Electronics 1 Presentation Part 3 (ME)

8/7/2019 Electronics 1 Presentation Part 3 (ME)

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Semiconductor Diode

Circuit Analysis

Prepared by: Armando V. Barretto



Load Line Analysis

• Below is a circuit which will be used to describe the operation of a diodecircuit.

• The diode in the circuit is forward biased.• The intersection of the characteristic curve of the diode and the load line

defines the current and voltage levels of the network.

Cathode (K)Anode (A)

ID

ID (mA)

VD

(V)

Quiescent Point (Q point)

Zener potential / voltage VZ

Zener region

Reversesaturationcurrent

ID (pA)

Load line

E R

E / R

Diodecharacteristiccurve

Q

EVDQ

IDQVR

VD



Load Line Analysis

• Applying Kirchhoff’s voltage law in the clockwise direction, the followingequation can be derived:

• Based on the equation above, if VD is equal to zero, ID = E / R.

• Also, if ID is equal to zero, VD = E.

• A straight line drawn between the two points defined when VD = 0 and ID = 0yields the load line of the network.

• The point of intersection between the straight line and the characteristic curve

of the diode is the point of operation of the network, and it is called theQuiescent (Q) point.

• The value of the current I D can be computed as:

• The above two equations can be solved simultaneously to determine unknown

parameters, although this process could be cumbersome.

RIVE 0VVE DDRD

R

VEI

D D

ID = IS (eVD /nVT-1)



Diode Circuit Analysis

• An easier approach is to use approximate analysis in which we assume that the

voltage across a diode is fixed once it is forward biased.

• For a forward biased diode, the voltage drop across the diode can be assumedto be:

0.7 volt for Silicon

0.3 volt for Ge1.2 volt for GaAs

• Example: With the circuit in the preceding slides having the followingparameters, determine the quiescent voltage and current of the diode.

E = 8 volts R = 4,700 ohms

volt0.7V

Ax101.5534700

0.7-8

R

V-EII

RIVE

Silicon),betoassumedis(diodevolt0.7isdiodethe

acrossdropvoltagethat theassumeweapproach,eapproximattheUsing

DQ

3DDQD

DD




• For a reverse biased diode, the diode can be assumed to be an open circuit.

• Example: For the circuit below, determine the current ID, voltage across the

diode and voltage across the resistor. E = 8 volts R = 4,700 ohms

Since the diode is reverse biased, the current ID is the leakage current which is very

small. The current is approximately equal to 0 A.

The voltage drop across the resistor (VR) is equal to 0 volt since the current

passing through it is equal to 0 A (V=IR).

The voltage drop across the diode (VD) is equal to the supply voltage (8 volts).


IDE R VR

VD

volts8V (0)(4700)V8 RIVE DDDD



Diode Circuit Analysis• Kircchoff’s voltage law is always applicable under any circumstances –

whether the voltages are dc, ac, pulses, or instantaneous values.• Example: For the circuit below, determine the current ID, voltage across the

diode and voltage across the resistor. E = 0.4 volt R = 4,700 ohms

Since the applied voltage (E) is less than 0.7 volt, it is not sufficient to turn on the

diode and the current will be approximately 0 A.

The voltage drop across the resistor (VR) is equal to 0 volt since the current

passing through it is equal to 0 A (V=IR).

The voltage drop across the diode (VD) is equal to the supply voltage (0.4 volt).


IDE R VR

VD

volt4.0V (0)(4700)V0.4 RIVE DDDD



Diode Circuit Analysis• Example: For the circuit below, determine the current ID, voltage across the

diodes and voltage across the resistor. E = 14 volts R = 4,700 ohms

Since the applied voltage (E) is greater than what is needed to forward bias the twodiodes, the two diodes will conduct, and the voltages across the diodes will be:

VD1= 0.7 v VD2 = 1.8 volts

The voltage drop across the resistor (VR) can be computed using Kircchoff’svoltage law as follows:

VR = E - VD1 - VD2 = 14 – 0.7 – 1.8 = 11.5 v

The current across the loop can be computed as:

IDE R VR

VD2

A10x2.4474700

5.11

R

VI 3R

D

VD1

D1 D2

D1 is a Silicon diode

D2 is a red LED




• Example: For the circuit below, determine the current passing through each

diode, current passing through R, voltage across the diodes, and voltage acrossR. E = 14 volts R = 4,700 ohms

Since the applied voltage (E) is greater than what is needed to forward bias the two

diodes, the two diodes will conduct, and the voltages across the diodes will be:VD1= 0.7 v VD2 = 0.7 v

The voltage drop across the resistor (VR) can be computed using Kircchoff’s

voltage law as follows:

VR = E - VD1 = 14 – 0.7 = 13.3 v

The current passing through R can be computed as:

The current passing through D1 and D2 computed as:

ID1 = ID2 = IR / 2 = 1.415 x 10 -3 A

E VD

A10x83.24700

3.13

R

VI 3R

R

D1 and D2 are Silicon diodesD1 D2R



Rectifiers Using Semiconductor

Diodes




Half Wave Rectifier Using Semiconductor Diode• A rectifier is a circuit which converts an alternating voltage into a pulsating dc

voltage.

• Rectifiers are used in power supplies to convert ac voltages to dc voltages.

• The power available from electric power companies are supplied using ac voltages

but many electrical and electronic devices need dc voltages, so there is a need to

convert ac voltages to dc voltages.

• A half-wave rectifier produces an output voltage during one-half (1800 ) of the input

ac signal. One half of the input signal is removed to establish a dc voltage.

• A half-wave rectifier using a semiconductor diode is shown below.

KA

IRL

AC Inputvoltage

(sine wave) RL

VO = output voltage(Pulsating DC voltage)

Half Wave Rectifier Using Semiconductor Diode

t

V

Input (Sine Wave)

t

V

Output (Pulsating DC)

Vp

Vp

Io




• In practically all rectifier applications, the peak voltage of the input signal is much greater than the minimum voltage required to forward biased the diode.

• During the positive half , the diode becomes forward biased when the ac inputsignal becomes 0.7 volt (for Si), and it conducts.

– The input voltage practically appears across R during the positive half .

– The peak voltage of the output signal is approximately equal to the peakvoltage of the input signal (Vp of input signal minus 0.7 v).

• During the negative half , the diode is reverse biased and does not conduct.Output voltage across R is equal to 0 volt.

• The average value (Vdc) of the input signal is equal to zero because thealgebraic sum of the positive half and negative half is equal to zero.

Vdc(input) = 0 volt (for sine wave input signal) = average voltage of input

• The peak voltage of the output of a half wave rectifier is approximately equal to the peak voltage of the input signal .

Vp(out) = Vp(input) = peak voltage of output signal




• If the voltage drop across the diode is ignored, the average value of the output voltage (Vo) and output current (Io) can be computed as:

rectifierwavehalf of currentoutputaverageR

Vp318.0

R

Vdc

Idc

rectifierwavehalf of tageoutput volaverageVp318.0Vp

Vdc

(1)(-1)-Vp210coscosVp

21

)cos(Vp2

1)(d)sin(Vp

T

1Vdc

LL

00

t t t

V

Vp

0 2

Half Wave Rectifier Output

Vdc

T

T = 2 = period

Vdc = average voltage

Vp = peak voltage



Half Wave Rectifier Using Semiconductor Diode• If the voltage drop across the diode is ignored, the effective or rms value of the

output voltage (Vo) and output current (Io) of a half wave rectifier can becomputed as:

rectifierwavehalf of currentoutputeffectiveorrmsR

Vp0.5

R

VrmsIrms

rectifierwavehalf of tageoutput voleffectiveorrmsVp0.52

VpVrms

2

)0sin(0

2

)2sin(

4

Vp

2

)2sin(

04

Vp

)(d)2cos(14

Vp

)(d2

)2cos(1

2

Vp

)(d)sin(2

Vp)(d)sin(Vp

T

1Vrms

LL

22

0

2

0

2

0

22

0

2

t t

t t t

t

t t t t

VVp

0 2

Half Wave Rectifier Output

T = 2 = period of output voltage

Vdc = average output voltage

Vp = peak voltage

rms – root mean squareT




• If there is no capacitor connected at the output of the rectifier,

– the maximum reverse bias voltage that will appear across the diode is

equal to the peak voltage of the input to the rectifier. This will occur

during the negative half of the input signal.

– the peak inverse voltage (PIV) or the peak reverse voltage (PRV) of thediode must be greater than the peak voltage of the input signal .

rectifier)of outputatconnectediscapacitorno(if VpPIV

KA

IRL

AC Inputvoltage

(sine wave) RL



t

V

Input (Sine Wave)

Vp

VD

t

V

Output (Pulsating DC)

Vp

Io




KA

IRL

AC Inputvoltage

(sine wave) RL

VO = output voltage(Almost pure DC voltage)

Half Wave Rectifier Using Semiconductor Diode And Capacitor Filter

Capacitor dischargingCapacitor charging

Output Voltage of Half Wave RectifierBefore Filtering

Output Voltage of Capacitor Filter Used WithHalf Wave Rectifier

Vp = peak voltageVp = peak voltage

+

-

signal)inputof half negative(during2VpVD

VD

Io



Full Wave Rectifier Using Center Tapped Transformer• A full-wave rectifier produces an output voltage during the entire cycle (3600 ) of

the input ac signal.

• A full-wave rectifier using two semiconductor diodes and a center tapped

transformer is shown below.

AC Inputvoltage

(sine wave)


Full Wave Rectifier Using Semiconductor Diode and Center Tapped Transformer

t

Voltage at Secondary

Vpsec

+

-

D1

D2

I-I-

I-

I-

I+

I+

I+

I-

I+

I-

Vp1/2sec

Vp1/2sec

tVoltage At Each Half (½)

of Secondary

Vp1/2secV1/2sec

V1/2sec

Vo = Output

(Pulsating DC)

Vp1/2sec

Io= output current

I+ = current during positive half

I- = current during negative half

RL

Io



Full Wave Rectifier Using Center Tapped Transformer• During the positive half of the voltage at the secondary of the transformer, D1 is

forward biased while D2 is reverse biased.– D1 conducts and current flows from the upper portion of secondary, to D1, to

RL, and then to the center tap of the transformer.

– The voltage at one half of the secondary appears across RL except for a small

voltage drop across D1.– The peak voltage at the output (Vpout) is almost equal to the peak voltage of

one half of the secondary voltage.

• During the negative half of the voltage at the secondary of the transformer, D2 is forward biased while D1 is reverse biased.

– D2 conducts and current flows from the lower portion of secondary, to D2, toRL, and then to the center tap of the transformer.

– The voltage at one half of the secondary appears across RL except for a smallvoltage drop across D2.

– The peak voltage at the output (Vpout) is almost equal to the peak voltage of one half of the secondary voltage.

• The average value (Vdc) of the input signal (voltage at ½ of secondary) is equal to zero because the algebraic sum of the positive half and negative half is equal to zero.

Vdc = 0 volt (for sine wave input signal)



Full Wave Rectifier Using Center Tapped Transformer• If the voltage drops across the diodes are ignored, the average value of the

output voltage (Vo) and output current (Io) can be computed as:

V

Vp1/2sec

0 2

Full Wave Rectifier Using Center Tapped Transformer Output

Vdc

T

T = = period of output voltage


Vp1/2sec = Vpsec /2 = peak voltage of 1/2 of secondary

Vpsec = peak voltage of whole secondary

rtransfomecenter tapusingrectifierwavefullof currentoutputaverage

RVp636.0

RVdc Idc

rtransfomecenter tapusingrectifierwavefullof tageoutput volaverage

Vp636.02Vp

Vdc

(1)(-1)-Vp1

0coscosVp1

)cos(Vp1

)(d)sin(VpT

1Vdc

L

sec1/2

L

sec1/2sec1/2

sec1/2sec1/2

0sec1/2

0sec1/2

t t t



Full Wave Rectifier Using Center Tapped Transformer

• If the voltage drops across the diodes are ignored, the effective or rms value of

the output voltage (Vo) and output current (Io) of a full wave rectifier using

center tapped transformer can be computed as:

ertransformcenter tapusingrectifierwavefullof currentoutputeffectiveorrms

R

Vp0.707

R2

Vp

R

VrmsIrms

ertransformcenter tapusingrectifierwavefullof tageoutput voleffectiveorrms

Vp0.7072

VpVrms

2

)0sin(02

)2sin(2

Vp2

)2sin(

02Vp

)(d)2cos(12

Vp )(d

2

)2cos(1Vp

)(d)sin(

Vp

)(d)sin(VpT

1

Vrms

L

1/2sec

L

1/2sec

L

1/2sec1/2sec

21/2sec

21/2sec

0

21/2sec

0

21/2sec

0

22

1/2sec

0

2

1/2sec

t t

t t t t

t t t t




• If there is no filter capacitor connected at the output of the rectifier, the peak

inverse voltage (PIV) or the peak reverse voltage (PRV) of the diode mustbe greater than approximately two times the peak voltage of half of the

secondary or the peak voltage of the whole secondary of the transformer.

rectifier)of outputatconnectediscapacitorno(if Vp2

rectifier)of outputatconnectediscapacitorno(if VpsecPIV

sec1/2

AC Inputvoltage

(sine wave)


+

-

D1

D2

I-I-

I-

I-

I+

I+

I+

I-

I+

I-

Vp1/2sec

V1/2sec

V1/2sec

Vp1/2sec

Io= output current

I+ = current during positive half

I- = current during negative half

RL

t


Vpsec

Vp1/2sec

t

Voltage At Each Half (½)

of Secondary

Vp1/2sec

Vo = Output(Pulsating DC)

Io




• If a capacitor filter is connected at the output of the rectifier, the output voltage

will be different.• The output voltage of a full wave rectifier and output voltage of a capacitor filter

connected to a full wave rectifier are shown below.

• The output voltage of the filter is a dc voltage with some ripple (ac variation).

• When the output of the rectifier is increasing, the capacitor is charging, and

when the output of the rectifier is decreasing, the capacitor is discharging

through the load of the filter (RL).

• If the capacitor has no load , the output of the capacitor filter will ideally be a

constant dc voltage, because it will not be discharging.

Capacitor DischargingCapacitor charging

Output Voltage of Full Wave RectifierBefore Filtering

Output Voltage of Capacitor Filter Used WithFull Wave Rectifier

Vp1/2secVp1/2sec




• Whether there is a capacitor connected or no capacitor connected to theoutput of the rectifier, the voltage that will appear across the diode when it is

reverse biased is equal to two times the peak voltage of half of the secondary(2Vp1/2sec ) or the voltage of the whole secondary (Vpsec).

• If there is a filter capacitor connected at the output of the rectifier, the peak

inverse voltage (PIV) or the peak reverse voltage (PRV) of the diode mustalso be greater than approximately two times the peak voltage of half of the

secondary (2Vp1/2sec ) or the peak voltage of the whole secondary (Vpsec) of the transformer.

rectifier)of outputatconnectediscapacitor(if Vp2

rectifier)of outputatconnectediscapacitor(if VpsecPIV

sec1/2



Full Wave Bridge Rectifier• A full-wave bridge rectifier using semiconductor diodes is shown below.

Full Wave Bridge Rectifier Using Semiconductor Diode

t


Vpsec

Vpsec

Vo = Output Voltage

(Pulsating DC)

Vo =

output

Voltage

D1D2

D3

D4

+

-

RL

Current flow during positive half of input signal

Current flow during negative half of input signal

Io =

output

current



Full Wave Bridge Rectifier• During the positive half of the voltage at the secondary of the transformer, D2 and

D3 are forward biased while D1 and D4 are reverse biased.– D2 and D3 conduct and current flows from the upper portion of secondary, to

D2, to RL, to D3, and then to the lower portion of the secondary of thetransformer.

– The voltage at the secondary appears across RL

except for a small voltage dropacross D2 and D3.

– The peak voltage at the output (Vpout) is almost equal to the peak voltage of the secondary of the transformer.

• During the negative half of the voltage at the secondary of the transformer, D1 and

D4 are forward biased while D2 and D3 are reverse biased.– D1 and D4 conduct and current flows from the lower portion of secondary, to

D4, to RL, to D1, and then to the upper portion of the secondary of thetransformer.

– The voltage at the secondary appears across RL

except for a small voltage dropacross D1 and D4.

– The peak voltage at the output (Vpout) is almost equal to the peak voltage of the secondary of the transformer.



Full Wave Bridge Rectifier• If the voltage drops across the diodes are ignored, the average value of the

output voltage (Vo) and output current (Io) can be computed as:

V

Vpsec

0 2

Full Wave Bridge Rectifier Output

Vdc

T

T = = period of output voltage


Vpsec = peak voltage of secondary

rectifierbridgewavefullof currentoutputaverage

RVp636.0

RVdc Idc

rectifierbridgewavefullof tageoutput volaverage

Vp636.02Vp

Vdc

(1)(-1)-Vp1

0coscosVp1

)cos(Vp1

)(d)sin(VpT

1Vdc

L

sec

L

secsec

secsec

0sec

0sec

t t t



Full Wave Bridge Rectifier

• If the voltage drops across the diodes are ignored, the effective or rms value of

the output voltage and output current of a full wave bridge rectifier can be

computed as:

rectifierbridgewavefullof currentoutputeffectiveorrms

R

Vp0.707

R2

Vp

R

VrmsIrms

rectifierbridgewavefullof tageoutput voleffectiveorrms

Vp0.7072

VpVrms

2

)0sin(02

)2sin(

2

Vp

2

)2sin(

02

Vp

)(d)2cos(12

Vp )(d

2

)2cos(1Vp

)(d)sin(Vp

)(d)sin(Vp

T

1Vrms

L

sec

L

sec

L

secsec

2sec

2sec

0

2sec

0

2sec

0

22

sec

0

2sec

t

t

t t t t

t t t t



Full Wave Bridge Rectifier• If there is a filter capacitor or no filter capacitor connected at the output of the

rectifier, the voltage that will appear across the reverse biased diodes will be approximately equal to the peak voltage of the secondary.

• If there is no filter capacitor connected at the output of the rectifier, the peakinverse voltage (PIV) or the peak reverse voltage (PRV) of the diode must be

greater than approximately the peak voltage of the secondary of thetransformer.

• If there is a filter capacitor connected at the output of the rectifier, the peakinverse voltage (PIV) or the peak reverse voltage (PRV) of the diode must alsobe greater than approximately the peak voltage of the secondary of thetransformer.

rectifier)of outputatconnectediscapacitorno(if VpsecPIV

rectifier)of outputatconnectediscapacitor(if VpsecPIV



Rectifiers

• If the voltage drops across the diodes are considered, the voltage drops

across the conducting diodes must be subtracted from the peak voltages in

the equations for the average voltage (Vdc) or rms voltages (Vrms) at the

output of the rectifier.

• Example: For a half wave rectifier with Silicon diode, the equation for the

average voltage if the voltage drop across the diode is considered will be:

• Example: For a full wave bridge rectifier with Silicon diodes, the equation

for the rms voltage (Vrms) if the voltage drop across the diodes is

considered will be:

0.7)(Vp0.318Vdc

)4.1(Vp0.707Vrms



Comparison of RectifiersHalf wave rectifiers:

• Advantage is only one diode is used .

• Disadvantages compared to full wave rectifiers are:

– lower rms output voltage and lower average output voltage

– Higher ripple

Full wave rectifier with center tapped transformer:

• Advantage over bridge rectifier is only two diodes are used.

• Disadvantages compared to bridge rectifier assuming the same voltages at

whole secondary are:– lower rms output voltage and lower average output voltage (1/2 of bridge

rectifier)

Full wave rectifier with center tapped transformer:• Advantage is higher rms output voltage and higher average output voltage

(assuming the same voltage at whole secondary).

• Disadvantage is four diodes are used.



Positive and Negative Voltages Power Supply• Below is a circuit which could be used to produce a positive and a negative

voltage using a transformer with center tap.

• During the positive half of the voltage at the secondary of the transformer, D1and D2 conducts, and C1 and C2 are charged with the polarity shown.

• During the negative half of the ac voltage at the secondary of the transformer,no diode conducts.

I2

0 volt

+ Output Voltage

- Output Voltage

+

-

-

+

D1

D2

C1

C2

P i i d N i V l P S l



Positive and Negative Voltages Power Supply

• Below is a circuit which could be used to produce a positive dc voltage and a

negative dc voltage using a transformer with center tap.• During the positive half of the voltage at the secondary of the transformer, D2

and D3 conducts, and C1 and C2 are charged with the polarity shown.

• During the negative half of the ac voltage at the secondary of the transformer,D4 and D1 conducts, and C1 and C2 are charged with the polarity shown.

0 volt

+ Output Voltage

- Output Voltage

D1 D2

D3 D4

+

-

-

+

C1

C2

P iti d N ti V lt P S l



Positive and Negative Voltages Power Supply

• Below is a circuit which could be used to produce a positive dc voltage and a

negative dc voltage using a transformer with no center tap.• During the positive half of the voltage at the secondary of the transformer, D1

conducts, and C1 is charged with the polarity shown.

• During the negative half of the ac voltage at the secondary of the transformer,D2 conducts, and C2 is charged with the polarity shown.

I2

0 volt

+ Output Voltage

- Output Voltage

+

-

-

+

D1

D2

D1 is conducting

D2 is conducting

C1

C2

T f



Transformers• Transformers are typically used to step up or step down the primary voltage, or

match a load impedance.• A transformer can increase or decrease the voltage or current levels on the

secondary depending on the turns ratio of the primary and secondary.

• A transformer can also increase or decrease the impedance of the load

appearing at the primary of the transformer depending on the square of the

transformer turns ratio.

• Turns ratio is equal to the number of turns in the primary divided by the

number of turns in the secondary.

• The turns ratio can be used to match the impedance of the load (RL

) to the

output impedance of the amplifier.

Transformers



Transformers• Assuming that the transformer is lossless (no power is dissipated in the

transformer), the following relationships can be derived:

RL =

R2

Transformer

N1:N2

V1 V2

secondaryatcurrentI

primaryatcurrentI

ertransformtheof impedanceloadRR

primaryat theRof impedancereflected

ertransformtheof primaryat theimpedanceinputR

secondaryat thevoltageaverageV

primaryat thevoltageaverageV

secondaryat thegepeak voltaV

primaryat thegepeak voltaV

secondaryat thevoltagermsV

primaryat thevoltagermsV

secondaryin theturnsof numberN

primaryin theturnsof numberN:where

ertransformof ratioturnsaR

R

I

I

V

V

V

V

V

V

N

N

2

1

L2

2

1

dc2

dc1

p2

p1

rms2

rms1

2

1

2

1

1

2

dc2

dc1

p2

p1

rms2

rms1

2

1

R1

I1

I2

2

2

2

1

2

1

2

1

22

11

2

1a

N

N

N

N

N

N

/IV

/IV

R

R

V l A d C R l i hi F A Si W



Voltage And Current Relationships For A Sine Wave• For a sine wave with a peak voltage = Vp, the rms voltage, rms current, average

voltage, average current, and peak current can be computed as:

measuredisvoltagewhich theacrossresistorR:where

wavesineaforcurrentoutputaverage

0R

0

R

Vdc Idc

wavesineaof tageoutput volaverage

volt0Vdc

measuredisvoltagewhich theacrossresistorR:where

wavesineaforcurrentoutputeffectiveorrms

RVp0.707

R2Vp

RVrmsIrms

voltagewavesineafortageoutput voleffectiveorrms

Vp0.7072

VpVrms

currentpeak R

Vp

Ip

AC voltage(sine wave) R

V

t

V

Voltage (Sine Wave)

Vp

I



Sample Problems

• Example: A half wave rectifier uses a step down transformer. The input to

the primary of the transformer is 220 volts rms and the turns ratio of thetransformer (N1:N2) is 10:1. Load resistance (RL) is 5,000 ohms. Nocapacitor is connected at the output of the rectifier. Determine thefollowing:

a. Peak voltage at the primary

b. Average voltage at the primary

c. Peak voltage at the secondary

d. RMS voltage at the secondary

e. Average voltage at the secondary

f. Peak voltage at the output of the rectifier (across RL)

g. RMS voltage at the output of the rectifier (across RL)

h. Average voltage at the output of the rectifier (across RL)

i. Peak current at the load

j. RMS current at the load

k. Average current at the load

l. Minimum PIV of the diode (without capacitor filter)

S l P bl



Sample Problems

a. Since the power outlet voltage is a sine wave, the input to the transformer

is assumed to be a sine wave and the peak voltage at the primary iscomputed as:

b. Since the voltage at the primary is a sine wave, the average voltage at theprimary is 0 volt.

c. The voltage at the secondary is also a sine wave. Using the turns ratio andthe peak voltage at the primary, the peak voltage at the secondary can becomputed as:

ertransformof primaryatgepeak voltavolts311.174 0.707

220

0.707

VrmsVp

voltagewavesineafortageoutput voleffectiveorrmsVp0.7072

VpVrms

secondaryat thegepeak volta

volts31.11710

311.174

N2

N1

priVpsecVp

ratioturnsN2

N1

secVp

priVp

S l P bl



Sample Problems

d. As computed above, the rms voltage across the secondary is 22 volts.

e. Since the voltage at the secondary is also a sine wave, the average voltageat the secondary is 0 volt.

f. The peak voltage at the load (R L ) is approximately equal to the peakvoltage at the secondary which is 31.117 volts.

wave)(sinesecondaryatgepeak voltavolts31.117

707.0

22

0.707

secVrmssecVp

secondaryatvoltagermsvolts2210

220

N2

N1priVrms

secVrms

ratioturnsN2

N1

secVrms

priVrms

:followsassecondaryat thevoltagermsthe

computingfirstbycomputedbealsocansecondaryat thegepeak voltaThe

S l P bl



Sample Problems

g. The rms voltage at the load is not the same as the rms voltage at the

secondary of the transformer and it can be computed as:

h. The average voltage at the load is not the same as the average voltage atthe secondary of the transformer and it can be computed as:

i. The peak current at the load can be computed as:

volts15.558)0.5(31.117secVp0.5Vp0.5RacrossVrms

rectifierwavehalf of tageoutput voleffectiveorrmsVp0.52

VpVrms

RLL

rectifierwavehalf of tageoutput volaveragevolts9.895(31.117)318.0secVp

Vdc

rectifierwavehalf of tageoutput volaverageVp318.0secVp318.0VpVdc RL

L3

L

RLRL Rthroughpassingcurrentpeak A10x6.223

5000

117.31

R

Vp Ip

Sample Problems



Sample Problems

j. The rms current at the load can be computed as:

k. The average current at the load can be computed as:

l. The minimum PIV of the diode (without capacitor) is equal to the peak voltage of the secondary, which is 31.117 volts.

L3

L

RLRL RthroughpassingcurrentrmsA10x112.3

5000

558.15

R

Vrms Irms

L3

L

RLRL RthroughpassingcurrentaverageA10x979.1

5000

895.9

R

Vdc Idc

Rectifiers



Rectifiers

• Example: A full wave rectifier is using a center tapped step down transformer.

The input to the primary of the transformer is 220 volts rms and the turns ratio of the transformer (N1:N2) is 10:1. Load resistance (RL) is 5,000 ohms. No capacitoris connected at the output of the rectifier. Determine the following: (Similar toprevious example except full wave rectifier with center tapped transformer)



c. Peak voltage at the whole secondary and peak voltage at ½ of secondary

d. RMS voltage at the whole secondary and rms voltage at ½ of secondary

e. Average voltage at the whole secondary and average voltage at ½ of secondary







l. Minimum PIV of the diode without capacitor

Sample Problems



Sample Problems




c. The voltage at the secondary is also a sine wave. Using the turns ratio andthe peak voltage at the primary, the peak voltage at the whole secondarycan be computed as:


220 0.707

VrmsVp


VpVrms

secondarywholeat thegepeak volta

volts31.11710

311.174

N2

N1

priVpsecVp

ratioturnsN2

N1

secVp

priVp

Sample Problems



Sample Problems

secondaryof 1/2atgepeak voltavolts15.5582

31.117

2

secVp Vp

wave)(sinesecondaryat wholegepeak voltavolts31.117707.0

22

0.707

secVrmssecVp

secondaryat wholevoltagermsvolts2210

220

N2

N1

priVrmssecVrms

ratioturnsN2

N1

secVrms

priVrms



sec1/2

d. As computed above, the rms voltage across the whole secondary is 22

volts. The rms voltage at ½ of secondary is one half the rms voltage of thewhole secondary and is equal to 22 / 2 = 11 volts.

e. Since the voltage at the secondary is also a sine wave, the average voltage at the whole secondary is 0 volt, and the average voltage at ½ of the

secondary is also equal to 0 volt.

Sample Problems



Sample Problems

f. Since the peak voltage that appears across the load (R L ) is approximately

equal to the peak voltage at one half of the secondary, the peak voltage at the load (R L ) is approximately equal to the peak voltage at ½ of the secondary and it is equal to 31.117 / 2 = 15.558 volts.

g. The rms voltage at the output of rectifier or at the load can be computed as:

h. The average voltage at the output of the rectifier or at the load can becomputed as:

ertransformcenter tapusingrectifierwavefullof tageoutput voleffectiveorrms

volts11)558.15(707.0Vp0.7072

VpVrms

1/2sec1/2sec

rtransfomecenter tapusingrectifierwavefullof tageoutput volaverage

volts9.895)558.15)(636.0(Vp636.02VpVdc

sec1/2sec1/2

S l P bl



Sample Problemsi. The peak current at the load can be computed as:



l. The minimum PIV of the diode (without capacitor) is equal to the peakvoltage of the whole secondary, which is 31.117 volts.

L3

L


5000

895.9

R

Vdc Idc

L3

L


5000

11

R

Vrms Irms

L3

L


5000

558.15

R

Vp Ip

Rectifiers



Rectifiers

• Example: A full wave bridge rectifier is using a step down transformer. The input

to the primary of the transformer is 220 volts rms and the turns ratio of thetransformer (N1:N2) is 10:1. Load resistance (RL) is 5,000 ohms. No capacitor isconnected at the output of the rectifier. Determine the following: (Similar toprevious two examples except full wave bridge rectifier)



c. Peak voltage at the secondary

d. RMS voltage at the secondary

e. Average voltage at the secondary







l. Minimum PIV of the diode without capacitor

Sample Problems



Sample Problems




c. The voltage at the secondary is also a sine wave. Using the turns ratio andthe peak voltage at the primary, the peak voltage at the secondary can becomputed as:


220

0.707

VrmsVp


VpVrms

secondary at thegepeak volta

volts31.11710

311.174

N2

N1

priVpsecVp

ratioturnsN2

N1

secVp

priVp

Sample Problems



Sample Problems

wave)(sinesecondaryat wholegepeak voltavolts31.117

707.0

22

0.707

secVrmssecVp

secondaryat wholevoltagermsvolts2210

220

N2

N1

priVrmssecVrms

ratioturnsN2

N1

secVrms

priVrms



d. As computed above, the rms voltage across the whole secondary is 22volts.

e. Since the voltage at the secondary is also a sine wave, the average voltage at the secondary is 0 volt.

Sample Problems



p

f. The peak voltage that appears across the load (R L ) is approximately equal to

the peak voltage at the secondary and it is equal to 31.117 volts.

g. The rms voltage at the output of rectifier or at the load can be computed as:

h. The average voltage at the output of the rectifier or at the load can becomputed as:

rectifierbridgewavefullof tageoutput voleffectiveorrms

volts22)117.31(707.0secVp0.7072secVpVrms

rectifierbridgewavefullof tageoutput volaverage

volts79.91)117.31)(636.0(secVp636.0sec2Vp

Vdc

Sample Problems



Sample Problemsi. The peak current at the load can be computed as:



l. The minimum PIV of the diode (without capacitor) is equal to the peakvoltage of the whole secondary, which is 31.117 volts.

L3

L


5000

79.19

R

Vdc Idc

L3

L


5000

22

R

Vrms Irms

L3

L


5000

117.31

R

Vp Ip

Sample Problems



p

• Example: The average (dc) current at the load of a half wave rectifier is

50 mA and the load resistance is 4,000 ohms. Determine:a. Average output voltage of the rectifier (or across the load)

b. Peak voltage at the output of the rectifier (or across the load)

c. RMS output voltage of the rectifier (or across the load)

d. Minimum PIV rating of the diode

rectifierof outputorloadatgepeak voltavolts628.93318.0

200

0.318

Vdc Vp

loadatvoltageaverageVp0.318Vdc

rectifierof tageoutput volaverageloadatvoltageaverage

volts200)000,4)(10x0(5 )(RIdcVdc

rectifierof outputatcurrentaverageRthroughpassingcurrentaverage

A10x05R

Vdc Idc

RL RL

RLRL

3LRLRL

L

3

L

RLRL

Sample Problems



)(Rloadat thevoltageeffectiveorrms

rectifierwavehalf of tageoutput voleffectiveorrms

volts314.4652

628.93

2

VpVrms

L

RL

The minimum PIV of the diode (without capacitor) is equal to the peak voltage

of the secondary or load, which is 628.93 volts.

Rectifiers



• Example:

Rectifiers



• Example:



Clippers


Clippers



• Clippers are networks which are used to remove or “clip” away a portion of

the input signal without distorting the remaining part of the input signal.• The input signal could be a square wave, sine wave or any other waveform.

• A half wave rectifier is a clipper which removes half of the input signal.

• The two general categories of clippers are series clipper and parallel clipper.

• A series clipper is one where the diode is in series with the load .• A parallel clipper is one where the diode is in a branch parallel with the load .

• The considerations in analyzing clippers are:

– Take careful note of where the output voltage is.

– Analyze which diodes are forward biased and which are reverse biased under certain conditions.

– Forward biased diodes may be considered short circuits while reverse biased diodes may be considered open circuits.

– DC voltages from batteries will add or subtract from the input signal and

the resulting voltages will determine the biasing of the diodes.• If the voltage drops across the diodes are considered , the voltage drops across

the conducting diodes must be subtracted from the peak voltages in theequations.

Clippers



• A series clipper which removes the negative half of the input signal is shown

below. This is also a half wave rectifier.

• During the positive half of the input signal, the diode conducts and the input

voltage appears across the load (RL).

• During the negative half of the input signal, the diode is reversed biased and

does not conduct. No current flows through the load and the output voltage isequal to zero.

ID

RL Output

Input (Sine Wave)

Output

Vp

Vp

V

Input

+

-

0 v

0 v

0 v

0 v

Clippers



• A series clipper which removes the positive half of the input signal is shown

below. This is also a half wave rectifier.• During the negative half of the input signal, the diode conducts and the input

voltage appears across the load (RL).

• During the positive half of the input signal, the diode is reversed biased and

does not conduct. No current flows through the load and the output voltage isequal to zero.

ID RL Output

Input (Sine Wave)

Output-Vp

Vp

V

Input

0 v

0 v0 v

0 v

Clippers



• A parallel clipper which removes the negative half of the input signal is

shown below.• During the positive half of the input signal, the diode does not conduct and

the input voltage appears at the output.

• During the negative half of the input signal, the diode is forward biased and it

conducts. The output voltage is the voltage across the diode which is almostzero volt.

OutputInput D1

t

V

Input

Input (Sine Wave)

Output

Vp

Vp

0 v

0 v0 v

0 v

Clippers

ll l l h h lf



• A parallel clipper which removes the positive half of the input signal is

shown below.• During the negative half of the input signal, the diode does not conduct and

the input voltage appears at the output.

• During the positive half of the input signal, the diode is forward biased and it

conducts. The output voltage is the voltage across the diode which is almostzero volt.

OutputInput D1

t

V

Input

Vp

Input (Sine Wave)

Output-Vp

Vp

0 v

0 v0 v

0 v

Clippers• A biased series clipper which removes a portion of the negative half of the input



• A biased series clipper which removes a portion of the negative half of the input

signal is shown below.• When the input signal is 0 volt, the diode is forward biased because of the 3 volt

battery. The diode conducts and approximately 3 volts appear at the output.

• During the positive half of the input signal, the input voltage adds up with the 3

volt battery and the diode is forward biased . The diode conducts and the sum of the input voltage and 3 volt battery appears at the output.

• During the negative half of the input signal, the diode remains forward biased

because of the 3 volt battery until the input voltage becomes approximately equal

to 3 volts. When the negative half of the input signal becomes approximately -3

volts or more negative, the diode is reversed biased and does not conduct. No

current flows through the resistor and the output voltage is equal to zero.

ID R Output

Input (Sine Wave)

t

10 v

Vp=10 v

Vp=10 v

Input

V= 3 volts

3 v

13 v

Output

Vp + V =10+3v

+

- 0 v

0 v0 v

Clippers• A biased series clipper which removes the negative half of the input signal and a



• A biased series clipper which removes the negative half of the input signal and a

portion of the positive half of the input signal is shown below.• When the input signal is 0 volt, the diode is reverse biased because of the 3 volt

battery. The diode does not conduct and output voltage is equal to zero.

• During the positive half of the input signal, the diode remains reverse biased

because of the 3 volt battery until the input voltage becomes approximately equalto + 3 volts. When the positive half of the input signal becomes more positive than

approximately 3 volts, the diode is forward biased and it conducts. Current flows

through the resistor and output voltage becomes greater than 0 volt. Peak voltage

is approximately equal to the peak voltage of input minus 3 volts (7 volts).

• During the negative half of the input signal, the diode is reverse biased . There is

no current and output is equal to 0 volt.

ID R Output

Input (Sine Wave)

t

7 v

Vp=10 v

Vp=10 v

Input

V=3 volts

3 v

Output

Vp-V = 10 v-3v

+

-0 v

0 v0 v

Clippers• A biased series clipper which removes a portion of the positive half of the input



A biased series clipper which removes a portion of the positive half of the input

signal is shown below.• When the input signal is 0 volt, the diode is forward biased because of the - 3 volt

battery. The diode conducts and approximately - 3 volts appear at the output.

• During the positive half of the input signal, the diode remains forward biased

because of the - 3 volt battery until the input voltage becomes greater than approximately + 3 volts. When the positive half of the input signal becomes greater

than approximately + 3 volts, the diode is reversed biased and does not conduct. No

current flows through the resistor and the output voltage is equal to zero.

• During the negative half of the input signal, the input voltage adds up with the - 3

volt battery and the diode is forward biased . The diode conducts and the sum of the

input voltage and -3 volt battery appears at the output.

ID R Output

Input (Sine Wave)

Vp=10 v

Vp=10 v

Input

V= 3 volts

-3 v

-13 v Output -Vp-V= -10-3v

0 v

0 v0 v

0 v

Clippers• A biased series clipper which removes the positive half and a portion of the



A biased series clipper which removes the positive half and a portion of the

negative half of the input signal is shown below.• When the input signal is 0 volt, the diode is reverse biased because of the 3 volt

battery. The diode does not conduct and output is equal to zero volt..

• During the positive half of the input signal, the input signal adds to the voltage of

the battery, the diode remains reverse biased and does not conduct. No currentflows through the resistor and the output voltage is equal to zero.

• During the negative half of the input signal, the diode remains reverse biased until

the input signal becomes more negative than approximately – 3 volts. When the

input signal becomes more negative than approximately – 3 volts, the diode

conducts, current flows through the resistor, and negative output voltage appears

at the output.

ID R Output

Input (Sine Wave)

Vp=10 v

Vp=10 v

Input

V= 3 volts

-7 v Output-Vp + V= -10 + 3v

0 v

0 v

0 v

0 v

Clippers• A biased parallel clipper which removes a portion of the positive half of the input

i l i h b l



signal is shown below.

• When the input signal is 0 volt, the diode is reverse biased . The diode does not

conduct and output voltage is equal to zero.

• During the positive half of the input signal, the diode remains reverse biased

because of the 3 volt battery until the input voltage becomes more positive than

approximately 3 volts. When the positive half of the input signal becomes more positive than approximately 3 volts, the diode is forward biased and it conducts.

Output voltage becomes equal to 3 volts because the diode is effectively a short

circuit when it conducts.

• During the negative half of the input signal, input signal adds up with the batteryvoltage, the diode is reverse biased and output voltage is equal to the input signal.

Input (Sine Wave)

Vp=10 v

3 v

Output

OutputInputV = 3 v

Vp=10 v

Input

Vp=10 v

-Vp = -10v

0 v

0 v0 v

0 vV = 3 v

Clippers• A biased parallel clipper which removes the positive half of the input signal and



p pp p f f p g

a portion of the negative half of the input signal is shown below.• When the input signal is 0 volt, the diode is forward biased and it acts as a short

circuit. The output voltage is equal to – 3 volts.

• During the positive half of the input signal, the diode remains forward biased and

the output voltage remains at – 3 volts because the diode acts like a short circuit.• During the negative half of the input signal, the diode remains forward biased

until the input signal becomes equal to approximately – 3 volts. When the input

signal becomes equal to or more negative than approximately – 3 volts, the diode

is reverse biased and it acts as an open circuit. The output voltage becomes equal

to the negative input signal.

Input (Sine Wave)

Vp=10 v

-V= -3 v

Output

OutputInputV = 3 v

Vp=10 v

Input

Vp=10 v

-Vp = -10 v

0 v

0 v0 v

0 v

Clippers• A biased parallel clipper which removes a portion of the negative half of the



p pp p f g f f

input signal is shown below.• When the input signal is 0 volt, the diode is reverse biased . The diode acts like an

open circuit and output voltage is equal to zero.

• During the positive half of the input signal, the input signal adds up with the

battery voltage, the diode remains reverse biased and acts like an open circuit.The output voltage is equal to the positive input voltage.

• During the negative half of the input signal, the diode remains reverse biased and

acts like an open circuit until the input signal becomes more negative than

approximately -3 volts. When the input signal becomes more negative than

approximately -3 volts, the diode is forward biased and it acts like a short circuit.

The output becomes equal to – 3 volts.

Input (Sine Wave)

Vp=10 v

OutputInputV = 3 v

Vp=10 v

Input

0 v

Vp =10 v

0 v

Output

-V = -3 v

0 v

Clippers• A biased parallel clipper which removes the negative half and a portion of the



positive half of the input signal is shown below.• When the input signal is 0 volt, the diode is forward biased and it acts as a short

circuit. The output voltage is equal to + 3 volts.

• During the positive half of the input signal, the diode remains forward biased and

the output is equal to + 3 volts until the input signal becomes equal to approximately + 3 volts. When the input signal becomes equal to or more positive

than approximately + 3 volts, the diode is reverse biased and it acts as an open

circuit. The output voltage becomes equal to the positive input signal .

• During the negative half of the input signal, the input signal adds up with the

battery voltage, the diode remains forward biased and the output voltage remains

at + 3 volts because the diode acts like a short circuit.

Input (Sine Wave)

Vp=10 v

+V = + 3 v

OutputInputV = 3 v

Vp=10 v

Input

0 v

0 v

Vp =10 v

Output

0 v

Clippers• A biased parallel clipper which removes a portion of the positive half and a

i f h i h lf f h i i l



portion of the negative half of the input signal is shown below.

• When the input signal is 0 volt, the diodes are reverse biased and they act like open circuits. The output voltage is equal to zero volt.

• During the positive half of the input signal, D2 remains reverse biased all throughout while D1 remains reverse biased until the input signal becomes more positive

than approximately + 3 volts. When the input signal becomes more positive than approximately + 3 volts, D1 is forward biased and it acts like a short circuit. Theoutput voltage becomes equal to + 3 volts.

• During the negative half of the input signal, D1 remains reverse biased all throughout while D2 remains reverse biased until the input signal becomes more negative

than approximately - 3 volts. When the input signal becomes more negative than approximately - 3 volts, D2 is forward biased and it acts like a short circuit. Theoutput voltage becomes equal to - 3 volts.

Input (Sine Wave)

Vp=10 v

OutputInput D1

3 v3 v

D2

Vp=10 v

Input Output

Vp=10 v

+ 3 v

- 3 v

0 v

0 v



Resistor – Capacitor (RC)Networks


RC Network

i



E

(charging

voltage)

R

1 k

C

4 F

eR

eC

10 v

+

-

S1

iC

+

-

eC

(volt)

Time (t)

3

6

9

t1 t2 t3

E = supply voltage

eR = instantaneous voltage across resistor R

ec = instantaneous voltage across capacitor C

When S1 is closed, capacitor C will be charged towards the supply voltage

E. As the capacitor voltage increases, the charging current decreases.

Capacitor voltage

Charging current

RC Circuit Operation



• The instantaneous voltage across resistor R iseR = E – ec

Where: eR = instantaneous voltage across the resistor R

E = supply voltage

ec = instantaneous voltage across the capacitor C

• The instantaneous current passing through R and C is

R

eEi

R

ei

CC

RC




• If the charge on the capacitor C is zero at the instant that the switch isclosed, then at t = 0, the current passing through R and C is

• This current causes the capacitor C to charge with the polarity illustrated,

so that at time t1, the capacitor voltage eC could be 3 volts. This alters eR

eR = E – ec

= 10 v – 3 v = 7 v

• Because C has accumulated some charge, eC is increased and the voltage

across R is reduced; thus the charging current through R is also reduced.

10mAkohm1

0v10i

R

eEi

C

CC




• Since the charging current is reduced , C is being charged at a lower ratethan before.

• After some longer time period, eC increases to 6 v., and the voltage across

R is

eR

= 10 v – 6 v = 4 v

• The charging current has now been reduced further, and even longer time period is required to charge C by another 3 volts.

• The capacitor does not receive its charge at a constant rate.

• When the voltage across C is increasing, the voltage across R is

decreasing, and the charging current is also decreasing.• C is charged at a rapid rate initially, and then the charging rate decreases

as the capacitor voltage increases.

mA4kohm1

v6v10iC




• The capacitor voltage follows the exponential equation

(ohms)resistorof resistanceR

(Farad)capacitorof ecapacitancC

(seconds)chargeof ntcommencemefromtimet

2.718constantlexponentia

(volt)capacitoron thechargeinitialEo

(volt) voltagechargingE

(volt)instant tatRacrossvoltagee

(volt)instant tatvoltagecapacitore:Where

Eo)(Ee

is Rresistortheacross voltagetheand

Eo)(EEe

R

c

RC

t -

R

RC

t -

c


When there is no initial voltage on the capacitor,



charged.iscapacitortheassmallerandsmallerbecomescurrentchargingtheas

tage,supply volthetochargedfullybecomeneverwillcapacitorThe

current)charging(initial R

EI:Where

Ii

R

Ei

R

)-E(1-Ei

R

e-Ei

iscapacitorandresistorhethrough tflowingcurrentousinstantaneThe

)-E(1e

capacitor)theof voltageeous(instantan 0)(EEe

capacitor)theof voltage(initial 0Eo

RC

RC

RC

C

RC

RC

t-

C

t-

C

t-

C

C

t-

c

t-

c


• When the capacitor becomes charged to a certain voltage Eo, and is



p g g ,

discharged through a resistor R (without a supply voltage), the following

equation can also be used to compute for the capacitor voltage at any time

during discharge:

(ohms)resistorof resistanceR

(Farad)capacitorof ecapacitancC

(seconds)dischargeof ntcommencemefromtimet

2.718constantlexponentia

(volt)capacitoron thechargeinitialEo

volt0(volt) voltagechargingE

(volt)instant tatvoltagecapacitore:Where

Eoe

Eo)(00e

Eo)(EEe

c

RC

t -

c

RC

t -

c

RC

t -

c


• Example: A 1 F capacitor is charged from a 6 v source through a 10



Example: A 1 F capacitor is charged from a 6 v source through a 10kohm resistor. If the capacitor has an initial voltage of – 3 volts,calculate its voltage after 8 ms.

Solution:

volts1.96e

3v))(v(6v6e

Eo)(EEe

c

k)(1uF)(10

ms8 -

c

RC

t-

c

RC Time Constant

Capacitor

voltageCapacitor Charging voltage (E)



eC

(volt)

time (t) in ms

4 v

6 v

8 v

4 ms2 ms 6 ms

vo tage

E =10 v

2 v

12 ms10 ms8 ms 14 ms 16 ms 18 ms 20 ms

90 % E99.3 % E

(at t = 5 RC)

63.2 % E

(at t = RC)

Initial Chargingcurrent voltage whilecharging

Regardless of the value of E, R and C, the capacitor is charged to 63.2 % E

when t = RC, to 90 % E at t = 2.2 RC, and to 99.3 % E at t = 5 RC.

In the diagram, it was assumed that E = 10 volts, R = 1 kohm, and C = 4 F.

Thus RC = 4 ms

(10 volts)

RC Time Constant



• The product RC is the time constant of the RC circuit. It is the time whenthe capacitor is charged to 63.2 % E, regardless of the value of E, R, and C.(assuming initial charge of C is zero)

• At time 5 RC , the capacitor is charged to 99.3 % E, regardless of the valueof E, R, and C. (assuming initial charge of C is zero).

• For practical purposes, it is assumed that the capacitor is fully charged or discharged at t = 5 RC .

• At t = RC, the current flowing through R and C is 36.8 % I . Where I is theinitial charging current which is equal to E / R.

• When a capacitor is charged from a DC voltage source through a resistor,the instantaneous level of the capacitor voltage may be calculated at anygiven time.

• There is a definite relationship between the time constant of the RCnetwork and the time required for the capacitor to charge to approximately

63 % and 99 % of the input voltage.

RC Time Constant

SquareV



Input

voltage

R

C

eR

eC

10 v

S1

iC when Cis charging

Output

(across resistor)

vi

wave

input

t

t

t

eR

for

RC= 10 PW

eR

for

RC=PW

eR

for

RC= PW / 10

t

PW

eR

eR

eR

During the time that input voltage is positive, C is

charged towards the peak voltage of the input signal.

The output voltage across R is equal to the inputvoltage minus the voltage across the capacitor. When

the input signal becomes 0 v, the capacitor discharges

through R , during which the current is in the reverse

direction, causing a negative going pulse across R.

iC when Cis discharging

V1 V2

V3 V4

V1

V2

V3 V4

V1

V2

V3

V4



Clampers


Clampers

• Clampers are networks which shifts a waveform to a different dc level without



changing the appearance of the waveform.• Clampers are typically constructed using diodes, capacitors and resistors.

• The dc shifts are typically provided by the capacitor which charges up to acertain dc voltage when the diode is conducting.

• Additional dc shifts can be introduced in a clamper network by using dc supplies.

• When the capacitor is charging, the resistor is typically shorted out and the RCtime constant is effectively 0 sec. The capacitor is charged instantaneously to theapplied voltage because the RC time constant is 0.

• When the capacitor is discharging, the resistor is not shorted out and the RC time constant of the resistor and capacitor combination must be high enough so

that the voltage drop across the capacitor is not significant while it is discharging.

– It is desired that the RC time constant is much greater than 5 times the discharging time of the capacitor (time when diode is not conducting).

– In the analysis of clampers, it is assumed that the voltage drop across the

capacitor while it is discharging is negligible.• The input signal could be a square wave, sine wave or any other periodical

waveform.

Clampers• The considerations in analyzing clippers are:

– Begin the analysis by taking note of which portion of the input signal is the



diode forward biased and the capacitor charged to a certain dc level.

– Assume that the capacitor is charged up instantaneously to the appliedvoltage when the diode is “on”. Typically, the resistor is shorted out whenthe diode is conducting and the capacitor is fully charged instantaneously to

the applied voltage because the RC time constant is effectively 0 second when the capacitor is charging.

– With the capacitor is charged, it is like a battery whose voltage will add to or subtract from the input signal.

– Determine the output signal while assuming that the capacitor is fully

charged when the diode is not conducting. It is assumed that the RC time constant when the capacitor is discharging is high enough that the voltageacross the capacitor does not drop significantly.

– Check if the shape of the output waveform is the same as that of the inputsignal except for the change in the dc level in the output waveform.

• In the succeeding analysis, approximate approach is used and it is assumed that a diode is shorted when forward biased and voltage across the diode is 0 v.

• When considering the voltage across the diode (when it is forward biased) isdesired, simply replace the diode with a battery with the corresponding forward

biased voltage ( 0.7 for Si, 0.3 for Ge).

Clampers• A Clamper which shifts the dc level from 0 volt to the negative peak voltage

(-Vp) of the input signal is shown below.



• During the positive half of the input signal, the diode conducts, the capacitoris charged to the peak voltage (Vp) of the input signal with the polarityshown, and the output voltage is approximately equal to zero volt becausethe diode acts almost like a short circuit. The capacitor chargesinstantaneously to the peak voltage of the input signal because the resistor iseffectively shorted out (RC time constant = 0).

• During the negative half of the input signal, the input signal which is –Vp adds up to the voltage across the capacitor, the diode is reverse biased and itacts like an open circuit. The output voltage becomes equal to two times the

negative of the peak voltage (-2Vp) of the input signal (voltage acrosscapacitor plus input voltage).

OutputInput D1

Square

wave

input

0 v

PW

Output

PW

+ -Vp

0 v

Vp

-2Vp

-Vp DC = 0 v

DC = -Vp-Vp

T

0 v

0 v

Clampers• A Clamper which shifts the dc level from 0 volt to the positive peak voltage

(+Vp) of the input signal is shown below.

i h lf di d d



• During the negative half of the input signal, the diode conducts, the capacitoris charged to the peak voltage (Vp) of the input signal with the polarityshown, and the output voltage is approximately equal to zero volt becausethe diode acts almost like a short circuit. The capacitor chargesinstantaneously to the peak voltage of the input signal because the resistor iseffectively shorted out (RC time constant = 0).

• During the positive half of the input signal, the input signal which is +Vp adds up to the voltage across the capacitor, the diode is reverse biased and itacts like an open circuit. The output voltage becomes equal to two times the

peak voltage (2Vp) of the input signal (voltage across capacitor plus inputvoltage).

OutputInputD1

Square

wave

input

0 v

PW

OutputPW

+-Vp

0 v

Vp

2Vp

-Vp DC = 0 v

DC = +VpVp

T

0 v

0 v

Clampers• A Clamper using a dc supply, which shifts the dc level from -2.5 v to + 10.5 v, is

shown below.

D i h i i f h i i l h i i l dd i h h



• During the negative portion of the input signal, the input signal adds up with thevoltage of the battery, the diode is forward biased and it acts almost like a short

circuit. The capacitor is charged to 13 volts (10 v + 3 v) with the polarity shown.The capacitor is charged instantaneously to 13 volts because the resistor iseffectively shorted out (RC time constant = 0). The output voltage is

approximately equal to 3 volts because the diode acts almost like a short circuit.

• During the positive portion of the input signal, the input signal (5 v) adds up tothe voltage across the capacitor (13 v) , the diode is reverse biased and it acts likean open circuit. The output voltage is equal to + 18 volts (5 v + 13 v).

OutputInput D1

Input

0 v

Output

+-13 v

0 v

5 v

18 v

-10 v

DC = -2.5 v

DC = 10.5 v3 v

3 v

0 v

0 v

Clampers• A Clamper using a dc supply, which shifts the dc level from -2.5 v to - 10.5 v, is

shown below.

• During the positive portion of the input signal the input signal adds up with the



• During the positive portion of the input signal, the input signal adds up with thevoltage of the battery (5 v + 3 v), the diode is forward biased and it acts almostlike a short circuit. The capacitor is charged to 8 volts (5 v + 3 v) with thepolarity shown. The output voltage is approximately equal to - 3 volts. Thecapacitor charges instantaneously to 8 volts because the resistor is effectively

shorted out (RC time constant = 0).

• During the negative portion of the input signal, the input signal (-10 v) adds upto the voltage across the capacitor (-8 v) , The voltage across the diode is 15 v(8v + 10 v – 3 v), the diode is reverse biased, and it acts like an open circuit.The output voltage is equal to -18 volts (-10 v - 8 v).

OutputInput D1

Input

0 v

Output

+ -8 v

0 v

5 v

-18 v

-10 v

DC = -2.5 v

DC = -10.5 v

-3 v

3 v

0 v

0 v

Clampers• A Clamper using a dc supply, which shifts the dc level from -2.5 v to - 4.5 v, is

shown below.

• During the positive portion of the input signal the voltage of the battery (3 v)



• During the positive portion of the input signal, the voltage of the battery (3 v)subtracts from the input voltage (5 v), the diode is forward biased and it actsalmost like a short circuit. The capacitor is charged to 2 volts (5 v - 3 v) with thepolarity shown. The output voltage is approximately equal to + 3 volts because

the diode is almost like a short circuit. The capacitor charges instantaneously

to 2 volts because the resistor is effectively shorted out (RC time constant = 0).• During the negative portion of the input signal, the input signal (-10 v) adds up

to the voltage across the capacitor (-2 v) , The voltage across the diode is 15 v(10 v + 2 v + 3 v), the diode is reverse biased, and it acts like an open circuit.The output voltage is equal to -12 volts (-10 v - 2 v).

OutputInput D1

Input

0 v

Output

+ -2 v

0 v

5 v

-12 v

-10 v

DC = -2.5 v

DC = -4.5 v

3 v3 v

0 v

0 v

Clampers• A Clamper using a dc supply, which shifts the dc level from -2.5 v to + 4.5 v, is

shown below.

• During the negative portion of the input signal the voltage of the battery (3 v)



• During the negative portion of the input signal, the voltage of the battery (3 v)subtracts from the input voltage (-10 v), the diode is forward biased and it actsalmost like a short circuit. The capacitor is charged to 7 volts (10 v - 3 v) withthe polarity shown. The output voltage is approximately equal to - 3 voltsbecause the diode acts almost like a short circuit. The capacitor charges

instantaneously to 7 volts because the resistor is effectively shorted out (RCtime constant = 0).

• During the positive portion of the input signal, the input signal (5 v) adds up tothe voltage across the capacitor (7 v) , The voltage across the diode is 15 v (5 v +7 v + 3 v), the diode is reverse biased, and it acts like an open circuit. The

output voltage is equal to + 12 volts (5 v + 7 v).

OutputInput D1

Input

0 v

Output

+- 7 v

0 v

5 v

12 v

-10 v

DC = -2.5 v

DC = 4.5 v

-3 v

3 v

0 v

0 v

Clampers• A Clamper using a dc supply, which shifts the dc level from 0 v to – 13 v, and

with a sine wave input is shown below.

• During the positive portion of the input signal the voltage of the battery (3 v)



• During the positive portion of the input signal, the voltage of the battery (3 v)adds up to the input voltage (10 v peak), the diode is forward biased and it actsalmost like a short circuit. The capacitor is charged to 13 volts (10 v + 3 v)during the + peak of the input signal with the polarity shown.

• During the negative portion of the input signal, the input signal (-10 v peak) adds up to the voltage across the capacitor (-13 v) , the diode is reverse biased,and it acts like an open circuit. The negative peak output voltage is equal to -23volts (-10 v - 13 v).

• The output voltage shifts by approximately - 13 volts.

OutputInput D1

Input

Output

+ -13 v

0 v

10 v

-3 v

-10 v

DC = 0 v

DC = -13v

-23 v

3 v

0 v

0 v



Voltage Multiplier Circuits



• Voltage multiplier circuits are used to step up the peak output voltage of therectified signal while employing a relatively low peak input voltage to therectifier (like lower transformer secondary voltage)



rectifier (like lower transformer secondary voltage).

• The multiplication of the peak rectified signal could be by a factor of 2, 3, 4 or more.

• Voltage multiplier circuits are typically made up of diodes and capacitors.

• In the analysis of voltage multiplier circuits, it is assumed that the capacitors are not discharged significantly, and the capacitor voltage does not changesignificantly while it is discharging.

Voltage Multiplier Circuits• A half wave voltage doubler circuit is shown below.

• During the positive half of the secondary voltage, D1 is forward biased and

current flows from the upper portion of the secondary, to C1, to D1, and thent th l ti f th d C1 i h d t th k lt f th



current flows from the upper portion of the secondary, to C1, to D1, and thento the lower portion of the secondary. C1 is charged to the peak voltage of the

secondary (Vp) with the indicated polarity. D2 is reverse biased during thisperiod and does not conduct..

• During the negative half of the secondary voltage, the capacitor voltage (Vp)

adds up with the secondary voltage. D2 is forward biased and current flowsfrom the lower portion of the secondary, to C2, to D2, to C1, and then to theupper portion of the secondary. C2 is charged to twice the peak voltage of thesecondary (2Vp) with the indicated polarity. D1 is reverse biased during thisprocess and does not conduct. The output voltage is twice the negative peak

amplitude of the secondary voltage (-2Vp).• PIV of diodes is 2Vp.

AC Inputvoltage

(sine wave)

+

-

D2Vp

D1

Output

= - 2Vp

+ -

C1

C2

Input

(secondary voltage)Vp=10 v

0 v0 v

0 v

- Vp= -10 v

- 2Vp= -20 v Output = - 2 Vp

I + half

I - half

Voltage Multiplier Circuits• A full wave voltage doubler circuit is shown below.

• During the positive half of the secondary voltage, D1 is forward biased and current

flows from the upper portion of the secondary, to D1, to C1, and then to the lowerti f th d C1 i h d t th k lt f th d (V )



ows o t e uppe po t o o t e seco da y, to , to C , a d t e to t e oweportion of the secondary. C1 is charged to the peak voltage of the secondary (Vp)with the indicated polarity. D2 is reverse biased during this period and does notconduct..

• During the negative half of the secondary voltage, D2 is forward biased and current

flows from the lower portion of the secondary, to C2, to D2, and then to the upperportion of the secondary. C2 is charged to the peak voltage of the secondary (Vp)with the indicated polarity. D1 is reverse biased during this period and does notconduct. The output voltage is twice the peak amplitude of the secondary voltage(2Vp).

• PIV of diodes is 2Vp.

AC Inputvoltage

(sine wave)

+

-

D2

D1

C1

C2

Input

(secondary voltage)Vp=10 v

0 v0 v

0 v

2Vp = 20 v Output = 2Vp

I + half

I - half

-

+

+

-

Output = 2Vp

Vp

Vp


• A circuit which could be used as a voltage doubler, tripler, and quadrupler isshown below.

• During the positive and negative alteration of the secondary voltage, the



g p g y g ,capacitors are charged with the indicated values and polarity.

• The outputs are taken at different points as indicated on the circuit.

AC Inputvoltage

(sine wave)

Vp

D1

+ -

C1

C2

D3D2 D4

C3

C4

2Vp+ -

2Vp+ - 2Vp+ -

Quadrupler (4Vp)Doubler (2Vp)

Tripler (3Vp)

Vp


For the voltage multiplier circuit in the preceding slide:

• During the first positive half of the secondary voltage, D1 is forward biased and current flows from the upper portion of the secondary to C1 to D1 and



g f p f y g , fand current flows from the upper portion of the secondary to C1, to D1, andthen to the lower portion of the secondary. C1 is charged to the peak voltage of

the secondary (Vp) with the indicated polarity. D2 and D4 are reverse biased during this process and does not conduct.

• During the first negative half of the secondary voltage, the voltage across C1(Vp) adds up with the secondary voltage. D2 is forward biased and currentflows from the lower portion of the secondary, to C2, to D2, to C1, and then tothe upper portion of the secondary. C2 is charged to twice the peak voltage of the secondary (2Vp) with the indicated polarity. D1 and D3 are reverse biased

during this process.• During the second positive half of the secondary voltage, the voltage across C2

(2Vp) and the secondary voltage add up while the voltage from C1 (Vp)subtracts from the sum of the two voltages. D3 is forward biased and currentflows from the upper portion of the secondary to C1, to C3, to D3, to C2 , andthen to the lower portion of the secondary. C3 is charged to twice the peakvoltage of the secondary (Vp) with the indicated polarity. D2 and D4 are

reverse biased during this process and does not conduct.


For the voltage multiplier circuit in the preceding slide (continuation):

• During the second negative half of the input signal, the secondary voltageadds up with the voltage across C1 and C3 while the voltage across C2



g g f p g y gadds up with the voltage across C1 and C3, while the voltage across C2subtracts from the sum of the voltages. D4 is forward biased and currentflows from the lower portion of the secondary, to C2, to C4, to D4, to C3, toC1, and then to the upper portion of the secondary. C4 is charged to twice

the peak voltage of the secondary (2Vp) with the indicated polarity. D1 andD3 are reverse biased during this process.

• During the succeeding alterations of the secondary voltage, the capacitorsare charged with the indicated voltage and polarity.

• The PIV of the diodes is 2Vp.

• Adding more diodes and capacitors will introduce more voltage multipliersin the circuit.



Zener Diode Applications


Zener Diode Used For Voltage Regulation

• A circuit using a zener diode for voltage regulation is shown below.



VZ = 5 volts

15 volts

Output Voltage

D1D2

D3 D4

+

-

C1

R1

Electronics 1 Presentation Part 3 (ME)

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Transcript of Electronics 1 Presentation Part 3 (ME)