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16-1aElectronicsSemiconductors

• They “collect” a positive electric charge on a small minority of the atoms.

• If a voltage is applied, the electron goes to the positive terminal.

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16-1bElectronicsSemiconductors

• If the semiconductor is doped with atoms that have three valenceelectrons, each dope atom forms three covalent bonds with itsneighboring Si or Ge atoms, resulting in one neighbor atom in thelattice with no atom to bond with.

• If a semiconductor is doped with atoms that have five valenceelectrons, each dope atom forms four covalent bonds with itsneighbors, resulting in one unshared electron in the dope atom,causing the dope atom to donate a free electron.

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16-1cElectronicsSemiconductors

• p-n Junction – when p-type and n-type doping occur next to eachother in the same crystal- Diffusion Current – free electrons from the n-type material combine

with the holes in the p-type material near the junction- Depletion Region – area near the junction

• Drift current – the potential difference creates an electric field thatpushes electrons back toward the n-type material from the p-typematerial

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16-1dElectronicsSemiconductors

Example 1 (FEIM):Which of the following is NOT true for intrinsic semiconductors?

(A) There are holes in intrinsic semiconductors.(B) There are free electrons in intrinsic semiconductors.(C) They make good insulators.(D) Increasing thermal energy increases their conductivity.

Intrinsic semiconductors will carry current, so answer (C) is not true.

Therefore, (C) is the answer.

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16-1eElectronicsSemiconductors

Example 2 (FEIM):In the depletion region of a semiconductor p-n junction, there

(A) is an electric field.(B) are more holes than outside the depletion region.(C) are more free electrons than outside the depletion region.(D) is current perpendicular to the current outside the depletion region.

Answers (B) and (C) are wrong because the depletion region has fewerholes and free electrons than outside the depletion region. Answer (D) isnonsense. However, there is an electric field.

Therefore, (A) is correct.

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16-1fElectronicsSemiconductors

Diode Symbol• P-type – anode• N-type – cathode

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16-2ElectronicsP-N Junction Biasing

• Forward biased• Reverse biased

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16-3ElectronicsDiode Characteristics

• Static forward resistance• Breakdown voltageFor an ideal diode in series witha voltage:

For an ideal diode with zeroresistance in the forward biasdirection and infinite resistance inthe reverse bias direction:

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16-4aElectronicsSpecial Diodes

Zener Diodes• They have a high doping concentration.• Avalanche – the effect of the e– in the depletion region accelerating

and colliding.• For an ideal Zener diode, Vo = 0, rf = 0, and ra = 0.

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16-4bElectronicsSpecial Diodes

Example (FEIM):When a Zener diode suffers breakdown, it

(A) is immediately destroyed.(B) behaves as a reversed biased ideal diode.(C) becomes an open circuit.(D) behaves as a voltage source.

Since the Zener diode is at the Zener voltage in the reverse biasdirection when it suffers breakdown, (D) is correct. Note that answers(B) and (C) are the same (just worded differently), so they both must bewrong.

Therefore, the answer is (D).

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16-5aElectronicsDiode Applications

Half-Wave Rectifiers• Half of a symmetric AC signal gets through• Used in AC-to-DC converters

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16-5bElectronicsDiode Applications

Full-Wave (Bridge) Rectifiers• Current is always going in the

same direction• Used in AC-to-DC converters,

and are more efficient thanhalf-wave rectifiers

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16-5cElectronicsDiode Applications

Clamping CircuitsOutput Voltage:Vout = Vin+ Vp - VmwhereVin = the input voltageVp = the clamping voltageVm = the maximum voltage of the input

For a clamping circuit output with a sinusoidal input:• Average Voltage: Vave = Vp – Vm

• RMS Voltage:

!

Vrms

=1

2

Vm

+Vp"V

m

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16-5dElectronicsDiode Applications

Base Clipper

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16-5eElectronicsDiode Applications

Peak Clipper

Valley Clipper

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16-5f1ElectronicsDiode Applications

Combined Clipper1. Valley clipper + peak clipper

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16-5f2ElectronicsDiode Applications

Combined Clipper (cont.)2. Two Zener diodes in series in the opposite directionThe ideal model for the Zener diode:

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16-5gElectronicsDiode Applications

Example (FEIM):What is the average current through the resistor in the rectifier shown?Assume ideal diodes.(A) 0 A(B) 0.76 A(C) 3.06 A(D) 4.80 A

This is a full-wave rectifier, so

Therefore, (C) is correct.

!

Vave

=2V

peak

"=

(2)(120 V)

"

!

Iave

=V

ave

R=

240 V

"

#

$ %

&

' (

1

25)

#

$ %

&

' ( = 3.06 A

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16-6a1ElectronicsOperational Amplifiers

• An electronic device used to perform mathematical operations onanalog signals.- Two inputs, one output, small current, and large gain

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16-6a2ElectronicsOperational Amplifiers

EIT8 Table 51.1

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16-6bElectronicsOperational Amplifiers

Example (EIT8):

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16-7a1ElectronicsInput Impedance

Example 1 (FEIM):What is the input impedance as seen by the source va of the followingcircuit?

(A) 5 kΩ(B) 7.5 kΩ(C) 10 kΩ(D) 12.5 kΩ

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16-7a2ElectronicsInput Impedance

The input voltage is 15 V. The input current is

The input impedance is the absolute value of the input voltage over theinput current. So the input impedance is

Therefore, (B) is correct.

!

iin

=15 V " 5 V

5 k#= 2 mA

!

Zin

=V

in

iin

=15 V

2 mA= 7.5 k"

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16-8aElectronicsAmplifiers

Therefore,

!

I1

=v

a"v

b

R1

!

I2

=v

o"v

b

R2

!

va"v

b

R1

=v

o"v

b

R2

!

vo

=R

2

R1

va

+ 1+R

2

R1

"

# $

%

& ' vb

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16-8bElectronicsAmplifiers

Noninverting Amplifiers• va = 0• v1 = v2

Since v2 is a voltage divider circuit of the operational amplifier output,

!

v2

=R

1

R2

+ R1

"

# $

%

& ' vout

!

vout

= 1+R

2

R1

"

# $

%

& ' vb

Since vb = v1 = v2, we can substitute and solve for vout:

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16-8cElectronicsAmplifiers

Inverting Amplifiers• vb = 0• v1 = v2 = 0

!

iin

=v

a

R1

Since v1 = v2 = 0,

input current:

current through the feedback resistor:

!

if

=v

out

R2

Since iin = –if,

!

va

R1

= "v

out

R2

vout

="R

2

R1

va

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16-8dElectronicsAmplifiers

Summing Amplifiers• Superposition theorem – currents

can be treated as independent forcestrying to push electrons into (or outof) node A.

For the noninverting amplifier:

!

"if

= i1+ i

2+ i

3+L

!

"v

out

Rf

=v

1

R1

+v

2

R2

+v

3

R3

+L

!

vout

= "Rf

v1

R1

+v

2

R2

+v

3

R3

+L#

$ %

&

' (

!

vout

= 1+R

2

R1

"

# $

%

& ' v1

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16-8eElectronicsAmplifiers

Integrating Amplifiers• Similar to inverting amplifier, feedback current has to be equal and

opposite to the input current.• The output voltage is the voltage across the capacitor.

Assume the initial voltage on the capacitor = 0; the voltageon the capacitor is:

Applying Ohm's law:

!

vout

=1

Ci dt"

!

vout

=1

RCv

indt"

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16-8fElectronicsAmplifiers

Differentiating Amplifiers• Feedback current has to be equal and opposite to the input current.• The output voltage is the voltage across the resistor.

!

Since v1

= v2

= 0: i = Cdv

in

dt

!

Substituting into vout

= "iR gives:vout

= "RCdv

in

dt

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16-8gElectronicsAmplifiers

Low-Pass FiltersThe output voltage divided by the feedback impedance is equal andopposite to the input voltage divided by the input impedance.

!

vout

vin

= "Z

f

Zin

= "1

ZinY

f

= "1

Ri

1

Rf

+ j#C$

% &

'

( )

!

vout

vi

="R

f

Ri1+ j#R

fC( )

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16-8hElectronicsAmplifiers

Example 1 (FEIM):What is the input impedance of the following ideal amplifier?

Therefore, (B) is correct.

!

(A) R1

(B) R3

(C)R

2

R1

+ R3

(D)R

1R

3

R1+ R

2

!

iin

= ia

+ i3

!

ia

= 0, because this is an ideal op amp.

!

Rin

=V

in

iin

=V

in

i3

= R3

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16-8iElectronicsAmplifiers

Example 2 (FEIM):What is the input impedance of the ideal amplifier shown?

Therefore, (A) is correct.

!

(A) R1

(B) R2

(C)R

2

R1

(D)R

1

R1+ R

2

!

Rin

=V

in

iin

!

iin

=V

in"V

a

R1

Va = 0 because this is an ideal op amp.Substituting yields:

!

Rin

= Vin

R1

Vin

"

# $

%

& ' = R

1

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16-8j1ElectronicsAmplifiers

Example 3 (FEIM):The 700 Hz signal shown is applied to the ideal amplifier circuit shown.What will the output signal be?

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16-8j2ElectronicsAmplifiers

Both the DC and AC part are multiplied by –3.Therefore, (B) is correct.

The input current and feedback currents must be equal and opposite, so:

!

iin

=V

in

3 k"= #i

feedback=#V

out

9 k"

!

Vout

Vin

="9 k#

3 k#= "3

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16-8k1ElectronicsAmplifiers

Example 4 (FEIM):Two AC signals V1 and V2 are to be combined such that

The following subtracting amplifier circuit is used. What must be thevalues of R1, R2, R3, and R4?

!

Vout

=3

2V

2"

5

2V

1.

!

(A) R1

= 2 k", R2

= 2 k", R3

= 5 k", R4

= 3 k"

(B) R1

= 2 k", R2

= 4 k", R3

= 5 k", R4

= 3 k"

(C) R1

= 4 k", R2

= 8 k", R3

= 10 k", R4

= 2 k"

(D) R1

= 5 k", R2

= 3 k", R3

= 4 k", R4

= 2 k"

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16-8k2ElectronicsAmplifiers

Solving for Vout yields:

!

Va

= V2

R4

R2

+ R4

"

# $

%

& '

So the possibilities narrow to (A),(B), and (C).Trying R1 = 2 kΩ in the otherrelationship yields

4R4 = 3R2Answer (B) fits because R4 = 3 kΩ,and R2 = 4 kΩ. Plugging in to confirm,

Therefore, (B) is correct.

!

i1

=V

1"V

a

R1

= "iout

= "V

out"V

a

R3

#

$ %

&

' (

!

Vout

=R

1+ R

3

R1

"

# $

%

& '

R4

R2

+ R4

"

# $

%

& ' V2

(R

3

R1

V1

=3

2V

2(

5

2V

1

!

R3

R1

=5

2

!

R1+ R

3

R1

"

# $

%

& '

R4

R2

+ R4

"

# $

%

& ' =

2 + 5

2

"

# $

%

& '

R4

R2

+ R4

"

# $

%

& ' =

3

2

!

Vout

=2 + 5

2

"

# $

%

& '

2

4 + 3

"

# $

%

& ' V2

(5

2V

1=

3

2V

2(

5

2V

1

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16-8l1ElectronicsAmplifiers

Example 5 (FEIM):A 30 mV sinusoidal signal must beinverted, amplified to 6 V (peak), andchopped at 4 V. If the following circuit isused, what are the values of R1 and R2,and the avalanche voltage of the Zenerdiodes Z? Assume –0.7 V forward biasvoltage drop and negligible dioderesistance.

(A) R1 = 1 kΩ, R2 = 20 kΩ, Z = 4 V(B) R1 = 1 kΩ, R2 = 200 kΩ, Z = 4 V(C) R1 = 2 kΩ, R2 = 400 kΩ, Z = 3.3 V(D) R1 = 2 kΩ, R2 = 800 kΩ, Z = 3.3 V

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16-8l2ElectronicsAmplifiers

The amplification without clipping is

This narrows the choices to (B) and (C).When |Vin| > 2 mV, the diodes will be forward biased and reversedbiased respectively, so the voltage across the two diodes in series willbe the Zener voltage plus the forward bias voltage. Thus, the Zenervoltage is 3.3 V.

Therefore, (C) is correct.

!

Vout

Vin

= "R

2

R1

="6 V

30#10"3

V

R2

= 200 R1