Discrete devices: FETs - · PDF fileField Effect Transistors ... •Bias Currents...

1

Discrete devices: FETsField Effect Transistors

Current passing from source to drain now controlled by

VOLTAGE at the gate

(rather than by CURRENT into the base as in a BJT).

Gate voltage either

ENHANCES or

DEPLETES the

conduction channel.

JFET = Junction FET

MOSFET = Metal Oxide

Semiconductor FET

MOSFETS have an

insulating layer at gate

so draw less current.

2

Discrete devices: FETs

Field Effect Transistors

There are FOUR kinds of MOSFETs:

Enhancement Mode: N type P type

HB-2 HB-1

M1

+V12V

Q72N3906

Q82N3904

+V12V

+V12V

+V12V

Q62N3906

Q52N3904

+V12V

Q4MTP2955

Q3MTP2955

Q2MTP3055

Q1MTP3055

R510k

R6100

R810k

R7100

R410k

R3100

R2100

R110k

HB-2 HB-1

M1

+V12V

Q72N3906

Q82N3904

+V12V

+V12V

+V12V

Q62N3906

Q52N3904

+V12V

Q4MTP2955

Q3MTP2955

Q2MTP3055

Q1MTP3055

R510k

R6100

R810k

R7100

R410k

R3100

R2100

R110k

Vgs

+-

Vgs +-

Increasing Vgs increases Id.

Depletion : Increasing Vgs decreases Id

Id

Id

HUF 75321IRF5305

4


Enhancement: N type P type

S1

L1

+V12V

+V12V

Q3MTP2955

R810k

R7100

MTP3055V

S1

L1

+V12V

+V12V

R810k

R7100

HUF 75321

IRF5305

5


MTP3055V

S1

L1

+V12V

+V12V

R810k

R7100

5V

GND

gds

HUF 75321

6


MTP3055V

S1

L1

+V12V

+V12V

R810k

R7100

5V

GND

HUF 75321

7


MTP3055V

S1

L1

+V12V

+V12V

R810k

R7100

If after closing S1 (and turning on the LED)

we cut the circuit at the red line:

1. The LED will stay lit

2. The LED will go off

HUF 75321

8


MTP3055V

S1

L1

+V12V

+V12V

R810k

R7100

5V

GND

Gate is floating, but lamp is still lit!!

Lecture 2 – Analog circuits

Seeing the light…..

I

t

IR detection

IR light

Vout

t

Noise sources:

• Electrical (60Hz, 120Hz, 180Hz….)

• Other electrical

• IR from lights

• IR from cameras (autofocus)

• Visible light

~ mV

Q1

OP805

Vout

+V

V19V

RL

What we want: 0 – 5 V DC signal representing the IR amplitude.

Analog circuits – filtering and

detection

IR

detect

DC

block

Amplify Filter

Peak

detect

Q1

OP805

Vout

+V

V19V

RL

Analog circuits – discrete devices: BJT

Application: light detection

Phototransistor:

Acts like BJT except charge carriers

generated by incident light add to the base

current.

In other words, Ic Incident light

Ic+

Vout = Ic * RL

+

Selecting RL ….

For the circuit below,

1) Smaller R increases the DC sensitivity to light

2) Larger R increases the DC sensitivity to light


detection

What is the result of the following:

+

-Z1

Z2

Z2/Z1= 3

Vout

Vout = 1)

2)

3)

4)


detection

IR

detect

DC

block

Amplify Filter

Peak

detect

Analog circuits – DC block

Capacitors:

• Open circuits for DC

• Short circuit at high frequencies


detection

IR

detect

DC

block

Amplify Filter

Peak

detect

Analog circuits: Op-amps

Ideal op-amp: +

-

+Vcc

-Vcc

VoutV+

V-Vout = K (V+ - V-)

K ~ 106 (at DC!)

With feedback to

limit gain:+

-

V+

V-

VoutV- = Vout

Vout = K (V+ - Vout)

Vout(1+K) = KV+

Vout=V+(K/(1+K))

Vout= V+

When negative feedback is

applied to an op-amp, V+ = V-

….if the amplifier is within its

operating range.

The next few slides will show some bad design choices

Analog circuits: Op-ampsEg: Inverting

amplifier.+

-V-

VoutV- = 0

I1 = Vin/Z1

Vout = 0 – Z2I1

Vout= - (Z2/Z1) Vin

Z1

Z2

Vin

I1

Eg 1: Z2 = 100kW

Z1 = 10kW Vout= - 10 Vin

Eg 2: Z2 = 100kW

Z1 = 1 W Vout= - 100,000 Vin !!

Not likely….

10x gain is a

“reasonable” value

Analog circuits: Real Op-amps

+

-V-

Vout

Z1

Z2

Vin

I1

Eg 2: Z2 = 100kW

Z1 = 1 W Vout= - 100,000 Vin !!

Several problems:

• I1 = 1A for Vin = 1 V !! (excessive load for upstream circuitry)

• Gain Bandwidth product ~ 3 MHz. This would limit the

bandwidth of the amplifier from DC up to 30 Hz (i.e. not a very

responsive system!).


Things to consider:

• Input impedance

• Gain Bandwidth product

• Bias Currents

• Voltage limitations

• Output current limitations

+

-V-

Vout

Z1

Z2

Vin

I1

Since V- is a virtual ground, input

impedance seen by Vin is Z1

Things to consider:

• Input impedance


• Bias Currents




+

-V-

VoutVin

Since Op-amp inputs source or sink very little

current (depends on type) , input impedance in this

case is very high. This is a commonly used buffer to

separate your low impedance circuit from a sensitive

source that you need to measure without drawing

current.

Things to consider:

• Input impedance


• Bias Currents



Analog circuits: Real Op-amps+

-V-

Vout

Z1

Z2

Vin

I1

Open loop gain (K)

Frequency 20

100

kHz

Gain-Bandwidth limit (Hz) = Gain * Max. Frequency = CONSTANT

Slew-rate is a similar

limit: it is a limit on

the rate of change of

output voltage

log K

log w

TL082: Gain*Bandwidth = 3 MHz

This means that at a gain of 100, Bandwidth is 30 kHz.


Things to consider:

• Input impedance


• Bias Currents



+

-V-

Vout

Z1

Z2

Vin

I1

Op-amp terminals can act as small current

sources. These Bias Currents can become

large error or offset voltages if the resistors

in the circuit are large.

Eg: 20 nA bias current * 10 MW = 200 mV!


Things to consider:

• Input impedance


• Bias Currents



Op-amp input voltages (V+, V-) must be at least a few

volts away from the power rails (+Vcc, -Vcc).

Applying input voltages equal or near the power rails

will cause the Op-amp to behave unexpectedly.

Rail-to-rail Op-amps are an expensive solution to this

limitation.

+

-

-Vcc

VoutV+

V-

+Vcc


Things to consider:

• Input impedance


• Bias Currents


• Output current / voltage limitations

Op-amp output terminals can only provide a few mA

of current. Motors, lamps and similar high current

devices cannot typically be driven by a normal OP-

amp. High power Op-amps exist that can provide

much higher current levels. Output voltage range is

also limited within a few volts of the power rails.

+

-

-Vcc

VoutV+

V-

+Vcc

Q: We wish to amplify a 100 kHz, 10mV p-p sine wave, as much as

possible. Which is the best circuit?

Vout+

TL082

9V

100kHz

V2-5m/5mV

9V

+

TL082

2k

100k100k

2k

-9V

Vout+

TL082

9V

-9V-9V

+

TL082

9V

100kHz

V2-5m/5mV

9V

+

TL082

10k

120k

10k

120k120k

10k

-9V

Vout+

TL082

9V

-9V-9V

+

TL082

9V

100kHz

V2-5m/5mV

9V

+

TL082

10k

100k

10k

120k120k

10k

1)

2)

3)


Application: amplifier stages

Vout

Vin

R6

1k

R5

30k

R41k

R330k

R21k

R110k

U3

TL082U2TL082

U1TL082

Total gain: 30*30*11 = 9900

Input impedance = high

Max Bandwidth = 100 kHz


detection

IR

detect

DC

block

Amplify Filter

Peak

detect

Analog circuits: Filters

To understand filters you should first understand the difference

between the TIME DOMAIN and FREQUENCY DOMAIN

Which is the correct match between the following time-domain (left)

signals and their Fourier Transforms (right) ?

a)

b)

c)

d)

i)

ii)

iii)

iv)

1) a-i, b-iv, c-ii, d-iii

2) a-ii, b-iv, c-iii, d-i

3) a-iii, b-iv, c-i, d-ii

4) a-iv, b-ii, c-iii, d-i

t f


Demo: Frequency generator and Spectrum analyzer

Frequency

Generator

Spectrum

Analyzer

ground

signal

Frequency

Generator

Spectrum

Analyzer

ground

Z1

Z2


Vin Vout

“Transfer Function” = Vout/Vin = H(w)

So: Vout (w) = H(w)*Vin(w)

This is all in terms of w since, in general,

impedances are functions of w.

Similar to voltage divider: except w dependent.

Zcap = 1/j wC

Zind = j wL

Zres = R

Frequency

Generator

Spectrum

Analyzer

ground

Z1

Z2


Vin Vout

Vout (w) = [Vin(w)/(Z1+Z2)]*Z2

So: H(w) = Z2/(Z1+Z2)

For resistors, this is just the well known

voltage divider: R2/(R1+R2)

21

2)(ZZ

ZH

+w

Frequency

Generator

Spectrum

Analyzer

Z1

Z2

Analog circuits: FiltersNow plug in a resistor and a capacitor:

VoutVin

R133 k

C1100 nF

100 k

50 nF

Z2 = 1/j wC

Z1= R

RCj

CjR

CjH

w

w

ww

+

+

1

1

/1

/1)(

Frequency

Generator

Spectrum

Analyzer

Z1

Z2


For low frequencies (small w), H = 1

For high frequencies (large w), H 0

This is a LOW PASS FILTERRCj

Hw

w+

1

1)(

VoutVin

R133 k

C1100 nF

f0 = 1/(2pRC) = 30 Hz

100 k

50 nF

At w = j/RC, (the pole of this function) H is infinite.

At the real value w = 1/RC, H begins to decrease in

amplitude.

Analog circuits: Transfer Functions

Bode plots: a graphical representation of frequency response on

logarithmic axes.

dBV = 20log10(|V|)

(20 is used instead of 10 so the

result will represent power ~ V2)

-3 dB = ½ as much power as 0 dB

Vout is 1/√2 of Vin at -3dB

Vertical axis:

log10(f)Log of frequency is used to ensure

linear plots from 1/f or 1/fn functions

Horizontal axis:

Pole: 1/(1+jw/w0)

Zero: (1+jw/w0)

-20 db/decade in amplitude after w0, -90 phase

+20 db/decade in amplitude after w0, +90 phase


RCjH

ww

+

1

1)(

-3dB, 30 Hz - 45 deg, 30 Hz

Bode Plot:

-20db/decade / pole

-90 deg / pole

Analog circuits: Transfer Functions

Bode plots: a graphical representation of frequency response on

logarithmic axes.

Pole: 1/(1+jw/w0)

Zero: (1+jw/w0)

-20 db/decade in amplitude after w0, -90 phase

+20 db/decade in amplitude after w0, +90 phase

)1)(1()(

1122

21

CRjCRj

CRjH

ww

ww

++-

Zero at w=0

Pole at w=1/(R2C2) Pole at w=1/(R1C1)

Analog circuits: Simple Pole

RCjH

ww

+

1

1)(

-3dB, 1/RC - 45 deg, 1/RC

Bode Plot:

-20db/decade

-90 deg

Analog circuits: Simple Zero

RCjH ww +1)(

+3dB, 1/RC +45 deg, 1/RC

Bode Plot:

+20db/decade

+90 deg


RCjH

ww

+

1

1)(

VoutVin

R133 k

C1100 nF

100 k

50 nF

Vout

Vin

Gnd

Ch1

Ch0

f0 = 1/(2pRC) = 30 Hz


VoutVin

R1

C1

How does this circuit affect the following waveform:

1)

2)

3)

4)

Analog circuits: Active Filters

C2R2

R1

C1

U1TL082

Z2

Z1

H(w) = - Z1/Z2

2

22

1

CjRZ

w+

Active Band Pass:

1

1

1

1 )1

( -+ CjR

Z w


C2R2

R1

C1

U1TL082

Z2

Z1

)1)(1()(

1122

21

CRjCRj

CRjH

ww

ww

++-

Active Band Pass:

Zero at w=0



)1)(1()(

1122

21

CRjCRj

CRjH

ww

ww

++-

Idealized Bode Plot:

Zero at w=0


w

20log(|H|)

0 dB

20log(R1/R2) dB

1/(R2C2) 1/(R1C1)

(1)

(2) (3)

(1)

(1)

(1)

(2)

(2) (3) 20db/decade


)1)(1()(

1122

21

CRjCRj

CRjH

ww

ww

++-

Idealized Bode Plot:

Zero at w=0


w

phase(H)

-180

1/(R2C2) 1/(R1C1)(1)

(2) (3)

(1)

(1)

(1)

(2)

(2) (3)

-90

-270


)1(

)1()(

22

11

CRj

CRjH

w

ww

+

+

Which best describes the amplitude

response (Bode plot) of the following

transfer function (R1C1 > R2C2):

1)

1/ R1C1

1/ R1C1

1/ R1C1

1/ R1C1

1/ R2C21/ R2C2

1/ R2C2 1/ R2C2

0db 0db

0db 0db

2)

3) 4)


detection

IR

detect

DC

block

Amplify Filter

Low

pass

Band Pass

C2R2

R1

C1

U1TL082 C2R2

R1

C1

U1TL082 C2R2

R1

C1

U1TL082

Use multiple stages to get steeper filter roll-offs…

Htot(w) = H1(w) * H2 (w) *H3(w) Remember –20dB/dec for each POLE

Debugging Circuits

Learn to systematically check your circuits:

• Power rails:

– Check that 15V is really 15V; if not, localize the component

that is shorting the power rail.

• Physical check:

– Check pinouts, missing/loose wires, etc.

• Isolate stages where possible

– Check output of stage 1 – if ok plug into stage 2 and see if

stage 1 output is degraded.

– If ok check output of stage 2 etc

• Keep wiring TIDY!

R1

VinVout

Vee-12V

Vcc

+12V

+

U2UA741

C1

D21N914

Vee

-12V

Vcc+12V

D1

1N914

+

U1UA741

DISCUSSION Q: What is wrong with this circuit, when

implemented with a TL082 OP-AMP?

Discrete devices: FETs - · PDF fileField Effect Transistors ... •Bias Currents...

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Transcript of Discrete devices: FETs - · PDF fileField Effect Transistors ... •Bias Currents...