Operational Amplifiers Magic Rules Application Examples Tom Murphy, UCSD tmurphy/phys121/phys121.html

Operational AmplifiersOperational Amplifiers

Magic RulesMagic Rules

Application ExamplesApplication Examples

Tom Murphy, UCSDhttp://physics.ucsd.edu/~tmurphy/phys121/phys121.html

Winter 2012

UCSD: Physics 121; 2012

2

Op-Amp IntroductionOp-Amp Introduction

• Op-amps (amplifiers/buffers in general) are drawn as Op-amps (amplifiers/buffers in general) are drawn as a triangle in a circuit schematica triangle in a circuit schematic

• There are two inputsThere are two inputs– inverting and non-inverting

• And one And one outputoutput• Also power connections (note no explicit ground)Also power connections (note no explicit ground)

+

2

3 4

76

divot on pin-1 end

inverting input

non-inverting input

V+

V

output

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The ideal op-ampThe ideal op-amp

• Infinite voltage gainInfinite voltage gain– a voltage difference at the two inputs is magnified infinitely– in truth, something like 200,000– means difference between + terminal and terminal is

amplified by 200,000!

• Infinite input impedanceInfinite input impedance– no current flows into inputs– in truth, about 1012 for FET input op-amps

• Zero output impedanceZero output impedance– rock-solid independent of load– roughly true up to current maximum (usually 5–25 mA)

• Infinitely Infinitely fastfast (infinite bandwidth) (infinite bandwidth)– in truth, limited to few MHz range– slew rate limited to 0.5–20 V/s

op amp: modelloop amp: modello

4

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Op-amp without feedbackOp-amp without feedback

• The internal op-amp formula is:The internal op-amp formula is:Vout = gain(V+ V)

• So if So if VV++ is greater than is greater than VV, the output goes positive, the output goes positive

• If If VV is greater than is greater than VV++, the output goes negative, the output goes negative

• A A gaingain of 200,000 makes this device (as illustrated of 200,000 makes this device (as illustrated here) practically uselesshere) practically useless

+

V

V+

Vout

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Infinite Gain in negative feedbackInfinite Gain in negative feedback

• Infinite gain would be useless except in the self-Infinite gain would be useless except in the self-regulated negative feedback regimeregulated negative feedback regime– negative feedback seems bad, and positive good—but in

electronics positive feedback means runaway or oscillation, and negative feedback leads to stability

• Imagine hooking the output to the inverting terminal:Imagine hooking the output to the inverting terminal:

• If the output is less than If the output is less than VVinin, it shoots positive, it shoots positive

• If the output is greater than If the output is greater than VVinin, it shoots negative, it shoots negative

– result is that output quickly forces itself to be exactly Vin

+Vin

negative feedback loop

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Even under loadEven under load

• Even if we load the output (which as pictured wants Even if we load the output (which as pictured wants to drag the output to ground)…to drag the output to ground)…– the op-amp will do everything it can within its current

limitations to drive the output until the inverting input reaches Vin

– negative feedback makes it self-correcting

– in this case, the op-amp drives (or pulls, if Vin is negative) a current through the load until the output equals Vin

– so what we have here is a buffer: can apply Vin to a load without burdening the source of Vin with any current!

+Vin

Important note: op-amp output terminalsources/sinks current at will: not likeinputs that have no current flow

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Positive feedback pathologyPositive feedback pathology

• In the configuration below, if the + input is even a In the configuration below, if the + input is even a smidge higher than smidge higher than VVinin, the output goes way positive, the output goes way positive

• This makes the + terminal even This makes the + terminal even moremore positive than positive than VVinin, making the situation worse, making the situation worse

• This system will immediately This system will immediately ““railrail”” at the supply at the supply voltagevoltage– could rail either direction, depending on initial offset

+

Vin

positive feedback: BAD

feedbackfeedback

9

A0

B

Vin VoutV’

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Op-Amp Op-Amp ““Golden RulesGolden Rules””

• When an op-amp is configured in When an op-amp is configured in anyany negative- negative-feedback arrangement, it will obey the following two feedback arrangement, it will obey the following two rules:rules:

– The inputs to the op-amp draw or source no current (true whether negative feedback or not)

– The op-amp output will do whatever it can (within its limitations) to make the voltage difference between the two inputs zero

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Inverting amplifier exampleInverting amplifier example

+

Vin

Vout

R1

R2

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Inverting amplifier exampleInverting amplifier example

• Applying the rules: Applying the rules: terminal at terminal at ““virtual groundvirtual ground””– so current through R1 is If = Vin/R1

• Current does not flow into op-amp (one of our rules)Current does not flow into op-amp (one of our rules)– so the current through R1 must go through R2 – voltage drop across R2 is then IfR2 = Vin(R2/R1)

• So So VVoutout = 0 = 0 VVinin((RR22//RR11) ) = = VVinin((RR22//RR11))• Thus we amplify Thus we amplify VVinin by factor by factor RR22//RR11

– negative sign earns title “inverting” amplifier

• Current is Current is drawn intodrawn into op-amp output terminal op-amp output terminal

+

Vin

Vout

R1

R2

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Non-inverting AmplifierNon-inverting Amplifier

+Vin

Vout

R1

R2

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Non-inverting AmplifierNon-inverting Amplifier

• Now neg. terminal held at Now neg. terminal held at VVinin

– so current through R1 is If = Vin/R1 (to left, into ground)

• This current cannot come from op-amp inputThis current cannot come from op-amp input– so comes through R2 (delivered from op-amp output)

– voltage drop across R2 is IfR2 = Vin(R2/R1)

– so that output is higher than neg. input terminal by Vin(R2/R1)

– Vout = Vin + Vin(R2/R1) = Vin(1 + R2/R1)

– thus gain is (1 + R2/R1), and is positive

• Current is Current is sourcedsourced from op-amp output in this example from op-amp output in this example

+Vin

Vout

R1

R2

Ref:080114HKN Operational Amplifier 15

+

vi

Ra

vov-

v+

Rf

+

vi

Ra

vov-

v+

Rf

R2R1

+

vi vov-

v+

Rf

+

vi

vov-

v+

Rf

R2R1

Noninverting amplifier Noninverting input with voltage divider

ia

fo v

RR

R

R

Rv ))(1(

21

2

Voltage follower io vv

ia

fo v

R

Rv )1(

Less than unity gain

io vRR

Rv

21

2


Ideal Vs Practical Op-AmpIdeal Practical

Open Loop gain A 105

Bandwidth BW 10-100Hz

Input Impedance Zin >1M

Output Impedance Zout 0 10-100

Output Voltage VoutDepends only on Vd = (V+V)

Differential mode signal

Depends slightly on average input Vc = (V++V)/2 Common-Mode signal

CMRR 10-100dB

+

~AVin

Vin Vout

Zout=0

Ideal op-amp

+

AVin

Vin Vout

Zout

~Zin

Practical op-amp

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Summing AmplifierSumming Amplifier

• Much like the inverting amplifier, but with two input Much like the inverting amplifier, but with two input voltagesvoltages– inverting input still held at virtual ground– I1 and I2 are added together to run through Rf

– so we get the (inverted) sum: Vout = Rf(V1/R1 + V2/R2)• if R2 = R1, we get a sum proportional to (V1 + V2)

• Can have any number of summing inputsCan have any number of summing inputs– we’ll make our D/A converter this way

+

V1

Vout

R1

Rf

V2

R2

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Differencing AmplifierDifferencing Amplifier

• The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider:

+

V

Vout

R1

R2

V+

R1

R2

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• The non-inverting input is a simple voltage divider:The non-inverting input is a simple voltage divider:– Vnode = V+R2/(R1 + R2)

+

V

Vout

R1

R2

V+

R1

R2

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• So So IIff = ( = (VV VVnodenode)/)/RR11

+

V

Vout

R1

R2

V+

R1

R2

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– Vout = Vnode IfR2 = V+(1 + R2/R1)(R2/(R1 + R2)) V(R2/R1)

+

V

Vout

R1

R2

V+

R1

R2

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– Vout = Vnode IfR2 = V+(1 + R2/R1)(R2/(R1 + R2)) V(R2/R1)

– so Vout = (R2/R1)(V V)

+

V

Vout

R1

R2

V+

R1

R2

23

Differentiator (high-pass)Differentiator (high-pass)

Vin

Vout

C

R

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Differentiator (high-pass)Differentiator (high-pass)

• For a capacitorFor a capacitor

• So we have a differentiator, or high-pass filterSo we have a differentiator, or high-pass filter

+

Vin

Vout

C

R

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Low-pass filter (integrator)Low-pass filter (integrator)

• IIff = V = Vinin/R, so C·dV/R, so C·dVcapcap/dt = V/dt = Vinin/R/R

– and since left side of capacitor is at virtual ground:

– and therefore we have an integrator (low pass)

+

Vin

Vout

R

C

esempio: 741esempio: 741

26


Frequency-Gain Relation• Ideally, signals are amplified

from DC to the highest AC frequency

• Practically, bandwidth is limited

• 741 family op-amp have an limit bandwidth of few KHz.

• Unity Gain frequency f1: the gain at unity

• Cutoff frequency fc: the gain drop by 3dB from dc gain Gd

(Voltage Gain)

(frequency)f1

Gd

0.707Gd

fc0

1

GB Product : f1 = Gd fc

20log(0.707)=3dB


GB ProductExample: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV

Sol:

Since f1 = 10 MHz

By using GB production equation

f1 = Gd fc

fc = f1 / Gd = 10 MHz / 20 V/mV

= 10 106 / 20 103

= 500 Hz

(Voltage Gain)

(frequency)f1

Gd

0.707Gd

fc0

1

10MHz

? Hz

Gain-Bandwidth product: GBP = Ao BW

Documents

Operational Amplifiers Magic Rules Application Examples Tom Murphy, UCSD tmurphy/phys121/phys121.html