OpAmp

Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002

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The Operational AmplifierThe Operational Amplifier


• Usually Called Op AmpsUsually Called Op Amps

• An amplifier is a device that accepts a varying input signal and An amplifier is a device that accepts a varying input signal and produces a similar output signal with a larger amplitude.produces a similar output signal with a larger amplitude.

• Usually connected so part of the output is fed back to the input. Usually connected so part of the output is fed back to the input. (Feedback Loop)(Feedback Loop)

• Most Op Amps behave like voltage amplifiers. They take an input Most Op Amps behave like voltage amplifiers. They take an input voltage and output a scaled version.voltage and output a scaled version.

• They are the basic components used to build analog circuits.They are the basic components used to build analog circuits.

• The name “operational amplifier” comes from the fact that they The name “operational amplifier” comes from the fact that they were originally used to perform mathematical operations such as were originally used to perform mathematical operations such as integration and differentiation.integration and differentiation.

• Integrated circuit fabrication techniques have made high-Integrated circuit fabrication techniques have made high-performance operational amplifiers very inexpensive in comparison performance operational amplifiers very inexpensive in comparison to older discrete devices.to older discrete devices.

• ii(+)(+), i, i(-)(-) : Currents into the amplifier on the inverting and noninverting lines : Currents into the amplifier on the inverting and noninverting lines

respectively respectively• vvidid : The input voltage from inverting to non-inverting inputs : The input voltage from inverting to non-inverting inputs

• +V+VSS , -V , -VSS : DC source voltages, usually +15V and –15V : DC source voltages, usually +15V and –15V

• RRii : The input resistance, ideally infinity : The input resistance, ideally infinity

• A : The gain of the amplifier. Ideally very high, in the 1x10A : The gain of the amplifier. Ideally very high, in the 1x101010 range. range.• RROO: The output resistance, ideally zero: The output resistance, ideally zero

• vvOO: The output voltage; v: The output voltage; vOO = A = AOLOLvvidid where A where AOLOL is the open-loop voltage gain is the open-loop voltage gain

The Operational AmplifierThe Operational Amplifier


+V+VSS

-V-VSS

vvidid

InvertingInverting

NoninvertingNoninverting

OutputOutput

++

__ii(-)(-)

ii(+)(+)

vvOO = A = Addvvidid

RROO

AARRii

The Four Amplifier TypesThe Four Amplifier Types


DescriptionGain

SymbolTransfer Function

Voltage Amplifieror

Voltage Controlled Voltage Source (VCVS)Av vo/vin

Current Amplifieror

Current Controlled Current Source (ICIS)Ai io/iin

Transconductance Amplifieror

Voltage Controlled Current Source (VCIS)

gm

(siemens)io/vin

Transresistance Amplifieror

Current Controlled Voltage Source (ICVS)

rm

(ohms)vo/iin

VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) Summary


Noninverting ConfigurationNoninverting Configuration

++

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vvinin

++++

--vvOO

vvidid

ii(+)(+)

ii(-)(-)

iiOO

iiFF

RRFF RRLL

RR11

ii11

vvidid = v = voo/A/AOLOL

Assuming AAssuming AOLOL

vvidid =0 =0

Also, with the Also, with the assumption that Rassumption that Rinin = =

ii(+)(+) = i = i(-)(-) = 0 = 0

__vvFF

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vv11

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__

vvLL

++

__

iiLL

Applying KVL the Applying KVL the following equations following equations

can be found:can be found:

vv11 = v = vinin

vvOO = v = v11 + v + vFF = v = vinin+ i+ iFFRRFF

This means that, This means that,

iiFF = i = i11

Therefore: iTherefore: iFF = v = vinin/R/R11

Using the equation to the left the output Using the equation to the left the output voltage becomes:voltage becomes:

vvoo = v = vinin + v + vininRRFF = v = vinin R RFF + 1 + 1

RR11 R R11


VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued

The closed-loop voltage gain is symbolized by AThe closed-loop voltage gain is symbolized by Avv and is found to be: and is found to be:

AAvv = v = voo = R = RFF + 1 + 1

vvinin R R11

The original closed loop gain equation is:The original closed loop gain equation is:

AAvv = A = AFF = A = AOLOL

1 + A1 + AOLOL

Ideally AIdeally AOLOL , Therefore A , Therefore Avv = 1 = 1

Note: The actual value of ANote: The actual value of AOLOL is given for the specific device and is given for the specific device and

usually ranges from 50k usually ranges from 50k 500k. 500k. is the feedback factor and by assuming open-loop gain is infinite:is the feedback factor and by assuming open-loop gain is infinite:

= R= R11

RR11 + R + RFF

AAFF is the amplifier is the amplifier

gain with gain with feedbackfeedback


VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryNoninverting Configuration ContinuedNoninverting Configuration Continued

Input and Output ResistanceInput and Output Resistance

Ideally, the input resistance for this configuration is infinity, but the a Ideally, the input resistance for this configuration is infinity, but the a closer prediction of the actual input resistance can be found with the closer prediction of the actual input resistance can be found with the following formula:following formula:

RRinFinF = R = Rinin (1 + (1 + AAOLOL)) Where RWhere Rinin is given for the is given for the

specified device. Usually Rspecified device. Usually Rinin is is

in the Min the M range. range.

Ideally, the output resistance is zero, but the formula below gives a Ideally, the output resistance is zero, but the formula below gives a more accurate value:more accurate value:

RRoFoF = R = Roo Where RWhere Roo is given for the is given for the

AAOLOL + 1+ 1specified device. Usually Rspecified device. Usually Roo is in is in

the 10the 10ss of of ss range. range.


VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Noninverting Configuration ExampleNoninverting Configuration Example

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__

vvinin

++++

--vvOO

vvidid

ii(+)(+)

ii(-)(-)

iiOO

iiFF

RRFF RRLL

RR11

ii11

__vvFF

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__

vv11

++

__

vvLL

++

__

iiLL Given:Given: vvinin = 0.6V, R = 0.6V, RFF = 200 k = 200 k

RR11 = 2 k = 2 k , A , AOLOL = 400k = 400k

RRinin = 8 M = 8 M , R , Roo = 60 = 60

Find: vFind: vo o , i, iF F , A, Avv , , , R , RinFinF and R and RoFoF

Solution:Solution:

vvoo = v = vinin + v + vininRRFF = 0.6 + 0.6*2x10 = 0.6 + 0.6*2x1055 = = 60.6 V60.6 V i iFF = v = vin in = 0.6 = = 0.6 = 0.3 mA0.3 mA

RR11 20002000 R R11 2000 2000

AAvv = R = RFF + 1 = 2x10 + 1 = 2x1055 + 1 = + 1 = 101101 = 1 = 1 = = 1 = 1 = 9.9x109.9x10-3-3

RR11 20002000 A AOLOL 101 101

RRinFinF = R = Rinin (1 + (1 + AAOLOL) = 8x10) = 8x1066 (1 + 9.9x10 (1 + 9.9x10-3-3*4x10*4x1055) = ) = 3.1688x103.1688x101010

RRoFoF = R = Roo = 60= 60 = = 0.015 0.015

AAOLOL + 1 9.9x10+ 1 9.9x10-3-3*4x10*4x105 5 + 1+ 1


VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting ConfigurationInverting Configuration

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__

RRLL

vvOO

++

--vvinin

++

__

RR11ii11

RRFFiiFFThe same The same

assumptions used to assumptions used to find the equations for find the equations for

the noninverting the noninverting configuration are configuration are also used for the also used for the

inverting inverting configuration.configuration.

General Equations:General Equations:

ii11 = v = vinin/R/R11

iiFF = i = i11

vvoo = -i = -iFFRRFF = -v = -vininRRFF/R/R11

AAvv = R = RFF/R/R11 = R = R11/R/RFF

Input and Output ResistanceInput and Output Resistance

Ideally, the input resistance for this configuration is equivalent to RIdeally, the input resistance for this configuration is equivalent to R11. .

However, the actual value of the input resistance is given by the However, the actual value of the input resistance is given by the following formula:following formula:

RRinin = R = R11 + R + RFF

1 + A1 + AOLOL

Ideally, the output resistance is zero, but the formula below gives a Ideally, the output resistance is zero, but the formula below gives a more accurate value:more accurate value:

RRoFoF = R = Roo

1 + 1 + AAOLOL

Note: Note: = R = R11 This is different from the equation usedThis is different from the equation used

RR11 + R + RFF on the previous slide, which can be confusing.on the previous slide, which can be confusing.Kristin Ackerson, Virginia Tech EEKristin Ackerson, Virginia Tech EESpring 2002Spring 2002

VCVS (Voltage Amplifier) SummaryVCVS (Voltage Amplifier) SummaryInverting Configuration ContinuedInverting Configuration Continued


VCVS (Voltage Amplifier)VCVS (Voltage Amplifier)Inverting Configuration ExampleInverting Configuration Example

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__

RRLL

++

--vvinin

++

__

RR11ii11

RRFFiiFF

Given:Given: vvinin = 0.6 V, R = 0.6 V, RFF = 20 k = 20 k

RR11 = 2 k = 2 k , A , AOLOL = 400k = 400k

RRinin = 8 M = 8 M , R , Roo = 60 = 60

Find: vFind: vo o , i, iF F , A, Avv , , , R , RinFinF and R and RoFoF

vvOO

Solution:Solution:

vvoo = -i = -iFFRRFF = -v = -vininRRFF/R/R11 = -(0.6*20,000)/2000 = = -(0.6*20,000)/2000 = 12 V12 V

iiFF = i = i11 = v = vinin/R/R11 = 1 / 2000 = = 1 / 2000 = 0.5 mA0.5 mA

AAvv = R = RFF/R/R11 = 20,000 / 2000 = = 20,000 / 2000 = 10 10 = R = R11/R/RFF = 2000 / 20,000 = = 2000 / 20,000 = 0.10.1

RRinin = R = R11 + R + RF F = 2000 + 20,000 = 2000 + 20,000 = = 2,000.05 2,000.05

1 + A1 + AOLOL 1 + 400,000 1 + 400,000

RRoFoF = R = Roo = 60= 60 = = 1.67 m 1.67 m

1 + 1 + AAOLOL 1 + 0.09*400,0001 + 0.09*400,000

Note: Note: is 0.09 because using is 0.09 because using different formula than abovedifferent formula than above

ICIS (Current Amplifier) SummaryICIS (Current Amplifier) Summary


Not commonly done using operational amplifiersNot commonly done using operational amplifiers

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LoadLoad

iiinin

iiLL

Similar to the voltage Similar to the voltage follower shown below:follower shown below:

Both these amplifiers have Both these amplifiers have unity gain:unity gain:

AAvv = A = Aii = 1 = 1

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iiinin = i = iLL

vvinin = v = voovvinin

++

__ ++

--vvOO

Voltage FollowerVoltage Follower

1 Possible 1 Possible ICIS ICIS

Operational Operational Amplifier Amplifier

ApplicationApplication

VCIS (Transconductance Amplifier) SummaryVCIS (Transconductance Amplifier) Summary


Voltage to Current ConverterVoltage to Current Converter

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LoadLoadiiLL

RR11ii11

vvinin

++

__

OROR++

__

LoadLoadiiLL

RR11ii11

vvinin

++

__vvinin

++

__


iiLL = i = i11 = v = v11/R/R11

vv11 = v = vinin

The transconductance, gThe transconductance, gmm = i = ioo/v/vinin = 1/R = 1/R11

Therefore, iTherefore, iLL = i = i11 = v = vinin/R/R11 = g = gmmvvinin

The maximum load resistance is determined by:The maximum load resistance is determined by:

RRL(max)L(max) = v = vo(max)o(max)/i/iLL


VCIS (Transconductance Amplifier)VCIS (Transconductance Amplifier)Voltage to Current Converter ExampleVoltage to Current Converter Example

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LoadLoadiiLL

RR11ii11

vvinin

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Given: vGiven: vinin = 2 V, R = 2 V, R11 = 2 k = 2 k

vvo(max)o(max) = = 10 V 10 V

Find: iFind: iL L , g, gmm and R and RL(max)L(max)

Solution:Solution:

iiLL = i = i11 = v = vinin/R/R11 = 2 / 2000 = = 2 / 2000 = 1 mA1 mA

ggmm = i = ioo/v/vinin = 1/R = 1/R11 = 1 / 2000 = = 1 / 2000 = 0.5 mS0.5 mS

RRL(max)L(max) = v = vo(max)o(max)/i/iLL = 10 V / 1 mA = 10 V / 1 mA

= = 10 k 10 k

Note: Note:

• If RIf RLL > R > RL(max)L(max) the op amp the op amp

will saturatewill saturate

• The output current, iThe output current, iLL is is

independent of the load independent of the load resistance.resistance.


VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter


iiFF = i = iinin

vvoo = -i = -iFFRRFF

rrmm = v = voo/i/iinin = R = RFF

++

__

iiFF

iiinin

RRFF

vvOO

++

--

VCIS (Transresistance Amplifier) SummaryVCIS (Transresistance Amplifier) SummaryCurrent to Voltage ConverterCurrent to Voltage Converter


• Transresistance Amplifiers are used for low-power Transresistance Amplifiers are used for low-power applications to produce an output voltage proportional to applications to produce an output voltage proportional to the input current.the input current.

• Photodiodes and Phototransistors, which are used in the Photodiodes and Phototransistors, which are used in the production of solar power are commonly modeled as production of solar power are commonly modeled as current sources.current sources.

• Current to Voltage Converters can be used to convert these Current to Voltage Converters can be used to convert these current sources to more commonly used voltage sources.current sources to more commonly used voltage sources.


VCIS (Transresistance Amplifier)VCIS (Transresistance Amplifier)Current to Voltage Converter ExampleCurrent to Voltage Converter Example

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iiFF

iiinin

RRFF

vvOO

++

--

Given: iGiven: iinin = 10 mA = 10 mA

RRFF = 200 = 200

Find: iFind: iF F , v, voo and r and rmm

Solution:Solution:

iiFF = i = iinin = = 10 mA10 mA

vvoo = -i = -iFFRRFF = 10 mA * 200 = 10 mA * 200 = = 2 V2 V

rrmm = v = voo/i/iinin = R = RFF = = 200200

Power BandwidthPower Bandwidth


The maximum frequency at which a sinusoidal output signal can be The maximum frequency at which a sinusoidal output signal can be produced without causing distortion in the signal.produced without causing distortion in the signal.

The power bandwidth, BWThe power bandwidth, BWpp is determined using the desired is determined using the desired

output signal amplitude and the the slew rate (output signal amplitude and the the slew rate (see next slidesee next slide) ) specifications of the op amp.specifications of the op amp.

BWBWpp = SR = SR

22VVo(max)o(max)

SR = 2SR = 2fVfVo(max)o(max) where SR is the slew rate where SR is the slew rate

Example:Example:

Given: VGiven: Vo(max)o(max) = 12 V and SR = 500 kV/s = 12 V and SR = 500 kV/s

Find: BWFind: BWpp

Solution:Solution: BW BWpp = = 500 kV/s500 kV/s = 6.63 kHz = 6.63 kHz

22 * 12 V * 12 V

Slew RateSlew Rate


A limitation of the maximum possible rate of change of the A limitation of the maximum possible rate of change of the output of an operational amplifier.output of an operational amplifier.

As seen on the previous slide, As seen on the previous slide, This is derived from:This is derived from:

SR = 2SR = 2fVfVo(max)o(max) SR = SR = vvoo//ttmaxmax

Slew Rate is independent of the Slew Rate is independent of the closed-loop gain of the op amp.closed-loop gain of the op amp.

Example:Example:

Given: SR = 500 kV/s and Given: SR = 500 kV/s and vvo o = 12 V (Vo(max) = 12V)= 12 V (Vo(max) = 12V)

Find: The Find: The t and f.t and f.

Solution: Solution: t = t = vo / SR = (10 V) / (5x10vo / SR = (10 V) / (5x1055 V/s) = 2x10 V/s) = 2x10-5-5 s s

f = SR / 2f = SR / 2VVo(max)o(max) = (5x10 = (5x1055 V/s) / (2 V/s) / (2 * 12) = 6,630 Hz * 12) = 6,630 Hz

f is the f is the frequency in frequency in

HzHz

Slew Rate DistortionSlew Rate Distortion


vv

tt

desired output desired output waveformwaveform

actual output actual output because of because of slew rate slew rate limitationlimitation

tt

vv

The picture above shows exactly what happens when the The picture above shows exactly what happens when the slew rate limitations are not met and the output of the slew rate limitations are not met and the output of the

operational amplifier is distorted.operational amplifier is distorted.

SR = SR = v/v/t = m (slope)t = m (slope)

Noise GainNoise Gain


The noise gain of an amplifier is independent of the amplifiers The noise gain of an amplifier is independent of the amplifiers configuration (inverting or noninverting)configuration (inverting or noninverting)

The noise gain is given by the formula:The noise gain is given by the formula:

AANN = R = R11 + R + RFF

RR11

Example 1: Given a noninverting amplifier with the resistance Example 1: Given a noninverting amplifier with the resistance values, R values, R11 = 2 k = 2 k and R and RFF = 200 k = 200 k

Find: The noise gain.Find: The noise gain.

AANN = = 2 k2 k + 200 k + 200 k = 101 = 101 Note: For the Note: For the

2 k2 k noninverting amplifier Anoninverting amplifier ANN = A = AVV

Example 2: Given an inverting amplifier with the resistance Example 2: Given an inverting amplifier with the resistance values, R values, R11 = 2 k = 2 k and R and RFF = 20 k = 20 k

Find: The noise gain.Find: The noise gain.

AANN = = 2 k2 k + 20 k + 20 k = 12 = 12 Note: For the Note: For the

2 k2 k inverting amplifier Ainverting amplifier ANN > A > AVV


Gain-Bandwidth ProductGain-Bandwidth Product

In most operational amplifiers, the open-loop gain begins In most operational amplifiers, the open-loop gain begins dropping off at very low frequencies. Therefore, to make the dropping off at very low frequencies. Therefore, to make the

op amp useful at higher frequencies, gain is traded for op amp useful at higher frequencies, gain is traded for bandwidth.bandwidth.

The Gain-Bandwidth Product (GBW) is given by:The Gain-Bandwidth Product (GBW) is given by:

GBW = AGBW = ANNBW BW

Example: For a 741 op amp, a noise gain of 10 k corresponds Example: For a 741 op amp, a noise gain of 10 k corresponds to a bandwidth of ~200 Hz to a bandwidth of ~200 Hz

Find: The GBW Find: The GBW

GBW = 10 k * 200 Hz = 2 MHzGBW = 10 k * 200 Hz = 2 MHz

Cascaded Amplifiers - BandwidthCascaded Amplifiers - Bandwidth


Quite often, one amplifier does not increase the signal enough Quite often, one amplifier does not increase the signal enough and amplifiers are cascaded so the output of one amplifier is the and amplifiers are cascaded so the output of one amplifier is the

input to the next.input to the next.

The amplifiers are matched so:The amplifiers are matched so:

BWBWSS = BW = BW11 = BW = BW22 = = GBWGBW where, BWwhere, BWSS is the bandwidth of all is the bandwidth of all

AANN the cascaded amplifiers and Athe cascaded amplifiers and ANN is is

the noise gainthe noise gain

The Total Bandwidth of the Cascaded Amplifiers is:The Total Bandwidth of the Cascaded Amplifiers is:

BWBWTT = BW = BWss(2(21/n1/n – 1) – 1)1/21/2 where n is the number of amplifiers where n is the number of amplifiers

that are being cascadedthat are being cascaded

Example: Cascading 3 Amplifiers with GBW = 1 MHz and AExample: Cascading 3 Amplifiers with GBW = 1 MHz and ANN = 15, = 15,

Find: The Total Bandwidth, BWFind: The Total Bandwidth, BWTT

BWBWSS = 1 MHz / 15 = 66.7 kHz = 1 MHz / 15 = 66.7 kHz

BWBWTT = 66.7 kHz (2 = 66.7 kHz (21/31/3 – 1) – 1)1/21/2 = 34 kHz = 34 kHz

Common-Mode Rejection RatioCommon-Mode Rejection Ratio


The common-mode rejection ratio (CMRR) relates to the ability of The common-mode rejection ratio (CMRR) relates to the ability of the op amp to reject common-mode input voltage. This is very the op amp to reject common-mode input voltage. This is very

important because common-mode signals are frequently important because common-mode signals are frequently encountered in op amp applications.encountered in op amp applications.

CMRR = 20 log|ACMRR = 20 log|AN N / A/ Acmcm||

AAcmcm = = AANN

loglog-1-1 (CMRR / 20) (CMRR / 20)

We solve for AWe solve for Acmcm because Op Amp data sheets list the CMRR value. because Op Amp data sheets list the CMRR value.

The common-mode input voltage is an average of the voltages that The common-mode input voltage is an average of the voltages that are present at the non-inverting and inverting terminals of the are present at the non-inverting and inverting terminals of the

amplifier.amplifier.

vvicmicm = v = v(+)(+) + v + v(-)(-)

22

Common-Mode Rejection RatioCommon-Mode Rejection Ratio


ExampleExample

Given: A 741 op amp with CMRR = 90 dB and a noise gain, Given: A 741 op amp with CMRR = 90 dB and a noise gain, AANN = 1 k = 1 k

Find: The common mode gain, AFind: The common mode gain, Acmcm

AAcmcm = = A ANN = 1000= 1000

loglog-1-1 (CMRR / 20) (CMRR / 20) log log-1-1 (90 / 20) (90 / 20)

= 0.0316= 0.0316

It is very desirable for the common-mode gain to be small.It is very desirable for the common-mode gain to be small.

Power Supply Rejection Ratio Power Supply Rejection Ratio


One of the reasons op amps are so useful, is that they can One of the reasons op amps are so useful, is that they can be operated from a wide variety of power supply voltages.be operated from a wide variety of power supply voltages.

The 741 op amp can be operated from bipolar supplies The 741 op amp can be operated from bipolar supplies ranging from ranging from 5V to 5V to 18V with out too many changes to the 18V with out too many changes to the

parameters of the op amp.parameters of the op amp.

The power supply rejection ratio (SVRR) refers to the slight The power supply rejection ratio (SVRR) refers to the slight change in output voltage that occurs when the power change in output voltage that occurs when the power

supply of the op amp changes during operation.supply of the op amp changes during operation.

SVRR = 20 log (SVRR = 20 log (VVss / / VVoo))

The SVRR value is given for a specified op amp. For the The SVRR value is given for a specified op amp. For the 741 op amp, SVRR = 96 dB over the range 741 op amp, SVRR = 96 dB over the range 5V to 5V to 18V.18V.


Problems and SolutionsCalculate the output voltage if V1 = –0.2 V and V2 = 0 V.

Solution:As v2 is zero so that branch can be neglected

i.e. 10k can be neglected

Now the circuit is simple non inverting ampand v0=(-Rf/R1)*v1=-(330k/33k)*(-0.2)

=2v


Problems and SolutionsDetermine the output voltage when V1 = –V2 = 1 V.

Solution:

Vo = (-RF/R1)V1 + (1+(RF/R1))V2'= -1 + (1+1)(20*V1/40)

= -2


Problems and SolutionsCalculate the input voltage if the final output is 10.08 V.

Solution:

Vo/Vin=(1+Rf1/R1)*(-Rf2/R2)*(-Rf3/R3)

10.08/Vin=2*1*30

Vin=0.168 V


Problems and SolutionsCalculate the output voltage if V1 = V2 = 700 mV.

Solution:G= -(500k/250k);

Vo1 (output of first stage op-amp)= -2 V1:

second stage op-amp has inputs as V2 and -(2*V1)

so Vo = - {[(-2*V1)*(500k/100k)]+(V2*(500k/50k))} (Summer)Vo = - {-10V1+10V1}

Vo = 0V;


Problems and SolutionsCalculate the input voltage for this circuit if Vo = –11 V.

Solution:Given circuit is inverting amp so gain is -(Rf/Rin)

vo/vin=-(100/10)-11/vin=-10=>

vin=1.1


Problems and SolutionsCalculate IL for this circuit.

Solution:

IL = -Vo/4k ;Vo = -(4k/2k)*10v;

IL = (20/4K) = 5 mA


Problems and SolutionsCalculate the output voltage if V1 = 0 V and V2 = 0.2 V.

Solution:

Vout=-Rf.V1/R1-Rf.V2/R2=-(330*0/33)-(330*0.2/10)

=-6.6V

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