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How to Derive the Differential seus amigos.
Amplifier Transfer Function The Instrumentation Amplifier (IA) resembles the differential amplifier, with the

How to Derive the Transfer main difference that the inputs are buffered by two Op Amps. Besides that, it is Privacy - Terms

Function of the Inverting designed for low DC offset, low offset drift with temperature, low input bias
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Summing Amplifier currents and high common-mode rejection ratio. These qualities make the IA very
How to Derive the Summing useful in analog circuit design, in precision applications and in sensor signal Tweet
Amplifier Transfer Function processing.

How to Derive the Inverting Why this website?

Amplifier Transfer Function
It's an Analog World by
Design

SPICE/EDA Tools

Semiconductor Manufacturers
How to Apply Thevenin’s
Electronics Websites
Theorem – Part 2. Nested
Notable Articles in Electronics Thevenin Sources Method
Design
Figure 1
Recent Posts
Sites I follow Figure 1 shows one of the most common configurations of the instrumentation
amplifier. Its clever design allows U1 and U2 operational amplifiers to share the RMS Value of a Trapezoidal
STMicroelectronics current through the feedback resistors R5, R6 and RG. Because of that, one single Waveform Calculator
resistor change, RG, changes the instrumentation amplifier gain, as we will see
Intel Why is the Op Amp Gain-
further. RG is called the “gain resistor”. If the amplifier is integrated on a single Bandwidth Product Constant?
The Verge monolithic chip, RG is usually left outside so that the user can change the gain as
How to Calculate the RMS
he wishes. One example of such instrumentation amplifier is Texas Instruments’
EDN Value of an Arbitrary
INA128/INA129.
Waveform
Analog Devices

To minimize the common-mode error and increase the CMRR (Common-Mode Design a Unipolar to Bipolar
New Products at Texas
Converter the Easy Way with
Instruments Rejection Ratio), the differential amplifier resistor ratios R2/R1 and R4/R3 are
Microsoft Mathematics
equal. (See The Differential Amplifier Common-Mode Error Part 1 and Part
EEWeb
2 for more on this matter.) Tenma 72-7745 Multimeter
Electronic Products Review

PCB Prototype Another potential error generator is the input bias current. Although, in most Open-loop, Closed-loop and
analysis, the input current into an Op Amp is considered zero, in reality that is not Feedback Questions and
the case. A small input current flows into the Op Amp inputs and is converted into Answers

voltage by the input resistors. If the resistors are not equal, the voltage difference How to Derive the RMS Value
between the two generates an offset, which is amplified and transmitted at the of a Sine Wave with a DC
You may also be
circuit output. Because of that, R1 is designed to be equal with R3. Similarly, R2 Offset
interested in …
equals R4.

How to Design a Circuit from Most Popular

its Transfer Function Graph
How do we derive the instrumentation amplifier transfer function?

Injecting AC into the DC It is well known that the instrumentation amplifier transfer function in Figure 1 is How to Derive the RMS Value
Power Supply Rail of Pulse and Square
Waveforms
How to Derive the Differential
Amplifier Transfer Function (1) How to Derive the RMS Value
of a Sine Wave with a DC
Derive the Transfer Function
Offset
of the Common Collector
Amplifier with Thevenin's when R5 = R6, R2 = R4 and R1 = R3. How to Derive the RMS Value
Theorem of a Triangle Waveform

A Summing and Differential

The proof of this transfer function starts with the Superposition Theorem. Let’s
An Op Amp Gain Bandwidth
Amplifier with One Op Amp make V2 zero by connecting the U2 input to ground, and let’s calculate Vout1 (see Product
Figure 2).
Solving the Differential How to Derive the
Amplifier - Part 1 Instrumentation Amplifier
Transfer…
How to Calculate the RMS
Value of an Arbitrary An ADC and DAC Least
Waveform Significant Bit (LSB)

The Transfer Function of the

Non-Inverting Summing…

How to Derive the Inverting

Amplifier Transfer Function

How to Derive the Differential

Amplifier Transfer Function
Figure 2
The Common-Collector
Amplifier Input and Output…
The calculation of Vout1 starts from the differential amplifier transfer function
shown in equation (2). U3 is in a differential configuration. If we note the voltage Popular posts by Top 10
levels at U1 and U2 outputs with V11 and V12 respectively, Vout1 can be written as plugin

(2) Categories

Analog Design
For the proof of equation (2) see The Differential Amplifier Transfer Function Calculators
on this website.
Differential Amplifier

To determine V11 and V12 we note that, if V2 is zero, the node between RG and R6 Education
is a virtual ground. This is because U2 sets its output at such a level, so that its
Electronic Circuits Examples
inverting input equals the non-inverting input potential. With this observation, one
would realize that U1 is in a non-inverting amplifier configuration, with its feedback Mixed-Signal Design
resistor network R5 and RG connected to a virtual ground. Operational Amplifier
Formulas
Therefore, V11 can be deduced from the non-inverting amplifier transfer function:
Opinion

Power Supply
(3) RMS

Summing Amplifier

In order to calculate V12, let’s observe that the current that flows through R5 and Superposition Theorem
RG, IG, is the same as the current through R6. That is because there is no other
Thevenin's Theorem
current path. The currents that flow into U1 and U2 inputs are too small to be
taken into consideration. Mathematically, we can write that the current through R5 Transistor Circuits
and RG equals the current through R6 as in equation (4). Since the node between
Waveforms
RG and R6 is at zero volts, V11 appears as a voltage drop on R5 and RG in series.
Also, V12 is the voltage drop on R6, forcing the output of U2 to be driven below
ground. Recent Comments

Adrian S. Nastase on Bipolar

(4) to Unipolar Converters Based
on a Summing Amplifier
Configuration
Therefore, V12 is
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loop, Closed-loop and
Feedback Questions and
(5) Answers

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Replacing V11 and V12 in equation (2), Vout1 becomes a Bipolar to Unipolar
Converter to Drive an ADC

Arvind Agrahari on Design a

(6) Bipolar to Unipolar Converter
to Drive an ADC

Aarohi on Open-loop, Closed-

After calculations, and taking into consideration that R5 = R6, the result for Vout1 loop and Feedback Questions
is as in equation (7). and Answers

Josef Burda on Bipolar to

Unipolar Converters Based on
(7) a Summing Amplifier
Configuration

Adrian S. Nastase on Design

For the second part of the Superposition Theorem, let’s restore V2 and let’s make a Bipolar to Unipolar
V1 zero. We will note the output voltage with Vout2, and with V21 and V22 the Converter to Drive an ADC
output voltage of U1 and U2 respectively (see Figure 3).
Adrian S. Nastase on
Injecting AC into the DC
Power Supply Rail

Adrian S. Nastase on
Summing Amplifier Calculator

Adrian S. Nastase on The

Common-Collector Amplifier
Input and Output Resistance
– The Proof

Figure 3

Vout2 depends on V21 and V22 in a similar manner as Vout1 in equation (2).

(8)

This time, U2 is in a non-inverting configuration, so that V22 can be written as a

function of V2 as in (9).

(9)

The circuit is symmetric, so we can write a similar equation for V21 and V22 as
equation (4) for V11 and V12.

(10)

and therefore, V21 is

(11)

Replacing V21 and V22 in equation (8) and after calculations, we find Vout2 as in
the following expression.

(12)

All we need to do now is to add Vout1 and Vout2 to find the instrumentation
amplifier transfer function. The result is given in equation (13).

(13)

Q. E. D.

>>> <<<

If we take a closer look at the instrumentation amplifier transfer function, we note

that, if RG is not connected and R2 = R1, the circuit gain becomes one. Changing
one single resistor, RG, results in large gain variations, so it gives the analog
designer flexibility in his application. This is the reason why the IC manufacturers
choose not to integrate RG on the monolithic chip, and also choose to make R1,
R2, R3 and R4 equal. As opposed to the differential amplifier, where the user has
to change at least two resistors to change the gain, in instrumentation amplifiers
one resistor does the job, bringing elegance and simplicity in the analog design.

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Related Posts

How to Derive the Differential Amplifier Transfer Function

How to Derive the Transfer Function of the Inverting Summing
Amplifier
How to Derive the Summing Amplifier Transfer Function
How to Derive the Inverting Amplifier Transfer Function

 Analog Design, Differential Amplifier, Superposition Theorem

 differential amplifier, instrumentation amplifier, op amp (opamp) formulas, operational
amplifier, proof, transfer function

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 How to Apply Thevenin’s Theorem – Part 1, Solving Circuits with
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26 thoughts on “How to Derive the Instrumentation

Amplifier Transfer Function”

arun
December 14, 2010 at 9:46 am | Reply

you did not solve equation number 6.how did u obtain equation 7 after
solving equation 6

Adrian S. Nastase
December 14, 2010 at 6:48 pm | Reply

Arun, here are the calculations:

First, factorize V1*(1+R5/RG),

Vout1 = (R2/R1)*(V1*(1+R5/RG)*(1+R6/(R5+RG)))

Then, introduce 1 in each fraction,

Vout1 = (R2/R1)*(V1*(RG+R5)/RG*(R5+RG+R6)/(R5+RG))

Simplify RG+R5
Vout1 = (R2/R1)*V1*(R5+RG+R6)/RG

And, because R5=R6,

Vout1 = (R2/R1)*V1*(RG+2R5)/RG

Distribute RG, and this is the final result:

Vout1 = V1*(R2/R1)*(1+2R5/RG)

Let me know if you need anything else.

Grant
February 17, 2011 at 4:43 am | Reply

I looked at the derivation for the transfer function of the differential

amplifier, as linked, but the transfer function proven on that page looks
nothing like equation 2. How did you derive equation 2 of this page from
the differential amplifier’s transfer function?

Adrian S. Nastase
February 17, 2011 at 5:57 am | Reply

Grant, the two equations are identical, if R1 = R3 and R2 = R4 as

stated two paragraphs above.

Equation (2) in this article is Vout1 = R2/R1 *(V11-V12).

Equation (1) in How to Derive the Differential Amplifier Transfer
Function is Vout = V1 * R2/(R1+R2) * (1+R4/R3) – V2 * R4/R3.

If R1 = R3 and R2 = R4 then
R2/(R1+R2) * (1+R4/R3) = R2/(R1+R2) * (1+R2/R1) = R2/R1,
and
R4/R3 = R2/R1

The inputs of the differential amplifier, which is the instrumentation

amplifier output stage, are V11 instead of V1 and V12 instead of
V2. We also note Vout with Vout1. Therefore, from the differential
amplifier transfer function, as applied to the instrumentation
amplifier output stage we get

Vout1 = V11 * R2/(R1+R2) * (1+R4/R3) – V12 * R4/R3 = V11 *

R2/R1 – V12 * R2/R1 = R2/R1 * (V11 – V12),

exactly as shown in equation (2).

luke
November 19, 2011 at 12:43 am | Reply

Thank you. I was looking at the same thing. This clarifies.

Great article by the way.

Hasrat Ali Anjum

November 25, 2017 at 1:43 pm | Reply

Ley us U3 non inverting terminal voltage Vp then

Vp=V11*R2/(R1+R2). {by voltage divider rule}
& Inverting terminal is connected R3 with V12 voltage
Now. R4=R2,R3=R1,
Apply superposition theorem
(1). Vp=0 then U3 act like a inverting amplifier
So Vout(1)’= –(R4/R3)V12,=–(R2/R1)V12,
(2) V12=0 then U3 act like a non-inverting amplifier so,
Vout(1)”=Vp*(1+R4/R3)=(1+R2/R1)Vp
=(1+R2/R1)(R2/R1+R2)*V11
Vout(1)” = V11*(R2/R1)
Vout1=Vout(1)’+Vout(1)”
=R2/R1*(V11–V12)

Robert
July 8, 2011 at 10:04 am | Reply

Hi,
If input voltages V1 and V2 are the same, does it mean that output
voltage equals zero volt?

Adrian S. Nastase
July 10, 2011 at 12:06 am | Reply

Yes, it will be zero. But nothing is a perfect zero in this Universe.

You will still have a few millivolts at the amplifier output due to
offset, or due to V1 and V2 not being perfectly equal.

Gilbert
November 14, 2011 at 8:13 am | Reply

Hi, if U3 is up side down, means R4 connects to ground and R2 connects

to Vout and U3 has the opposite sign. Will all the equation be not
changed?
for example, will the equation 2 become Vout1=R2/R1(V12-V11)?
Thank you

Adrian S. Nastase
November 15, 2011 at 5:13 am | Reply

The notations are just a convention. Your U3 being turned upside

down, is the same as saying “let’s call the upper transistors R3 and
R4 and the lower transistors R1 and R2, and let’s switch V11 and
V12 labels between them”. The resistor ratio is the same, since
R4/R3 = R2/R1. Because we switched V11 and V12, then, yes,
Vout1 = R2/R1 (V12-V11).

htsh
October 6, 2012 at 12:00 pm | Reply

Adrian, In fig 2 applying KCL at node between Rg and R6, the current
direction should be towards that node. Current should flow out from both
opamps. Only then will equation 10 be valid, right?

Adrian S. Nastase
October 10, 2012 at 8:36 pm | Reply

No, not right. The current that flows from U1 output through R5
and RG is the same current that flows through R6 and into the
output of U2. Current does not flow out from both Op Amps. If
flows out from U1 and into U2 when V1 is greater than V2 as in
figure 2. Equation 10 refers to figure 3 not 2. In figure 3, V2 is
greater than V1 and current flows from U2 and into U1.

Suleiman
April 25, 2015 at 10:28 am | Reply

what is the significance of output voltage in the instrumentation amplifier?

Adrian S. Nastase
April 26, 2015 at 9:22 pm | Reply

As equation 13 shows, Vout is directly proportional with the

difference between the amplifier two inputs.

ashish burande
October 4, 2015 at 6:57 am | Reply

how to design an instrumentation amplifier to get 2v output from 1 and

0mv input with designing step

Adrian S. Nastase
February 25, 2016 at 6:27 am | Reply

I think my article shows that. You need to choose an

instrumentation amplifier (go to digikey.com) and look in the data
sheet for the transfer function. Should be similar with what I
describe here. You need to calculate a resistor value to set the
gain. 1 mV is a small signal. You need to choose a low noise
amplifier with low offset.

saif
January 19, 2016 at 7:55 pm | Reply

How to decide the value of the resistor R1,R2,R3,R4,R5,R6 ?

What I know the value should be the same.
Is it if we put the too high or too small it will affect the gain ?
What is the best range value for the resistor, because my input is in mV
from the Wheatstone Bridge.
and for the Vout VALUE, is it we need to evaluate by our own value to
calculate the value of RG?

Adrian S. Nastase
January 29, 2016 at 11:16 pm | Reply

Look at the last paragraph of this article. Choose all resistors

equal, with a value of 1kohm to 10kohm, and then calculate RG to
give you the desired gain. The gain is shown in Eq 1.

Saif
February 16, 2016 at 10:04 pm | Reply

I use 200kohm for every resistors. Is the value make sense

?

I am now in the process of designing signal conditioning

circuit for thermistor.
The temperature range is between 0-100 deg C.
So I make the maximum temperature which is 100 deg C as
maximum output voltage which is 5V.

The value for V1 measured is 131.35mV

the value for V2 measured is 27.41mV

Then I calculate using your equation by substitute the Vo as

5V
and I find the value of RG is about 8491ohm.

Is my calculation is correct or not ?

Saif
February 16, 2016 at 12:43 am | Reply

Hello.
Very helpful articles.
Im in the process of design my signal conditioning circuit for thermistor. I
do need this amplifier since the output from Wheatstone Bridge is in mV.
Is it make sense the resistor I used for this amplifier is all 200k ohm ?
Is it too big ?

Adrian S. Nastase
February 25, 2016 at 5:58 am | Reply

I would use 10kohm resistors.

Marwah
May 1, 2016 at 8:47 am | Reply

How to drive common mode gain of the first stage?

Adrian S. Nastase
May 19, 2016 at 5:35 am | Reply

I don’t understand this question. You need to reformulate it.

Tasnim Taleeb
October 27, 2017 at 7:34 pm | Reply

Prove that the gain of the INA 126 amplifier is equal to ? ?

?? ????/??

john
November 16, 2017 at 2:38 pm | Reply

hello,how to design an intrumentations amplifers to satisfy a fixed

differential voltage gain of Af=500? please reply me as soon as possible

Adrian S. Nastase
November 30, 2017 at 10:47 pm | Reply

You can use INA126 (Texas Instruments). RG is the gain resistor.

With RG = 162 ohms, 1% tolerance, the gain is 500.

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