# OpAmp circuit

#### azim85bd

Feb 3, 2013
3
Hi All,

Would you please suggest how I can make a division circuit using OpAmp. I can make addition and subtraction, but bit confused how to make a division operational circuit.

Thanks.

#### Laplace

Apr 4, 2010
1,252
The common method for making an analog divider is to obtain a multiplier IC from a place such as Analog Devices and use it in the feedback loop of an op amp . See MT-079: Analog Multipliers, http://www.analog.com/static/imported-files/tutorials/MT-079.pdf for an introduction.

But there are other design approaches. http://www.etasr.com/index.php/ETASR/article/view/239/137 is an example. Attached is a facsimile of the first page of this report, "A Novel Voltage Divider Circuit".

Another article, "Analog divider uses few components", implements an analog divider with a 555 timer. The file size is small enough to be attached.

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• 12212-Figure_1.pdf
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#### Electrobrains

Jan 2, 2012
259
Addition and subtraction are in fact the same thing (just adding negative number(s)).
So are multiplication and division (multiplying the reciprocal (1/X)).

The simplest possible multiplication (division) circuit, is a normal Operational Amplifier circuit, with the feedback or feed-forward resistor made variable.
Use a digital potentiometer, controlled by a small uC in place of R1 in the below circuit!
Placing the potentiometer at R2 makes a strait multiplier.
If multiplying AND dividing, the reciprocal / non reciprocal could be adapted in software for the same circuit.

Gain = - R2 / R1 , Vout= -Vin * (R2 / R1)

Also, for instance a JFET transistor can work as a voltage controlled resistor.

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#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
I think the OP wants to divide one voltage by another voltage. That's what Laplace has assumed as well. He has given a good answer.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Another approach (in theory) is to have a pair of circuits what have logarithmic and exponential response to input voltages.

Use them to take the log of your two inputs, subtract them, then perform the exponential function on them.

In practice, I'm not sure that this approach would be very effective. I would agree that Laplace has provided the best answer so far.

#### Electrobrains

Jan 2, 2012
259
I think the OP wants to divide one voltage by another voltage. That's what Laplace has assumed as well. He has given a good answer.
I think Laplace's circuit looks very interesting, but it has a 555 with a high impedance, first order filtered PWM signal as an output!
Ripple, high source impedance and slow response might be issues. Buffering and additional filtering might be needed (there are already 4 RC-filters in that circuit!).

The idea with a digital potentiometer would allow for a much faster and cleaner signal, with respect to the straight OP input voltage (dividend). Added of course the flexibility of the uC.

The divisor voltage would have to be collected on an A/D or comparator pin of a micro controller (or right away produced in software). Then translated into the right bit-length and format. Then serially output on 2 or 3 other uC pins to set the digipot.

Of course the divisor would also be subject to slow response time and digital quantification, but I suppose that would seldom matter.
As an example where I would use such circuit would be in an analog measure instrument, with accurate scaling of the input range (dividing the input signal entering the instrument).

In fact the simplest division circuit I can think of (without operational amplifiers) would be the digital potentiometer itself. Setting the ratio of a potentiometer would divide the voltage over it in the way the programmer wishes (you could program any mathematical function).

The second solution I mentioned, with the JFET, is a very simple pure analog solution. Here I found the simplest possible division OP-circuit from an old National Semiconductor Application Note:

A more accurate, modern but still simple solution would be this multiplier circuit. It uses MOSFETs instead of JFETs:

Vout=(R2/R1)*v1*v2/Vref

Using Vref as input (and v2 fixed) would give the desired division function. Connecting v1 and v2 together, would give a squaring circuit on top of it.

For deeper explanation and also an example of Steve's log approach (as multiplier), see the Wikibook site itself:
http://en.wikibooks.org/wiki/Electronics/Analog_multipliers

#### Electrobrains

Jan 2, 2012
259
For deeper explanation and also an example of Steve's log approach (as multiplier), see the Wikibook site itself:
http://en.wikibooks.org/wiki/Electronics/Analog_multipliers
p.s.
The Multiplication Circuit based on the logarithm rule log(A*B) = logA + logB in the Wikibook link, should easily be possible to modify to a Division Circuit.
Insert an inverter after one of the logarithmic OP-Amps, before the resistor to the summing OP-Amp.
The input of that branch will be the Divisor!
Mathematically: log(A/B) = logA - logB

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