# Circuit question -about current source circuit

Discussion in 'General Electronics Chat' started by deme, Dec 3, 2015.

1. ### deme Thread Starter New Member

Mar 24, 2014
11
1
The following circuit is presented here. Basically it is a circuit that keeps constant current through the OUT terminals even if the resistance of the 'Device Under Test' changes. The current value can be varied (4 - 20mA) using the VR1 potentiometer. The author makes a good job explaining the functionality but I have one question.

What is the purpose of R3?

As I understand it, the voltage at the positive input of the op-amp (pin 3) has to vary from 160mV up to 800mV and this is achieved with the R1, R2, VR1 and R4 resistors. So why is R3 there?

Thank you.

2. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
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to set the 4ma point.

3. ### deme Thread Starter New Member

Mar 24, 2014
11
1
You mean the 160mV? Isn't that what R4 is doing?

Thanks.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
All resistors in this circuit (R2, R3, R4, VR2) are there to "set" the voltage to 0.8V and 0.16V. And if you remove R3 voltage will change.

absf likes this.
5. ### AnalogKid Distinguished Member

Aug 1, 2013
4,685
1,297
You are correct that a different set of values for R2, VR1, and R4 would make R3 unnecessary. Without running all the math, it might be that things work out to standard resistor values more easily with R3 in there. Assuming a relatively high U1 input impedance, R2 and VR1 are in parallel with R3, and you can ignore R1 when calculating the voltages at the ends of VR1.

ak

absf likes this.

Nov 19, 2015
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7. ### hp1729 Well-Known Member

Nov 23, 2015
2,084
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The circuit seems to work pretty good without R3, or D1.

8. ### deme Thread Starter New Member

Mar 24, 2014
11
1
Thank you all for your replies. In case someone wants to see a simulation of this circuit, I drawn the circuit in falstad.com simulator here.