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Reference question: understanding constant current source circuit

With respect to the linked question, I am seeing a variation in the load current as I change the voltage \$V_2\$ to and fro from \$10 V\$ to \$20 V\$. Is it expected or am I operating the device in it's undefined region?

Sonder
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2 Answers2

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Here's the circuit linked in the question: -

enter image description here

It uses an LT6231 op-amp and, the front page of the data sheet tell us this (inside the red box): -

enter image description here

I am seeing a variation in the load current as I change the voltage V2 to and fro from 10V to 20V

You are exceeding the supply voltage of the device. Don't do this without expecting problems. Generally though, even if you weren't exceeding the supply voltage maximum, power supply rejection on all op-amps isn't infinite.

Andy aka
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  • Dear Sir, i over saw this. Still the variation is there when i change the input from 9 V to 12 V. still the change in current is present. – Sonder Feb 21 '20 at 11:46
  • Because the input voltage that is relevant for controlling the current through the 100 kohm resistor (R1) is between positive rail and non-inverting input of the op-amp. There are many variations on this type of circuit and you have chosen a type of circuit where the applied voltage (to keep good current regulation) is between top positive rail and the op-amp input. – Andy aka Feb 21 '20 at 12:20
  • See [this](https://electronics.stackexchange.com/questions/350491/constant-current-circuit-with-transistor/350493#350493) and [this](https://electronics.stackexchange.com/questions/103787/maintaining-a-constant-current-with-varying-loads/103794#103794) and [this](https://electronics.stackexchange.com/questions/196017/high-side-constant-current-source/196026#196026) and [this](https://electronics.stackexchange.com/questions/207671/conditioning-a-voltage-using-constant-current-for-an-adc-range/207699#207699) for other ideas and examples. – Andy aka Feb 21 '20 at 12:33
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enter image description here

You would not likely make the circuit this way in real life, because the current is dependent on the difference between the supply voltage V2 and V1. Ignoring base current and op-amp non-idealities, Iout = (V2-V1)/R1.

You should have something like a two-terminal shunt reference (or a negative regulator if you are that sort of designer) hanging off of V2 so that the voltage between V2 and the non-inverting terminal is constant.

In your simulation, just move V1(-) to the non-inverting input and V1(+) to connect to V2(+).

For example:

enter image description here

Spehro Pefhany
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