0
\$\begingroup\$

I want to build something like this. This circuit controls 0-20mA but the circuit have one big problem. If the circuit load e.g 100 Ohm changes to e.g 200 Ohm, then less current is going to flow. To solve that issue, I need to tune in the feedback for the operational amplifier by turning on the potentiometern

enter image description here

So I have an idea! What if I measure the current flow over a small resistor e.g 25 Ohm. I then send the voltage difference back as a feedback signal. If I get +3.3v as the reference value, then I should have 20mA as current flow in the circuit.

Like this. But even if I tune in current flow to 20mA when I have +3.3v as reference, the current flow in the circuit WILL change if I change the circuit load.

enter image description here

Question:

How should I build up this feedback circuit so when I have +3.3V as reference input signal, then 20mA will flow through the 25 ohm resistor, no matter what arbitrary load we have?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ This can be easily done with a Howland current source with suitable references. 0 mA looks like an open circuit, Are you sure you want that? Look under Op Amp examples and swap - Vin+ with Vin - and use a single supply (Rail to Rail Op Amps can't drive 1k pots well, ... except Falstad's they have infinite current so you model them with Ro inside the loop \$\endgroup\$ Mar 8 at 0:38
  • \$\begingroup\$ Do you require one side of the load to be grounded? \$\endgroup\$ Mar 8 at 0:44
  • \$\begingroup\$ I don't know.... \$\endgroup\$
    – MrYui
    Mar 8 at 6:56
3
\$\begingroup\$

If your load resistor Rload can be "floated" so that neither end is grounded, the circuit becomes very simple.

As you say, a current-sense resistor (120 ohm) is used to sample transistor emitter current. Transistor collector current is very slightly smaller. Opamp must be of "rail-to-rail" type, because output voltages and input voltages can range anywhere from 0 - 3.3V:

schematic

simulate this circuit – Schematic created using CircuitLab


However, if one end of your load must be grounded, the circuit becomes a little more complex.

\$\endgroup\$
0
\$\begingroup\$

What you need is a constant current source, but you've built a constant voltage source!

The easiest way forward here is typically building a constant current sink out of your opamp + a transistor (search for "constant current sink opamp"), connect your sensor /variable load between supply voltage and that sink, and measure the voltage at the sink.

\$\endgroup\$
2
  • \$\begingroup\$ No sensor here. \$\endgroup\$
    – MrYui
    Mar 8 at 6:54
  • \$\begingroup\$ That's why I referred to "variable load" there, too! \$\endgroup\$
    – mmmm
    Mar 8 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.