Electrical – How Replica Feedback bias circuit rejects supply noise


I attached the circuit diagram along with a short description of its working principle. However, it's not clear how the circuit rejects supply noise. It will be great if someone can brief about what exactly happens when-

  1. The supply voltage goes up
  2. The supply voltage goes down.


Best Answer

This is a current sink/source based on the load resistor and the control voltage. If the load is a 1.00 K ohm resistor and the control voltage is 1.00 volts, the current sourced is 1.00 mA. It is a linear scaled loop, so the limit is the positive voltage rail (usually +15) minus the mosfet drop voltage.

The current is buffered by the mosfets, but you cannot put more than +12 volts into the op-amp so with 1 K that is a 12 mA limit. With a 100 ohm load you multiply that by 10, so 120 mA is the limit. 10 ohms may be your finite limit as it would top out at 1.20 amps. Zero current is possible with this circuit.

If Vcc is raised to the op-amps limits then your control voltage is limited to Vcc - 2 volts, due to the output stage of the op-amp. You may be able to push the cntrl voltage close to Vcc but not likely closer than 1/2 volt. There are single power supply op-amps like the OP220EZ which will work up to 30 VDC. To prevent ringing when changing loads a 22 ohm resistor at the op-amp output will isolate it from the mosfets high gate capacitance.

Any resistor you put at the op-amp output will slow down the loop correction time. Do not use a resistor over 1K ohm or a major change in load will take many uS to correct.

The circuit is a servo-loop, so does not only it regulate current but small ripples in Vcc are ignored. That is they are cancelled out by the same feedback loop. That is why constant current sources and sinks are used very often. For voltage references, current mirrors, amplifier voltage sources, some battery chargers, etc.