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How to tune this current buffer?

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I recently came across this circuit for a current buffer, and I've been trying to understand how part of it works. When working properly, this circuit should output the same amount of current being input.

I think I understand the output stage of the circuit: that appears to be a voltage-controlled high side current source, but I'm confused about the circuit that drives the output stage.

After simulating the circuit in spice, my version works "okay", but not awesome. I think I need to tune the values of R3, R4, and R5, but I'm not sure how. Also, I don't understand why there is a diode, but I think it's important and changing the diode used greatly changes the output of the circuit.

Assuming D1 is important, how would I go about picking an appropriate diode?

Spice file and picture of circuit is attached.

I agree the circuit looks messy and its gain because of no feedback is dependent upon setting a number of resistor values just right, not a good solution. The diode is important as it matches the Vbe drop across Q3 to help linearise the circuit. There are other ways to do the same thing, it would be best to choose one of them.

Here is a simple example, I am not proposing you use it, but it is much simpler as it depends upon current mirrors and matched transistors, the emitter resistors help deal with any mismatch between the matched pairs. The traces are the input current and output current.


--- Quote from: AJ528 on May 25, 2023, 04:51:38 pm ---Assuming D1 is important, how would I go about picking an appropriate diode?

--- End quote ---
Diode D1 is to track the VBE of Q3, so usually that would be a diode connected Q3, and in the same package.
Usually you would also pick Q1,Q2 in the same package, as their VBE need to nominally track.
Change that in spice to be a Diode connected BC856B

--- Quote from: AJ528 on May 25, 2023, 04:51:38 pm --- I think I need to tune the values of R3, R4, and R5, but I'm not sure how.

--- End quote ---
Spice lets you change things easily.

The current gain here is the ratio of  R5:R1 x ratio of R3:R4

A simple circuit like this will not track well at low currents, or over a wide range of currents, and has modest PSRR - but that may be fine for your needs.

David Hess:
Q1 and Q2 operate as emitter followers, meaning that the voltage at their emitters follows the voltage at their bases.  The voltage at the base of Q1 comes from current Iin through resistor R1.  This voltage is duplicated at the emitter of Q2 because the Vbe voltages of Q1 and Q2 cancel.

The voltage at the emitter of Q2 causes a current through R4, which is duplicate at Q4's collector.  This current flows through D1 and R3.  The Vbe of Q3 is the same as the voltage across D1, so the voltage across R3 is the same as the voltage across R5.  This voltage across R5 causes a current which shows up at the emitter of Q4.

So it is indeed a current to current converter, but the current input to ground is equal to the current out from the positive supply to ground.  The ratios of the various emitter resistors control the current gain.

For best accuracy the various Vbe voltage should be matched, including D1.  Nothing can be done about Q1 and Q2 since they are a PNP and NPN transistor, however D1 could be the base-emitter junction of a BC856B to closely match Q3.  With discrete transistors their Vbe voltages would need to be matched manually but this is not difficult.  D1 and Q3 could be a dual matched transistor.


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