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I want to use an opamp in a current sink configuration in order to control the emitter-base current of a PNP transistor via a digital-analog pin from a microcontroller.

The circuit is as shown below.

Now, normally, I would use an inexpensive single-channel, single-supply SOT-23 opamp like the MIC6211YM5-TR that can drive 50mA and allows a supply voltage of up to 32V.

In this setting, I can supply the opamp from the same voltage as the load R_LOAD, in this case ca. 9V DC.

Such opamps are extremely hard to buy these days, while similar opamps that work up to 5.5 DC supply are more available.

Here is my question:

If I change the Vcc of the opamp to 5V while the supply voltage for the load (at the emitter of Q28) is still 9V, will any current flow through the output of the opamp even if the DAC voltage is zero?

In other words, how will the opamp output of a current sink behave when its supply voltage and the supply voltage of the load differ?

Will the opamp output be at 5V or just be a high-impedance sink to ground?

If it was at ca. 5V, could a current would flow from +9V via Q8 emitter->base to the +5V supply (maybe even compromising that supply voltage and / or harm the LDO regulator)?

Edit 1: I was also considering the use of an old-fashioned TL431, which is widely available (also in SOT-23), as per the attached sketch below. The downside of the TL431 is that (1) the internal reference voltage is set to ca. 2.5V, so the effective DAC range is only from 2.5...3.3V, and (2) it is unclear to me how reliable the TL431 would switch off the PNP transistor.

Circuit (corrected)

Alternative with TL431

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  • \$\begingroup\$ Thanks for pointing this out! I messed up the pins when extracting the circuit from a larger schematic. Just fixed. \$\endgroup\$ Commented May 18, 2022 at 13:32

2 Answers 2

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First, let's take a look at the output section of an op amp:

schematic

simulate this circuit – Schematic created using CircuitLab

It's not exactly the same as above but the idea is the same: The op amp has a push-pull output.

As you can see, the output section is basically a buffer i.e. unity-gain amplifier. So, unless the output hits the positive supply rail, changing supply voltages will not change the current flowing through the output because the output voltage will not change.

The negative supply input is tied to ground through a 100R resistor and the voltage drop across it is fed back to the non-inverting input. The current flowing through it will be too close the output sink current (neglecting the bias currents of the input section of the op amp as they are in uA or even nA range).

More importantly, the op amp is running in open loop configuration, so any insignificant change at one of the inputs will create a significant change at the output because of the very large open loop gain of the op amp. Even if you ground both inputs through small resistors there'll be an offset error between the inputs (can be seen as Vos in the datasheets) so this will be multiplied by the open loop gain and reflected to the output. In your case, there's a huge risk of output saturation (hit either rail) even if the DAC voltage is near zero.

What I'm trying to say is, that you may not be able to observe a change because the output will hit either rail. Maybe you should think about building a closed loop circuit.

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  • \$\begingroup\$ Thanks! But I do not fully understand you argument that it was an open-loop circuit, because the current flowing through R79 depends on the sum of the currents running to the OpAmp's output via R81 and the E-B of Q28. The voltage at pin 2 of the opamp is fed back to the positive input at pin 1. \$\endgroup\$ Commented May 18, 2022 at 13:51
  • \$\begingroup\$ My circuit is btw. based on circuit B in maximintegrated.com/en/design/technical-documents/app-notes/3/… \$\endgroup\$ Commented May 18, 2022 at 13:52
  • \$\begingroup\$ @MartinHepp sorry, missed that part. Okay then, what I said in the 1st paragraph is your answer. \$\endgroup\$ Commented May 18, 2022 at 14:01
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Something like this can help, some checks must be done.
Feedback need to be used ...

enter image description here

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  • \$\begingroup\$ Thanks! So basically your simulation shows that it should work, doesn't it? \$\endgroup\$ Commented May 18, 2022 at 14:25
  • \$\begingroup\$ Quasi sure. I have try changing R3 -> 500 Ohm, no change in characteristics. If R3 too high, -> saturation below 50 mA. \$\endgroup\$
    – Antonio51
    Commented May 18, 2022 at 14:44
  • \$\begingroup\$ Tried your first circuit with MAX4162 ... very "weird" (?). \$\endgroup\$
    – Antonio51
    Commented May 18, 2022 at 18:29
  • \$\begingroup\$ Could you elaborate re "weird" please? Also, could you please explain your rationale for adding Q1 and Q3 in your simulation? Thanks! \$\endgroup\$ Commented May 18, 2022 at 20:11
  • \$\begingroup\$ Q1 and Q3 form a "Widlar" current source. So, if you impose the current through Q1, then the current through Q2 is "proportional" to the imposed current ic(Q1)= V2/75 Ohm (if Q1 and Q2 are the same, will show later). Also, I read the Maxim note and tried "circuit B" ... not been successful until now. \$\endgroup\$
    – Antonio51
    Commented May 18, 2022 at 20:56

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