The other day I built a basic buck power supply on a breadboard from scratch (i.e. sawtooth generator, comparator, zener power supplies for control circuitry and references, feedback loop, and low-pass filter for the power output).

It got me thinking - could any off-the-shelf power supply chip be made into a current source using the following arrangement of the feedback?


simulate this circuit – Schematic created using CircuitLab

For the sake of example here, this would be a buck converter with an internal 1V FB reference which has been designed as a 1A current source. Rather than the traditional voltage divider feedback loop for voltage feedback, this uses a 0.01\$ \Omega \$ feedback resistor with some generic amplifier block (could be an IA, could have an active low-pass filter as well for noise/anti-aliasing etc.) with a gain of 100x.

Once the voltage is increasingly ramped until the current through the resistor hits 1A, the feedback voltage will reach 1V - and so the power supply will maintain the voltage (and therefore current) at that level.

Obviously, buck, boost, or buck-boost would have to be chosen appropriately to ensure the output voltage was correctly reachable, over the desired current compliance range.

Any first-glance reasons why this wouldn't work?


EDIT: Decided to simulate it with a basic linear tech buck part. Seems to work (note, this has a reference of 2.21V). Obviously as some have pointed out, the op-amps will affect the transfer function in the feedback loop. I believe as long as the correct op-amps were chosen (high-frequency, high CMRR, proper filtering), this is a viable solution however. Feel free to disagree, but.

enter image description here

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    \$\begingroup\$ The devil is in the details. While the concept is fine, it is entirely possible that the closed feedback loop will be unstable. It depends on the frequency of the current sense amp and the internals of the regulator. \$\endgroup\$ Jul 5, 2018 at 23:30
  • \$\begingroup\$ The circuit as shown can't work since the +/- inputs of the opamp would switch the output at very low levels ..perhaps just a few uV. You need a way to detect both the output voltage and the output current. \$\endgroup\$ Jul 5, 2018 at 23:49
  • \$\begingroup\$ @JackCreasey, since the gain is given on the schematic, we can assume the amplifier symbol doesn't indicate an op-amp. \$\endgroup\$
    – The Photon
    Jul 6, 2018 at 1:36
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    \$\begingroup\$ Or, to get a more explicit "proof" of stability, break the feedback loop and do an AC simulation to find the phase margin. \$\endgroup\$
    – The Photon
    Jul 6, 2018 at 1:42
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    \$\begingroup\$ @JackCreasey - What? The feedback voltage is proportional to current, as sensed by the 22.1 mohm resistor. With a gain of 100, this gives the required 2.21 volts operation at the FB pin for a current of one amp. It most certainly is a constant current source. \$\endgroup\$ Jul 6, 2018 at 4:26

1 Answer 1


There are a few problems with using a DC to DC converter as a 'power supply'.

The first one is stability, most DC to DC converters are only stable across a certain range of voltages.

The second one is dropout, most DC to DC converters drop out on the low end of the voltage range. The LT1076 is stable across this range, from the datasheet

Buck Converter with Output Voltage Range of 2.5V to 50V

So conceivably you could use the current limiting function but after 2.5V it would drop out.

There is a schematic that I've seen from linear (that I can't find right now) If I remember right it uses the LT3042 (or simmilar LDO) to drop the remainder of the voltage down to get the full voltage output range.


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