Unexpected behaviour of current sensing circuit with op-amp

Question

I would like to use the schematic below to measure the output current of a buck converter (0-10 A) and convert it to a voltage of 0-3.3 V which is read by a microcontroller. During startup I sometimes run into the problem which I have simulated below.

When the voltage at the sense resistor drops to near 0 V, the op-amp (LM324) output jumps to the supply voltage of 24 V and stays there. Since the input voltage is close to 0 V I would expect some undesired behaviour like this, but I don't understand why it does not 'recover'? What would be the best way to resolve this?

Answer 1

Looking at your schematic, I can't see what Q1 is supposed to do, but it seems as if the inverting nature of Q1 should make the loop feedback negative.

However, one thing that stands out is that it is easily possible for Q1's base potential (the op-amp output) to exceed its collector potential, and forward bias the base-collector junction. If that happens, then collector potential will "follow" the base, and feedback becomes positive.

I think that's why your op-amp locks up in a high state.

Since I don't know what Q1 is supposed to achieve here, the best answer I can give you is a suggestion which wouldn't suffer from this problem, but doesn't use Q1:

^{simulate this circuit – Schematic created using CircuitLab}

That's just a basic differential amplifier, with gain of 33. It behaves like this:

Answer 2

Thanks all for your comments/anwsers! I seem to have resolved my problem by replacing Q1 with a mosfet and I also added some capacitive feedback (see below). I updated my prototyping board with these changes as well and the microcontroller indeed seems to return the actual output current of my buck converter.

Answer 3

If you must use a high-side current monitor, you could try a circuit like the one I have simulated below. It won't be accurate for input voltages near zero, but it is within 1% from 5 V to 35 V, for which I use a 15 V sine wave with 20 V DC offset.

The output voltage between R2 and R4 (a and b) equals the voltage on the 100 mΩ shunt R5 (Vin and Vout). It also works with a 10 mΩ shunt, but the error increases to about 5%-8% for common-mode voltages of less than 6 V. If you do this, you may need to tweak the values to get the performance you desire, and you will still need an amplifier with a gain of 30 to get the required signal to your ADC.

I was able to achieve about 1% error down to a range of 1 V to 39 V with a 10 mΩ shunt by using 975 Ω for R1 and R3, and 1 MΩ for R7.

I simulated the OP's circuit using an AD820 and 2N2222, and the output was basically slammed to the 24 V Vcc rail.

OK. Here is a practical circuit, although it also has some unwanted behavior, which might be due to the simulator. For one thing, the voltage across the shunt resistor should not be distorted (or go below zero).

I was not really satisfied with the last circuit, so I came up with the following, using a current mirror and a single op-amp. I also used a step function to test it for a range of power supply voltages. This still needs very close matching of the differential amplifier resistors:

I simulated the circuit provided by @emdura, and found problems with it. However, I used most of the same components and made some changes that work properly. The source follower MOSFET is not really necessary, but is provided in case higher output current is desired. At least this design does not depend on the value of the load resistor.

Answer 4

Real op-amps have offset. You'll have to add the worst-case offset to the input plus a small margin. Then it won't saturate. But that will require a slightly different design.

You have two choices:

1. Use a purpose-made high sensing current shunt amplifier. Those usually are easy to interface straight to 3.3V ADC inputs.

   The upside: one chip. The downside: there are no industry standard pinouts/specs, so if the part is not available, you either have to wait or redesign.

2. Use a discrete design.

   The upside: if you can't buy the parts, the world is ending and you don't need to worry about it anymore :) The downside: more drift, but performance can be improved by using better op-amps in the same package.

There isn't much to applying high-side sensors if you follow their datasheet and/or app notes. So we'll explore a discrete design.

OA1 copies scaled version of the input current onto R6. I2 driving R2 adds an offset voltage that compensates for worst-case offsets in OA1 and elsewhere in the system. I2 would be e.g. an LM334, or any other current source with compliance voltage range from 10 to 25V.

OA2 mirrors the current from R6 onto R9 and R11.

OA3 buffers the output voltage.

D1-R10 establish about 4V of headroom from the top rail, so that the op-amp inputs are within their common-mode range.

V1, V2 and V3 are used to model op-amp offset voltages.

^{simulate this circuit – Schematic created using CircuitLab}

The output voltage waveform is shown below. The load is driven with a 1kHz 0-10A sine wave current waveform. The top and bottom curves are for the worst case input offset voltage in OA1, since that offset dwarfs the others.

This circuit always recovers from disturbances even if they saturate the outputs.

To improve performance, use op-amps with more gain, better linearity and offset drift. The requirements for the op-amps are modest:

• OA1, OA2
  1. Input common mode range from (V-) to (V+)-2V.
  2. Output voltage range: (V-)+2V to (V+)-2V.
• OA3 - if powered from 24V (eg LM324):
  1. Input common mode range from (V-) to (V+)-2V.
  2. Output voltage range: (V-) to (V+)-5V.
• OA3 - if powered from 5V:
  1. Input common mode range: (V-) to (V+)-2V.
  2. Output voltage range: (V-) to (V+)-2V.
• OA3 - if powered from 3.3V:
  1. Input common mode range: rail-to-rail
  2. Output voltage range: rail-to-rail

R12 ensures LM324 performance with output close to 0V. It should be removed for other op-amp families, unless the datasheet dictates otherwise.