Input Protection for an Electronic Load

Question

I'm currently working on a dummy load designed for testing small power supplies less than about 10V. I'd like to provide reverse input protection to prevent the op-amps from being damaged. Because the load has to measure the input voltage (constant power and resistance operation), the voltage drop across the input protection needs to be as small as possible.

I've thought about using a P-channel MOSFET, but that requires that the input voltage be above the MOSFET's Vgs threshold, which would be above the minimum voltage I want this load to be able to test (under ~1V ideally). Also, this would cause large voltage drops and power dissipation when the input voltage is near the MOSFET's Vgs threshold.

Another solution I've seen is a zener with a fuse/PTC. PTCs are too slow to trigger and fuses have to be replaced, so that doesn't seem to be a great solution.

The simplest solution that I can think of is a single diode, but this is also not ideal because of the rather large voltage drop (min. of about .3v with a schottky).

Here is my current circuit:

^{simulate this circuit – Schematic created using CircuitLab}

I appreciate any ideas you all have to offer!

EDIT: I forgot to mention that this IS NOT powered from the power supply under test. It's going to be battery powered so I'm thinking 2 3v coin cells in series regulated to 5v.

Also, I don't think I was clear about what I meant by using a P-Channel MOSFET: https://www.circuitlab.com/circuit/bka32t/p-channel-protection/

Answer 1

For reverse-voltage protection, you could consider (conceptually) something like this. Diodes are the inherent body diodes of the N-channel MOSFETs, not discrete components.

Each input could go negative as much as one (body) diode drop so some biasing may be necessary depending on the amplifier/comparator, perhaps a resistor to ground on each input of the comparator and a resistor to +5 to bias the input above ground. Some amplifier/comparators may require only a divider network.

Anyway, if Vx is negative, then the output of U1 is ~0V so Q2 stays off, and Vx can be as high as the breakdown voltage of Q2 without causing damage (assuming R4, R5 are high enough to avoid damage to U1). If Vx is positive more than a few mV, then Q2 turns on fully, adding only milliohms to the circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

You'll want to put a 10k resistor in the path from the FET source to the -ve input of the op-amp. You don't want that directly connected there, and you're only sensing voltage. However, be careful of the input bias on this op-amp, it will causes a voltage drop across the 10k and add error, so choose a resistor that is low enough to not add significant error to your voltage sense, or a precision opamp with low input bias currents.

You also want to put a 10 ohm up to 100 ohm between the opamp output and the gate of the FET. The op amp doesn't want to drive the capacitive gate and the 10-100ohm there along with the gate capacitance forms a pole that stabilizes the loop. Some people like to add extra capacitance from the gate to GND if the loop still oscillates and they need it even slower.

For reverse voltage protection, perhaps you can use the current sense resistor as it is there, and either a depletion mode FET (if you can work out the Vgs issues) or simply a relay. Note that when the load is connected in reverse the body diode of the MOSFET is forward biased, and the sense resistor and MOSFET is in the current path. This will be high current and you can sense that, and you can also sense it's in reverse by the polarity across the sense resistor.

Furthermore, you cannot power these from the LOAD. you must have independent power for the opamps. You can share a ground, that's all.

For a more advanced design, you can float the output section and power that from a separate supply. Closing the loop from the MOSFET to the op-amp would be via opto-isolators or signal transformers in both directions.

Answer 3

If you have op amps in your dummy load then you cant power them from the power supply your testing if it is only providing 1 volt. This has to mean that your dummy load has an external supply and if it has then what is the problem using that to generate sufficient negative voltage to turn your p ch fet on? Maybe I'm missing something here?