2
\$\begingroup\$

I have a circuit that tests DC PSUs. They are typically unregulated and unsmoothed, up to 50V. It performs tests to establish I/V curves, down to about 1V, and up to about 3A. I want a sub-circuit that protects the testing circuit from accidental reverse-polarity connection of the PSU under test, without dropping too much voltage. (i.e., a rectifier is definitely not ideal).

The testing circuit has +5V, -12V and +12V supplies which can be used for the protection sub-circuit.

What circuit would achieve this? Ideally I would use an N-channel MOSFET given its particularly low On-Rds. This question comes close to the answer, but it will not work for a PSU voltage of less than about 5-10V, or more than about 20V. (Although a potential divider is suggested for use for PSUs above 20V). I have tried scribbling some circuits but am not getting anywhere. Is what I'm asking even possible?

\$\endgroup\$
2
\$\begingroup\$

Consider using an LM311 comparator powered from the available +/- supplies and driving an SSR.

Divide the input voltage down by (say) 11:1 as shown. That gives a 45mV to 4.5V signal at the inverting input of U1 for a 0.5 to 50V input. The non-inverting terminal is held at about 23mV, so input voltages greater than +360mV (allowing for 10mV \$V_{OS (MAX)}\$) will turn the SSR on. Zero or negative input voltages leave the SSR off.

The circuit shown presents a small resistive (11K) load across the input. If that's objectionable, an additional high-impedance buffer could be added ahead of the comparator.

schematic

simulate this circuit – Schematic created using CircuitLab

A suitable SSR might be the 60V-rated CPC1907B, which will look like a ~0.06 ohm resistor when it is on and is okay for 3A (6A rating) (a bit more resistance worst-case as it warms up).

enter image description here

The same principle could be used with the comparator driving a heftier SSR, or a mechanical relay, which would allow even lower series resistance. It's also permissible to parallel a few CPC1907Bs with the input LEDs either driven from individual 1K resistors or connected in series with single a lower value series resistor.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.