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I'm working on a circuit I have a buffer amplifier with a low value pull-up on the output. I need to protect the op-amp because there could be high voltage on the pull-up pin.

The circuit normally works at 5V but there is a possibility that the output of the op-amp can see up to 15V. The op-amp would get damaged. I can't simply use a diode or other common techniques because this over-voltage could last minutes plus because of the low pull-up on the output, I need very low output impedance. Any suggestions would be appreciated.

What I need is somehow to detect an over-voltage on the op-amp output (I'm thinking of using a zener or comparator) and then disconnect the op-amp circuitry from the output.

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  • \$\begingroup\$ Please provide a circuit. Why do you have a pull-up at the output of a buffer amplifier - that is unusual? \$\endgroup\$ – Kevin White Oct 25 '17 at 0:36
  • \$\begingroup\$ I updated the post with a basic schematic (sorry did this on my Mac, new to LTSpice on Mac). The pull-up is required for the microcontroller. There is another side to the micro and the pull-up needs to be there. It's not normally 47 Ohm, usually somewhere around 100 Ohms. \$\endgroup\$ – aiq25 Oct 25 '17 at 2:35
  • \$\begingroup\$ So.. can your micro tolerate 15V? What else is there beyond opamp and micro that can cause 15V? Kevin probably asked for a complete circuit. Why do you need such low pull-ups? \$\endgroup\$ – Wesley Lee Oct 25 '17 at 2:38
  • \$\begingroup\$ Okay I will get a complete setup tomorrow. I think that will help a lot, sorry about the lack of info. There are diodes on the input side of the micro to clamp beyond 5V. \$\endgroup\$ – aiq25 Oct 25 '17 at 3:01
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    \$\begingroup\$ @aiq25 - that is very unusual requiring a very strong pull-up into an analogue input. Most opamps will not be able to drive that 47 (or 100) ohm load. Can you explain why you need the pull up? A simple diode to the +5v rail could avoid the output being pulled too high. Could you also show a schematic under the fault condition (15v) that you are protecting against? \$\endgroup\$ – Kevin White Oct 25 '17 at 3:05
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When you put 15V into a "very low impedance", one of two things will happen.

  1. The low impedance will sink the current from the voltage source, and the voltage source will be dragged down to a low voltage. In this case, you will have a large current, and it's the large current you need to detect to save your circuit.

  2. The voltage source "wins", and basically destroys the circuit to get to 15v. In this case, it's too late for protection.

Other than the instructive insights here, the point is that "something that reacts to 15V" isn't really what you're after.

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