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Below is an example circuit: enter image description here

I have some circuits based on opamps similar to the ones above. I want to prevent a situation if one powers supply rails Vcc and Vee reverse way. Normally this would damage the circuit.

Now I want to add a simple interface component or circuitry across one of the supply lines of the opamp which would protect this circuit for reverse polarity situation without contributing much noise.

What is in general used such cases to prevent reverse polarity for the opamp supplies?

edit:

Which way of placing the diode below is better for the purpose of reverse polarity?: For single supply opamps

enter image description here

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  • \$\begingroup\$ How they could reverse. Let's for example have a multimeter. How can the polarity in the instrument opamp gets reversed? \$\endgroup\$
    – Marko Buršič
    Commented Aug 28, 2017 at 8:36
  • \$\begingroup\$ Opamps are ICs and all ICs have ESD protection meaning they have reverse biased diodes across between the supply connections. These could perform the function you need provided that you limit the maximum supply current. I think 100 mA should be a safe value. You could add an extra reverse protection diode across the supply pins as well. For non-critical applications you could add (Shottky) diodes in series with the supply lines. \$\endgroup\$
    – Bimpelrekkie
    Commented Aug 28, 2017 at 8:38
  • \$\begingroup\$ @MarkoBuršič If one wires Vcc to negative and Vee to positive terminal of power supply. That was to prevent reverse wiring of supply rails to power supply. Please see my edit also. I tried to make it more clear what I plan to use but not sure \$\endgroup\$
    – user1245
    Commented Aug 28, 2017 at 8:43
  • \$\begingroup\$ Are you trying to bomb-proof a tutorial kit so that the students can't accidentally (or intentionally) destroy the chips? If so, then pop that information into the question. \$\endgroup\$
    – Transistor
    Commented Aug 28, 2017 at 8:45
  • 2
    \$\begingroup\$ I am no wiser. It does not answer my question. \$\endgroup\$
    – Transistor
    Commented Aug 28, 2017 at 8:47

1 Answer 1

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There are various solutions to this problem. They range from diy circuits to IC's built for this specific purpose.

The most basic circuit uses series diodes with every supply rail. This does mean you lose power in the diodes as they have a voltage drop. It might not be a problem for you if you are dealing with very low-power devices that draw almost no current and power is not a constraint.

You can also use a parallel diode with a very small series resistor. This parallel diode would be reverse biased in normal operation, but become forward biased on reverse-power condition. In this case, the diode will shunt the rails. This does mean there will still be the forward voltage of this diode on your rails, and this could still pose an issue.

The better methods is to use series MOSFETs. They will turn on if the power is connected correctly, and turn off when it is reverse. The advantage is that they can have a far lower voltage drop. You can use a simple resistor devider or such to wire this up, but there are also ICs that can do this for you.

Simple example of the series FET would be like in the example on this hackaday page:

enter image description here

More information can be found in this TI appnote

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  • \$\begingroup\$ Why is Shottky type preferred ? \$\endgroup\$
    – user1245
    Commented Aug 28, 2017 at 9:23
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    \$\begingroup\$ @Matthew That is why you put the body diode such that is forward biased for normal operation. If the battery is connected the wrong way, it will be reverse biased. This is a circuit I and others have verified many times, so I would have trouble accepting it being "useless" \$\endgroup\$
    – Joren Vaes
    Commented Aug 28, 2017 at 9:34
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    \$\begingroup\$ @user134429 Shottky is generally prefered because the forward voltage drop is lower. This means that the power lost in the circuit is lower compared to a regular silicon PN diode \$\endgroup\$
    – Joren Vaes
    Commented Aug 28, 2017 at 9:35
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    \$\begingroup\$ @JorenVaes I'm sorry. It was my mistake. I was distracted when I was looking at your post. I was completely wrong. \$\endgroup\$
    – Matthew
    Commented Aug 28, 2017 at 9:45

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