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I am designing a circuit to scale an input voltage to the range of an analog to digital converter pin on a microcontroller, and also provide electrical protection for the pin.

I put together the following circuit which consists of a resistor voltage divider, a Zener diode, and a rail to rail op amp (single ended, powered by the microcontroller's digital supply voltage: 3.3V).

Will this circuit function as intended to both divide the voltage and provide over-voltage protection? I expect an input voltage range of 0V-6V but want to be prepared for voltages outside this range.

enter image description here

The idea was to divide the voltage in half before hitting the op-amp input, and to include a Zener diode for additional protection against both high voltage and negative voltage in a single component. I then realized that the voltage in this circuit might not be divided, but instead be shunted through the Zener immediately, leading to clipping of the input voltage rather than voltage division.

Which behavior will I get in this scenario, clipping or voltage division, and why? If I get clipping, what modifications would you recommend to get my intended behavior?

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2 Answers 2

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You are right about the clipping behaviour of the zener diode, and about the dividing behaviour of the resistors, and your design will do what you require.

In fact, you can test this behaviour in the simulator, to be sure:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

As AnalogKid pointed out, the clipping won't be sharp, but that's of no concern as long as the output signal is linear within the region of interest to you, and doesn't ever reach a value that could damage the following stage (the opamp, in your case).

As you can see, the output potential is half of the input, is restricted to a maximum of about 3.6V, and is also "accidentally" clipped at -0.6V or so, when the zener diode is forward biased, thereby adding protection against negative input potentials.

If you really need to buffer this output with the opamp, then you also have the benefit of the opamp's own output being clipped to the supply rails, of +3.3V and 0V.

According to the MCP6002 datasheet, it will tolerate input voltages extending 1V beyond either supply rail, which means that with a 3.6V zener diode as shown, the opamp's inputs will never be exposed to damaging potentials outside that range of -1V to +4.3V.

From the datasheet you can find out the opamp's common-mode input voltage range, with supply potentials of 0V & 3.3V. It is surprisingly good, at -0.3V to +3.6V. Your clamped/divided signal could possibly deviate outside this zone of "guaranteed behaviour", if it descends below -0.3V. If that happens then you can't trust the opamp output to be what you expect, but it won't be damaged.

Your design is good to go.

There's an alternative though. Since you have a +3.3V reference, you can clamp your divided signal to it with regular or schottkey signal diodes:

schematic

simulate this circuit

enter image description here

The response is very similar to your original design, except it has no need for a zener diode, instead relying on regular diodes to limit excursions beyond the power supply rails. This will be sufficient to protect your opamp.

I remind you again that the opamp's output cannot possibly extend beyond its own supply potentials, and is guaranteed to stay between 0V and 3.3V, regardless of its inputs.

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I would do two things.

  1. Your ideas on how a cheap opamp can protect the uC, and a zener diode would protect the opamp from both over and under voltages, are correct. However, IRL, the zener diode conduction knee is not very sharp. To guarantee that the diode has no effect on a 3 V signal, it would have to be rated for at least 4 or 5 V, leaving the opamp input at least partly unprotected. Instead of the zener, use two high speed signal diodes (1N914, 1N4148, etc.) from the non-inverting input to Vcc and GND. Shottkey diodes would have a lower clamp voltage, but also higher leakage current that might mess up the divider ratio.

  2. Replace the connection from the output to the inverting input with a resistor equal to the parallel equivalent of the two resistors in the input divider; in your case, 49.9 K. This minimizes input stage offset errors.

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