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I am having some trouble solving an input protection issue for a design that has to measure the voltage of an input between ~10V two 60V DC via a microcontroller ADC. The issue is the micro may be unpowered but the input could remain anywhere between 10V to 60V. This would cause the ESD protection diodes in the micro to conduct which would be bad.

I have come up with the following circuit that simulates but I'm concerned that the Vgs of M3 is close to the -20V limit. In the circuit VIN can be between 10V and 60V, VCTRL would be from the micro to enable M3 during normal operation. If the micro is unpowered then M5 is turned off because R8 pulls the gate to ground. M4 is used for reverse polarity protection.

Am I way off base with this design?

Thanks,

60V Input with protection

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  • \$\begingroup\$ Back to back FETs are a useful way. In the past I have used a similar approach, generating an isolated gate voltage, which can be done e.g. with a small photovoltaic isolator: electronics.stackexchange.com/a/567019/237061 \$\endgroup\$
    – tobalt
    Dec 23, 2021 at 6:09
  • \$\begingroup\$ Why not simply clamp VAdc to the rails with schottky diodes? \$\endgroup\$ Dec 23, 2021 at 11:49
  • \$\begingroup\$ What is the frequency of the signal you want to measure, and the source impedance? If frequency is low, you can use a pretty high input impedance, which makes protection a lot simpler. \$\endgroup\$
    – bobflux
    Dec 23, 2021 at 12:52

4 Answers 4

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I have come up with the following circuit that simulates but I'm concerned that the Vgs of M3 is close to the -20V limit.

Then use a Zener diode to protect gate-source just like you have done for M4 (the reverse protection circuit). Put the zener across R7 and make R11 similar in value to R6 (50 k).

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This is a good place to use an inverting op amp. The input pin is always controlled to a safe voltage during operation, and low-threshold clamp diodes can be safely employed in the knowledge that they will be at zero volts when a measurement is taken.

schematic

simulate this circuit – Schematic created using CircuitLab

The op amp will require + and - power supplies (+/- 5V, for instance). The 'output' will be - V/20, so instead of 10 to 60V, expect -0.05V to -3.0V; a second op amp, or some resistor divider and pullup voltage, will bring this into your ADC range. R3 and C1 can be omitted (R3 set to zero ohms).

At 60V, the 100k resistor will take under one milliamp, less than a tenth watt. The Schottky clamp diodes will keep the terminal voltage on the amplifier below protection-diode damage thresholds.

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    \$\begingroup\$ +1 Nice and very safe variant. Changes are needed if low noise and/or high input impedance are desired. \$\endgroup\$
    – tobalt
    Dec 23, 2021 at 6:05
  • \$\begingroup\$ You should NEVER use a dual supply op-amp to feed a single polarity ADC input. You must guarantee that the input signal is uni-polar and limited to the ADC input range. For the schematic as shown the input to the ADC would only work for negative excursions of the signal ...yet it shows a positive excursion signal. \$\endgroup\$ Dec 23, 2021 at 18:40
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60V DC through your resistor network become around 2.8V DC which is within the safety input range of your microcontroller. I don't know what your microcontroller is, but most of their inputs become Hi-Z when they're not powered.

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  • \$\begingroup\$ Yes most inputs are Hi-Z as you say but they also contain ESD diodes that are connected to the power rails that allow voltage and current to be present on these rails when the micro is not powered. \$\endgroup\$ Jan 5, 2022 at 17:19
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Input protection design for a microcontroller ADC

Zener diodes or any form of diode usually provides poor protection for an ADC input.

You need to provide:

  1. Accurate voltage clamping for the ADC signal.
  2. Overvoltage and polarity clamping for ADC input limits (below the intrinsic diode limits).
  3. Zero effect on signal input network in linear range.

For complete reference see this and this answer,

I'll repeat the protection circuit here for clarity:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the op-amp provides accurate clamping AND input overvoltage protection to about +/-200V. The circuit is high impedance so would meet your 100K Ohm requirement in your circuit.

If you need to provide ESD protection, then this should be applied direct at the input (shown in the first answer link). Be aware that there is some leakage current so at high input impedance may be significant within your signal range.

The TLV600x would typically be powered by the same VCC/VDD used by your MCU. If you take no action to address this then an overvoltage on the input will attempt to power up you op-amp and MCU from the signal input if the VCC/VDD is missing. Typically this will NOT be a problem as the MCU will require many mA of supply current to reach it's reset level detector voltage level. If you really want to address this problem then you have to separately power the front end protection op-amp.

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