I have this circuit for having measurements like with a multimeter diode function.

The ADS1115 result is reported to PC software and stored in a database for later reference comparison.

I could protect the ADC INPUT by Zener diode and op-amp buffer. but if the user measures two points carrying voltage higher than 3.3V, the 3.3V regulator will destroyed.

How can I protect my circuit against that scenario?

I want to keep the 3.3V at the measurements probes so I could not use a diode which will lower down that voltage at the probes end.

Circuit diagram

  • 1
    \$\begingroup\$ How high a voltage could be applied to the terminals? In other words, what is an acceptable maximum voltage to protect against and what current could be sourced from that voltage. Without these figures it is guesswork. \$\endgroup\$
    – Andy aka
    Nov 25, 2022 at 12:39
  • \$\begingroup\$ Are you sure you can't put a diode in series with R1? It looks like the 3.3V voltage doesn't need to be terribly precise anyway. (In fact why even bother with the regulator? seems you could just as well use 5V for this measurement) \$\endgroup\$
    – user253751
    Nov 25, 2022 at 12:47
  • \$\begingroup\$ The porotype was without any protection and produced several boards after many references have been token. After ADC burns reports I decided to modify the circuit by adding protection. The reason I do not want to change the 3.3V is that I do not want to have measurements offsets after many references have been taken on the first version especially that the offset is not linear which can be fixed by software. \$\endgroup\$
    – Husam
    Nov 25, 2022 at 13:22
  • \$\begingroup\$ The expected high voltage over terminals is not more than 5 volts. but still need more tolerance till 12Volts. \$\endgroup\$
    – Husam
    Nov 25, 2022 at 13:25

2 Answers 2


I'd replace the 3.3 volt regulator with a current source (op-amp, transistor and a couple of resistors) fed directly from the 5 volt rail. I'd put that current source in series with a protection diode also. Normally, for these types of measurement a 1 mA current source will be adequate; that is typical of what most multimeters generate when measuring diode forward voltage: -

enter image description here

The major benefit of using a 1 mA current source is that the voltage you measure (via the op-amp) is directly related to the equivalent resistance of the diode at 1 mA. If you design the current source carefully you won't need to compensate in software for nuances in the 3.3 volts (or 5 volts).


It's probably acceptable to put as much as 5V, perhaps 5.5 on the output of the HT7833 regulator with 5V on the input. Let's say you want to protect against +/-24V being applied.

You can put a TVS or zener diode from the output of the regulator, slightly higher voltage than 3.3V, say 3.9V or 4.3V. Applying 24V through 3.3K will conduct only about 6mA and the resistor will dissipate less than 200mW.

If you need to protect against higher voltages, such as mains voltage (120VAC or 240VAC) the same zener should work, but the resistor will be dissipating a lot of power (eg. with 240VAC it would be around 17W, so the resistor would have to be huge, and cooking the resistor will directly harm the accuracy of your measurement). Perhaps a fuse and TVS could be added, but fuse resistance is not zero.

An extra bit of insurance g would be to add a blocking diode (Schottky) on the input of the HT7833. That will prevent much current from flowing backwards through the regulator. There's plenty of headroom for the regulator.

  • \$\begingroup\$ Using blocking diode (Schottky) on the regulator input would be nice idea, the 5V circuit power comes from PC USB. Also protecting the Regulator output with Zener and TVS would be extra protection. I will prototype that against worst scenarios \$\endgroup\$
    – Husam
    Nov 25, 2022 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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