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I am using a voltage divider to measure input voltage, a zener diode was added to protect opamp in case of voltage spike.

schematic

simulate this circuit – Schematic created using CircuitLab

I was told that the leakage current of zener will reduce measurement accuracy and advised to use ESD diode below:

schematic

simulate this circuit

How is the design better? Doesn't ESD diode also have the leakage current?

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    \$\begingroup\$ WHy not include 2 datasheets and check yourself? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 7 '19 at 5:52
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    \$\begingroup\$ If you need low leakage you can also use FET transistor as diode for protection: electronics.stackexchange.com/questions/396985 \$\endgroup\$ – Rokta Aug 7 '19 at 6:30
  • \$\begingroup\$ What input over-voltage might you expect? What values for the resistors and what op-amp are you considering? \$\endgroup\$ – Andy aka Aug 7 '19 at 6:57
  • \$\begingroup\$ ANY diode will have a leakage current, what counts is that the leakage current is such that it is low enough. So do calculations to determine what leakage current is acceptable, then choose a diode with a similar or smaller leakage current. Next time when you're "told something" don't just assume that that's it, get to the bottom of it, understand WHY it is like that. Maybe you were being told nonsense so be sceptical and ask "Why?" \$\endgroup\$ – Bimpelrekkie Aug 7 '19 at 7:03
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You can use an OA that has ESD protected inputs. You need to add a current imiting resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

If you use very high impedance resistor divider, then the current can't be higher than 0.2 micro amp (5 giga resistor). The input ESD protection will easily limit the voltage. But you have to be aware that resistor divider has also a capacitance, so it is RC divider not a pure resistor divider.

Now if we look the capacitive part of the divider, you can see that output voltage is inversely proportional to the filter capacitor. If not placed, you could get very large spike on OA input. The capacitive divider takes its place when switching events occurs on the measured rail.

schematic

simulate this circuit

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    \$\begingroup\$ Some additional info: all modern opamps have ESD protection so practically any opamp will work (it will have the build-in protection). If you make Rlimit such that no more than about 10 mA can flow, your opamp should be well protected. I agree with this answer in that adding external diodes isn't needed as long as Rlimit limits the current sufficiently. \$\endgroup\$ – Bimpelrekkie Aug 7 '19 at 7:09

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