I heard that a capacitor can protect a DC power line from a transient voltages by ESD event. Unfortunately I don't understand how it works in detail. It would be appreciated if someone could answer the following questions.

For the questions I selected an example product whose specs are 0.01uF, 25V.

(1) basic principle
The transient voltage is bypassed by the capacitor. This protection works for either positive or negative transient. A negative transient voltage is bypassed from GND to VCC through the capacitor so that the potential difference between VCC and GND remains small. Am I correct?

(2) maximum voltage that a capacitor withstands
The example product has a voltage rating of 25V. This would mean that the capacitor can withstand up to 25V. Then is it possible for the product to protect power lines from a 8kV voltage peak?

(3) resulting voltage on the power line
TVS diodes have a rating called clamping voltage. For a capacitor can you assume that the clamping voltage is just the DC power voltage?

(4) Would such capacitor protect 3.3V DC power line successfully against any kind of ESD events that include human body model, charged device model?

  • \$\begingroup\$ Most of the ESD energy goes towards charging the capacitor \$\endgroup\$ Commented Jul 10, 2018 at 23:29

2 Answers 2


ESD will charge the capacitor. Here is an LTspice model for an ESD source.

The notable part of this model is that you are charging a 150pF and 8pF source then discharging through a R and L; the total amount of charge related to the ESD event is limited due to this. Which is why a capacitor that can respond with sufficient speed can absorb the ESD event without creating an excessive rise in voltage.

If you want to be careful, I would suggest simulating your ESD level (the link includes a initial condition for each IEC level) against the model for the capacitance you want to include. That should capture the total rise you will see.


The current Human Body Model (HBM) for Electrostatic Discharge (ESD) events at 8kV is usually 30A for ~2ns (some are worse, but generally people test for 6kV or 8kV). Then there is an additional 10A peak for roughly 30ns. But let's talk about the first one.

Thats 30*2e-9= 60e-9 Coulombs of static discharge

If the capacitor is charged up to 3.3V there is ~500e-9 Couloumbs in the capacitor

$$ \frac{1}{2}\frac{Q^2}{C}= \frac{1}{2}QV^2$$


$$ \frac{Q}{C}= V^2$$

So if you have 540e-9 Coulombs, and throw in the an additional 60e-9 Coulombs and 0.01uF for the capacitance you get ~8V which could kill your electronics. What I am trying to show is a capacitor is not a great way to handle ESD events, a better way is a TVS diode. The best way is to design your product to shunt any potential ESD event away from the electronics via a chassis or guard.

If you do have inputs to the outside world on your PCB, use a protection scheme (YMMV depending on your source, and sink impedance or what type of sensor you have). The most important concepts are the diodes and the series resistance.

enter image description here

Source: https://www.digikey.com/en/articles/techzone/2012/apr/protecting-inputs-in-digital-electronics


Example of TVS for power: enter image description here Source: https://e2e.ti.com/support/power_management/led_driver/f/192/t/467628?TPS92551-and-ESD-Protection

  • \$\begingroup\$ (1) Is it recommended to use TVS diodes to protect power lines? (2) Is it okay to put 1kohm in the power line? Doesn't it reduce the input voltage? \$\endgroup\$
    – Nownuri
    Commented Jul 18, 2018 at 4:04
  • \$\begingroup\$ You can use TVS for power, the series resistance would need to be low, if anything at all, if you used 1kohm you wouldn't be able to run a circuit more than a few 10uA's because of the voltage drop in the resistor. See edit \$\endgroup\$
    – Voltage Spike
    Commented Jul 18, 2018 at 5:22

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