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I have a question about this circuit, using Zener diode as a protection for mosfet. I've found some related topics but didn't found any circuit which uses the same configuration.Schematic

Can anyone suggest me that, Is this the correct approach?

The gate voltage is +10V maximum, 0V minimum and VPP1 and VNN1 are +100V and -100V respectively.
Zener voltage for the diode is 15V.

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    \$\begingroup\$ Well, to protect the gate of the MOSFET from over-voltage, the Zener voltage rating should be lower than the MOSFET Vgs absolute max rating (from the MOSFET datasheet). There should possibly be a resistor in series with the outputs from the IC to make sure the Zener clamp current is not excessive. You can start with zero Ohms and only install a larger value later if needed. You will need to monitor Vgs with an oscilloscope after you build boards and make sure everything is OK. \$\endgroup\$ – mkeith Nov 4 '17 at 17:40
  • \$\begingroup\$ Thanks, Sorry I forgot to mention but Vgs of the mosfet is 30V. \$\endgroup\$ – Embedded Nov 4 '17 at 17:50
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Let's look at just Q2.

When you first power the device, OUTB will either be 0V or 10V, let's say 0V. VNN1 will be -100V. This will make the gate 100V with respect to the source. Most MOSFETs won't be able to handle that and will burn out (or blow up), or they would if it weren't for the Zener. The Zener will effectively clamp Vgs at 15V, which the MOSFET WILL be able to handle. Once the capacitor C35 is charged everything will reach steady state, the MOSFET will see a Vgs of 0V and everything will work normally.

The Zener also acts as a regular diode, clamping Vgs at -0.7V or thereabouts, instead of allowing it to reach -10V; as it would do if OUTB stayed on for a long time, then was switched off.

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Here is a fragment from one of my project where I use mosfet as reverse polarity protection (for battery input):

enter image description here

As someone said in the comment, the Zener voltage has to be lower than the maximum voltage of mosfet gate (VGs).

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