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The project I want to build is shown below. I want to operate the P-channel mosfet on and off with variable frequency from 20-100Hz and I choose P-channel to do this so that the +350V will not be present on the load when the control circuit is off.

I want also the high power supply to the mosfet and the load to be separated from the control circuit.

My questions are:

  1. Will this circuit works? please give me suggestions or revisions.
  2. What if the load is disconnected or fails and becomes "open circuited" - will be high voltage present on the drain or will the mosfet still switch on and off in that case?

I am new to electronics.

enter image description here

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    \$\begingroup\$ The "GS" in V_GS(th) is very important. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 11 '14 at 6:16
  • \$\begingroup\$ This type of circuit is often called a high-side switch or high-side driver. It's a fairly standard circuit configuration, and you may find example schematics via an internet search. \$\endgroup\$ – John Honniball Nov 11 '14 at 9:11
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As Ignacio is trying to say, you need to pull the gate of the FET up to the 350V supply rail to turn it off. This makes the VGS ~= 0V. Which means the FET is off.

To turn it on, you need to reduce the voltage on this gate, perhaps the full 350V is not healthy for it. Perhaps drop the voltage at the gate to +330V to turn it on. This will make the VGS = -20V which is plenty. But always check datasheets.

This is a very high voltage switching circuit and your BJT will need to sustain the 350V as well, like I said. I suggest an opto-coupler to isolate the signal from the 350V supply.

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  • \$\begingroup\$ Not to mention that V_GSS on the IXTR16' is only +/-20V, so cue the smoldering crater where the die used to be. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 11 '14 at 6:40
  • \$\begingroup\$ An opto to isolate the signal AND a floating supply to ensure the Opto output & the Gate-Source are references near 350V rail. \$\endgroup\$ – JonRB Nov 11 '14 at 12:32
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No, this will not work, it will first destroy the mostfet, and then (depending on whether the mosfet gate-source failed open or short) destroy the 2n3906 and possible the rest of the circuit.

The reason is that the gate of the mosfet MUST be within +/- 20 V of its source. You switch it between -350 and -338, which is WAY too much.

If it is possible to connect the ground of your 555 circuit to a different voltage you could drive the mosfet (I would prefer an N type) directly. Note that this makes touching the circuit dangerous. Otherwise use a standard level shifter with a high-voltage transistor. Or one that uses an opto-coupler: isolation from 350V is A GOOD THING.

The fact that you are unaware of such basic principles and yet you are building a 350V switch makes me a bit nervous about your future health. But Darwinistic selection has its place. Leave us an address where we can send the award.

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I would like to add to Kyran's comment. You can consider using SiHfrc20 offered by Vishay. It has a Vds = 600 V and you could drive your load at around 1.3 A at Tc = 100 Deg. C.

When the load is open, it will be able to handle the voltage. However, I propose that you should be take some feedback to avoid the stress on the MOSFET.

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