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schematic

simulate this circuit – Schematic created using CircuitLab

Gentlefolk, I am attempting to build a circuit to switch power through a P-CHANNEL MOSFET device. I believe I am on the right track, but I appear to be frying my NPN transistors (2N2222) as soon as the NPN base goes high (5v). The smell is pretty putrid and I am probably knocking a year off my life every time I breathe in the fumes. Not quite sure what I am doing wrong here. R1 is simply a dummy load resistor. In the production circuit, this is where the equipment sits.

I am simply trying to use the NPN to drag the MOSFET base to ground which should turn on the MOSFET and allow current to pass. There should only be 17mA through R2 which shouldn't be enough to fry the device. Any idea what I'm doing wrong here?

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    \$\begingroup\$ Thanks for your help guys. Have fixed the problem, which in the end turned out to be a toasted MOSFET. The busted MOSFET was in turn, toasting my NPN's. Replaced the NPN with an N-CHANNEL MOSFET and replaced the P-CHANNEL MOSFET with a new one. Dropped the pull-up resistor back to 1k. All working a treat. \$\endgroup\$
    – Mark
    Commented May 27, 2014 at 16:07
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    \$\begingroup\$ Thank you for the feedback, Mark. It's always good to know that when we hit the nail on the head (or miss entirely). \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 27, 2014 at 20:22

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Either the MOSFET has an internal gate-source short or it is wired up wrong (or, see Joe Hass's answer in case you're actually applying +5 directly to the base).

If you want to be able to apply +5 without the 1K resistor you show, you can use an N-channel MOSFET rather than a BJT.

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  • \$\begingroup\$ BJT? Is a Gate-Source short a feature of the MOSFET design or a dodgy MOSFET? Need to use PChannel due common grounds everywhere in the equipment load. \$\endgroup\$
    – Mark
    Commented May 27, 2014 at 13:04
  • \$\begingroup\$ Not a feature. ;-) If you expose the gate to an excessive voltage wrt the source even for an instant (rating is +/-20V, but it might take 50V or 70V to actually kill it), the virtually perfect (but very thin) gate oxide insulation ruptures and the MOSFET is toast. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 27, 2014 at 13:10
  • \$\begingroup\$ OK - just snipped the MOSFET gate lead off the board. NPN is no longer smoking. Fixed one problem :-) Have changed the collector resistor to 20k (from 1k) NPN collector is tied to ground whatever the state of the NPN base - so looks like this one is fried. Checked the MOSFET gate - there is definitely a goodly amount of voltage appearing on the MOSFET gate - 16.8v which explains why it's toasting the NPN. Is this expected behaviour for a MOSFET gate? Should I throw a few ohms between the NPN collector and the MOSFET gate to limit current through that path? \$\endgroup\$
    – Mark
    Commented May 27, 2014 at 13:23
  • \$\begingroup\$ The 1K is okay. You'll slow down the MOSFET by using 20K (so it will get hotter during rapid switching) but it will still work. OTOH, the MOSFET is not okay. If you've confirmed the wiring, it's trash. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 27, 2014 at 13:30
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If you are really bringing the base of the NPN transistor to 5 V then you are destroying the base-emitter PN junction with overcurrent. In a circuit like this you wouldn't expect the voltage at the base of the transistor to exceed about 0.8 V. Having said that, the circuit you provided doesn't quite match your description. The circuit shows a 1 kilohm resistor between an inverter and the transistor base, which should work just fine. You really don't need more than 1 mA of current into the base to easily get 17 mA from collector to emitter, so you could probably increase the resistor value to 2.2 k or 4.7 k. However, the 5 V signal should be applied at the inverter output, not directly to the NPN base.

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  • \$\begingroup\$ the action of grounding the inverter input causes the inverter output high - which you probably know already. High at that point in the circuit is 5V. The MOSFET side of the circuit is 16.8v - do we think this is the appropriate way to do this? Is the 2n2222 appropriate for this application? \$\endgroup\$
    – Mark
    Commented May 27, 2014 at 13:07
  • \$\begingroup\$ I understand quite well how an inverter works, but what you said was "the NPN base goes high (5v)". If you really did what you said you did then you destroyed the NPN transistor. \$\endgroup\$
    – Joe Hass
    Commented May 27, 2014 at 13:19
  • \$\begingroup\$ A 2N2222 or 2N3904 would be fine for this application, if used correctly. \$\endgroup\$
    – Joe Hass
    Commented May 27, 2014 at 13:26
  • \$\begingroup\$ ok. yes I get that the NPN is fried. I just don't get why? the NPN base can take 5v no problem - it appears to be a voltage on the MOSFET gate that is the problem. Can you suggest a solution? \$\endgroup\$
    – Mark
    Commented May 27, 2014 at 13:33
  • \$\begingroup\$ No, it is not true that you can apply 5 V to the NPN base with "no problem". For a true 2N2222 the rated breakdown voltage of the BE junction is 5.0 V so you are right at the maximum limit. However, what you really need to worry about are the base current with \$V_{BE} = 5\$V and the resulting power dissipation of the NPN transistor. \$\endgroup\$
    – Joe Hass
    Commented May 27, 2014 at 13:55

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