1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I have a FQP27P06 P-Channel MOSFET that I am trying to drive. My source to my motor is 12v. From my understanding, I should be able to turn this "ON" with -10v (so I should be able to use 1-2V or either put the gate to ground?). I have tried everything and I can not get this thing to conduct. Here is my schematic. What do I need to do to get this to work? I've wasted two days on this.

schematic

simulate this circuit

\$\endgroup\$
  • 1
    \$\begingroup\$ Is your schematic accurate? Because you don't show your 12 volt supply being grounded. Otherwise, the most likely explanations are a dead MOSFET or simply connecting the FET wrong. Your circuit should work. \$\endgroup\$ – WhatRoughBeast Jul 3 '18 at 19:59
  • \$\begingroup\$ Oh, yes, the left side of the 12v supply goes to ground. \$\endgroup\$ – mScientist Jul 3 '18 at 20:04
  • \$\begingroup\$ If I probe from my drain to ground, I DO get 12v, but my motor does not turn. I can disconnect my motor and hook straight to the 12v battery and it does turn... \$\endgroup\$ – mScientist Jul 3 '18 at 20:13
  • \$\begingroup\$ If the left side of the 12 V source is grounded then it should be shown on the schematic. Note that he left side of the 12 V source is the positive. (The arrow points in the direction of increasing voltage.) I suspect that you have inserted it backwards. If you have deliberately connected it with positive ground then D1 and D2 are forward biased and you have a short-circuit on your 12 V supply. \$\endgroup\$ – Transistor Jul 3 '18 at 20:26
  • \$\begingroup\$ So try a different FET, and check to make sure you know which pins are which. Note, for instance that the gate is not the middle pin. \$\endgroup\$ – WhatRoughBeast Jul 3 '18 at 20:39
2
\$\begingroup\$

There are several things wrong here:

enter image description here

  1. The voltage source is backwards. From the diodes, it seems you are using a normal negative-ground system. However, you have the negative output of the supply connected to the supply rail.

  2. The other end of the supply is dangling. That won't allow any current to flow, and the supply is effectively not there. The positive end of the supply should go to your supply rail, and the negative end to ground.

  3. Grounding the FET gate makes no sense. That will permanently turn on the FET. If you plan to do that, might as well replace it with a wire.

Updated to new schematic

This now looks correct for what is shown here, although the change of R1 from 10 kΩ to 100 Ω makes no sense. That's going to cause 120 mA of quiescent current when the gate is pulled low. That's ridiculous. 10 kΩ was a good value for what you are trying to show. Perhaps you got some bad advice, but changing it was otherwise silly.

Since you didn't provide a link to the datasheet, we can't verify that the transistor is what you say it is and that it is used suitably.

If the motor still doesn't turn on, then maybe the FET is blown, you have its pins connected wrong, or there is a broken wire someplace. With the supply connected backwards as you originally had it, it's quite possible that the diodes and/or FET are blown.

Verify that there is 0 V on the FET gate and 12 V on its source. The drain should then be at 12 V. If not, the FET is blown or installed backwards. If there is 12 V on the drain, then that 12 V isn't getting to the motor somehow, or the motor is dead.

When debugging, break the problem into little individual problems, and verify each individual piece is working. On a simple circuit like this, all you need is a voltmeter to verify the voltages.

\$\endgroup\$
  • \$\begingroup\$ I have no idea what #1 means, I am very new to this. The left side of the 12v supply goes to ground (not shown). I am grounding the FET gate just to see if this schematic works. I am going to drive it with a transistor or an opto once I get the schematic to work. \$\endgroup\$ – mScientist Jul 3 '18 at 20:21
  • \$\begingroup\$ I edited the schematic and it is show above in the original post. \$\endgroup\$ – mScientist Jul 3 '18 at 20:50
  • \$\begingroup\$ The symbol used for the source means actually current source (although the quantity given is volts). A voltage source symbol contains +/- signs. \$\endgroup\$ – Curd Jul 3 '18 at 21:25
0
\$\begingroup\$

The above schematic does work. I don't know why I couldn't do it on a breadboard at all. I soldered it to a test board and it works fine. I guess I need to improve my breadboard skills or by a better one instead of the cheap ones. Thanks for all the input.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.