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I am trying to use an NMOS(2N7000BU) to "turn off" the LM350 voltage regulator. I have an OR gate going into the gate of the mosfet, so either input will turn it off. The hope was to ground the adjust pin on the 350 so there is no output. When I first put this on a breadboard it worked, but now when I try it the voltage out stays around 1V, which is "off" enough for my application. Removing the voltage from the gate of the mosfet doesn't turn the 350 back on, which it should. I have also tried it without the 10k resistor between gate and source. Any help is much appreciated! enter image description here

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    \$\begingroup\$ The minimum voltaje you can get is 1.2 V, shorting pin 1 to ground, check datasheet. When you remove the gate voltage and don't recover the output adjusted voltage, have you measured pin1 to ground? \$\endgroup\$
    – Bravale
    Aug 10, 2023 at 20:44
  • \$\begingroup\$ You could have just used a couple of diodes onto the gate of the MOSFET to get the OR function. It still won't turn the LM350 off completely though. \$\endgroup\$
    – Finbarr
    Aug 10, 2023 at 21:18
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    \$\begingroup\$ The LM350 internal voltage reference is 1.25 V. This is the lowest possible output voltage unless you pull the ADJ pin below GND. \$\endgroup\$
    – AnalogKid
    Aug 10, 2023 at 22:09

3 Answers 3

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Simplest thing to do (if you can redesign)

  1. Use a regulator with an enable pin. Next best choice:
  2. Use a P-channel FET as a load switch. You can use this in the regulator's input path or output path to turn it on or off. See this other stackexchange article about using a P-Channel FET as a load switch. p-channel MOSFET switch
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The two NPNs are working as emitter-followers, so the emitter will be ~0.7V below whatever the base is at. It may be that even when the NPNs are off they are leaking enough that they're still turning on the FET.

My recommendation? If your intention is to have an OR for turn-off, just use two MOSFETs in parallel, with your 'A' and 'B' inputs controlling each FET's gate. As follows (simulate it here):

enter image description here

You could also do the same thing with a pair of NPNs.

Regardless, your output voltage will go no lower than ADJ (pin 1) + 1.2V. You said that's ok, I would beg to differ. Consider using a load switch instead, or a regulator with a separate enable pin.

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Your resistance of 750 ohms is much higher than the 240 ohms maximum on the datasheet. Then you must calculate the new resistance for the voltage adjust pot.

The inputs of both NPN transistors must be 0.5V maximum for the Mosfet to turn off. Do it like this: LM350

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