I have a 6 volt supply that I need enable/disable using another 3 volt supply.

If I use a PNP Darlington set-up like this: PNP setup Then I get just under 6 volts at the Output node.

My problem is that I don't have any PNP transistors but I do have several NPN transistors and I was hoping that setting up a NPN Darlington would give me similar results, but building something like this yields about 2 volts at the Output node:

NPN setup

My question is; is there a circuit I can make using NPN transistors (2N2222 to be exact) what will give me a voltage close to V2 voltage when V1 is enabled?

  • \$\begingroup\$ Is your load really only 10k? Or what's the required current? \$\endgroup\$ – jonk Aug 28 '18 at 19:01
  • \$\begingroup\$ @jonk My load is a 6 volt aquarium-pump, I think it's more than 10k but I don't know the details. I can potentially swap the 6 volt supply for a 12 volt if that makes a big difference. \$\endgroup\$ – bornander Aug 28 '18 at 19:04
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    \$\begingroup\$ It's hard to know what to recommend without knowing the required current compliance. Is it possible that you could just measure it? Or does it have a rating on it that tells you "watts"? \$\endgroup\$ – jonk Aug 28 '18 at 19:09
  • \$\begingroup\$ @jonk Current should be less than 250mA. \$\endgroup\$ – bornander Aug 28 '18 at 19:16
  • \$\begingroup\$ Are you controlling it via an MCU output pin? Or what exactly is providing the 3 V controlling voltage? \$\endgroup\$ – jonk Aug 28 '18 at 19:19

Try this circuit:


simulate this circuit – Schematic created using CircuitLab

The reason your circuit doesn't work is because you're using an "emitter follower" circuit. Basically what it does is the transistor maintains a voltage at the emitter that is the same as that at the base, minus a diode drop. Using a darlington in this case doesn't help because the transistor will not conduct if the base is lower than the emitter.

Note that the voltage at the collector of the transistor will not be 6 volts when it is on (you don't want that anyway), however the load will have very close to 6 volts across it when the transistor is switched on.

Edit: as mentioned by @jonk, your micro's output pin does not source enough current to drive this transistor enough to switch your pump. You have a couple of options here, you could buffer the output of your microcontroller through a buffer like a 74HC34, or you could use a darlington configuration with 2 2n2222 transistors as shown here.

  • \$\begingroup\$ The OP says the load is a motor so (1) you need a kickback diode around the load and (2) the 2N2222 may not handle enough current -- especially at motor start or motor stall. \$\endgroup\$ – Brock Adams Aug 28 '18 at 19:18
  • \$\begingroup\$ Also, assuming \$250\:\text{mA}\$, there won't be sufficient base drive with this circuit. The std-drive I/O is only rated at \$500\:\mu\text{A}\$ and the high-drive (only 3 I/Os) is rated at \$5\:\text{mA}\$ max. \$\endgroup\$ – jonk Aug 28 '18 at 20:45
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    \$\begingroup\$ @jonk To be fair, when I posted the answer this information wasn't available. \$\endgroup\$ – C_Elegans Aug 29 '18 at 0:34
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    \$\begingroup\$ Which is why I was asking questions before attempting an answer. \$\endgroup\$ – jonk Aug 29 '18 at 2:04
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    \$\begingroup\$ @bornander no it wouldn’t. With the PNP transistor, you need to bring the base up to the positive supply to turn it off, which you can’t do with only 3 volts. So you wouldn’t have been able to turn your pump off \$\endgroup\$ – C_Elegans Aug 29 '18 at 13:11

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