I have a simple circuit using an NPN as a switch using the BCW71 (https://www.mouser.es/datasheet/2/308/BCW71-D-1802620.pdf), as seen in the following picture:

NPN switch

My issue is that when I put 3V3 at the base, I get 2.6V at the emitter, so I have a voltage drop of almost 13V at the transistor. When I put 0V at the base the emitter is zero. Am I missing something?

Edit: R23 goes to the gate of an N-channel MOSFET.

  • 2
    \$\begingroup\$ Yes. Your base voltage needs to be 0.7 V higher than the emitter. You have built an emitter follower which will give out 3.3 -0.7 V. You need an PNP transistor for this configuration. \$\endgroup\$
    – winny
    Aug 18, 2020 at 8:33
  • \$\begingroup\$ thanks! which configuration should I use in order to keep on using an NPN? \$\endgroup\$ Aug 18, 2020 at 8:36
  • \$\begingroup\$ @winny You need an PNP transistor for this configuration Actually an NPN and a PNP would be needed, similar to this: electronics.stackexchange.com/questions/12972/… \$\endgroup\$ Aug 18, 2020 at 8:43
  • \$\begingroup\$ @Bimpelrekkie Yes, of course. I assume you understood what I meant:-) \$\endgroup\$
    – winny
    Aug 18, 2020 at 9:49
  • 2
    \$\begingroup\$ IMHO just replace it all with a gate driver IC. It will work by definition with any MOS and any application (PWM?) , it's going to take less board area saving two resistors and can't be significantly more expensive. \$\endgroup\$
    – carloc
    Nov 6, 2021 at 8:25

2 Answers 2


Your circuit is working as it should. You can visualize your circuit in this way:

Vbe = 0.7V (Assumed; the datasheet specifies that it is between 0.6 and 0.75)
VR16 = 1000 * Ib but Ib is very small so we can say that the voltage is nearly zero.

VR24 = 3.3 - VR16 - Vbe

Thus VR24 should very close to 2.6V


simulate this circuit – Schematic created using CircuitLab

This configuration is not useful to drive an N channel MOSFET (I am assuming you are using the MOSFET as a power switch and the MOSFET is N channel.)

What to use is a different question but, assuming the MOSFET is a switch, I can suggest you some quick approaches.

  • Use a common emitter configuration (The logic will be inverted though.)
  • Use a PNP transistor.
  • Use a P channel MOSFET with a transistor in common emitter configuration (That will have a direct logic behavior.)

Yup. The only way for there to be more than about 0.7V between the base and the emitter is for the transistor to be broken.

Put another way, the emitter is always about 0.7V below the base when current flows through the emitter. The base is at 3.3V, so the emitter will be at about 2.6V.

Since you are driving a MOSFET gate with this, you need to rearrange things slightly.

Like this:


\$Y\$ goes to your 3.3V signal. \$\overline{Y}\$ goes to R23 and then to the gate of your MOSFET.

This circuit inverts the driving signal. A high on the base will put a low on the gate of the MOSFET.

You can either invert the logic driving \$Y\$ in code, or add another NPN in front of this one to invert the signal.

  • \$\begingroup\$ thanks! R23 is connected to the gate of a MOSFET. I edited the question accordingly. \$\endgroup\$ Aug 18, 2020 at 8:38
  • \$\begingroup\$ thank you, I will try this. \$\endgroup\$ Aug 18, 2020 at 9:23

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