I have a problem with a circuit that should be simple but has caused me some annoying problems.


I need a switch that closes when the power is off and opens when the power is on. My solution doesn't work and I feel stupid.

Desired effect:

  • When power is on, QNP should act as an open switch. In zero-input state, Collector and Emitter voltage should be (almost) zero. A 12V PWM signal on either Collector or Emitter side should leave the other side unaffected.

  • When power is off, QNP should act as a closed switch. A 12V PWM should be conducted from the Emitter side to the Collector side with minimal voltage drop.

My Problems:

I chose to use a PNP Bipolar Transistor, namely a MMBT3906, which worked in all my simulations using CircuitLab and Circuit Simulator.

But once I got my circuit produced I found a strange problem, where I would measure some voltage (1.2V) on the emitter of the PNP transistor, even though it should have been off (base at 12V). It seems like voltage "leaks" from the base to the emitter.

I tried with a blank board using only the components for this circuit and the problem persisted. I tried changing the transistor multiple times. I tried using a PMOS BSS84P with similar problems.

Screenshot from CircuitLab Simulation

The use cases can be seen here: https://www.circuitlab.com/circuit/mgt5u8/qnprequirementcircuits/

I have many more simulations I can post if you want.

Thank you in advance for your help.

  • \$\begingroup\$ Your circuit is wrong; why both emitter and collector are grounded? \$\endgroup\$
    – ilkhd
    Commented May 28, 2015 at 17:42
  • \$\begingroup\$ When power is on, is it correct you provide a dc voltage where your posted diagram shows ground connected to RaT1-2? When power is off, do you actually short that node to ground, or just disconnect it or disable the supply that is connected there? What kind of supply is it? \$\endgroup\$
    – The Photon
    Commented May 28, 2015 at 17:43
  • \$\begingroup\$ Sorry for the confusion. RaT1-2 and RaT1-3 signify the load on the emitter and collector. You can take a look at the CircuitLab link for the different scenarios. When power is off, SW12 is open and RPD-1-2 should act as pull-down. \$\endgroup\$
    – Johis
    Commented May 28, 2015 at 19:51

3 Answers 3


Page 2 of this datasheet tells the story.

enter image description here

Under absolute maximum ratings it states that the maximum reverse voltage of base and emitter is 5V. You have exceeded this by using 12V and although the base-emitter region is probably still intact (due to the 100k resistors in your design) you can't expect leakage current to be insignificant.

Don't expect simulation models to go beyond (or even get near) real-world issues due to absolute maximum ratings - models expect you to obey rudimentary principles and this is one of them.

Try a 5V control signal.

Also, when power is off you won't be able to conduct much signal thru from emitter to collector because you are relying on the base-emitter junction to be foward biased and this requires at least +0.7 volts on the emitter to do so.

  • \$\begingroup\$ Just started to write basically the same thing. No point now but I'll give you the graphic I made. \$\endgroup\$
    – The Photon
    Commented May 28, 2015 at 18:05
  • \$\begingroup\$ Transistor is PNP; -0.7v \$\endgroup\$
    – ilkhd
    Commented May 28, 2015 at 18:09
  • \$\begingroup\$ @ilkhd yes it is and you are reverse biasing the emitter base region - the spec says (courtesy of @thephoton) the junction limit is when the emitter is 5V below the base voltage (hence the negative sign). You, potentially have the emitter 12V below the base! \$\endgroup\$
    – Andy aka
    Commented May 28, 2015 at 18:17
  • \$\begingroup\$ That was a very enlightening answer, thank you. I originally used a P-MOSFET, but it had some other issues and I must have glanced too fast over the datasheet of the transistor after the simulations told me it should work. Do you have any suggestions on how to solve the problem? \$\endgroup\$
    – Johis
    Commented May 28, 2015 at 20:04
  • \$\begingroup\$ @Johis consider using a P channel JFET - basically it is naturally conducting from source to drain when the gate voltage = the source voltage. To turn it off you take the gate voltage higher than the source. The following article is useful: learningaboutelectronics.com/Articles/P-channel-JFET also note that these types of device are used in a lot of quality analogue gates (which is basically what you are trying to build) - Spehro's answer alludes to these but the 4066 uses MOSFETs which aren't as good. \$\endgroup\$
    – Andy aka
    Commented May 28, 2015 at 20:12

When I look at the datasheet for a 2N3906 manufactured by Fairchild (https://www.fairchildsemi.com/datasheets/2N/2N3906.pdf), I see on page 3 that it has an Emitter-Base breakdown Voltage of -5.0V.
So if your SW12 switch is closed then you're applying 12V to the base of your 2N3906.
Its emitter is tied to ground through RaT1-2, so you're effectively applying -12V to the 2N3906's base-emitter junction - far in excess of the datasheet's specified breakdown limit of -5V.
So what happens?
As you see, the base-emitter junction 'breaks down' and starts to conduct - so you see a voltage at the emitter since there's now current flowing through it.
You measure 1.2V at the emitter, so I'd assume the same voltage drop across RQNP-1-2.
So in actual fact your 2N3906's b-e junction is breaking down at about 9.6V (12 - 1.2 - 1.2).

  • \$\begingroup\$ Oh lol - 3 answers all saying the same thing posted within a few minutes of each other :P \$\endgroup\$
    – brhans
    Commented May 28, 2015 at 18:08
  • \$\begingroup\$ Last in first out! \$\endgroup\$
    – Andy aka
    Commented May 28, 2015 at 18:15
  • \$\begingroup\$ I guess I'd better upvote them all :) \$\endgroup\$
    – bitsmack
    Commented May 28, 2015 at 18:30
  • \$\begingroup\$ Thank you, such an obvious thing I overlooked. Any suggestions on how to achieve what I want? \$\endgroup\$
    – Johis
    Commented May 28, 2015 at 22:58

This is expected.

Edit: It's not leakage- you're breaking down the E-B junction in reverse by exceeding the breakdown voltage (typically rated at 5V with actual breakdown 6~9V). A p-channel MOSFET might give you what you want since the gate is insulated. There are also (somewhat rare) 'symmetrical' transistors that have equal E-B and C-E breakdown voltages (in the olden days they were used to make analog switches, but now mostly MOSFET transmission gates have replaced them). There are also some (somewhat rare) transistors such as the 2SA1252 which have guaranteed Vebo breakdown voltage > 12V.

It looks like you want an analog switch function. You could use 1/4 of a CD4066 which would act as I think you want (except with inverted control input- you could use another 1/4 of the chip with a resistor as an inverter).

  • \$\begingroup\$ How exactly? Base at zero potential (+12); both junctions are reverse biased. \$\endgroup\$
    – ilkhd
    Commented May 28, 2015 at 17:53
  • \$\begingroup\$ @ilkhd See edit above. Base-emitter reverse breakdown is usually less than 12V by a few volts. \$\endgroup\$ Commented May 28, 2015 at 18:03
  • 1
    \$\begingroup\$ A timely edit methinks LOL \$\endgroup\$
    – Andy aka
    Commented May 28, 2015 at 18:04
  • \$\begingroup\$ @Andyaka Well you may laugh, and out loud, sir. \$\endgroup\$ Commented May 28, 2015 at 18:05
  • \$\begingroup\$ @Andyaka Now I look like a moron \$\endgroup\$
    – ilkhd
    Commented May 28, 2015 at 18:06

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