Switch using 2N 3906 transistor

Thread Starter


Joined May 14, 2014
I am trying to use one 2N3906 transistor as a switch. I have a control signal that vary between 0v and 3,3 Vdc at base, but the transitor not open when the voltage at base reach the 3.2 V. I try with several resistor values but the transistor never off.

Would somebody help me indicating me what must be the configuration and the correct resistors values?


Joined Feb 24, 2006
With a PNP transistor correctly configured as a switch you need to connect the emitter through a pullup resistor to Vcc, and the base to a pullup resistor to Vcc. Aseries resistor from the contro signal to the base and the collector to the grounded load.

With the control signal at 3.3V, the transistor is off and the grounded load is undriven. With the control signal at GND, current flows through the base pullup resistor and the series resistor to pull the base to 3.3V - 0.7V = 2.6V. Noe the transistor is on and the load is pulled up to Vcc - Vce = 3.3 - 0.1 = 3.2V determined by the load resistance and the emitter pullup resistor.

Thread Starter


Joined May 14, 2014
I probed with 600 ohm series resistor at base, a 10 kohm pullup resistor from base to VCC( 12V) and 1k pullup resistor from colector to Vcc , but transistor not off.

Thread Starter


Joined May 14, 2014
Thank you but I must use necessarily one terminal of v2 to gnd because is comes from a board which reference voltage is Ground.


Joined Jun 5, 2013
Show us the circuit you are using. If your circuit is the standard PNP switching circuit and the voltage you are switching is > 3.3V then the transistor will be on at 0V and also on at 3.3V. If this is the case, you will need a level shifter.

Edited: or better yet, use a low-side switch with an NPN transistor.

Last edited:


Joined Aug 1, 2013
I probed with 600 ohm series resistor at base, a 10 kohm pullup resistor from base to VCC( 12V) and 1k pullup resistor from colector to Vcc , but transistor not off.
The connections you describe are not correct for either a PNP nor an NPN transistor. Please post your schematic so we can see what you are doing.



Joined Nov 30, 2010
I second that. Your words do not make sense. Nobody can guess what, "v2" means.
At least you told us you have 12 volts.

Is this what you're doing?



Joined May 9, 2014
Your failure to post a schematic has many people wondering about the legitimacy of your problem or a need for attention.

You have 2 possible solutions, use the 2N3906 and wire it correctly or use a 2N3904 and wire it correctly.

Here's 3 example circuits that I have used on countless occasions without any issues and are known to work.

I've got this in my cheat-sheet for easy reference but using IC's has become my preferred solution.

The most commonly used transistor switch is the PNP variety shown in Figure 1. The secret to making a transistor switch work properly is to get the transistor in a saturation state. For this to happen we need to know the maximum load current for the device to be turned on and the minimum HFE of the transistor. For example, if we have a load that requires 100MA of current and a transistor with a minimum HFE of 100, we can then calculate the minimum base current required to saturate the transistor as follows:

Maximum Current Required = 100MA
Supply Voltage = 12 Volts

R1 = Supply Voltage / ( Maximum Current Required / Minimum HFE * 1.3 )
R1 = 12 / (.1 / 100 * 1.3)
R1 = 9230.7 or 10K for nearest standard value.

Resistor R2 is not essential to this circuit but is generally used for stability and to insure that the transistor switch is completely turned off. This resistor insures that the base of the transistor does not go slightly negative which would cause a very small amount of collector current to flow. The value of this resistor is not critical but a value about 10 times R1 is normally chosen. For this circuit we will calculate R2 to be 10 times R1 as follows:

R2 = 10 * 10000
R2 = 100K

To turn on our transistor switch all that is needed is to short resistor R1 to the negative ground.

While PNP transistors are normally used for a negative ground configuration, it is possible to use a NPN transistor if a positive ground configuration is desired as indicated in Figure 2. The calculation of resistor values is identical to the PNP version. However, in the NPN transistor, R1 must be shorted to the positive end of the supply to turn the switch on.

While our transistor switch can easily replace many mechanical relays, it does have a few drawbacks. The maximum design current must not be exceeded or the output voltage will be reduced. A short circuit of the output will overheat and destroy the transistor in many cases. Although the transistor is in saturation when turned on, about .3 volts is lost through the collector to the emitter of the transistor. We must also insure that the maximum power dissipation of the transistor is not exceeded. We can calculate the power dissipation by multiplying the current by .3 volts. In the case of 100 MA, the transistor must be able to withstand 30 milliwatts (.3 times .1).​