# PNP switching voltages

I am trying to use a TIP2955 PNP as a simple switch for an amplifier circuit. (I know TIP2955 is overkill, but I have so many laying around). The amp requires 12V. My logic is 5V. PNP pinouts:

* Emitter connected to +12V
* Collector connected to amp input
* Base connected to logic circuit through a 1K resistor.


When the logic is disconnected the PNP is off. However, when I apply 5V to the base (via the 1K), the PNP switches on.

What is the required ratio between the emitter and base voltages? From a little testing, it seems this ratio is about 1/2. That is, the base voltage must be at least 1/2 the emitter voltage for the PNP to be in off state. This is unfortunate, as I need to control the 12V switch from 5V.

If you really want to go with the TIP2955 I would suggest you need an extra transistor (NPN type) to interface it with the logic circuit. Something like this.

TR1 switches 'ON' when the input voltage is greater than 0.6V. R3 controls the base current. With the value given this will be about (12 - 0.6)/100 = 114 mA. The current gain of the TIP2955 is between 20 and 70. R2 ensures that TR2 is fully OFF when TR1 is turned OFF.

• I was thinking about this, but hoping to avoid more soldering. Soldering on, hopes off. Thanks! :-) – Pål Thingbø Jul 6 '14 at 11:10
• I think R3 is a bit small... And maybe he can get away with it without adding a transistor. – Vladimir Cravero Jul 6 '14 at 13:14
• @VladimirCravero My assumption was that the amplifier would be taking a couple of amps and the minimum gain of the tip2955 was 20 giving a minimum base current of 100mA, hence the value of R3 chosen. Also, to turn the TIP2955 OFF the base voltage must be less than 0.6V from the 12V supply. Buffering the 5V logic output make sense in terms of current and voltage swing. – JIm Dearden Jul 6 '14 at 16:16
• I am sorry, you are perfectly right. I've not been paying enough attention when I read in the last couple of days, sorry. – Vladimir Cravero Jul 6 '14 at 16:17