PNP and NPN question

Thread Starter


Joined Mar 3, 2011
I'm having a hard time completely understanding PNP and NPN transistors. I know that PNP is meant to be pulled low and has a common anode and NPN is meant to be pulled high and has a common cathode. But why, for instance, could I not just use an NPN transistor in this circuit?:
Why do we need an NPN pulling a PNP low? Why can't we have the NPN emitter to the motor?

I'm not doing anything in particular, I'm just trying to understand this.



Joined Feb 12, 2009
Both these transistors are running in common-emitter mode.

This way, when the first transistor is switched on with a positive signal on the input, current flows INTO the base of the NPN transistor. This causes a the NPN transistor to switch on.

This causes a much larger current to flow OUT of the base of the PNP transistor. This causes an even larger current in the collector of the PNP transistor.

Two stages of current amplification.

Important note: current flows INTO the base of the NPN. It flows OUT of the base of the PNP.

"Conventional" current flow - from positive to negative. The actual electrons flow the other way but don't worry about that here!

Thread Starter


Joined Mar 3, 2011
Ok so in this schematic it is not necessarily important that one is NPN and one is PNP. It's only using two because the motor PNP requires a higher current(saturation?). So they could both be NPN if the motor transistors base was connected the the first transistors emitter instead of collector? Have I got it right?


Joined Jul 17, 2007
In that circuit, they used NPN/PNP so that they could both be saturated switches.

One glaring omission is that there is no current limiting resistor between the base of the PNP and the collector of the NPN. This means that the NPN transistor will run very hot, and will likely burn up.

You might use an NPN as an emitter follower to source more current to the base of an NPN that is used as a saturated switch. You would still need to limit the current flowing into each of the bases, or risk overheating the transistor(s). In the case of the 1st NPN, you should use a resistor on the base AND another on the collector in lieu of a resistor between the emitter and base of the 2nd transistor.

It's much easier to calculate the base resistors for the NPN/PNP pairs, as they are both saturated switches.
For the PNP transistor, it's base resistor needs to be sized to allow 1/10 of the desired collector current.
The NPN transistor needs it's base resistor sized to allow 1/10th of it's desired collector current, which is the base current of the PNP.
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Joined Mar 24, 2008
Agreed, follow the current path. There is a straight shot between the two diode junctions and the power supply. I've used similar circuits to reverse which power supply is coupled to the output of some very low current circuits (with moderate current outputs).

Take a look at the dual transistor driver on this circuit. I believe it does what you want.

MOS 555 Long Duration Red LED Flasher

Thread Starter


Joined Mar 3, 2011
I'm not working on anything, I'm just trying to understand transistors.

Could you explain why two NPNs could not both be saturated? I'm at a very basic level here. Just trying to understand why you would choose one or the other in whatever case. Thanks for your help so far.


Joined Mar 24, 2008
The function of the circuit I showed is simple enough, it is to use the other power supply rail instead of the one a conventional single common emitter design would use.

You need to figure out how to post your circuits here. It's isn't hard, many people use hand drawings along with either a digital camera or a flat bed scanner. I use M/S Paint with a set of templates to draw circuits or parts. Thing is, schematics are the language of electronics, where words will not suffice.

Different transistors in different configurations do different things. For example, a Darlington Pair meets your description of two NPN transistors, but there are many more configurations out there.

Same thing with a Sziklai Pair, which uses a NPN and a PNP.

Both of these pairs make what is occasionally called a super transistor. They look like a single transistor with exceptionally large gain, but with some minor differences.

I suspect you are wanting basics, to try to understand how transistors work. You will still run into the basic configurations, because they are also fundamental to how transistors work, and more importantly, how to use them.

Here is a good start...


Potato Pudding

Joined Jun 11, 2010
That circuit is probably fine. The resistor on the base of the NPN is critical.

It is a Sziklai pair which is a configuration used something like how Darlington pairs are used.

No I am wrong. The emitter of the NPN would need to be tied to the collector of the PNP to make a Sziklai pair.

It might still work but the base current of the NPN would be critical. If it is oversaturated the circuit fries.

The NPN needs to be used to control the base current of the PNP.
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Joined Jul 3, 2008
Could you explain why two NPNs could not both be saturated?
Think about what happens if you have an NPN with the collector tied to the top upper rail power supply (call it Vcc). In order for the transistor to turn on, Vbe (voltage drop from base to emitter) needs to be about 0.7 volts. The highest voltage available for Vb (the base voltage) is Vcc, which means that the emitter voltage is less than Vcc-0.7 volts. Now think about what Vce is. Vce must be greater than 0.7 Volts. However, a transistor in saturation will have Vce less than 0.3 volts typically. Hence, the transistor is not saturated and you do not have the capability to put all available voltage onto the load.

With a PNP with it's emitter on the top rail, you are free to fully saturate the PNP transistor and deliver nearly the full power supply voltage to the load. This is because the collector voltage is free to go above the base voltage.
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Joined Nov 30, 2010
Another person saying the same thing in a different way:

One reason this circuit is used is that the first emitter will always be at a lower voltage than the input. If you need full available voltage to get to the load you do this "double invert circuit" to get the same direction of voltage applied to the load, but with higher voltage and current than just connecting the original signal to the load or just connecting the first emitter to the load.

You can't just substitute an NPN for the PNP because the current would be going the wrong way to turn it on.