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Topic: NPN & PNP transistors (Read 3244 times) previous topic - next topic

What is the difference between NPN & PNP transistors?
Does NPN transistors activate by connecting its base to Vcc? and connecting the base of a PNP transistor to ground will activate it?
(NPN: collector-positive; emitter-negative; base-positive || PNP: collector-negative; emitter-positive; base-negative;) Correct?
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johnwasser

My limited understanding is:

An NPN transistor acts as a "low-side switch", connected between the load and Ground.  A voltage greater than 0 on the base causes a small current to flow through the base-emitter junction and that allows a larger current to flow from the load to the collector.

A PNP transistor acts as a "high-side switch", connected between power and the load.  A voltage GREATER THAN THE POWER RAIL on the base causes a small current to flow through the base-emitter junction and that allows a larger current to flow from the power source to the load.  Because the PNP switch requires a base voltage higher than the power rail it is not as easy to use for power switching.
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Magician

Quote
A voltage GREATER THAN THE POWER RAIL on the base

I find that explanation quite confusing. Bipolar transistor is current amplifier, you apply small current trough base-emitter junction and you getting much bigger trough emitter-collector circuit.
The major difference from other kind of amplifiers ( magnetic, for example http://en.wikipedia.org/wiki/Magnetic_amplifier) is that polarity of the signal at the base and power source has to be observed, it doesn't work with AC.


johnwasser


Quote
A voltage GREATER THAN THE POWER RAIL on the base

I find that explanation quite confusing. Bipolar transistor is current amplifier, you apply small current trough base-emitter junction and you getting much bigger trough emitter-collector circuit.


I think the problem is that when the transistor conducts the voltage at the Emitter terminal becomes the same as the voltage at the Collector (let's say +5V).  In order to keep the transistor turned on you need, as you say, current flowing through the Base-Emitter junction.  To get a current to flow from Base to Emitter you need the voltage at the Base to be greater then the voltage at the Emitter.  I think that means you need the Base to be at a voltage greater than +5V.  That's just my limited understanding.
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retrolefty

Quote
I think that means you need the Base to be at a voltage greater than +5V.  That's just my limited understanding.


No, the emitter base voltage just needs to be higher then the forward voltage drop of the base/emitter junction, typically .7vdc. It's just like a led where there is no current flow until the anode/cathode voltage exceeds the forward voltage drop rating, say 1.5vdc for a common red led.

But as already posted you don't control the base emitter junction with voltage, but rather current, so speaking of base voltage is somewhat incomplete. You could apply 10 vdc to a base emitter junction, but if applied through a series 1 meg ohm resistor the transistor would not 'turn on'.

Lefty


winner10920

A npn transistor as a switch gets 5v at the base, emitter to ground and the collector gets the load, effectivetly switching low side
A pnp transistor as a switch gets 0 volts at the base, emmiter to 5volts and the collector gets the load, effectively switvhing the high side

johnwasser

This site explains a lot: http://www.w9xt.com/page_microdesign_toc.html

Of particular interest are:

Part 7 - Using Transistors to Switch Higher Power Loads
Part 8 - Using PNP Transistors as High Side Switches
Part 12 - Switching Higher Voltage Loads with PNP Transistors

From Part 8:
Quote

One additional thing you need to be careful with PNP high side switches is the voltage used to drive the load.  Normally it is best to use the same voltage to drive the load that is used to power the microcontroller.  Consider the following.:  Suppose the load voltage is +12V and the microcontroller is running at 5 volts.  Assume the output pin is high, at 5V, and the base resistor is 1000 ohms.  Base current would be calculated by
Ib =  (Vcc - Vp0 - Vbesat)/R1 =( 12 - 5 - .7)/1000 = .0063A = 6.3ma

Unless the transistor has exceptionally low gain, it will be turned on even though with the output pin high it should be turned off.  Worse yet, the microcontroller output pin will be seeing more than 5V, which greatly exceeds the usual limitation of the microcontroller supply voltage plus 0.3V.  The microcontroller would likely be damaged in this situation.  Part 12 discusses ways of driving a PNP transistor when the load must be driven with a higher voltage than the microcontroller.
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