# Difference between NPN & PNP transistors

Discussion in 'General Electronics Chat' started by YorksireDave, Dec 5, 2015.

1. ### YorksireDave Thread Starter New Member

Dec 9, 2010
5
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I've read many descriptions of how transistors work but apart from understanding that the current flow is reversed, I just don't get why or in what circumstances one would use either type.

Is the following correct ?

For a PNP transistor to turn on - used as a simple switch - the voltage at the base must higher or equal to the voltage at the base
For an NPN transistor to turn on - used etc - the voltage at the base must be LOWER than the voltage at the emitter

So, with an NPN with 12v at the base and 12v at the emitter its off but when you drop the voltage down to say three volts on the base it turns on. With a PNP with 12v at the emitter and 12v at the base its hard on. Drop the base voltage down to 3v and it turns off.

I apologise for apparently trivialising such a wonderful object, but I really do want to start to understand and my simple mind works best with the kind of 'which means that' simplicity of everyday life...

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2. ### #12 Expert

Nov 30, 2010
16,659
7,304
Too much reading! You have bipolar transistors and mosfets mixed up.
Start with the basic number line from math: negative numbers are less than positive numbers, and zero is in the middle.

For each polarity of a transistor, there is a reversed polarity transistor. For each mosfet, there is one version that is normally on and one that is normally off, and each of them has a reversed polarity version. I think in terms of electron flow, but this website has declared that we are going to use, "conventional" flow so I will be writing this as if something positive flows, and that is called current.

Start with bipolar transistors (because that's where we started in historical terms), the NPN and the PNP.
The bipolar transistors are remembered by their middle letter. In case I miss some repetitive verbiage, almost all the sentences here imply, "with respect to the emitter". For an NPN, P means positive, and when the base is positive with respect to the emitter, the transistor moves toward, "on". The base-emitter junction is like the gate of the valve (transistor). Yes, the word, "valve" is British for "vacuum tube", but transistors are just as much a valve as a vacuum tube is.

So, with an NPN transistor, you usually apply a voltage to make the collector voltage more positive than the emitter, at which point, nothing happens. Then you apply some positive voltage to the base, through a resistor, and current starts flowing from the base to the emitter, and a lot more current starts flowing from the collector to the emitter. Then you add resistors to modify and limit the amount of current that is going to flow.

So, with a PNP transistor, you usually apply a voltage to make the collector voltage more negative than the emitter, at which point, nothing happens. Then you apply some negative voltage to the base, through a resistor, and current starts flowing from the emitter to the base, and a lot more current starts flowing from the emitter to the collector. Then you add resistors to modify and limit the amount of current that is going to flow.

Now, the mosfets. With an N-channel mosfet, N means a negative voltage will pinch off the current flow, so I think of an N-channel mosfet as an N-pinch mosfet. The basic arrangement of voltages is like an NPN transistor in that you generally start with a more positive voltage on the drain with respect to the source. Your usual enhancement mode mosfet is normally off, much like an NPN transistor. Zero current, base to emitter with an NPN transistor resembles zero volts gate-to-source with an enhancement mode mosfet. As you apply a more positive voltage to the gate (with respect to the source) the mosfet starts allowing current to flow from the more positive drain to the source. The big differences here include the fact that a bipolar transistor needs current flow for the gate(base) to turn the transistor on, but a mosfet needs more positive voltage on the gate to turn the transistor on.

For the P-channel mosfet, reverse several of the words and you have that one categorized in your head.

Then we get to the depletion mode mosfet. Again, start with an N-channel mosfet. It is called depletion mode because it is normally "on" and you must apply some negative voltage (with respect to the source) to turn it off. Reverse the polarities and you have the P-channel depletion mode mosfet categorized.

Then we get to j-fets. They use a reverse biased P-N junction for their control mechanism. j-fets are normally "on" and that current is called IDss which means, Current through the Drain with gate Shorted to Source (Zero Vgs). The N-channel j-fet normally operates similar to the NPN with a positive voltage supplied to the drain and the Volts Gate-to-Source controlling the current flow, Drain to Source. More negative voltage on the gate pinches off the flow.

There is also a P type j-fet.

This is a good start. There is plenty left to discuss, and the other denizens of AllAboutCircuits will contribute plentifully. If I got something backwards, they will correct me.

Last edited: Dec 5, 2015
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3. ### hp1729 Well-Known Member

Nov 23, 2015
2,071
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Close. For an NPN transistor to turn on the base must be about 700 mV more positive than the emitter. For a PNP to turn on the base must be about 700 mV more negative than the emitter. While conducting this 700 mV will vary only slightly, only a few 100 mV between fully off and fully on. The more current you draw through the base the lower the resistance will be between the emitter and collector.
Why use one over another? What voltage do you have for an input and what voltage do you have for the output? Do you want an applied voltage to turn in on or a ground in to turn it on?
Does that help?

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4. ### #12 Expert

Nov 30, 2010
16,659
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Getting to quantities now. A bipolar transistor actually works clear down to the pico amp range, but that isn't usually useful. I have attached a graph in which I measured the collector current from 1 nanoamp to 1 milliamp as the Volts, base-to-emitter changed from 0.24 volts to 0.63 volts. When you try to work at currents less than 1 milliamp, the impedance of the circuit is so high that you are in danger of picking up stray voltage radiated by the power wires in your house, to the AM radio band, and right on up in the frequency domain. People get the idea that 600 millivolts to 700 millivolts (base to emitter) is the working range of a bipolar transistor because 1 ma to 10 ma is a useful current range for a small bipolar transistor. The other end of the spectrum is described by something like an MJ2955 where figure 8 shows a Vbe as high as 1.8 volts when you're trying to run 10 amps through a transistor as big as a silver dollar.

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5. ### #12 Expert

Nov 30, 2010
16,659
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A mosfet, on the other hand, is a voltage operated device. Some are called, "Logic Level" because they can allow significant drain current with less than 5 volts from Gate-to-Source. The older mosfets needed voltages in the range of 5V to 10V to reach their maximum current flow. I don't think there is any such thing as a mosfet that can work in the Vgs range of zero to 1 volt, but progress keeps happening, and I could become "wrong" at any moment.

A j-fet always works in the pinch region. An N-channel j-fet will usually have its gate more negative than the source because they are naturally "on" and you modulate them with a less positive voltage on their gate, generally in the zero to negative 4 volt range. j-fets were manufactured with their IDss in the range of 5 uA to 150 ma, the last time I checked. You just choose one that conducts more current than you need and throttle it down to where you want it to operate.

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6. ### WBahn Moderator

Mar 31, 2012
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Read that statement again and ask what it means for the voltage at the base to be higher than the voltage at the base. Is that even possible?

I think you meant that the voltage at the base must be higher or equal to the voltage at the emitter. This is almost correct except that it is the description for an NPN transistor, not a PNP. Furthermore, for any appreciable current to flow, the voltage at the base must be one diode voltage drop (typically 0.5 V to 0.7 V) higher than the voltage at the emitter.

Closer (but still with the transistor type reversed). The voltage at the base of a PNP transistor must be about one diode voltage drop lower than the voltage at the emitter.

Go back and revisit these descriptions in light of what you have learned. Also, with bipolar transistors you will generally not be able to forward bias the base-emitter junction with much more than one diode voltage drop without destroying the device.

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7. ### KeepItSimpleStupid Well-Known Member

Mar 4, 2014
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Take a look here http://efxkits.com/blog/difference-between-npn-and-pnp-transistor/ and here https://www.google.com/url?sa=t&rct...sistor&usg=AFQjCNHtUMB1VRIaeThkJN3gah-hMalDqg

One thing to remember,is the way the emitter is drawn. In the NPN, conventional current will flow from the Base to the emitte rin the direction of the arrow on the emitter.

Ben Franklin made a mistake, so we have "conventional current" from positive to negative. Electrons, go the other way. It matters in electro chemistry and somewhat in solid state physics.

Also note, as an aside, Power is positive for "power dissipated. That means for analysis our 4 MW power plant is -4 MW.
Solar cells may be another area where current doesn't have the correct sign.

8. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
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Here is a simulation of two simple transistor circuits powered from 12V; Circuit A employs an NPN while its complementary circuit B employs a PNP. It shows what happens to the voltage at the output of both circuits as the input voltage V(in) sweeps from 0V to 12V.

A uses an NPN transistor as a simple common-emitter inverter (amplifier, switch), where as the input voltage V(in) increases, the voltage at the collector decreases at a faster rate. Note the plot of V(a), Red, plotted as Volts on the Y axis vs the input voltage along the x-axis, also in Volts.

Note that the collector voltage is at 12V as the input sweeps upward from 0V to ~0.6V, and by the time the input reaches ~1.1V, the collector voltage V(a) has fallen to near 0.3V. As the input sweeps upward from 1.1V to 12V, the collector voltage V(a) falls a bit more toward ~0.1V.

B uses a PNP transistor as a simple common-emitter inverter (amplifier, switch), where as the input voltage V(in) decreases, the voltage at the collector increases at a faster rate. Note the plot of V(b), Green, plotted as Volts on the Y axis vs the input voltage along the x-axis, also in Volts.

Note that the collector voltage is at 0V as the input sweeps downward from 12V to ~11.4V, and by the time the input reaches ~10.7V, the collector voltage V(b) has risen to near 11.7V. As the input sweeps downward from 10.7V to 0V, the collector voltage V(b) rises a bit more toward ~11.9V.

Do you see the complementary symmetry in these behaviors?

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9. ### YorksireDave Thread Starter New Member

Dec 9, 2010
5
2
#12 Therin lies the problem. YOU understand it. To YOU it all fits and flows. To ME, all you've unwittingly done is add to my confusion ;-)

10. ### YorksireDave Thread Starter New Member

Dec 9, 2010
5
2
A genuine thank you to everyone for trying to help this old fool...

#12. Unfortunately, you've confused me even more, but I do now understand (I hope) that transistors are operated by current levels and MOSFETS operated by voltage levels. So, some real progress. Thanks.

WPBahn. According to some of your colleagues one can use a voltage divider and get less voltage on the base. Is that wrong? Thanks for the rest it has clarified things for me somewhat. Unfortunately when you spoke of forward biasing I'm afraid the fog decended and you lost me ;-(

KISS. TBH, I'm not sure what you are getting at. My apologies.

MikeML. Ahh. The clouds are parting... So, with an NPN you apply current to the base to turn it off. With a PNP, you apply current to turn it on.
Can I run this by you all?

I understand the base resistor is there to deliver an appropriate current to the base (does that mean base current MUST be limited?). What I'm not sure about is why R1 & R6 are there! Is it to simply limit current flow and to stop whats effectively a short? Where do you connect a load? Is it always at the point being measured (A or B)?

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11. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
You would need to sketch a schematic of what you are talking about in order for me to answer it. Yes, you can use a voltage divider to get less voltage on the base (depending on the circuit), but if you do it the wrong way you merely turn off the transistor.

The reason that I lost you when I mentioned forward biasing is because you are trying to jump too far forward without laying the necessary foundation. Before you start even thinking about transistors, you need to become comfortable with diodes. They are much easier to understand and concepts like forward and reverse bias and forward voltage drop will make a lot more sense if you start there.

Wrong. In either case you need base current in order to turn on the transistor. The collector current is a multiple of the base current. No base current, no collector current. In an NPN the current flows into the base (using conventional current) while in a PNP the current flows out of the base.

In general you need to make sure that the base current is limited somehow (the same goes for the collector current for that matter). There are several ways to do it depending on the circuit being used. A base resistor is a common way, though often not the best way if you are trying to make an amplifier and not just a switch. R1 and R6 are there because, if they weren't, then the outputs would simply be tied to the power supply rails and would be at a fixed voltage that the transistor couldn't change (unless the internal resistance of the source is large enough). In some circuits, they also serve to limit the current, particularly if the transistor is being used as a switch.

I HIGHLY recommend you take a step back and get familiar with diodes and diode circuits.

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12. ### kyka New Member

Jun 7, 2015
23
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Imagine that R1 and R6 were not there (or are simply 0). What would be the output?

The location of the load depends on the topology used. There are three basic transistor topologies and this is a common-emitter because the emitters are connected to a constant voltage. Since the input is applied to the base, the only point left for the load is the collector.

13. ### #12 Expert

Nov 30, 2010
16,659
7,304
But, yes, for me it fits and it flows.
I tried to start simple and show that the inverse behavior and reversed voltages work for each kind of transistor because that's how I sort them in my mind. That's why the paragraphs keep getting smaller as I go through each type of transistor. That's also why there are dozens of people here to answer questions. You might get a dozen different answers that all say basically the same thing, but one of them lights up in your mind. If I failed, that's OK. My style fits some people, and it doesn't fit some other people. You are responsible for choosing which person speaks your language and address that person for details. The main thing to know right now is, This is a conversation. It flows both ways. Ask a small question and get a small answer if that's what floats your boat. You started with a rather large question and got a rather large answer. You are free to create a different question now that you have a few examples of how each person communicates.

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14. ### Techno Tronix Member

Jan 10, 2015
140
10
The main difference between the two types of transistors is that holes are the more important carriers for PNP transistors, whereas electrons are the important carriers for NPN transistors. Then, PNP transistors use a small base current and a negative base voltage to control a much larger emitter-collector current.

15. ### hp1729 Well-Known Member

Nov 23, 2015
2,071
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Nice drawings. What software do you use for that?

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18. ### signal6271 New Member

Dec 1, 2015
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I agree with New. It is confusing, due to the fact that we all will have a different understanding/interpetation of the same device. Keep looking and trying to find a possible understanding .

19. ### Sinus23 Member

Sep 7, 2013
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465

I tried to make it as simple as I could by just using NPN''s since the original poster was wondering how a transistor as a switch works

Notice the collector-emitter voltage drop. In the circuit on the left I tied the base to the ground and the voltage drop across the collector to emitter is almost the same as the supply voltage. The 1K Ohm resistors voltage drop is trivial.(You would have to have a pretty decent meter to measure it) But when I supply current to the base the voltage from collector to emitter becomes trivial and most of the voltage drop is across the 1K ohm resistor.

Then if you check the current flowing through the R1 to the negative terminal of the battery, you can see that there is almost none(I say almost none because there might be such a small current that it could be disregarded in most cases) But in the second circuit the base gets supplied with enough current to saturate the transistor so the voltage drop across the collector emitter becomes really small so the current through Rc becomes almost, I=supply/Rc

This is a gross simplification and the datasheet for different transistors will give you more detailed information. But for me it was the CE voltage drop that make the idea of a transistor as a switch somewhat click

20. ### Sinus23 Member

Sep 7, 2013
178
465
After reading through the posts I had forgotten the name of the thread...Here are the same circuits using PNP's