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This is a simple Class A amplifier using an NPN transistor:

Class A NPN Amplifier

Is there a way this can use a PNP transistor, instead (with only 0V and +Vcc, not 0V and -Vcc)? Why or why not?

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  • \$\begingroup\$ Not sure this is a bad enough question to warrant the downvote it got, but why would you want to use a PNP when an NPN works perfectly as it is? \$\endgroup\$ – Polynomial Feb 12 '13 at 11:49
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    \$\begingroup\$ Because I want to know if it's possible, and if it isn't, why it can't be. 'Just use an NPN' is maximally unhelpful. \$\endgroup\$ – Ehryk Feb 12 '13 at 11:52
  • \$\begingroup\$ I would imagine that you probably could build a similar amplifier with a PNP, but it wouldn't look like this, and would therefore cease to be a common emitter amplifier. Though don't take my word for it - I'm no good with analogue stuff. You might want to expand your question to specify why you want to know. Academic curiosity is a perfectly valid reason. In the meantime, you get my upvote because I'd be interested to know too. \$\endgroup\$ – Polynomial Feb 12 '13 at 11:55
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    \$\begingroup\$ If you want a non-inverting amp, then you are asking the wrong question. Any PNP equivalent would also be inverting. If you need non-inverting, look at common collector (AKA emitter follower) which has current gain but voltage gain=1, or common base which has voltage gain but current gain=1. And an NPN will do either of these. \$\endgroup\$ – Brian Drummond Feb 12 '13 at 12:08
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    \$\begingroup\$ Common emitter has both voltage gain and current gain. Technically, it has current gain; but the high output impedance allows you to translate that into voltage gain by using a high load resistor. \$\endgroup\$ – Brian Drummond Feb 12 '13 at 13:28
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Any amplifier that can be made with an NPN BJT can also be made with a PNP. Whether its inverting or not really depends on how the output is interpreted. More on that later.

To convert this NPN common-emitter amplifier to a PNP common-emitter amplifier, just mirror the entire thing except for the supply voltages top-for-bottom:

PNP common emitter amplifier

Equivalently, you can just swap the NPN transistor for a PNP transistor, and replace +Vcc with -Vcc. However, the convention is to draw schematics with higher voltages at the top, so the resulting schematic would look a bit funny.

Remember that voltages are relative. The only important thing with a PNP common emitter amplifier is that the emitter is at a higher voltage than the other terminals, the base is approximately 0.6V lower than the emitter, and the collector will be lower than the emitter, with how much lower controlled by the base current and load.

If we call the highest voltage in the circuit "ground", then we can have negative voltages. Or, we can call the lowest voltage "ground" and have positive voltages. We can even pick a voltage in the middle and have both. Or, we can ignore ground altogether and talk about the voltage "drop" or "across" a component or between any two points in the circuit.

This is the same circuit, just with a different notion of "ground", which is completely irrelevant to the operation of the circuit, only our discussion of it:

with ground at top

Really, the terminal labeled "output" is a voltage somewhere between the other two terminals. Is it inverting? Well, are we considering the signal to be the output relative to the higher voltage at the top, or the lower voltage at the bottom?

Let's get rid of the names which the electricity doesn't know about, and draw the whole circuit, with a power supply and all:

without labels

There is no "ground" and there is no "Vcc"; there's just a battery with two terminals, one with a higher potential than the other. We can call them what we like; the circuit doesn't care, as long as there's that voltage difference there.

There is also no "output" terminal, but instead we have two voltage differences, either of which could be considered the "output": \$V_a\$ and \$V_b\$.

When the input \$V_{in}\$ is low, this forward-biases the transistors base-emitter junction more, turning it on more, making it effectively look like a smaller resistor and allowing more current in \$R_L\$. By Ohm's law, if the current over a resistor increases, so too will the voltage across it. Or, you can think of \$R_L\$ and the transistor as making a voltage divider. Either way, you can see than when \$V_{in}\$ goes down, \$V_b\$ goes up.

\$V_a + V_b\$ must be equal to the battery voltage, since they are in parallel with the battery. So if \$V_b\$ is going down, \$V_a\$ must be going up to make up the difference.

So is it inverting? I can't say! Is \$V_a\$ or \$V_b\$ the output?

What if \$V_{in}\$ is connected between the + side of the battery and C1, instead of the - side of the battery and C1, as it is now? Then is this an inverting amplifier, or not?

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  • \$\begingroup\$ 1) The output signal is coming out underneath the PNP, above RL, correct? 2) The reason I was clear about +Vcc and 0V is that I'm using an Arduino where GND and +5v are clearly labeled, where calling +5v '0v' and GND '-5v' seems odd; though I realize conceptually they represent the same thing. 3) What software did you use to make the diagrams? 4) Thank you! \$\endgroup\$ – Ehryk Feb 12 '13 at 13:12
  • \$\begingroup\$ @Ehryk forget about the names. The electricity can't see those labels anyway. See edits. Schematics were drawn with gschem. \$\endgroup\$ – Phil Frost Feb 12 '13 at 13:48
  • \$\begingroup\$ So both PNP and NPN transistors in this design are non-inverting with respect to the higher voltage potential/emitter side, and inverting with respect to the lower voltage potential/collector side? \$\endgroup\$ – Ehryk Feb 12 '13 at 15:05
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    \$\begingroup\$ @Ehryk my advice would be to become comfortable with how this circuit works. Ask a new question with a finer point if you need help. Once you understand how the circuit works, the invertingness will be easy to figure out. \$\endgroup\$ – Phil Frost Feb 12 '13 at 15:15
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    \$\begingroup\$ @PhilFrost A common emitter circuit does not necessarily have its emitter on the lower voltage side, that is only true when using a NPN transistor. All "common emitter" tells us is that both input and output signals share the emitter as a reference. Strictly speaking, a common emitter doesn't have an \$\text{R}_\text{E}\$ \$\endgroup\$ – jippie Feb 12 '13 at 18:38

protected by Community May 8 '16 at 16:38

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