Why is the signal inverted in a common emitter amplifier?
-
3\$\begingroup\$ Do you understand how a transistor works? \$\endgroup\$– Ignacio Vazquez-AbramsJun 10, 2015 at 14:53
-
3\$\begingroup\$ Have you ever looked at the schematic for a common emitter amplifier? The output is taken from the collector. What happens as you increase the base drive? What happens when you decrease it? \$\endgroup\$– R DrastJun 10, 2015 at 14:54
6 Answers
Why signal is inverted in a common emitter amplifier?
In simple terms: -
- If base voltage (input) increases, base current increases.
- If base current increases then collector current increases
- For a fixed collector resistor, when collector current increases, collector voltage (output) decreases
Therefore output is inverted to input.
-
\$\begingroup\$ You talking about collector "ac" current here, right? \$\endgroup\$– emaJun 10, 2015 at 15:36
-
\$\begingroup\$ ....collector voltage decreases because the voltage DROP across the collector resistor increases. \$\endgroup\$– LvWJun 10, 2015 at 15:48
-
\$\begingroup\$ is it possible to get a non-inverted output by taking output voltage across the collector resistor? \$\endgroup\$ Mar 11, 2017 at 3:16
-
\$\begingroup\$ Not with a constant dc supply voltage. \$\endgroup\$– Andy akaMar 11, 2017 at 10:30
In the common-emitter amplfier shown above, a DC bias point is set up by a voltage divider made up of R\$_{1}\$ and R\$_{2}\$. So with no signal coming in, the transistor will be partially on as shown in the graphic for the output (Q point).
If the input signal is negative, this will drive the transistor off, and when the transistor is completely off the output will be equal to V\$_{CC}\$ since there is no current flowing through R\$_{L}\$.
If the input signal is positive, this will drive the transistor on, and at its maximum saturation, the output will be equal to:
$$V_{out}= \frac{V_{CE} + VR_{E}}{VR_{L} + V_{CE} + VR_{E}}$$
which will go towards zero for a large value of R\$_{L}\$ compared to R\$_{E}\$, e.g. 2200 Ω and 220 Ω.
So as you can see by the graphs on the left and right side, the output is the inverse of the input.
I think @Andyaka may have missed a subtlety in question in your comment on his answer.
simulate this circuit – Schematic created using CircuitLab
Figure 1. (a) The circuit described in your question. (b) The circuit described in your comment.
Is it possible to get a non-inverted output by taking output voltage across the collector resistor?
The answer to this is "yes" if you don't mind it not being referenced to ground. We use this in the commen NPN LED driver. The LED lights when the input is high and turns off when the input is low. In this sense the logic is not inverted even though the collector voltage level with respect to ground is.
The best answer for that question would depend on what level of abstraction you are using for the transistor (what model etc.).
Take the hybrid-\$\pi\$ model:
The inversion is because of the direction of the current source is the reverse of the vbe, so an increase in vbe results in decrease of vce.
-
1\$\begingroup\$ Spehro Pefhany, with all respect: is this an explanation? One could ask: Why is the polarity of the current source as shown? More than that, can the direction of a current source be the reverse of a voltage source? \$\endgroup\$– LvWJun 10, 2015 at 15:46
-
\$\begingroup\$ @LvW Well, you can go to various models that are closer to the physics and always ask "why" once again and go deeper. At some level you'll have to invoke a deity or admit lack of knowledge. The hybrid-pi model voltage-controlled current source is a decent starting point for a model useful for analyzing small signal behavior of a BJT so I used that. There are worse ways to 'understand' transistor behavior. \$\endgroup\$ Jun 10, 2015 at 16:02
-
\$\begingroup\$ P.S. Specifically, the current-controlled current source is a poor one for analog circuit design (the key word in the original question being 'amplifier' not switch. It's better to think of the base current as a kind of defect, as Barrie Gilbert (inventor of the eponymous translinear cell used in many analog ICs) suggests. \$\endgroup\$ Jun 10, 2015 at 16:29
-
1\$\begingroup\$ Perhaps I was not clear enough: My only point was that such an eqivalent small-signal circuit can only reflect the properties of a device (BJT) but not explain! And the question was "WHY..." \$\endgroup\$– LvWJun 10, 2015 at 20:12
Couldn't find a subscript in the editing tools so here's an image from Word. This is what I understand from working through chapter 4 of "Learning the Art of Electronics", Hayes T:
-
\$\begingroup\$ What do you mean by "suffix". Do you mean subscript? e.g. VC? If so you're looking for MathJAX. On this site use
$$
for an equation on its own line and\$
for inline. \$\endgroup\$ Mar 4, 2018 at 21:57
It seems no one here was actually capable of explaining the quantum mechanics regarding why the signal is inverted, and one of you went as far as saying a BJT configured for simple class-A shuts off, but this is most definitely not correct. In its quiescent state, it is "listening" and the input signal modulates the D.C. current and we say it "rides the D.C." or "blends with the D.C."
Because the P-type semiconductor material is mostly positive in charge but not all the holes are occupied, it bucks the D.C. current away, and attracts it when it swings negative, and thus the output waveform is inverted. BJTs sound cleaner to me than FETs and MOSFETs, and I think part of that is some natural compression from the leakage current taming transients and sounds that are a natural part of a guitar's decay but undesired.
-
1\$\begingroup\$ Who is "we" in "we say it "rides the DC" or "blends with the DC""? I've never heard that before--certainly not a common turn of phrase among anglophone EEs. I'm also having a very hard time understanding what you mean by "bucks the current away, and attracts it when it swings negative". You might want to clarify in more precise language. \$\endgroup\$– HearthMar 13 at 20:26