# The common-emitter amplifier - CORRECTION

Discussion in 'Feedback and Suggestions' started by mrchen, Oct 6, 2013.

1. ### mrchen Thread Starter New Member

Oct 6, 2013
7
1
In "The common-emitter amplifier" article:

the phrase:

"output voltage is inversely proportional to the input signal strength. "

IS INCORRECT becase, by definition, A and B are inversely proportional when AxB=Const.

Actually Vout is (approximately, taking into account DC offsets) directly proportional to Vin with a negative multiplier, i.e.
Vout = Const x Vin, where Const is negative.

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2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Inversely in this context means inverting. I believe the paragraph to be correct as is.

Not verified.

3. ### MrChips Moderator

Oct 2, 2009
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3,451
I agree.

The output is proportional to the input but out of phase with the input, i.e. decreases as the input increases.

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4. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
Isn't it true that the circuit shown will invert the input signal, but in a non-linear fashion due to the dead-zone before the transistor actually begins to turn on?
Can that be considered truly (inversely) proportional?

5. ### Wendy Moderator

Mar 24, 2008
20,772
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Transistors in general are operated within their linear region, the two extremes being saturation and cutoff and considered outside this example.

6. ### WBahn Moderator

Mar 31, 2012
18,079
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I disagree.

The output is directly proportional to the input voltage, it's just that the proportionality constant happens to be negative.

The statement that Y is directly proportional to X means that Y = kX where X is the "constant of proportionality". There is no constraint that says that k has to be greater than zero.

If you want to make the inverting nature more explicit, then you can also say that Y is directly proportional to the negative of X.

But to say that Y is inversely proportional to X means that Y = k/X or that XY is a constant. And, again, that constant can be negative.

Or perhaps this might make it more obvious:

Is the force between two charged particles directly or inversely proportional to the square of the distance between them?

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7. ### Georacer Moderator

Nov 25, 2009
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I agree with the OP.

When I read that A and B are inversely proportional, I understand that $A \cdot B = c \in \Re$

On the other hand, in the inverting amplifier case, we have $A / B = -c, c \in \Re^+$

Last edited: Oct 7, 2013
8. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
I don't think that c even has to be a member of the reals.

After all, the voltage across an impedance in the frequency domain is directly proportional to the current through that impedance and the impedance is, in general, complex.

9. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I made an error in my post. I have corrected it now.

Regarding your remark, yeah, I guess you are right. I just didn't want to think too hard on the generalization onto the Complex Realm.

10. ### Dcrunkilton E-book Co-ordinator

Jul 31, 2004
416
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Thanks mrchen,

For pointing out this unfortunate choice of words.

Changed from:

An important quality to note here about the common-emitter configuration is that the output voltage is <italic>inversely proportional</italic> to the input signal strength.

to:

An important quality to note here about the common-emitter configuration is that the output voltage is <italic>inverted</italic> with respect to the input signal.

at ibibliolorg. And, credited to mrchen in the Contributors List

Dennis

Last edited: Feb 1, 2014