The common-emitter amplifier - CORRECTION

mrchen

Joined Oct 6, 2013
7
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.

Wendy

Joined Mar 24, 2008
22,582

Common emitter amplifier develops voltage output due to current through load resistor.

With the solar cell darkened (no current), the transistor will be in cutoff mode and behave as an open switch between collector and emitter. This will produce maximum voltage drop between collector and emitter for maximum Voutput, equal to the full voltage of the battery.

At full power (maximum light exposure), the solar cell will drive the transistor into saturation mode, making it behave like a closed switch between collector and emitter. The result will be minimum voltage drop between collector and emitter, or almost zero output voltage. In actuality, a saturated transistor can never achieve zero voltage drop between collector and emitter because of the two PN junctions through which collector current must travel. However, this “collector-emitter saturation voltage” will be fairly low, around several tenths of a volt, depending on the specific transistor used.

For light exposure levels somewhere between zero and maximum solar cell output, the transistor will be in its active mode, and the output voltage will be somewhere between zero and full battery voltage. An important quality to note here about the common-emitter configuration is that the output voltage is inversely proportional to the input signal strength. That is, the output voltage decreases as the input signal increases. For this reason, the common-emitter amplifier configuration is referred to as an inverting amplifier.
Inversely in this context means inverting. I believe the paragraph to be correct as is.

Not verified.

MrChips

Joined Oct 2, 2009
24,436
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.
I agree.

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

tubeguy

Joined Nov 3, 2012
1,157
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?

Wendy

Joined Mar 24, 2008
22,582
Transistors in general are operated within their linear region, the two extremes being saturation and cutoff and considered outside this example.

WBahn

Joined Mar 31, 2012
26,398
Inversely in this context means inverting. I believe the paragraph to be correct as is.

Not verified.
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?

Georacer

Joined Nov 25, 2009
5,182
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 $$A / B = -c, c \in \Re^+$$

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WBahn

Joined Mar 31, 2012
26,398
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 $$A \cdot B = -c, c \in \Re^+$$
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.

Georacer

Joined Nov 25, 2009
5,182
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.

Dcrunkilton

Joined Jul 31, 2004
422
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

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