# small signal gain of transistor

#### ntetlow

Joined Jul 12, 2019
52
I'm trying to work out the gain of the transistor in the attached circuit. I must be doing something wrong, could someone please tell me what error I'm making?
The Q point I'm using is 1.25mA which gives a gm of 0.0481 (1.25/26).
The peak small signal voltage I get as 0.46v.
This gives a collector current of 11mA (0.46 * 0.0481 * 0.5). This is way too much as the small signal gain from Ltspice is showing as 0.9mA.

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#### LvW

Joined Jun 13, 2013
1,466
Please, would you be so kind and tell us the meaning and the derivation of the expression (0.46 * 0.0481 * 0.5) ?
It seems that you are doing "something wrong" - however, without information about the circuit and the corresponding formulas used it is impossible to give you an answer.
As far as the last sentence is concerned - do you really think that the small-signal gain is given in milli-amps?

#### ntetlow

Joined Jul 12, 2019
52
Please, would you be so kind and tell us the meaning and the derivation of the expression (0.46 * 0.0481 * 0.5) ?
It seems that you are doing "something wrong" - however, without information about the circuit and the corresponding formulas used it is impossible to give you an answer.
As far as the last sentence is concerned - do you really think that the small-signal gain is given in milli-amps?
Hello,
thanks for replying. Sorry I didn't make things clear. In the expression the 0.46 is the voltage at the small signal peak voltage at the transistor base (the voltage above the q point voltage). The 0.0481 is the gm calculated, and the 0.5 is the multiplier factor required. This all courtesy of Basic Audio Amplifier (ecircuitcenter.com) .

#### MisterBill2

Joined Jan 23, 2018
12,424
It will be useful if you show the equation that you have used to calculate the gain.
And that LTspice file is useless to those of us who do not use simulators. Besides that, simulators are at best approximations of how a circuit may function.

Also, there is no mention of the circuit configuration, is it common emitter or common collector??

#### Audioguru again

Joined Oct 21, 2019
4,798
Are you blind? Does the +12VDC have any signal on it?
Please lookup Emitter-Follower then try the circuit with a 10uF capacitor.

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#### MisterBill2

Joined Jan 23, 2018
12,424
Are you blind? Does the +12VDC have any signal on it?
Please lookup Emitter-Follower then try the circuit with a 10uF capacitor.
INDEED!! My computer is blind, it does not see files with a ".asc" extension. And I did not follow that second link.
And still I am asking about the formula the TS used to calculate the gain.

#### crutschow

Joined Mar 14, 2008
30,116
simulators are at best approximations of how a circuit may function.
In most cases, it's a pretty good approximation.

#### crutschow

Joined Mar 14, 2008
30,116
The Q point I'm using is 1.25mA which gives a gm of 0.0481 (1.25/26).
The gm value of a emitter follower circuit has only a small effect on its current gain.
If put a large capacitor across the emitter resistor, then the gm will determine the AC gain.

#### MisterBill2

Joined Jan 23, 2018
12,424
Once again I am asking the TS to show the formula, WITH UNITS that they used to calculate the gain.
and putting a "large" capacitor across the emitter resistor will tend to reduce the output voltage swing and effectively reduce the gain.
and is it voltage gain or current gain or maybe POWER gain that is in question here.

#### Audioguru again

Joined Oct 21, 2019
4,798
The gain calculation above assumes the transistor is common emitter, not common collector.
A capacitor parallel with the emitter resistor in a common emitter transistor increases the voltage gain and increases the even-harmonics distortion.

Here is the small signal voltage gain that is slightly less than 1 because the input capacitor is too small to pass 40Hz without the loss produced by it.

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#### LvW

Joined Jun 13, 2013
1,466
Hello,
thanks for replying. Sorry I didn't make things clear. In the expression the 0.46 is the voltage at the small signal peak voltage at the transistor base (the voltage above the q point voltage). The 0.0481 is the gm calculated, and the 0.5 is the multiplier factor required. This all courtesy of Basic Audio Amplifier (ecircuitcenter.com) .
It seems that the question is related to a long-tailed pair (diff. amplifier).
Perhaps the questioner has lost some interest.

#### MisterBill2

Joined Jan 23, 2018
12,424
Why not use the correct formula for calculation of the current gain in a common collector amplifier circuit? Certainly it is more common to use the hfe or the HFE terms to calculate the gain.
And once again, a string of numbers without units has very little value in trying to find the error.
Please be advised that very few of the participants in this thread are able to see into other folks minds and know what they are thinking. My mind-reading ability is exceptionally poor.

#### ntetlow

Joined Jul 12, 2019
52
It seems that the question is related to a long-tailed pair (diff. amplifier).
Perhaps the questioner has lost some interest.
Hello,
No, I have n't lost interest in the problem though I'm close to getting a blinding headache with it.
I repeat that my info source is Basic Audio Amplifier (ecircuitcenter.com) , trying to work our gm for the transistor I would draw your attention to the phrase
' The transconductance of Q1 and Q2 is gm = Ic1 / VT = 0.5 mA / 26mV = 0.0192 (A/V). The transconductance of the input stage, gm1 = Ib(Q3)/V(2,3), is double to that of the basic amplifier (1/2*gm) thanks to the current doubling of the current mirror Q8, Q9' extracted from the above website.
I have also attached my Ltspice simulation for the above.

Thank you.
ps for voltage V(2,3) the simulation would have V(n010) - V(n012).

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#### LvW

Joined Jun 13, 2013
1,466
So you are dealing with the whole multi-stage amplifier?
trying to work our gm for the transistor I would draw your attention to the phrase
' The transconductance of Q1 and Q2 is gm = Ic1 / VT = 0.5 mA / 26mV = 0.0192 (A/V). The transconductance of the input stage, gm1 = Ib(Q3)/V(2,3), is double to that of the basic amplifier (1/2*gm) thanks to the current doubling of the current mirror Q8, Q9' extracted from the above website.
Transconductance gm for Q1 and Q2 is OK - but what do you mean with "input stage"? Perhaps Q3? And Q3 has the transconductance gm1 (your notation)?
What is really your problem? It is not clear at all.

#### ntetlow

Joined Jul 12, 2019
52
So you are dealing with the whole multi-stage amplifier?

Transconductance gm for Q1 and Q2 is OK - but what do you mean with "input stage"? Perhaps Q3? And Q3 has the transconductance gm1 (your notation)?
What is really your problem? It is not clear at all.
First of all, it's not my "input stage" , this is a quote from the passage in the website,
My aim is to corroborate what is written in the website. It is to ideally find the same results as them
That the gm is 0.0192. That the result for their IC1 is the same as in my simulation etc.

#### MisterBill2

Joined Jan 23, 2018
12,424
I have observed over the years that not everything presented on the internet is correct, and even some of what may be somewhat correct is applied properly. Some sites are good, and some are not. And some are rather useless.

#### crutschow

Joined Mar 14, 2008
30,116
The gain of a single transistor is much different than a differential pair, since the second transistor significantly changes the emitter impedance that the first transistor sees.

#### Audioguru again

Joined Oct 21, 2019
4,798
In your new amplifier circuit with the differential input transistors, the output has a DC offset voltage because the input transistors have different base currents because the resistances of Rin1 and Rf2 are different.

#### LvW

Joined Jun 13, 2013
1,466
The transconductance of the input stage, gm1 = Ib(Q3)/V(2,3), is double to that of the basic amplifier (1/2*gm) thanks to the current doubling of the current mirror Q8, Q9' extracted from the above website.
May I remeind you that any transconductance is defined as the OUTPUT current divided by the INPUT voltage?