# Little to no gain in common emitter amplifier

#### goldenthyme

Joined Apr 18, 2021
7
Hey guys, it's me again. I'm trying to build a simple collector emitter amplifier with current source and Vbe multiplier(used for later) for an audio amplifier.

It is driving a 40kohm load in this simulation.

Here is the input (green) and the output (blue):

If I'm using a current source instead of a collector resistor, how exactly do I bias the amplifier to have a certain amount of gain? I'm really lost as to how to adjust it.

#### crutschow

Joined Mar 14, 2008
32,861
How did you come up with such an odd circuit?
Did you calculate any of the circuit values or are they just guesses.

Since you have a perfect current source I1 providing the collector current, Q2 does nothing.
What is Q2 supposed to do?

With a -35V emitter voltage source and a 22 ohm emitter resistor, you are pulling about 1.5A through Q1's base-emitter junction which will totally saturate the transistor.

You can't use an ideal current source for the collector as it will bias it either to a very high voltage or saturation.

Here's an example of how to bias a common-emitter amplifier.

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#### LvW

Joined Jun 13, 2013
1,669
While speaking of a "simple emitter amplifier" I recommend to design at first a really simple stage with 4 resistors and one transistor.

#### Danko

Joined Nov 22, 2017
1,735
Hey guys, it's me again. I'm trying to build a simple collector emitter amplifier with current source and Vbe multiplier(used for later) for an audio amplifier.
Instead of current source, connected in series with collector, here collector current of Q1 is stabilized by parallel connected repeater (Q2, Q3).
Q1 voltage gain is 4000.

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#### goldenthyme

Joined Apr 18, 2021
7
The design is based on one in a book by Bob Cordell. Here is the full amplifier design:

I'm trying to replicate the voltage gain stage, which is Q4,5,6,7. Any suggestions as to how my diagram differs from his circuit?

#### Danko

Joined Nov 22, 2017
1,735
It differs by bias:

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#### goldenthyme

Joined Apr 18, 2021
7
Ok, so I managed to get the transistor bias properly but it doesn't seem to play well with a non ideal current source. Here's the full circuit I drew up with an ideal current source:

With the non ideal current source in place (the sub circuit on the right) I get lots of distortion:

Why would the ideal current source behave nicely but the non ideal one change the signal so severely? I thought the 2 transistor feedback source was fairly stable.

#### Danko

Joined Nov 22, 2017
1,735
Connect your power supplies 35 V in right order (polarity), and this circuit will work.
Pay attention to ground points.

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#### goldenthyme

Joined Apr 18, 2021
7
Connect your power supplies 35 V in right order (polarity), and this circuit will work.
Pay attention to ground points.
View attachment 236352
I'm still seem to be getting some type of clipping:

Forgive for my ignorance but I'm a bit of novice with circuits...

Does it have something to do with the fact I am driving a 40kohm load and not the output stage?

#### Ian0

Joined Aug 7, 2020
8,378
That's because you are running the whole think with no feedback. A long-tailed pair driving a current mirror, followed by a common emitter amplifier will have a gain of around 100dB.
You have also connected both ends of R12 to ground.
There's a very good example in Linsley-Hood's book "Audio Electronics".

#### Ian0

Joined Aug 7, 2020
8,378
Sorry - wrong book - it’s “Valve and Transistor Audio Amplifiers”

#### Danko

Joined Nov 22, 2017
1,735
I'm still seem to be getting some type of clipping:
Does it have something to do with the fact I am driving a 40kohm load and not the output stage?
40k load is shorted to ground.

You can start from working source of amplifier.
Voltage gain G = (R2 + R3) / R2 = (1 + 19) / 1 = 20.

Diode D1 may be replaced with 3 in series connected diodes 1N4148. In this case R5 value should be changed to 560 Ω.

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#### Veracohr

Joined Jan 3, 2011
771
In case you’re not aware, the Vbe multiplier isn’t part of the voltage amplifier stage, even though it’s in the collector of that stage. It’s there for the output stage.