# Transistor amplifier gain

#### GhostLoveScore

Nov 27, 2016
71
Hello

I am experimenting with transistor amplifier and I was wondering - transistor voltage gain seems to be Rc/Re for this schematic

So, is there any difference between putting Rc=20k, Re=2k and Rc=2k, Re=0.2k. The gain should be the same, only the collector current is higher in second example. For strictly voltage amplifier, would the two configuration be the same?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,507
Yes, the gain will be approximately the same in both cases.

Can you think of some possible negative side effects of the higher current version? What if you used 20 ohms and 2 ohms?

And what about the positive effects? Do you understand output impedance?

#### GhostLoveScore

Nov 27, 2016
71
Yes, the gain will be approximately the same in both cases.

Can you think of some possible negative side effects of the higher current version? What if you used 20 ohms and 2 ohms?

And what about the positive effects? Do you understand output impedance?

With the common emitter configuration output impedance is Rc. Lower output impedance is good for driving the load. I don't really know why would we want amplifier with high output impedance.

With low values of Rc and Re I don't see any negative side effects other than higher heat dissipation in transistor. If the current is within transistor specifications it should behave the same as at any other current.

For example, my RF amplifier is using BF199 transistor, voltage divider bias, Rb1= 600k, Rb2= 50k, Rc=20k and Re=1k. It is biased as calculated, around 6-7 V. Quiescent collector current is 0.3mA. I don't know if that's a good design, with that small collector current.

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,507
With low values of Rc and Re I don't see any negative side effects other than higher heat dissipation in transistor. If the current is within transistor specifications it should behave the same as at any other current.

Would this result in a higher quiescent current drain? Could that be a bad thing?

Also, look at the current gain of the transistor with respect to collector current. Do you notice anything as the collector current rises? What impact would this have on the base current and hence the voltage divider which biases the base? And what would this do to the input impedance?

#### Audioguru

Sep 24, 2016
3,650
A transistor can have the emitter resistor bypassed with a capacitor or it can have no emitter resistor. But the gain is not infinite because the transistor has an internal emitter resistance that is "the emitter current number in mA divided by 26". This resistance is in series with an actual unbypassed emitter resistor.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,507
But the gain is not infinite

Yes, those formulae are somewhat simplified and only work for Gv << β and Re >> re, and DC (or at least low (audio) frequency).

#### LvW

Apr 12, 2014
604
But the gain is not infinite because the transistor has an internal emitter resistance that is "the emitter current number in mA divided by 26". This resistance is in series with an actual unbypassed emitter resistor.

To me, it is a severe misconception to think that re=1/gm would be a resistance belonging to the emitter.
Such a view results from a false understanding for the transistors working principle.
The bipolar transistor acts as a voltage controlled curent source - and the relation between input and output is characterized by the slope of the transfer characteristiv Ic=f(VBE). Hence, the transconductance is gm=d(Ic)/d(VBE).
This transconductance appears in the formula for the voltage gain: A=-Rc/(RE+1/gm)=-Rc/(RE+re) .
That means: For convenience. in the formulas we are using the inverse transconductance 1/gm=re (which is a transresistance).
This quantity re connects the output current and the input voltage, Therefore, it is not a resistance belonging to the emitter.

#### LvW

Apr 12, 2014
604
So, is there any difference between putting Rc=20k, Re=2k and Rc=2k, Re=0.2k.

Yes - there is a difference. Without mentioning other points (power consumption, operational point, collector current,...) it is the correct gain formula which gives the difference:

Gain A=-Rc/(RE+1/gm).

The transconductance gm is the slope of the Ic=f(VBE) characteristic and can only be neglected if RE>>1/gm.
The value of gm can simply be found by differentiating the given function and results in gm~Ic/Vt (temperature voltage Vt~26mV at room temperature).

#### GhostLoveScore

Nov 27, 2016
71
Would this result in a higher quiescent current drain? Could that be a bad thing?

Also, look at the current gain of the transistor with respect to collector current. Do you notice anything as the collector current rises? What impact would this have on the base current and hence the voltage divider which biases the base? And what would this do to the input impedance?

Of course it would increase quiescent current. It can be a bad thing if the circuit is running on batteries. For the same amplification factor it would draw more current - depleting the battery much sooner.

First thing that I notice is that as base current is increasing, the collector-emitter voltage becomes less and less linear in active region. It doesn't have that "knee" after which Vce is not dependent on collector current. I have to think about the rest of the questions, I don't have answer to them right now.

#### GhostLoveScore

Nov 27, 2016
71
Yes - there is a difference. Without mentioning other points (power consumption, operational point, collector current,...) it is the correct gain formula which gives the difference:

Gain A=-Rc/(RE+1/gm).

So, for larger emitter resistor 1/gm has little effect on gain while for smaller emitter resistor 1/gm has larger effect - it reduces gain?

#### Audioguru

Sep 24, 2016
3,650
The gain is calculated with the collector resistor (parallel with the load resistance) divided by the emitter resistor in series with the internal emitter resistance (0.026/emitter current). If the gain is AC then the calculation uses the unbypassed emitter resistor value.
Then a higher emitter resistor value results in less gain.

Yes, the 1/gm is a pretty small resistance so it has a small effect on gain when the emitter resistor value is high.
When there is no emitter resistor or its resistance is low then 1/gm provides a high gain.

#### LvW

Apr 12, 2014
604
Yes, the 1/gm is a pretty small resistance so it has a small effect on gain when the emitter resistor value is high.
When there is no emitter resistor or its resistance is low then 1/gm provides a high gain.
Without emitter degeneration resistor RE (or when it is bypassed for AC) the gain is A=-gm*Rc. If this value is considered to be "high" depends on the DC current Ic and the collector resistor Rc.

Example (Ic=1mA and Rc=2kohms): A=-(Ic/Vt)*Rc=(1E-3/25E-3)*2E3=-80.

#### GhostLoveScore

Nov 27, 2016
71
I have one more side question - I am messing with this circuit

I didn't build that part on the left of the R5/R6 that's for audio. I added FET buffer and after that BF199 amplifier on the Q2 collector. I calculated and biased it properly. But when I connect emitter capacitor, operating point changes, goes from 6V to 1.5V or similar. That's without the input to the transistor base. I am guessing that AC is getting in somewhere and messing the biasing when emitter capacitor is not there?

FET buffer

amplifier

#### Audioguru

Sep 24, 2016
3,650
If you built it on a solderless breadboard then the messy wires all over the place are picking up all kinds of interference that are overloading the transistor and causing it to rectify them.
On my FM transmitter I needed to add a little 100pF filter capacitor between base and emitter of the mic preamp transistor.

Nov 27, 2016
71

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