# Increasing the gain of a BJT amplifier without affecting dc bias

Reading through my textbook, I came across a basic common-emitter amplifier that drives a load. Please have a look at the circuit, and input/output waveforms shown in the image below.

How would one increase the gain without affecting the DC bias? From my point of view, the value of RE would determine this. Decreasing this resistance would increase the base current, thereby increasing the collector current. Am I correct in my reasoning? I would appreciate some guidance.

• In your setup,the gain is expressed as -Rc/rE,because the capacitor shunts RE. – Daniel Tork Sep 11 '16 at 10:23
• That is the voltage gain – Daniel Tork Sep 11 '16 at 10:31
• It would help to know what RL is, and the frequency range of interest. There's one obvious change you can make, but whether it's relevant depends on the additional info. – Brian Drummond Sep 11 '16 at 12:32
• Are you asking this because of the analogue electronics tutorial due today at UCT? – user123382 Sep 11 '16 at 13:10
• In the following Rl == Rc. Vl == V_Rc. AND Re = transistor internal emitter resistance. || Gain = VdcRc x 38.4 [!!!!!!] - Really. So here gain = [12-8.2] * 38.4 = 145.9. This is because gain = Rl/Re and Re = 26/Ie . For Ie ~~= Ic you get V_Rl = Il.Rl = Ie.Rl. Play with that and you end up with the "magic" Gain = 38.4 x Vl | SO the DC collector bias point is crucial in setting the gain. – Russell McMahon Sep 11 '16 at 13:16

Yes, you are right, decreasing $R_\textrm{E}$ would increase gain, but this would violate your constraint "without changing DC bias".

In that circuit in-band voltage gain is basically $A_\text{v}=-g_\text{m}R_\text{C}$
-it should be RC||RL, but the latter is not given, I assume is open circuit-

Then $g_\text{m}=I_\textrm{C}/V_\text{T}$ and finally $I_\text{C}\approx I_\text{E}=V_\text{E}/R_\text{E}$ as long as we do not drive transistor out of active region.

So putting all toghether $$A_\text{v}\approx - \frac{V_\text{E}}{V_\text{T}}\,\frac{R_\text{C}}{R_\text{E}}= -\frac{2.13\,\text{V}}{26\,\text{mV}}\times\frac{1\,\text{k}\Omega}{560\,\Omega}\approx -145$$ with your numbers

and indeed lowering $R_\textrm{E}$ increases gain BUT you have to consider also that this is surely going to reduce output dynamics (i.e allowed output voltage swing), everything in life is a tradeoff.

Also, generally, take care to verify reduced $R_\textrm{E}$ will extra load base bias voltage divider with changes in $V_\textrm{E}$ too (your circuit has quite heavy divider though).

Finally, note as we don't need any BJT parameters (e.g. $\beta$) to make rather accurate gain calculations, that's a well-engineered circuit!

• Thanks so much for your answer! I haven't really covered too much content on transistors, so I am not too sure of the gain you have calculated in terms of $g_m$. So, what parameters of the circuit could be changed to increase gain without changing the DC bias? You stated that changing RE would violate the constraint. What else can I vary along with RE in order to keep the constraint? – abruzzi26 Sep 11 '16 at 13:46
• So to vary gm (I believe that problem is for someone who knows what it is) you have to vary Ic and hence bias point: forbidden. The other term in the gain expression (its knowledge required) is Rc||RL, if you vary Rc you change bias point again: forbidden. So what remains? – carloc Sep 11 '16 at 14:23
• From what I see, RE remains? – abruzzi26 Sep 11 '16 at 15:42
• @abruzzi26 He and I are saying that it can't be done - while RE is the external emitter resistor, what matters and what we are dealing with is what I called Re = internal emitter resistance of the actual transistor. The external RE is bypassed for AC by C2 so does not affect AC gain. Gain then becomes Rc/Re. But what I said and he says is that Re varies with mA in Re and in transistor and in Rc = all the same current (almost Ib slightly affects this). When you re arrange this and look here you find Gain and bias are hard-related. Alas. – Russell McMahon Sep 11 '16 at 23:09