# Common emitter: maximize effective voltage gain

My homework is to design npn common emitter amplifier, with minimum output aplitude=5V, powered by +26V and -20V with generator resistance=40kOhm and output resistance=10kOhm. I have calculated effectice voltage gain formula and its derivative. Know that collector current Ic doesnt mean as much as collector resistor Rc value, but still I have no idea how to design(calculate) collector and than emitter resistor values. I even have no idea where should I start. Please guide me

some general drawing, no values:

gain is expressed as:

RL is parallel Ro and Rc. Rc as big as possible, but when transistor is cut-off, there makes a divider (does it?), so RL max is 42kOhm, RL=39kOhm (for e24 series). Than I got stuck.

• What is the input signal Vpk? (I gather its source impedance is 40k.) What is the required gain? Is it fixed? Or are you free to let it be "anything?" (Your grounded C20 means you don't have a specific gain and that you will have distortion needing NFB to repair.) Which brings up: How linear does the output have to be, relative to the input? Etc. Your R38 is too high to meet your output resistance requirement, already. This suggests profound confusion on your part. Finally, do you have to use this exact topology? – jonk Apr 19 '18 at 19:56
• Welcome to EE.SE! This appears to be a homework question. As such, you need to show us your work so far, and explain which part of the question you're having trouble with. For future reference: Homework questions on EE.SE enjoy/suffer a special treatment. We don't provide complete answers, we only provide hints or Socratic questions, and only when you have demonstrated sufficient effort of your own. Otherwise, we would be doing you a disservice, and getting swamped by homework questions at the same time. See also here. – Dave Tweed Apr 19 '18 at 20:10
• @jonk source resistance is 40kOhm, output is 10kOhm resistor. I do not have any requirements for gain. It's just effective gain that must be as high as possible. – kamillabaarnowak Apr 19 '18 at 20:15
• It's beyond the scope of this site to do a complete tutorial on CE amplifiers and how to design with them. First read your textbook and everything you can find with Google. Maybe go sit with your professor during office hours. Then give it another try and if you're stuck and have specific questions post another question here. – John D Apr 19 '18 at 20:19
• @kamillabaarnowak Are you aware of what happens when you apply C20? As the collector current varies, so does the dynamic emitter resistance and therefore also the gain. So you are able to accept substantial variable gain over the span of a single cycle? Also, again, are you stuck with this topology? Why did you select it? – jonk Apr 19 '18 at 20:20

The gain of the common emitter amplifier is not determined by the collector resistance, but rather by the Q point, where the collector resistance takes into account the load. What happens is that if you increase the collector resistance, the base-emitter resistance $r_\mathrm{e}$ increases to compensate for the same Q point.
Under the assumptions that the collector current is equal to the emitter current, the emitter resistor is perfectly bypassed by the emitter capacitor, and the transistor is in small signal operation, the gain is $$|A_\mathrm{v}| = \frac{R_\mathrm{c}}{r_\mathrm{e}} = \frac{(V_\mathrm{cc} - V_\mathrm{Q}) / I_\mathrm{Q}}{k_\mathrm{B}T / (q I_\mathrm{Q})} = \frac{q}{k_\mathrm{B}T} (V_\mathrm{cc} - V_\mathrm{Q}) \approx 40(V_\mathrm{cc} - V_{Q}),$$ where $I_\mathrm{Q}$ is the quiescent current, $V_\mathrm{Q}$ is the quiescent voltage across the collector, $r_\mathrm{e}$ is the base emitter resistance, $k_\mathrm{B}$ is the Boltzmann constant, $T \approx 290\ \mathrm{K}$ is the absolute temperature, $q$ is the electron charge, and $V_\mathrm{cc}$ is the positive power rail.