# Why does adding a capacitor increase voltage gain? [duplicate]

I'm trying to get a bit of intuition when dealing with transistor amplifiers. I recently watched this tutorial, and wanted a bit more elaborate explanation about why adding the cap circled in blue greatly increases gain.

I am aware that caps block dc signal, so adding the cap shorts the ac signal to ground, but why does that drastically increase the gain?

## marked as duplicate by Andy aka, PeterJ, laptop2d, Michael Karas, Dmitry GrigoryevAug 3 '17 at 12:05

• Draw the small signal equivalent circuit and compare the formula for AC gain. Note how C1 basically shorts Re above a certain frequency. – Bimpelrekkie Jul 29 '17 at 20:35
• Very similar question: electronics.stackexchange.com/questions/297098/… – Big6 Jul 29 '17 at 20:50
• Is the idea that the emitter voltage follows the base intuitive to you? If so here's a simple explanation: Wiggle the base up. The emitter wiggles up. The emitter current is then the ratio of this wiggle to the impedance of the resistor in parallel with the capacitor. The bigger you make the capacitor, the smaller this impedance, and thus the bigger the resulting emitter current for a given wiggle. This emitter current flows through the 680 ohm resistor. The bigger the current, the bigger the voltage drop across the 680 ohm. – rusty_old_jfet Jul 29 '17 at 21:00
• You also might want to have a look at this question: electronics.stackexchange.com/questions/271061/… – Bimpelrekkie Jul 29 '17 at 21:30

With a transistor having reasonable current gain, configured as a common emitter amplifier, the AC gain can be approximated by the ratio of the collector resistor to the emitter resistor. The capacitor across the emitter resistor greatly reduces the effective emitter resistance at frequencies where the capacitive reactance is less than the emitter resistor. Note that the input signal, Vin, basically appears across this emitter resistor, Re, so the emitter current is given by Vin/Re. This current, again for a reasonably high transistor current gain, is the same as the collector current. Thus the output signal is this current multiplied by the collector resistor, Rc, which is Vin(Rc/Re). Hence, as already mentioned, the voltage gain is Rc/Re. The capacitor reduces the effective value of Re, hence increasing the AC gain.

• Did you want to expand this a little? The reason I ask is that there's no mention of the AC impedance of the capacitor at likely frequencies (in this case, the value shown is almost certainly needs to be made larger to avoid this question) and there is no mention of little-re ($r_e\approx \frac{1}{g_m}$) which in this case happens to be a significant factor. – jonk Jul 29 '17 at 21:00
• I agree with Jonk's comment and would like to add that in my view approximations is not something I would use to explain circuits to beginners. An approximation can only be done under certain conditions which are often not recognized/understood by beginners so using an approximation often steers them in the wrong direction ! First priority for beginners is to fully understand what happens (analyze the circuit) without any approximations. In response to your question I'd ask: what is a reasonable current gain ? and when you say a number I'd ask Why ? – Bimpelrekkie Jul 29 '17 at 21:27
• @Bimpelrekkie, assuming 'ideal' conditions when learning something new is a valid pedagogical approach, otherwise beginners get so bogged-down with the minutiae they lose sight of the objective and lose interest. – Chu Jul 30 '17 at 8:12
• @Chu True and False, I think. No one-size-fits-all, bright line here. Sometimes, it is the very best thing to start someone off with where they are headed even if you don't belabor it right away. For example, I think its a good thing to tell someone that Sue may see a magnetic field as a result of charge motion in an experiment while Bob may not see a magnetic field at all looking at the exact same experiment if Bob is in a different frame of reference. BEFORE talking about relativity. It helps to know beforehand that what you are learning is already wrong (sometimes) before you even start. – jonk Jul 31 '17 at 4:21
• @Chu Yes, I get class size issues. Wide distribution of talents, skills, etc. (I taught at the largest university in my state of Oregon, with my class sizes being about 75.) I guess I just wanted to clarify things by pointing out that what is and isn't minutiae depends upon the topic, the direction ahead, and the target audience. In the case of this specific OP's question, it's hard to know. But I think Bimpelrekkie and you are both simultaneously right. "Develop enough added model detail to know what is being simplified and why." That ignores still deeper modeling. So you both may be right. – jonk Jul 31 '17 at 18:03

Lets examine the amplifier in 2 pieces: voltage-to-current converter, driving a current-to-voltage converter.

simulate this circuit – Schematic created using CircuitLab

Without the capacitor the transconductance of the NPN is degenerated by the emitter resistor Re. You can show that the degenerated transconductance is

$$g_{m,degen} = \dfrac{g_m}{1+g_mR_e} \simeq \dfrac{1}{R_E}$$

The gain of a CE amplifier with emitter degeneration is approximately

$$A_V = -g_{m,degen}R_C \simeq - \dfrac{R_C}{R_E}$$

With the bypass capacitor, in the passband the transconductance is unaffected by Re as its bypassed (AC grounded). Leaving you with the raw $g_m$ of the NPN (much larger). Which has an approximate gain of, $$A_V = -g_mR_C$$