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I solved the circuit and found these values for each one. But I wanted to know:

Which capacitor has more influence in the low frequency and in the amplification?

Does RE have impact in amplification?

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    \$\begingroup\$ Welcome to SE EE. Since you "solved" the circuit already, you can easily re-calculate what happens if you for example double the value of \$R_E\$. Does doubling \$R_E\$ impact the DC bias solution? So does gm change? \$\endgroup\$ – Bimpelrekkie Jun 11 at 7:02
  • \$\begingroup\$ I only had some values that were missing, so firstly I solved them, and now I wanted some more information(the questions I asked). \$\endgroup\$ – Enn Jun 11 at 9:20
  • \$\begingroup\$ It is a relatively simple task to find the various 3-dB corner frequencies (wi) of the three RC combinations. You only have to compute the corresponding time constants (TAU)i=1/wi. The TAU values are the products of Ci with the connected R combinations. \$\endgroup\$ – LvW Jun 11 at 9:56
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Gain is determined by the ratio of the collector resistance (actually parallel combination of RL & RC) to the emitter resistance (RE + re).

Where re = 25mV/Ie.

With C3 in place, RE is shorted out as far as the signal is concerned and so the emitter resistance just equals re thereby increasing the gain substantially. RE then just has an effect on the DC bias conditions.

Each of the three capacitors combines with its associated resistance to form three high pass filter. If the three filters all had the same cut-off frequency (f = (1/(2*pi*RC)) then the output would be down 9dB at this cut-off frequency because the three filters are effectively cascaded. To put the response down 3dB at the same cut-off frequency all three capacitors would need to be doubled in value.

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