I have designed the common emitter (see figure) based on the specifications (A=20, Rout=50 Ohm).
Now I want to choose the values of the capacitors, the specifications are Fl=200KHz and Fh=1MHz (at least).
For the lower cutt-off frequency I use the method of "short-circuit time constant", I analyze one capacitor per time, I find the equivalent resistor and so I impose the specific and I find the value of the capacitor.
I have uploaded the figures about the small signal models used to find the three capacitors.
Some parameters:
Vcc=+6 V, Vee= -6 V, Beta=170, Gm=0.4 S, Ic=10.04 mA, rbe= 420 Ohm, ro=7968 Ohm.
I choose to give the 10% of the weight to Ci and Co and 80% to CE because the resistor seen from Ce is the smallest.

About Ci:
Req1 = R1 || R2 || rbe = 360.9 Ohm
Ci = 1 / (2*pi*Fl*Req1*0.1) = 22.05 nF

About Co:
Req2 = RC || ro + R5 = 427.8 Ohm
Co = 1 / (2*pi*Fl*Req2*0.1) = 18.6 nF

About Ce:
Req3 = Re || (rbe + R1 || R2)/(Beta+1) = 16.6 Ohm
Ce = 1 / (2*pi*Fl*Req3*0.8) = 60 nF.

Note that I don't have included any generator resistance because this common emitter is only one stage of a bigger project and the signal will come from an operational amplifier. So if I simulate exactly this circuit I obtain an Fl=1MHz instead of Fl=200KHz, numerically it is like if there is a generator resistor Rg=50 Ohm in parallel to R1 and R2 OR like there is a factor beta=170 that divide the R1 || R2 on the calculations of Ce.
Further if I repeat the calculations of all the capacitors including a generator resistor Rg=50 Ohm, Ci and Co are about the same, Ce will be 360 nF and if I simulate the circuit without the Rg but with Ce=360 nF (so designed like the Rg is here) I obtain Fl=185 KHz so a good approximation of the 200 KHz.

Can someone help me about this mystery? Does PSpice include a resistor on the AC generators? Or what may be?

 

As far as I know, the generator output impedance in PSpice is zero.

Normally the coupling and emitter bypass capacitors are made much larger than needed so that they provide no signal attenuation at the lowest frequency of interest, (unless you are trying to filter out the lower frequencies).
Your calculations appear to be for a -3dB rolloff for each capacitor at the lower frequency (200kHz).
That give a lot of rolloff at 200kHz (My simulation shows a 16db rolloff from 1MHz to 200kHz).
Try increasing all the caps by a factor of 10 and see if that's better.

 

Thanks for your response.
Yes, my calculations are for the -3dB lower cut-off frequency.
I agree with you that increasing all the capacitors by a factor of 10 leads to a decrease of the lower cut-off frequency by about the same factor 10 that is a more desirable situation. However in my case the project is academic and it is required to satisfy (exactly) the specifications.
Furthermore make the capacitors bigger solves the problem but it doesn't explains why with my calculations I get a simulation result different from the theoretical analysis, that is what intrigues me.
"That give a lot of rolloff at 200kHz (My simulation shows a 16db rolloff from 1MHz to 200kHz)." do you mean with Ce = 60 nF? The problem is here, on the theoretical analysis this should be -3db but it is not, and with Ce = 360 nF it works but it is not the theoretical results.
How to explain it?

 

How to explain it?

First all the rolloffs add together.
So if each contributed -3dB at 200kHz, then the output rolloff would be -9dB.

In addition I think the Ce calculation is incorrect.
I think your calculation is for the gain being 3dB above the minimum gain (with no capacitor across Re), not 3db below the maximum (AC short across Re).

 

On the calculations of the capacitors, there are the factors 0.1 for Ci and Co and 0.8 for Ce that are like a share quote and this should lead to 200 kHz total.

It is interesting the note about the calculation of Ce but I don't get what you mean, can you please be more precise?
My calculation are made after short the independent voltage source and the other capacitors (Ci and Co) and finding the resistor seen from Ce.

 

With Rs = 0Ω the signal source will "short" R1 and R2 therefore Ce pole is equal to :

\(Fp = \frac{1}{2 \pi (R_E||r_e) C_E} \approx \frac{0.16}{r_e*C_E} \approx \frac{0.16}{\frac{442 \Omega }{171} *60nF } \approx 1MHz \)

We also have a zero at

\(Fz = \frac{1}{2 \pi R_E* C_E} \approx \frac{0.16}{R_E*C_E}\approx 8kHz \)

So the get -3dB around 200kHz we need Ce ≥ 0.16/(200kHz * 2.5Ω ) = 320nF

 

...........
It is interesting the note about the calculation of Ce but I don't get what you mean, can you please be more precise?

The main problem seems to be the value of rbe you used.
rbe = Vt/Ie where Vt is the junction thermal voltage of 26mV at room temperature.
Your circuit has an emitter current of about 10mA, giving rbe= 2.6Ω (not the 420Ω you used).
So when the emitter impedance to ground is comparable to that, the gain will be reduced by 3dB.
This occurs for a Ce of about 260nF at 200kHz in my LTspice simulation, as shown below, for Ce values of 1mF (maximum gain of ≈22.5dB) and 260nF (gain of ≈19.5dB).


`

 

With Rs = 0Ω the signal source will "short" R1 and R2

YESS!! This is the error I was looking for! Thanks you!

crutschow, for Rbe I mean the \( r_{\pi}=\frac{\beta}{g_m}\) and not the \( r_e=\frac{1}{g_m}\approx \frac{V_T}{I_C} \), without R1 and R2 on my calculations, \( \frac{r_{be}}{\beta+1}=r_e=2.45 \Omega \) and that leads to your analysis and results.

Thanks everyone.

 

Yes, the should be re not rbe in my post.