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The transfer function of the following circuit is (I read A(s) on the solution of my professor):

enter image description here

$$A(s)=A_{CB} \frac{s \space \omega_{pC2}}{(s+\omega_{pC1})(s+\omega_{pC2})}$$

where Nevertheless is the pole of \$C_i\$ and \$A_{CB}\$ is the midband gain. Why there is the pole \$\omega_{pC2}\$ at numerator?

\$C_1=1\mu F\$ and \$C_2=100pF\$

\$\omega_{pC1}<\omega_{pC2}\$

Please show me where \$\omega_{pC2}\$ at numerator comes from. I applied the seen resistance method; this allows me to obtain \$\omega_{pC2}, \space \omega_{pC2}, \space A_{CB}\$. Given these 3 parameters, I want to write A(s) like A(s) above.

Thank you very much in advance.

Gennaro Arguzzi
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  • Some qestions: (1) What is the meaning of "pole of C1"?; (2) If Acb is the midband gain, the remaining frequency-dependent part of A(s) should be identical to "1" at w=SQRT(w1*w2). But this is not the case! What is your expression for the midband gain? – LvW Nov 11 '16 at 11:18
  • Hi @LvW, (1) $\omega_{pC1}$ is the pole due to the presence of C1. (2) If I write the A(s) in Bode form, I obtain $$A(s)= \frac {A_{CB} \space s}{\omega_{pC1} \space \left(1+ \frac {s}{\omega_{pC1}}\right) \space \left(1+ \frac {s}{\omega_{pC2}}\right)}$$ where $$\frac{A_{CB}}{\omega_{pC1}}=-43.83 dB$$ is the value of magnitude in w=1. This result is ok, I verified it in LTSpice. – Gennaro Arguzzi Nov 11 '16 at 11:27
  • The expression of midband gain is: $$A_{CB}=-\frac{h_{fe}R_C}{h_{ie}+R_E(h_{fe}+1)}=-4.717$$ – Gennaro Arguzzi Nov 11 '16 at 11:38
  • This expression neither contains C1 nor RB. This cannot be fully correct - unless its contribution is so small that you have neglected it. Did you neglect it from the beginning or based on the result for ACB? – LvW Nov 11 '16 at 12:35
  • The C1 capacitor is a short circuit --> ACB is not affected by RB//Vin; C2 is an open circuit. – Gennaro Arguzzi Nov 11 '16 at 13:12
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    In this case, you would have a frequency-independent gain stage. Why do both capacitors appear in the diagram (and in your "pole frequencies") ? It is not clear what you are asking for. – LvW Nov 11 '16 at 13:46
  • Hi @LvW, ACB is frequency-independent gain because is evaluate with C1 short and C2 open. – Gennaro Arguzzi Nov 11 '16 at 17:32
  • OK - so you calculate the frequency dependent gain formula (bandpass response), but you do NOT use this formula for finding the midband gain. Instead, you are using another simplified model (without capacitive influences) for this task. To me, this is not a very "consistent" way. Nevertheless, what is your problem now? – LvW Nov 12 '16 at 09:47

1 Answers1

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At first, I will try to answer your main question (concerning the numerator) based on a basic RC-Bandpass.

The transfer function is H(s)=sR1C1/[1+s(R1C1+R2C2+R1C2)+S²R1R2C1C2

As you can see, it is the TIME CONSTANT of the first RC element which appears in the numerator. But this is not the inverse of one of the two pole frequencies. Instead, we have two REAL pole frequencies (on the negative real axis in the s-plane). These pole frequencies are computet by setting the denominator equal to zero.

(And, of course, the result will be NOT simply w1=1/R1C1 and w2=1/R2C2. )

The midband gain can be found by dividing numerator and the s-element of the denominator Ao=R1C1/(R1C1+R2C2+R1C2). From this expression one can see which role the expression in the numerator plays (it determines the midband gain - together with the other two parts).

schematic

simulate this circuit – Schematic created using CircuitLab

LvW
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