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enter image description here Calculate the limit frequency \$fg\$ of the attenuation \$v_u =\frac{ u_a} { u_e}\$ at the current source I (for the value given in example), so that for \$f>> fg\$ exist ideal behavior. Note: First consider each capacitor individually by setting the impedance 0 and consider which of the resulting frequencies is relevant.

My problem:

I solved first part of example for \$C_2 \mapsto \infty\$, but I am having problem with another part.

\$C_1 \mapsto \infty\$

\$v_u =\frac{ u_a} { u_e}= \frac{r_D || (\frac{1}{i \omega C_2} +R_L)}{r_D || (\frac{1}{i \omega C_2} +R_L) +R} =\frac{r_D(1+iR_L\omega C_2)}{i\omega C_2(r_DR_L+r_DR+RR_L)+r_D+R}=\frac{r_D}{r_D+R}\frac{X}{1+iY}\$

So I dont know how to get the last step, how to write it in that form, because \$ Y=\frac{1}{\omega_g}\$ and after that I can find \$f_g\$ easy.

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  • \$\begingroup\$ Hint: the output of the current source is defined as being constant regardless of the load. \$\endgroup\$ – Dwayne Reid Mar 12 '17 at 14:58
  • \$\begingroup\$ Maybe it will be easier for you if you will try to find the "resistance" seen from C2 capacitor terminals. Rth = RL + rd||(RG+R). And \$Fg = \frac{1}{2\pi R_{th} C_2}\$ \$\endgroup\$ – G36 Mar 12 '17 at 15:01
  • \$\begingroup\$ Upon reflection of the diode resistance, I think the signal is so attenuated that peak diode current is almost double the mean DC 10mA resulting in a very low resistance and high attenuation (e.g. Ua=1% of 3V) with high distortion of sine wave making the cut-off frequency rather ill defined. -3dB or x% THD. just looking at rD avg = ( ΔVd / ΔId ) I expect this to make the answer quite non-linear with compressed positive sine more than negative sine and not a simple clamp result. \$\endgroup\$ – Sunnyskyguy EE75 Mar 12 '17 at 18:42
  • \$\begingroup\$ A better question is what does output signal look like at 100Hz CW and 1MHz with 100Hz AM? \$\endgroup\$ – Sunnyskyguy EE75 Mar 12 '17 at 18:49
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Assuming that the questioner really wants to ignore Rg's effect on the frequency response and starts from Ue:

Replace Ue, R and the diode with their Thevenin equivalent:

enter image description here

The most band limited case is with the highest attenuation, when rd is the smallest, practically =0 Ohm for a small signal AC.

In the high attenuation case the RC highpass filter RL & C2 is the decisive part. Solve the fg-candidate from RL and C2.

If you know something spesific about the diode, you can calculate the rd, In that case you can consider this to be a RC highpass filter that has the resistance = RL + Parallel(R,rd)

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