# How do I calculate limit frequency of the attenuator circuit?

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.

• Hint: the output of the current source is defined as being constant regardless of the load. – Dwayne Reid Mar 12 '17 at 14:58
• 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 $Fg = \frac{1}{2\pi R_{th} C_2}$ – G36 Mar 12 '17 at 15:01
• 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. – Tony Stewart Sunnyskyguy EE75 Mar 12 '17 at 18:42
• A better question is what does output signal look like at 100Hz CW and 1MHz with 100Hz AM? – Tony Stewart Sunnyskyguy EE75 Mar 12 '17 at 18:49