Question

01. Find appropriate values for the coupling capacitors Cai and Co and the bypass capacitor Cs for a CS amplifier as shown in figure below for which Rg = 4.7 M2, Rp =R_ = 15 ks2, Rsig = 100 k22, and gm=1 mA/V. For the amplifier, it is required to have fi at 100 Hz and that the nearest break frequency be at least a decade lower. a. Find the value of the capacitance Ccı and Cc and the bypass capacitor Cs? ZR cc MHD ŽR

Answer 1

solution: 'rinen Ru=4.7M2, RD=RL = 15kr Rsig= looks and gm= 1 mal f is at 100Hz. exon wie Rue cs cs= bypass Capacitar viss C., & Ccr are Coupling capacitat The above circuit can be reduced to Single Time Constant (STC) circults that are Composed of one capacitance and one reactance. To find the finae Constant and the break in frequencies associated with each of the w ee following capacitors, we have to follow iter Bollѕи, дws 2 -

③ we have to reduce Veig to Jeso. 0 Take one capacitar at a time and short circuit the other two capacitors. & For each capacitor calculate the Resistance seen between its terminals. - Ccil RG + Rsig.) Ren, wi- cc, (Rut Rsig) (Break or corner frequence . Of High Pass tybe very as Ws = // = fm RD § (break frequency of High Pass typi rum moring or a coz (RD+R() feca trebate Wi- / = cca (Rot R2). Lue "o break dirawe M ( break frequency of high pars wa RL type) W s is generelly higher (because of grm) than w, and we oo Ws is the highest frequency break point which will determine the lower 3 dB

karJ 5L. hod/decade ode 40 dB/ducts 68 se/ ready! dochen fe=fs= Zaca I'm to t3 tits (43) (log scales) = 1x10-3 = 16 UF 2TX fa 20x100 For nearest break frequency to be at least a decade lower (as given in question) d=f3=1049. (4.8 - 100 Hz = 1043) TO- 2EXCc, (0.142844.78X106 T = 100 me » Can = = 3.32 nF 27x19.8710 9.6 π and 10 = Te Can (15+1597/05 Cc2= GTX 105 X 105 1 X100F 67 0.53 MF -

To find the value of coupling and bypass capacitor, above circuits has to be reduced to single time constant circuits (STC) which consists of single resistance and capacitance. To constitute STC we have to follow the procedure as given below:

1. Reduce input voltage (Vsig to 0)

2. Take one capacitor at a time and short circuit the other capacitors.

3. Find the equivalent resistance across the terminals of the capacitor.

Ws will be the highest frequency breakpoint which will determine the lower 3dB frequency.

Final values of capacitances are :

CS= 1.6 $\mu$ F , CC1= 3.32 nF and CC2= 0.53 $\mu$ F