Question

Hand Calculation: 1. With Cs, calculate Am. . . BW and GB for Ri=100 KS2 and R-10 k2 assuming gm1.04 mA/V and Cos=CGD=CB 1.8 pF. Try to deduce the trade off between gain and bandwidth from these results. 2. From the results obtained in step 1. try to realize the effect of RL on the AM and the bandwidth 3. Without Cs, calculate Am. fi. fu, BW, and GB when Ru=100 k22. 4. From the results obtained in step 1 and 3. try to recognize the effect of C, on the Am and the BW. 5. Record your results in Table 2.

i don't know what you mean ,but the question can be answered withing the given information

Answer 1

SOLO steps: for For De analysis Ac sources (shorted voltage coureus) (ohen circuited for current sources). grounded & DC roures are For 2 AC Ac analysis analyns fi= 27.6-15) X166 x (1k+ 300k 1l 100) Wis ci (Rsigt RillR2) to= 13.96 HZ (104.3k 21 x 0 15x10 W2 - (RP+ R62. 2 to = 10.17 HZ Bw = total = 3.79 HZ

Gain = RbIRL) → Frequene = -1.04 mm (4.3k X100k wp.sk AV = – 4.2876 More Gain with out as: (RolIRL) AV = isted a 19m+ Rs. st Ar = – 1.790 with out es? gain reduced compared to with 'cs'. Bypass capacitor cs'is used to stabilire the circuit it increases gain in ac analyies. In ac analysis es stabilizes urmit.

és is Frequency due to Sn 27T CSC RS 1 Yom) b= 277. 1000 (1-3K/LIK) f. = 28-158 HZ