Question

Q #2 (Total 5 marks): Using an ideal op-amp model for the following circuit, a) Find the voltage gain at 300 MHz frequency. [3 mark] b) What is the value of Vo(t) if the input voltage is Vn(t) 5 sin(10t) V (assuume initial charge on the capacitors is zero). 2 mark] C1 20nF Cz 10nF in

Answer 1

1)
X_c = 1/jwc = -j/wc = -j/2 pi f c f = 300 MHZ, x_c_1 = = j/2 pi times 3 times 10^8 times 20 times 10^-9 = -j(0.026) ohm X_C_2 = -j(0.053) ohm 0 - V_in/1k + 0 - V_0/-j 0.053 + 0 - V_0/-j 0.026 = 0 V_0/V_in = -1800/57.32 j = -17.44 j Vertical-bar V_0/V_in Vertical-bar = 17.44 = A leftarrow Gain \r\n X_c = 1/jwc = -j/wc = -j/2 pi f c f = 10, x_c_1 = -j/2 pi times 10 times 20 times 10^-9 = -j(795 times 10^3) x_c_2 = -j(1591 times 10^3) ohm 0 - V_in/1k + 0 - V_0/x_c_1 + 0 - V_0/x_c_2 = 0 V_0 = -V_in(530 times 10^6 j) V_0 = -2650 times 10^6 sin(10 t + 90 degrees) v(+) = v(-) = 0, virtual short is valid because opamp is ideal and in negative feedback. in both part i applied kcl at V(-) terminal of the op amp. you have to use calculator to do the this complex
2)v(+) = v(-) =0, virtual short is valid because opamp is ideal and in negative feedback.

in both part i applied kcl at V(-) terminal of the op amp.

you have to use calculator to do the this complex calculations  

and j gives +90 deg phase shift.

this op amp is inverting op amp thats why gain and Vo is in negative value