(a) For the inverting amplifier circuit to the right of node B in Fig 1.2, what
ID: 2085637 • Letter: #
Question
(a) For the inverting amplifier circuit to the right of node B in Fig 1.2, what is the expected closed-loop gain (as measured from node B to node D)? What is the input resistance viewed from the right of node B? (b) Consider the basic inverting op amp shown in Fig 1.2, with a 91 mV peak-to-peak signal applied at node B. For an ideal op amp, what signals would you measure at nodes C and D? (a) Using an ideal op amp, design a non-inverting amplifier with a gain of +11 V/V, low currents in the associated resistor network, but no resistor larger than 10 k ohm. (b) What is the gain of the circuit in Fig 1.3, from node B to node D, with R_1 shunted by a resistor of equal value? With R_2 shunted likewise? With R_2 shorted? Calculate the expected gains for individual inputs A, D, F to output C, of the circuit shown in Fig 1.4.Explanation / Answer
1 a closed loop gain is -r2/R1=-10
At c we will have a constant DC level at ground and at d we will have a peak to peak signal of 910mv which is out of phase of the input
2 a use the same circuit in 1.3 but change the resistor values, use R1=10000/11.1 ohms. So the gain will be (10+(10/11.1))/(10/11.1)*vin/1.1=121/10*vin/1.1=11*vin
2b R1 shunted gain is 21 as gain is (R1+R2)/R1= 10+0.5/0.5=10.5*2=21
R2 shunted is (1+5)/1=6
So R2 shunted gain is 6
If R2 is shorted then gain is 1 as R2 is zero then
3 input at a gain is -10 as b is shorted so -r2/R1
Input at d then gain is 1 as voltage at b is equal to that at e is vin/11 and c to b gain is 11 so net 11/11=1
Input at f then gain is 10 as similar to that of input at d but now the voltage at e and b is vin/1.1 so 11/1.1=10
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