The circuit (Figure. 3.86) of a cascade of op-amp circuits illustrate the reason

Question

The circuit (Figure. 3.86) of a cascade of op-amp circuits illustrate the reason why op-amp realizations of transfer functions are so useful. Find the transfer function relating the complex amplitude of the voltage mu out(t) to the source. Show that this transfer function equals the product of each stage's transfer function. What is the load impedance appearing across the first op-amp's output? Figure 3.87 illustrates that sometimes "designs" can go wrong. Find the transfer function for this op-amp circuit (Figure 3.87), and then show that it can't work! Why can't it?

Explanation / Answer

Let Vo1 be the voltage at the output of the first opamp The circuit is an inverting amplifier, its gain is given by AV1 = - Z2 / Z1
The second circuit is also an inverting amplifier, its gain is given by AV2 = - Z4 / Z3 The overall gain of the two stage amplifier is the product of the individual stage AV = AV1 * AV2 AV = -(Z2 / Z1 )(-Z4 /Z3 ) Vout / Vin = Z2Z4 / Z1Z3
AV1 = Vo1 / Vin AV2 = Vout / Vo1 [Vo1 / Vin][Vout / Vo1] = Vout / Vin
b) Due to virtual ground, the impedance Z3 will be in parallel with the feedback impedance Z2 Thus the impedance at the output of the first opamp will be Z = Z2|| Z3
c) In the given circuit
The series impedance Zs(s) = 1000 + (1/10*10-9s) The feedback impedance Zf (s) = 4700 || (1/10-6s)
The circuit is in the inverting mode, the transfer function is given by T(s) = Vout(s) / Vin(s) T(s) = - Zf  (s) / Zs(s) T(s) = - [ 4700 || (1/10-6s) ] / [1000 + (1/10*10-9s) ] T(s) = - [4700 *10-8s] / [ (1+4700*10-6s)(1+10-5s)]



Note:The custom equation editor is not working]


The overall gain of the two stage amplifier is the product of the individual stage AV = AV1 * AV2 AV = -(Z2 / Z1 )(-Z4 /Z3 ) Vout / Vin = Z2Z4 / Z1Z3
AV1 = Vo1 / Vin AV2 = Vout / Vo1 [Vo1 / Vin][Vout / Vo1] = Vout / Vin
b) Due to virtual ground, the impedance Z3 will be in parallel with the feedback impedance Z2 Thus the impedance at the output of the first opamp will be Z = Z2|| Z3
c) In the given circuit
The series impedance Zs(s) = 1000 + (1/10*10-9s) The feedback impedance Zf (s) = 4700 || (1/10-6s)
The circuit is in the inverting mode, the transfer function is given by T(s) = Vout(s) / Vin(s) T(s) = - Zf  (s) / Zs(s) T(s) = - [ 4700 || (1/10-6s) ] / [1000 + (1/10*10-9s) ] T(s) = - [4700 *10-8s] / [ (1+4700*10-6s)(1+10-5s)]



Note:The custom equation editor is not working]


AV = -(Z2 / Z1 )(-Z4 /Z3 ) Vout / Vin = Z2Z4 / Z1Z3
AV1 = Vo1 / Vin AV2 = Vout / Vo1 [Vo1 / Vin][Vout / Vo1] = Vout / Vin
b) Due to virtual ground, the impedance Z3 will be in parallel with the feedback impedance Z2 Thus the impedance at the output of the first opamp will be Z = Z2|| Z3
c) In the given circuit
The series impedance Zs(s) = 1000 + (1/10*10-9s) The feedback impedance Zf (s) = 4700 || (1/10-6s)
The circuit is in the inverting mode, the transfer function is given by T(s) = Vout(s) / Vin(s) T(s) = - Zf  (s) / Zs(s) T(s) = - [ 4700 || (1/10-6s) ] / [1000 + (1/10*10-9s) ] T(s) = - [4700 *10-8s] / [ (1+4700*10-6s)(1+10-5s)]



Note:The custom equation editor is not working]



AV1 = Vo1 / Vin AV2 = Vout / Vo1 [Vo1 / Vin][Vout / Vo1] = Vout / Vin
b) Due to virtual ground, the impedance Z3 will be in parallel with the feedback impedance Z2 Thus the impedance at the output of the first opamp will be Z = Z2|| Z3
c) In the given circuit
The series impedance Zs(s) = 1000 + (1/10*10-9s) The feedback impedance Zf (s) = 4700 || (1/10-6s)
The circuit is in the inverting mode, the transfer function is given by T(s) = Vout(s) / Vin(s) T(s) = - Zf  (s) / Zs(s) T(s) = - [ 4700 || (1/10-6s) ] / [1000 + (1/10*10-9s) ] T(s) = - [4700 *10-8s] / [ (1+4700*10-6s)(1+10-5s)]



Note:The custom equation editor is not working]


Thus the impedance at the output of the first opamp will be Z = Z2|| Z3
c) In the given circuit
The series impedance Zs(s) = 1000 + (1/10*10-9s) The feedback impedance Zf (s) = 4700 || (1/10-6s)
The circuit is in the inverting mode, the transfer function is given by T(s) = Vout(s) / Vin(s) T(s) = - Zf  (s) / Zs(s) T(s) = - [ 4700 || (1/10-6s) ] / [1000 + (1/10*10-9s) ] T(s) = - [4700 *10-8s] / [ (1+4700*10-6s)(1+10-5s)]



Note:The custom equation editor is not working]

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