The circuit of an amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1.

Question 3

The circuit of an amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1.

Fig. 7.1

(a) By reference to the properties of an ideal op-amp,

(i) explain why point P is referred to as a virtual earth, [4]

(ii) derive an expression, in terms of the resistances R1 and R2, for the gain of the amplifier circuit. [4]

(b) In the circuit of Fig. 7.1, the ratio R2 / R1 is 4.5.

The variation with time t of the input potential VIN is shown in Fig. 7.2.

Fig. 7.2

                                                                                           

On Fig. 7.2, show the variation with time t of the output potential VOUT. [3]

 [Total: 11]

Reference: Past Exam Paper – November 2017 Paper 41 & 43 Q7

Solution:

(a)

(i)

{For an ideal op-amp,} the gain of the amplifier is very large           

V⁺ is connected directly to earth (at zero potential)                           

For the amplifier not to saturate,                                                        

the difference between V^- and V⁺ must be very small           or V^- must be equal to V⁺ {or else, the op-amp’s infinitely large gain would lead to saturation}

(ii)

{For an ideal op-amp,} the input impedance is infinite.                     

(So,) current in R1 = current in R2                                                                        

{From Ohm’s law: I = V / R}

(VIN – 0) / R1 = (0 – VOUT) / R2                                                                                

{VIN / R1 = – VOUT / R2 }

(gain =) VOUT / VIN = – R2 / R1                                                                                 

(b)

graph: correct inverted shape (straight diagonal line from (0,0) to a negative potential, then a horizontal line, then a straight diagonal line back to the t-axis at the point where VIN = 0)    

{Gain = - R₂ / R₁ = - 4.5

Gain = V_OUT / V_IN = - 4.5

V_OUT = - 4.5 V_IN }

horizontal line at correct potential of (–)9.0 V

{When V_IN = 3 V, V_OUT = - 4.5 × 3 = - 13.5 V

But the supply voltage is ± 9V, so the V_OUT saturates at - 9 V.

That is, when V_IN = 3 V, V_OUT = - 9 V}

both ends of horizontal line occur at correct times (coinciding with when VIN = 2.0 V)

{We need to find at which value of V_IN the output becomes – 9 V.

V_OUT / V_IN = - 4.5

- 9 / V_IN = - 4.5

V_IN = - 9 / -.45 = 2 V

So, when V_IN = 2 V, V_OUT = - 9 V

But as V_IN increases from 0 V to 2 V, the value of V_OUT also increases (with a negative sign) from 0 V to – 9 V.}