A-LevelPhysicsD.C. circuits9702/42 · Oct/Nov 2017 · Q710 Marks

(b) An amplifier circuit incorporating an ideal operational amplifier (op-amp) is used to amplify the output of a microphone. The circuit is shown in Fig. 7.1. [Figure 7.1] When the potential at point P is 48mV, the output potential difference VOUT is 3.6 V.

(a)(i)State what is meant by feedback,[2]
(a)(ii)two benefits of negative feedback in an amplifier circuit. 1. 2.[2]
(b)(i)1.the gain of the amplifier circuit, gain = ..............................[2]
(b)(i)2.the resistance of resistor R. resistance = .............................. Ω[2]
(b)(ii)State and explain the effect on the amplifier output when the potential at P exceeds 68 mV.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[1] (a)(i) (part of) the output is combined with the input
[1] (a)(i) reference to potential/voltage/signal
[2] (a)(ii) increased (operating) stability / increased bandwidth/range of frequencies over which gain is constant / less distortion (of output) Any 2 points.
[1] (b)(i)1. gain = 3.6/(48 × 10⁻³)
[1] (b)(i)1. = 75
[1] (b)(i)2. gain = 1 + RF / R
[1] (b)(i)2. 75 = 1 + (92.5 × 10³)/R
[1] (b)(i)2. R = 1300 Ω
[1] (b)(ii) for 68 mV, gain × VIN = 5.1 (V) or output voltage would be greater than the supply voltage
[1] (b)(ii) amplifier would saturate (at 5.0 V) or output voltage = 5.0 (V)

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