The circuit for an inverting amplifier incorporating an ideal operational amplifier (op-amp) is shown in Fig. 7.1. [Figure 7.1]
- (a)(i)For the circuit of Fig. 7.1: explain why point P is known as a virtual earth[3]
- (a)(ii)calculate the gain of the amplifier.[2]
- (b)When the op-amp is saturated, the potential difference across the LED is 2.3 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 30 mA.[3]
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Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [1] (amplifier) gain is very large/infinite
- [1] for amplifier not to saturate, V+ = V- or feedback (loop) ensures V+ = V-
- [1] V+ is at earth/0 V so V-is (almost) at earth/0 V
- [1] gain = (-)5200/800 or (−)5.2/0.80
- [1] = -6.5
- [1] (at saturation,) VOUT = 5 V
- [1] p.d. across R = 5 – 2.3 = 2.7 (V)
- [1] resistance = 2.7/(30 × 10⁻³) = 90 Ω
- or
- [1] Rdiode = 2.3/0.030 = 77 Ω
- [1] Rtotal = 5.0/0.030 = 167 Ω
- [1] Rresistor (= 167 – 77) = 90 Ω
- or
- [1] Rdiode = 2.3/0.030 = 77 Ω
- [1] 77/(Rresistor + 77) × 5 = 2.3
- [1] Rresistor = 90 Ω
- or
- [1] Rdiode = 2.3/0.030 = 77 Ω
- [1] Rresistor = 77 × (2.7/2.3)
- [1] Rresistor = 90 Ω
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