Q#14 (Past Exam Paper – November 2015 Paper 23 Q5)
(a) The I-V characteristic of a semiconductor diode is shown in Fig. 5.1.
Fig. 5.1
(i) Use Fig. 5.1 to explain the variation of the resistance of the diode as V increases from
zero to 0.8 V. [3]
(ii) Use Fig. 5.1 to determine the resistance of the diode for a current of 4.4 mA. [2]
(b) A cell of e.m.f. 1.2 V and negligible internal resistance is connected in series to a semiconductor diode and a resistor R1, as shown in Fig. 5.2.
Fig. 5.2
A resistor R2 of resistance 375 Ω is connected across the cell.
The diode has the characteristic shown in Fig. 5.1. The current supplied by the cell is 7.6 mA.
Calculate
(i) the current in R2, [1]
(ii) the resistance of R1, [2]
(iii) the ratio
power dissipated in the diode
power dissipated in R2
[2]
Solution:
(a)
(i)
{Note that the gradient of the graph does not give the resistance.
To obtain the resistance, we calculate the ratio V/I for that point.}
Resistance = V / I
{Initially, even though there is a p.d. across the semiconductor, the current is zero. This indicates that the resistance is very high at low voltages.}
(Initially,) there is a very high/infinite resistance at low voltages
Then, the resistance decreases as V increases
(ii)
{for a current of 4.4 mA,} p.d. from graph 0.50 (V)
{R = V/I}
Resistance = 0.5 / (4.4×10-3) = 110 (114) Ω
(b)
(i)
{The p.d. across each loop is equal to the e.m.f. in the circuit.}
{I = V/R}
current (= 1.2 / 375) = 3.2×10-3 A
(ii)
{I + 3.2 mA = 7.6 mA giving I = 7.6 – 3.2 = 4.4 mA}
current in diode = 4.4×10-3 A
{For the middle loop in the circuit, R = V / I
where V is the sum of p.d. across the loop (= e.m.f.),
I is the current in the middle loop
R is the total resistance in the middle loop}
total resistance = 1.2 / (4.4×10-3) = 272.7 (Ω)
{as calculated in (a)(ii), when current = 4.4 mA, resistance of diode = 114 / 113.6 Ω}
{total resistance = resistance of diode + R1}
resistance of R1 = 272.7 – 113.6 = 160 (159) Ω
OR
{from the graph, when current = 4.4 mA, p.d. across diode = 0.5 V
So, p.d. across R1 = 1.2 – 0.5 = 0.7 V}
p.d. across diode = 0.5 V and p.d. across R1 = 0.7 V
{R = V/I}
resistance of R1 = 0.7 / (4.4×10-3) = 160 (159) Ω
(iii)
power = IV or I2R or V2/ R
{Note that, when calculating the power dissipated, V, I and I should be across the component being considered respectively.}
{using P = VI, ratio = IV in diode / IV in R2}
ratio = (4.4 × 0.5) / (3.2 × 1.2)
OR
{using P = I2R,}
ratio = [(4.4)2×114] / [(3.2)2×375]
OR
{using P = V2/R, ratio = (0.52 / 114) / (1.2 / 375)}
ratio = [(0.5)2 × 375] / [114 × (1.2)2]
ratio = 0.57
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