Use Fig. 5.1 to explain the variation of the resistance of the diode as V increases from zero to 0.8 V.

Question 14

(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]

Reference: Past Exam Paper – November 2015 Paper 23 Q5

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