The circuit diagram used to study the characteristic curve of a zener diode is connected to variable power supply (0 - 15 V) as shown in figure. A zener diode with maximum potential V z = 10 V and maximum power dissipation of 0.4 W is connected across a potential divider arrangement. The value of resistance R P connected in series with the zener diode to protect it from the damage is ________________ Ω.

Solution

Physics sir · 3 steps

Find the largest current the zener can survive, then let R_P drop whatever the supply gives beyond 10 V.

1. Safe current limit. The zener holds V_z = 10 V across itself and may dissipate at most 0.4 W. So the largest current it can carry is Iₘₐₓ = P/V_z = 0.4/10 = 0.04 A.
2. Worst case. The diode is in most danger when the supply is turned up fully to 15 V. The zener still takes only 10 V, so the remaining 15 - 10 = 5 V must fall across R_P.

   R_P = (Eₘₐₓ - V_z)/Iₘₐₓ = (15 - 10)/0.04

3. So. R_P = 5/0.04 = 125 Ω. With this resistor the current can never exceed 0.04 A, so the answer is 125.

The common slip: 250. Dividing the zener voltage instead of the extra voltage, 10/0.04, gives 250. The resistor only takes 15 - 10 = 5 V.

The idea to keep. For a protective series resistor, use the maximum supply voltage. The resistor takes the extra voltage Eₘₐₓ - V_z while carrying the zener's maximum safe current P/V_z.

Answer from the NTA final answer key. Solution by PrepWiser.