The voltage of a 9V battery falls from 9V when it is new to about 7V near the end of its life. To get reasonable life out of a 9V battery, any circuit powered from it should be able to operate at 7V. To prevent the performance from changing as the voltage of the battery falls, design a circuit, that draws 10mA, to run off a 6V, and to control the voltage supplied to it with a 6V zener diode, which can operate with as little as 2mA through it. R = 82Ω
When the battery is new, what current flows from it?
Attempt:
I tried to work out the resistance of the zener:
Rz = Vz / Iz
= 6 / 0.002
= 3000Ω
I = Iz + Iload
Vin max = 9V
I = Vin max / R + Rz
= 9 / 82 + 3000
= 2.92mA
The correct answer was 36.6mA, but I cant see what I did wrong.
When the battery is new, what current flows from it?
Attempt:
I tried to work out the resistance of the zener:
Rz = Vz / Iz
= 6 / 0.002
= 3000Ω
I = Iz + Iload
Vin max = 9V
I = Vin max / R + Rz
= 9 / 82 + 3000
= 2.92mA
The correct answer was 36.6mA, but I cant see what I did wrong.