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.
Maybe you can draw a schematics. It's will be much easier for you to understood whats happen in the circuit. And forget about 2mA and diode resistance 3000Ω. Just apply KVL to 82 ohm resistor
Sorry but I havent learnt KVL. I thought about saying: I = V / R = 9 / 82 = 0.11A but that doesnt give the right answer. If the battery is fresh then V = 9V surely should be used and there is only one resistor and so it should be used to calculate the current of the fresh battery?
KVL is Kirchoff's voltage law, And for you circuit we have: Vin = V_R + V_Zener = I*R + 6V then I = (9V - 6V)/82 = ??
Well, you assume that Zener diode current is always equal to 2mA. Which is not true. Zener diode need a minimum current level to flow through it to generate the Zener voltage. And in your example minimum current level is equal 2mA. And the minimum current will flow through the diode when load current reaches its maximum value. So when Vin=9V and I_load = 10mA the Zener diode current is equal: (36.6mA - 10mA) = 26.6mA And when Vin = 7V then : I_Rs = (7-6V)/82 = 12mA so diode current is equal: 12m - 10mA = 2mA So every think is ok and Zener work fine. And also you have to remember that Zener diode resistance is not constant. Diode resistance change with Zener diode current.