# Are my zener concepts wrong?

If a battery of 60v with internal resistnace 4kohm (in series) is connected to a zener diode of v_z=10v in parallel and a load resistance of 2kohm in parallel too, what will be the currents in different circuits. since zener diode will breakdown at 10 v the load voltage will be max 10 v so as 60 v is > 10 v the currrent in load must be 5ma and the remaining 50 v will be across internal resis. so the current in that branch will be 12.5 ma and using kirchoff's rule the remaining current must be 7.5. but this is wrong. I don't understand why. this is surely due to conceptual mistake. simulate this circuit – Schematic created using CircuitLab

• Would Electrical Engineering be a better home for this question? – Qmechanic Mar 13 '15 at 6:50
• Looks fine to me. Total current from your source is 12.5mA. When zener is in zener breakdown, the potential between anode and cathode is 10V. Voltage across load is 10V, so all the currents line up. – efox29 Mar 13 '15 at 7:00
• @efox29 answer is 12.5,5,7.5 (I battery,I zener,I load) – RE60K Mar 13 '15 at 7:05
• What's your question ? – efox29 Mar 13 '15 at 7:07
• @efox29 my answer is 12.5,7.5,5 – RE60K Mar 13 '15 at 7:07

The voltage across your load is 10V (this is assuming that zener is in breakdown). So ohms law says that the current through the resistor is 5mA.

You're total current from the source is the current through your series resistor (4kohm). Again, if we assume zener breakdown, then the voltage across that series resistor is 50V. So Ohms Law says that current through that resistor is 12.5mA.

The remaining current is the current that goes through the zener.

The reason I say "assuming zener breakdown", is because a zener will go into zener breakdown when the reverse current is greater than the zener knee current. You're knee current was not specified. If the current through the zener is less than the knee current, then the zener is in reverse bias, and the calculations have be done over but this time assuming that the zener is simply a reverse biased diode.

You are correct. Your book or whoever told you that the load current was 7.5mA is wrong. $V1 = 60V$

$V2 = 10V$

$It = \frac {V1 -V2}{R1} =12.5mA$

$Il = \frac{V2}{R2} = 5mA$

$Iz = It - Il = 7.5mA$