# Zener diode circuit When I tried to find current flowing through 1.2k ohm resistor, I am finding 8.33 mA but there comes the problem: I am finding the current through the 1k ohm resistor is 6 mA. What am I doing wrong? Is my method wrong? I am trying to find current with this formula:

$$Vss-Vz/R1= 16-10/1000 = 6mA$$

If the zener diode has the same voltage as R2, R2 must have 10/1200= 8.3mA, right?

• What's the voltage across the 1.2k ohm R without the zener in the circuit? – George Herold Oct 14 '14 at 18:24

I believe the question you're asking is: "How can the 1.2k resistor have 8.3mA going through it if there is only 6mA going through the 1k resistor."

The answer is that there isn't 8.3mA going through the 1.2k resistor.

The zener diode will break down and conduct for any voltage above 10V. The misleading thing about this circuit is that the voltage at the top of the zener diode never reaches 10V. The voltage divider made up of the two resistors puts the voltage in the middle at about 8.7V. As a result, there's not enough voltage to cause the zener diode to break down. Except for a small amount of reverse leakage current, the diode does not conduct.

For the purposes of circuit analysis, the zener diode can be ignored. It has very little effect on the circuit. That said, the current through both resistors is just: $$I=\frac{V}{R}=\frac{16V}{1k+1.2k}=7.27mA$$

• I'm curious: Why would Vz be given as 10V in the diagram? It's just the reverse voltage specification of the zener diode, rather than what's being measured across it? – JYelton Oct 14 '14 at 18:32
• @JYelton, I interpreted the "V_z=10V" as specifying it's a 10V zener diode. Without more context from the OP, I think that's a safe assumption. – Dan Laks Oct 14 '14 at 18:37
• @JYelton, by the way, I think you win the award for fastest edit ever on this answer. – Dan Laks Oct 14 '14 at 18:39
• ı forgot the add picture thanks by the way ı get it now :D – Kemil Kemil Oct 14 '14 at 18:44
• I was studying this question, considering providing an answer, then your excellent answer appeared! I just edited one tiny thing. :) – JYelton Oct 14 '14 at 19:50