# Circuit Analysis of a circuit w/ Zener Diode.

#### guyvsdcsniper

Joined Sep 19, 2023
9
Attached is the problem and my work.

I am using circuit analysis to find the power dissipated in a Zener Diode. I used the loop method to solve for $$I_1$$

It seems like my error is during this process. I view crossing the cathode (-) up toward the anode (+) as going uphill / going toward a positive potential. For that reason I should get a +20V, just as I would get a +25V for traveling up the DC source.

This leads to the wrong answer, apparently it should be -20V for the diode, based on the fact the answer is given in the question (2W) and my answer is 18W.

Any help understanding why the same conventional signs dont apply for Diodes as they would EMFs ?

#### dl324

Joined Mar 30, 2015
16,688
It seems like my error is during this process.
That's your problem. You have a voltage drop across the resistor and a voltage drop across the zener. The polarity of both drops is the same.

#### WBahn

Joined Mar 31, 2012
29,865
For a diode that is operated conventionally, we generally assign the current polarity as being positive when the current is going from anode to cathode. By the passive sign convention, this makes the anode the positive terminal and the cathode the negative terminal.

We could certainly stay with this assignment when working with Zener diodes, but we generally don't. The reason is that these diodes are usually operated with a reverse current and sticking with the prior assignment would mean that all of the Zener voltages would have to be specified as being negative and so would the current flowing through the Zener. So, by convention, we flip both signs and talk about a positive Zener voltage, understanding that this is the voltage at the cathode relative to the anode and that the current is positive if it is likewise flowing the anode to cathode. The reason is simple -- humans are not good at dealing with negative values, or at least not as good at dealing with those as with positive values, so we try to adopt conventions that avoid us having to do so as much as possible.

#### BobTPH

Joined Jun 5, 2013
8,664
We know the voltage drop across the Zener is 20V. So what is the voltage across the 50Ω resistor? (hint: Kirchoff’s voltage law)

Now what is the current? (hint: Ohm’s law)

Now, what voltage do we multiply the current by to get the total dissipation?

#### MrAl

Joined Jun 17, 2014
11,268
Attached is the problem and my work.

I am using circuit analysis to find the power dissipated in a Zener Diode. I used the loop method to solve for $$I_1$$

It seems like my error is during this process. I view crossing the cathode (-) up toward the anode (+) as going uphill / going toward a positive potential. For that reason I should get a +20V, just as I would get a +25V for traveling up the DC source.

This leads to the wrong answer, apparently it should be -20V for the diode, based on the fact the answer is given in the question (2W) and my answer is 18W.

Any help understanding why the same conventional signs dont apply for Diodes as they would EMFs ?

View attachment 304556
Hi,

Yeah, it looks like you marked the polarity of the zener incorrectly as others have mentioned.

Zener diodes are a little different than regular rectifier diodes. Some of them can conduct in either direction, but the way they are most often used is in a circuit where they conduct when their cathode is positive with respect to their anode. That is when they work at their rated zener voltage. This means your equation for I1 has to change a little bit.