Zener Diode Usage

Hey Folks,

I’m new to electronics and I’m trying to learn more. I’m trying to get a good understanding of how a zener diode works and was hoping people could help me out a little.

In the attached image there are two example of a circuit with a zener diode that has a zener voltage of 5v.

My understanding is that in example #1 the voltage to the load would be 35v and in example 2 the voltage would be 5v. Is this correct?

By the way, the 40v is mostly for purposes of illustration - I will be using a circuit like example #2 that will probably have an voltage supply in the 9v range and I want to limit the input to an arduino pin to 5v.

Thanks in advance for any help.

1. You have to account for the voltage drop across R1, that is going to decrease the voltage seen in both cases.

2. Things in parallel have the same voltage drop (which is exactly what you have in case 2), doing a simple KVL (Kirchhoff's Voltage Law, drawing a simple loop from V+ to V- or Ground to see the way the voltage is dropped across various components), you can see how this is so (in case you needed proof beyond the simple concept that things in parallel have the same voltage, just as things in series share the same current).

As I stated in 1), the voltage in both cases you presented is entirely dependent on what R1 is (as well as your load, but you didn't even take R1 into account).

Note: The easiest ways to analyze an electrical circuit are using KVL's and KCL's, some people say the "Loop Rule" and the "Junction Rule" respectively, but they are Kirchhoff's Voltage and Current Laws, I advise looking them up. There are also other various methods, Mesh, which is very useful, as well as Nodal Analysis (which is more complicated, but allows you to solve any circuit using a system of equations). Keep in mind all these things are for hand calculating results, which it doesn't seem most people do.

Thanks. I will do some reading on KVL and KCL.

If I understand your answer, in the case of example 2, the answer is that the voltage at the load would be 5v since things in parallel have the same voltage drop. This is assuming that more than 5v is supplied to the zener diode (and that I haven't supplied so much voltage as to destroy it) then the voltage to the load would also be 5v since it is in parallel with the zener.

As for example #1, what if we remove R1? Is my assumption correct that in this case the zener would reduce the voltage to the load by 5 volts and therefore 35 volts would be supplied to the load?

By the way I like your Bene Gesserit quote - it has been a bit of a motto of mine since reading Dune about 30 years ago.

The load voltage in the #1 will depend on the Zener's current (Iz where you get Vz).

40V = Iz * R1 + Iz * Rload + Vz

#2: the voltage on the load will be 5V (Vz) provided the Iz is sufficient enough to get Vz=5V.

wrecks135:
Thanks. I will do some reading on KVL and KCL.

If I understand your answer, in the case of example 2, the answer is that the voltage at the load would be 5v since things in parallel have the same voltage drop. This is assuming that more than 5v is supplied to the zener diode (and that I haven't supplied so much voltage as to destroy it) then the voltage to the load would also be 5v since it is in parallel with the zener.

As for example #1, what if we remove R1? Is my assumption correct that in this case the zener would reduce the voltage to the load by 5 volts and therefore 35 volts would be supplied to the load?

By the way I like your Bene Gesserit quote - it has been a bit of a motto of mine since reading Dune about 30 years ago.

Hehe, yeah, great books, just about to finish up reading the 5th one, Heretics. There is some seriously good philosophy out of those books.

You are correct in your assumption of case 2, and with case 1, if you remove R1, it will perform as you have assumed.

Pito above provided the necessary equations to show the relationships between the voltages/currents/resistances.

Evill Mad Scientist talks about using a Zener as a power supply:

This is not necessarily a good power supply for all purposes– the resistor limits how much current can be drawn. It is also not necessarily a precision voltage reference; the voltage will depend on the amount of current drawn. (That is to say, for the voltage to be steady, the load driven by that reference voltage must be consistent.) The voltage also depends upon the temperature.

Just something I saw that I thought I would add.

Bigred, Thanks for this comment and caution. This won’t be a primary power supply.

This is just part of an “emergency” external power supply for an home built perfboard Arduino treasure box with an electronic locking mechanism. There will be an internal 9v power supply but I wanted a way of “jump starting” the system and automatically opening it from the outside if the internal batter dies or if you simply forget the access code. When you supply power to the external positive and negative leads the Arduino will power up and Pin 1 will go high, thereby automatically opening the box. I’ve gotten bits and pieces of the answer from various places and I have attached a diagram for my overall plan.

In order to save battery power I’m powering my perfboard Arduino from a 5v switching power module that can take in anywhere from 7v to 28v and output 5v with up to 92% efficiency.

Any thoughts or advice on this overall plan would be greatly appreciated!