# Zener diode circuit question

#### nothing909

Joined Nov 5, 2016
54 So basically, as the question says, you need to calculate the current in the zener and load when the supply voltage is 10V and then again when it's 15V.

FOR 10V SUPPLY

Voltage across resistor = 10-5 = 5v

Supply current = 5/1000 = 5mA

Voltage across load = 5V as Zener and load are in parallel

Current through load = 5/1000 = 5mA

Current through Zener = Supply current – Load current = 5 – 5 = 0

FOR 15V SUPPLY

Voltage across resistor = 15-5 = 10v

Supply current = 10/1000 = 10mA

Voltage across load = 5V as Zener and load are in parallel

Current through load = 5/1000 = 5mA

Current through Zener = Supply current – Load current = 10 – 5 = 5 mA

This doesn't look correct. I've never did a question with zener diodes before and I'm confused. What is it I'm doing wrong?

#### MrChips

Joined Oct 2, 2009
21,325

#### bertus

Joined Apr 5, 2008
20,567
Hello,

It depends on how exact you take the zener voltage.
The schematic states a zener of 5.1 Volts.

Bertus

#### dl324

Joined Mar 30, 2015
10,943
Welcome to AAC!
FOR 10V SUPPLY

Voltage across resistor = 10-5 = 5v
Supply current = 5/1000 = 5mA
Voltage across load = 5V as Zener and load are in parallel
Current through load = 5/1000 = 5mA
Current through Zener = Supply current – Load current = 5 – 5 = 0
How did you determine the voltage across R1?

Other than using 5V for the zener voltage and not showing your work, I don't see anything wrong.

#### nothing909

Joined Nov 5, 2016
54
Yea, I was supposed to use 5.1V instead of 5V. So if I redo the calculations, I get this:

FOR 10V SUPPLY

Voltage across resistor = 10-5.1 = 4.9v

Supply current = 4.9/1000 = 4.9mA

Voltage across load = 5.1V as Zener and load are in parallel

Current through load = 5.1/1000 = 5.1mA

Current through Zener = Supply current – Load current = 4.9 – 5.1 = -0.2 mA

FOR 15V SUPPLY

Voltage across resistor = 15-5.1 = 9.9v

Supply current = 9.9/1000 = 9.9mA

Voltage across load = 5.1V as Zener and load are in parallel

Current through load = 5.1/1000 = 5.1mA

Current through Zener = Supply current – Load current = 9.9 – 5.1 = 4.8 mA

For the 10V supply I'm getting -0.2mA for the zener. For the load I'm getting 5.1mA. How is it possible to get 5.1mA for the load if I've only got a 4.9mA supply?

#### dl324

Joined Mar 30, 2015
10,943
Yea, I was supposed to use 5.1V instead of 5V. So if I redo the calculations, I get this:

FOR 10V SUPPLY
Now your calculations for the 10V supply are wrong. Didn't check for 15V.

#### WBahn

Joined Mar 31, 2012
25,911
You are assuming that (with a 10 V input) that the Zener has 5.1 V across it because you are assuming that the Zener is reverse biased by enough to place it in reverse conduction. That's fine for the purpose of carrying out the analysis, but then you have to answer the question of whether your assumption was valid. If it wasn't, then you need to step back and reduce the analysis using a different assumption for the state of the Zener.

#### crutschow

Joined Mar 14, 2008
25,116
How can the load current be higher than the supply current when in input is 10V?
You are making an incorrect assumption about the voltage across the zener.
A zener can only sink current, it can't supply current.

#### MrChips

Joined Oct 2, 2009
21,325
Actually, a typical 5.1V 1W zener such as 1N4733A is spec'd at 50mA.

Hence using 5V in your calculation would not be terribly wrong. Besides, it makes the calculations easier.

#### dl324

Joined Mar 30, 2015
10,943
Hence using 5V in your calculation would not be terribly wrong. Besides, it makes the calculations easier.
Except that by using 5V in the first 10V supply calculation, it wasn't clear if he was using the zener voltage or the divider voltage to do the current calculation.

#### MrChips

Joined Oct 2, 2009
21,325
Except that by using 5V in the first 10V supply calculation, it wasn't clear if he was using the zener voltage or the divider voltage to do the current calculation.
It didn't matter.

5V across R1.
5V across R2.
0mA into the zener.

Any current > 0mA through the zener means there is no longer 5V across R2 and hence across the zener.

#### dl324

Joined Mar 30, 2015
10,943
It didn't matter.
It did for the OP because when he did the calculation using 5.1V for the zener voltage, it showed he made the wrong assumption and didn't adjust for it and recalculate the load voltage using the resistive divider.

#### MrChips

Joined Oct 2, 2009
21,325
It did for the OP because when he did the calculation using 5.1V for the zener voltage, it showed he made the wrong assumption and didn't adjust for it and recalculate the load voltage using the resistive divider.

5V across R2.
Still 5V across R1. Not 10 - 5.1V.

#### dl324

Joined Mar 30, 2015
10,943
For the 10V supply I'm getting -0.2mA for the zener. For the load I'm getting 5.1mA. How is it possible to get 5.1mA for the load if I've only got a 4.9mA supply?
This is Homework Help so you need to do most of the work. Give us your theory for what the circuit is doing, and why, and we can try to nudge you in the right direction.

#### dannyf

Joined Sep 13, 2015
2,197
as the question says,
the general design approach is to design the zener so that its output impedance is sufficiently low, vs. the load, so as to connecting the load makes minimal impact on the zener's output current.

I typically use 10x so as "sufficiently low". At 5v, the current on the load is 5ma. So I would design the current through the zener to be 10x 5ma = 50ma.

That would indicate R1 = (10v-5v)/50ma = 100ohm.

At this point, you will need to calculate the power dissipation over the zener: 50ma * 5v = 250mw. So a 1w zener is needed.

The basic approach works for the 15v Vcc case as well.

BTW, you may not be able to find the right parameters sometimes -> zener regulators aren't design to drive such low loads.