Overvoltage protection circuit specification. Zener diodes... what wattage?

Thread Starter

ianjuk

Joined Oct 10, 2020
14
I'm awaiting guidance from mods as to where to post this question but i'm looking for help asap so hoping some kind soul can help me out. I am aware this has likely been answered many times so apologies for duplication.

1. I have a wind turbine whose output I am testing.
2. I am using a data logging device that measures voltage of between 0-30v.
3. The wind turbine, is frequently generating more than 30v.
4. I need to build an overvoltage circuit that will protect the data logging instrument when the voltage goes above 30v, though realistically the cap could be between 24-30v for the purpose of my research.
5. I'm thinking zener diode is the way to go, with or without a fuse.
6. It's likely that at 30v+ the turbine is generating somewhere around 10a (guesstimate)

So my questions:

1. What wattage diode would I need, would 300w be sufficient? What would happen if the voltages go over eg: 400w?
2. Do I need a fuse and if so what size?

Many thanks for any help/direction.
 

Alec_t

Joined Sep 17, 2013
11,730
I need to build an overvoltage circuit that will protect the data logging instrument when the voltage goes above 30v
What load do you have on the turbine, apart from the data-logger? Do you have the data-logger spec? I'd be surprised if the zener needed to pass more than a few mA for pure over-voltage protection of a logging device.
 

jjw

Joined Dec 24, 2013
555
You need a series resistor to the zener.
If you know the input impedance of the data logger and it is relatively high ( ~100kohm or more) , the wattage of the resistor and the zener will be reasonable.

Edit: I assumed that the logger has a separate power supply not depending of the windmills output voltage.
 
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Thread Starter

ianjuk

Joined Oct 10, 2020
14
What load do you have on the turbine, apart from the data-logger? Do you have the data-logger spec? I'd be surprised if the zener needed to pass more than a few mA for pure over-voltage protection of a logging device.
At the moment, no load. Adding a load will mess with the data and therefore research. I’m doing this across 3 turbines so i’d have to ensure the loads were identical.
 

Thread Starter

ianjuk

Joined Oct 10, 2020
14
You need a series resistor to the zener.
If you know the input impedance of the data logger and it is relatively high ( ~100kohm or more) , the wattage of the resistor and the zener will be reasonable.

Edit: I assumed that the logger has a separate power supply not depending of the windmills output voltage.
Small bbattery. This is the data logger:
What current does your Data Logger inputs draw?
This is the data logger i’m using

https://www.peaktech.de/productdetail/kategorie/datenlogger/produkt/peaktech-5186.html
 

MrChips

Joined Oct 2, 2009
22,067
Your data logger is battery operated. You do not need protection except from over-voltage on the input.

Use a x2 attenuation voltage divider at the input using two resistors of equal value, for example, R1 = R2 = 1000Ω. Now the input will be good for 0-60VDC.

Multiply all your readings by 2.

1602333456234.png
 

Thread Starter

ianjuk

Joined Oct 10, 2020
14
Your data logger is battery operated. You do not need protection except from over-voltage on the input.

Use a x2 attenuation voltage divider at the input using two resistors of equal value, for example, R1 = R2 = 1000Ω. Now the input will be good for 0-60VDC.

Multiply all your readings by 2.

View attachment 219216
Thanks for this. Yes battery powered. I want to be sure that when the voltage goes above 30v it doesn’t fry the data logger, so limiting the input voltage to 30v max but I don’t want to affect the input below 30v. I’m interested at the voltages produced all the way from 0-30v.

will your solution cover those bases?
 

MrChips

Joined Oct 2, 2009
22,067
Thanks for this. Yes battery powered. I want to be sure that when the voltage goes above 30v it doesn’t fry the data logger, so limiting the input voltage to 30v max but I don’t want to affect the input below 30v. I’m interested at the voltages produced all the way from 0-30v.

will your solution cover those bases?
Yes, it does.

If you do not want to have to multiply your readings by 2, then replace R2 with a 30V zener, any wattage.
Here are some part numbers:
1N725
1N972
1N5256
1N4643
1N4751


1602335160142.png
 

Thread Starter

ianjuk

Joined Oct 10, 2020
14
Super, I was on the right track. Thank you and to everyone here who chipped in with A’s and guidance. Very grateful indeed.
 

jjw

Joined Dec 24, 2013
555
Yes, it does.

If you do not want to have to multiply your readings by 2, then replace R2 with a 30V zener, any wattage.
Here are some part numbers:
1N725
1N972
1N5256
1N4643
1N4751


View attachment 219219.
At 40V input the zener power with 1kohm resistor is 300mW.
1N725 has max power of 250mW, 1N972 400mW, 1N5256 500mW
1N4643 and 1N4751 are 1W
Use the 1W parts or higher value of R1, if it does not reduce the accuracy too much.
 

Alec_t

Joined Sep 17, 2013
11,730
It's likely that at 30v+ the turbine is generating somewhere around 10a (guesstimate)
If you don't have a load connected there won't be any current to speak of; just the few(?) microamps of current flowing through the input impedance of the logger.
 

Tonyr1084

Joined Sep 24, 2015
5,196
If you don't have a load connected there won't be any current to speak of; just the few(?) microamps of current flowing through the input impedance of the logger.
The zener needs to be protected from the current. Hence, a resistor will be needed.

The logger may only use microamps, but the value of the resistor still needs to be calculated so as to protect the zener from excessive current.

A 30V zener will only conduct when voltages reach or exceed the 30V (± its tolerance). Therefore, if the generator is putting out 50V and the zener is rated for 1 watt then Ohms Law needs to be applied. I'm using 1 watt as a reference point. Since the zener conducts at and above 30V then 50V - 30V = 20V. So 20V at 1W equals 50mA (or 0.05A). 20V ÷ 0.05A = 400Ω. That will conduct 1 watt when the generator is putting out 50 volts. If the generator goes higher - say 75V then:
1W ÷ (75V - 30V) = 0.0222•••A or (22mA)
45V ÷ 0.022A = 2045Ω. 2KΩ would be required to limit the wattage to 1 watt. Putting a 400Ω resistor through that voltage will mean the zener has to dissipate [45V ÷ 400Ω = 0.1125A. 45V x 0.1125A =] 5.0625W. You'll burn up the zener if it's rated for 1 watt and you push 5 watts through it. In this second case you would need a 2KΩ resistor to limit the current to slightly over 1 watt at 75V output. Using a higher wattage rated zener would improve matters.

Since your data logger won't be seeing more than 30V (due to the presence of the zener) it will be safe. Unless the zener burns out and goes open. Then at 75V, 2KΩ the logger could see 38mA. If your logger can handle that then no problem. If not - you'll need to replace the logger.
 
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