- Joined Oct 10, 2020
Thankyou for this, it's added a lot of clarity.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.
So going with a 30v 1w zener and a 1KΩ resistor should be enough to limit excess voltage of between 30v and 50v, but if I suspected (not likely) that a peak of 75v was possible, i'd need to up the resistor to 2KΩ?
Have I got that right?