Diodes used for input protection - How to protect over voltage - Page 1
Forum section: electronics / beginners

author: pigtwo
author_rank: Regular Contributor
timestamp: July 11, 2018, 05:55:54 pm
content: Hello all,I've run into a situation where I thought I understood something simple but now I realize I'm missing something.Say I'm trying to add some input protection to a circuit. I've commonly seen two reverse biased diodes connected to ground and the supply rail connected to the input. See the attached picture for an example.For the diode connected to ground this makes sense to me. But I'm looking at a circuit designed by someone else where they connected other diode to the supply rail. The supply rail is generated from a linear regulator. I was finding that the over voltage protection diode didn't seem to be working. Instead, when a voltage over the supply rail voltage was put on the input the over voltage protection diode would conduct and actually raise the supply rail voltage. After thinking about it this makes sense because I don't believe linear regulators can sink current.So from this I conclude that you can't use input protection diodes like this on supplies that are generated from linear supplies. Is this correct? Also what is the appropriate way to protect from over voltage in a design with only linear supplies?Thank you!

author: TimFox
author_rank: Super Contributor
timestamp: July 11, 2018, 06:52:30 pm
content: You can connect the top end of the upper diode to a Zener diode of less than the positive supply voltage, and add an appropriate resistor from the positive supply to maintain a small current in the Zener to ensure that the upper diode is reverse-biased during normal operation. Component values are left as an exercise for the reader and depend on the maximum “normal” input voltage.

author: langwadt
author_rank: Super Contributor
timestamp: July 11, 2018, 06:58:26 pm
content: you need a series resistor to limit the current protect the diodes, if that resistor also limits maximum current low enough injecting it into the supply is not a problem, else you will need to use a zener to ground instead

author: TimFox
author_rank: Super Contributor
timestamp: July 11, 2018, 07:14:01 pm
content: Yes, I didn’t notice the lack of the resistor at the left. If the device input can tolerate the very large resistor shown after the diodes, then a similarly large resistor to the left can be used. A single Zener will conduct in both directions (negative to ground through the PN diode and positive to ground through the Zener). This is not used for fast or critical inputs, due to high capacitance and leakage current of the Zener.

author: pigtwo
author_rank: Regular Contributor
timestamp: July 11, 2018, 07:47:12 pm
content: @TimFox Using a Zener in that configuration makes sense. Or as you and langwadt mention putting a reverse biased zener to ground also makes a lot of sense.@langwadt I forgot to mention that in my situation the input current is limited which is why there is no series resistor. But I believe the results are the same with either a current limited source or a series resistor. The reverse biased Zener solution is actually really obvious. I kind of felt like I was missing something easy when I was writing it up.Thank you for the explanation.As an aside question, I feel like I see this configuration all the time where one diode is connected to a supply for protection. Will a supply generated from other means(buck converter for example) be able to sink current and thus allow the use original configuration? I would think not given the configuration but I see this frequently so I'm not sure. Maybe most of the time the current is limited enough such that driving the supply won't increase the voltage.

author: langwadt
author_rank: Super Contributor
timestamp: July 11, 2018, 10:31:05 pm
content: Quote from: pigtwo on July 11, 2018, 07:47:12 pmSome synchronous buck converters can sink current, but usually the input current is low enough that it isn't need. if it is high current you'll have other problems, the energy has to go somewhere

author: David Hess
author_rank: Super Contributor
timestamp: July 11, 2018, 11:30:17 pm
content: Usually the load on the positive supply is enough to keep it in regulation when current is injected through the input protection diode. If that is not the case, then other arrangements need to be made like using a zener shunt or transistors to divert the current from the positive supply to ground.

author: JS
author_rank: Frequent Contributor
timestamp: July 12, 2018, 01:05:43 am
content: Bypass caps would handle fast spikes, a zenner won't let the rail to go too high. TVS ICs have all you need in many different configs, look at them and their application notes to understand this better. Sometimes zeners are polarized all the time so they kick in faster, sometimes they are left there wating for the spike.JS

author: RobertHolcombe
author_rank: Frequent Contributor
timestamp: July 12, 2018, 02:24:01 am
content: You've already had a lot of responses, but I wanted to add this link as I found it a good reference for what you're askinghttps://www.digikey.com.au/en/articles/techzone/2012/apr/protecting-inputs-in-digital-electronics