# Zener Diode Wattage Suggestions

Discussion in 'The Projects Forum' started by mattd860, Feb 25, 2016.

1. ### mattd860 Thread Starter Member

Jan 9, 2016
57
2
Can someone please help explain how to determine the appropriate wattage for zener diodes, specifically for the circuit below? Input voltage is 5.0v dc and I'm trying to limit current to ~25mA max so what should the wattage be for the zener diode? Also - If I was trying to limit current in a 5.0v dc circuit to 150mA with a 33ohm resistor, what should the wattage be for the ZD?

2. ### ISB123 Well-Known Member

May 21, 2014
1,239
525
Why would you want to limit the output voltage of MCU pin when its already working at 5V.

P=I x V

3. ### mattd860 Thread Starter Member

Jan 9, 2016
57
2
Just trying to protect the pins from transients.

4. ### mattd860 Thread Starter Member

Jan 9, 2016
57
2

So do I need to match the wattage of my circuit? I.E. if power = 125mW does that mean I go with a 125mW 5.1v zener diode? Or should I higher? I'm not trying to calculate wattage for my circuit, I'm just trying to figure out which 5.1v zener diode to use. Thanks

5. ### WBahn Moderator

Mar 31, 2012
17,460
4,701
Then put protection diodes to the supplies. Put one diode to ground and the other to Vcc. Orient both so that they are reverse biased while the output pin is within it's intended voltage range.

6. ### ronv AAC Fanatic!

Nov 12, 2008
3,224
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You have to decide how large a transient you want to protect against.

7. ### WBahn Moderator

Mar 31, 2012
17,460
4,701

If you try to limit the current in a 5 V circuit by putting a 33 Ω resistor in series with what ever it is driving, then when the load is pulling 50 mA you will only be delivering 3.35 V to it. Is that what you want?

The needed rating of the zener is independent of the load since the two have nothing to do with each other. If your input voltage exceeds ~5.1 V then the load will have 5.1 V across it (and the series resistor) but you will get uncontrolled current flow in the zener and will blow it very quickly.

This is not a proper role for a zener, at least not in the circuit as shown.

Just put a fuse in place of your resistor and ditch the zener altogether. If you really have transient spike issues, then deal with them using protection diodes or other suitable protection circuit.

Here's one that might work for you, and it's only 35 cents from Digikey -- you can almost certain beat that price elsewhere.

http://www.digikey.com/product-detail/en/bel-fuse-inc/3JQ 150-R/507-1639-ND/1010035

8. ### mattd860 Thread Starter Member

Jan 9, 2016
57
2
Thanks for the clarification. I was just trying to incorporate some of the features found here -http://www.rugged-circuits.com/ruggeduino-1 - because I'm adapting my custom board (with an arduino processor) to work in an automobile environment.

9. ### WBahn Moderator

Mar 31, 2012
17,460
4,701
If you read their rationale for what they are doing, you can see that they are dealing with specific threats (and not others).

In particular, that resistor is limiting the short circuit current, which is a fault condition, to 150 mA. This IS causing a voltage drop that might be unacceptable in many situations and therefore they provide for a means of bypassing it. The zener is protecting against threats coming from the connector pin in conjunction with the resistor and it basically serves the same purpose as the two diodes I recommended. It does not provide protection against threats coming the other way, but that should be fine.

As for sizing the zener, that depends on how long and how big the transient is that you are protecting against. For short transients (call it less than a second), the power rating doesn't matter too much since it won't have time to heat up. To really play it safe, assume that you have a hard, permaent fault that would apply 15 V (well above the 13.8 V full battery charge in a typical car) to the terminal pin. That would result in a current of (15 V - 5 V)/220 Ω (I'm doing some rounding here) which would be about 45 mA, let's call it 50 mA. So the zener would be dissipating about 250 mW. So you might use a 0.5 W zener. If that is too large, physically, I wouldn't be too troubled by using a 1/4 W or even less given the unlikeliness of such a persistent fault.

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10. ### ebeowulf17 Active Member

Aug 12, 2014
672
78
Great explanation!

The only thing I would add is that there's a difference between trying to provide broad, universal GPIO protection like theirs (with inevitable compromises) and providing specific protection for a dedicated input or output.

If you're building a circuit around a microcontroller for a specific project and you know what each pin will be used for, you can choose better protection circuits.

I have limited knowledge and experience, but I think the example you posted looks good for nearly any input scenario. If the pin will be an output, then it depends on what the output is driving. If you're simply lighting an indicator LED, the odds are good that you can size a resistor suitable for limiting normal LED current that also provides adequate short circuit protection. But if the output needs more power or versatility, you might be better off building an output stage with a transistor or MOSFET.

11. ### mattd860 Thread Starter Member

Jan 9, 2016
57
2
Actually the VIN is only 5V and all inputs come from the same 5v dc source because I'm using a 12v SPST reed relay as a switch. So I think the only thing I really need to protect is the VIN source which I have done. I'm going to remove the 5.1v 1W diodes and 220ohm resistor.

12. ### ebeowulf17 Active Member

Aug 12, 2014
672
78
Sounds ok to me, as long as the wires from MCU to relay2 are reasonably short and safe from the risk of induced noise.

I am curious about the value of R2. How much current does that relay1 coil need? 5.1k seems high to me, but I could be wrong. Depends on required coil current and on the characteristics of T1.

13. ### ian field Distinguished Member

Oct 27, 2012
4,333
772
The traditional way is to clamp input pins to the supply rails with a pair of ordinary diodes.

Most MOS devices have internal diodes, but they're usually rated for about 10mA max - adding external diodes can take a bigger hit without damage.

Most often the datasheet says something along the lines; any pin must not exceed the supply rails by more than 0.7V - a Zener doesn't guarantee that unless you select Vz well under Vcc/Vdd.

Schottky barrier diodes have a lower Vf, but they tend to have higher gate capacitance.

14. ### #12 Expert

Nov 30, 2010
16,033
6,545
Yes, R2 seems wrong. Should probably be another 220.
and add a diode across the coil so it doesn't kick back at the transistor T1.

15. ### Dodgydave Distinguished Member

Jun 22, 2012
4,796
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Cant see the point of D3,D4, and there are no bemf diodes across the relays.

16. ### ebeowulf17 Active Member

Aug 12, 2014
672
78
Oh yeah, forgot the flyback diode. Good catch!

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Jan 9, 2016
57
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18. ### #12 Expert

Nov 30, 2010
16,033
6,545
You are mistaken. T1 grounds all the current through the relay coil, not just some of it.

19. ### mattd860 Thread Starter Member

Jan 9, 2016
57
2
I think I was misunderstood. T1 grounds all the current through the relay coil when the gate receives a very small amount of current from the IC. I wanted to limit the current to T1 since it doesn't need much. I'm trying to limit current to or from the relay.

20. ### #12 Expert

Nov 30, 2010
16,033
6,545
That is likely. The usual way to use a transistor as a switch is to give its base 1/10th of its collector current. If you are trying to limit the current through the coil, giving the base of T1 less current will accomplish that.

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