dissipation in a zener diode

Discussion in 'General Electronics Discussion' started by flippineck, Feb 18, 2017.

1. flippineck

341
15
Sep 8, 2013
I am experimenting with an unstabilised DC power supply (solar panel array rated at 44V open circuit) which in practise seems to fluctuate up to about 45V open circuit in operation. Could I ensure the output stays at or below 40V by connecting a suitably high power & heatsinked 40V zener diode across it? Like a DO5 stud mount type?

If so.. when the voltage rises above 40V, how would I appraise the actual power dissipation in the zener?

The panels short circuit output is listed at 15A. Would I use straightforward P=IV?

That would seem to say the diode could experience 15 x 45 = 675W above 40V

However the conduction of the diode above 40V, would start to drop the PSU's output voltage so would the actual diode dissipation under full supply output actually be lower?

The highest rating I can find for such zeners seems to be about 50W.

Should I use some form of dummy load in series with the zener, and if so, would that mean I should subtract a figure to reflect the dummy load's own voltage drop, from the value chosen for the zener?

2. AnalogKid

2,641
765
Jun 10, 2015
For this discussion, lets assume an ideal zener diode with a brick wall response curve. At 39.999 V it is an open circuit, and at 40.000 V it conducts with a dynamic resistance as low as 2.67 ohms. That's the Ohm's Law equivalent of 40 V / 15 A. But the panel's output current at 0 ohms and into a 40 V zener are two very different things. I think you'll find that the peak current into a 40 V "load" is less than 15 A. Whatever that current is, the power dissipation in the diode is 40 times it.

BTW, this is a very inefficient way to regulate the voltage out of a solar panel, and will shorten the life of the panel.

ak

3. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,852
Jan 21, 2010
The dissipation is V*I.

If you have a datasheet for the panels it might have a V vs I curve at a nominal 100% solar irradiance. From that you can determine a current at 40V.

Let's say it's 5A. That will give you a power dissipation of up to 200W.

It may be very expensive to get a zener with that rating.

However, some wire wound resistors rated at a total power of (say) 300W, with total resistance of about 5 ohms could be used with a shunt regulator. The semiconductors would run cooler and the relatively cheaper resistors would be dissipating most of the power.

Last edited: Mar 3, 2017
4. Petkan

19
2
Feb 9, 2011
flippineck

Why would you need to clamp a solar panel. What is wrong if it reaches 45V?
The appropriate question is how to use the solar panel (say to charge a battery).
Solar panels are neither ideal voltage sources, nor ideal current sources. The voltage under load declines almost linearly with load current up to a point, then rapidly declines.
A PV panel with 45V open load voltage may drive miserable current into a 40V zener.
Note that power zeners could be made by small zener and power BJT (NPN CE across, Zener from + to base). There are many Buck regulators tolerating up to 60V input voltage, that can step down to say 12V.

Last edited by a moderator: Mar 4, 2017
5. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,852
Jan 21, 2010
I think the reason is to limit the max voltage so as not to exceed the input voltage limit of some downstream device.

In this case, the low current at the zener voltage is a good thing.

Connecting a buck or boost dc-dc converter to a source like a solar panel can lead to a situation where the system gets stuck in a mode where the solar panel is supplying it's max current at a low voltage. This may require the load to be removed in order to reset things to normal operation. This is one reason we have MPPT regulators.