1
\$\begingroup\$

I am attempting to simplify a schematic by using a zener diode as a regulator for a small radio. At peak, the radio will consume about 50mA @ 3.3V, and in its low power sleep state, virtually nothing (a couple uA). In a previous experiment, where a Zener diode with 1mA Iz and 400mW Pd was used, the voltage would shoot up far above 3.3V. After doing some of the analysis, something is still not quite clicking as to whether or not I would experience voltage fluctuations now while using this particular zener (20mA Iz, 0.5W).

If my voltages at Vz are basically 4.13V and 12V absent the diode on the high and low spectrum of the load draw, respectively, and both would trigger the zener ON, what are some things to keep in mind that might cause this zener to deregulate as mentioned? As far as I know, I'm currently looking at ~200mW dissipation at radio idle, but could come up short of Iz by about 12mA, providing only 8mA of available current when the radio is maxed out.

enter image description here

Thanks

\$\endgroup\$
2
\$\begingroup\$

If you want something that just works, use a linear voltage regulator. They're available for pennies. If you want less power dissipation, use a switching regulator.

If you want to use your circuit, the things to ensure are that the value of Rs is small enough to provide enough current at 3.3V, and physically large enough to dissipate all the heat that it generates, ~450mW by my calculation (12v-3.3v/50mA). Your zener also needs to be able to dissipate the required power, which will be ~160mW (3.3v*50mA). Your voltage regulation probably wont be great, but your radio can probably deal with it (although if you post what the radio actually is that would help). You should put a capacitor across the output of your voltage regulator.

Overall, 600-700mW is quite a lot of power to be continuously dissipating in a small space. You need to account for the temperature rise that this will cause. The issue with this circuit, compared to a linear (series) regulator, is that it always consumes the same amount of power, which is the maximum amount. A series linear regulator would be a much better fit for this application, as presumably the 50mA draw doesn't happen much of the time (although even if it happens 99% of the time, a series regulator is better).

Try your circuit out here.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ I agree on the linear regulator bit; especially in the context of this being in a confined space, which it is. I had only been talking about dissipation at the diode, but that is definitely quite the reminder of the heat! I was also trying to understand what made the other one fail, so part of that may have driven the circumstance of this question. And I assume that when a zener does not get its ideal Iz, the voltage can ramp up or down accordingly. \$\endgroup\$ May 27 '20 at 2:35
  • 1
    \$\begingroup\$ @user8585939 the point of a zener diode is that it has a steep IV curve. That means that for large variations of current, the voltage variation should be small. At any current between 1ma and 100ma, the voltage should probably be around 3V-3.5V (assuming no temperature rise). \$\endgroup\$
    – BeB00
    May 27 '20 at 2:40
  • \$\begingroup\$ The quiescent current of a recently searched LDO seems impressive; on an LP2950 3v3, even at loads drawing 50mA, the "overhead" current will likely average out to be only 1mA when factoring intermittent packet communication. On the Zener, if at least 50mA must always be active to achieve the regulated zener drop (and also enable providing downstream power to achieve the desired functionality), it's looking like this could be up to a 50x reduction in dissipation. Pinch me if that seems way off. \$\endgroup\$ May 27 '20 at 14:44
  • \$\begingroup\$ LDO quiescent currents will routinely be in the 10's and 1's of uA. Its entirely believable that your average current could be 1ma or less, espceially with radio, where you wont be transmitting the majority of the time. \$\endgroup\$
    – BeB00
    May 27 '20 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.