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Is it possible to use two zener diodes in series back to back to replace a diac?


I've accidentally burnt my dimmer with a short. It is a common dimmer with a triac/diac/pot of the simplest configuration (the triac is a BTA16). After replacing the triac, this still does not work, and I'm almost sure the problem is the DB3 diac (breakdown voltage at 32 V). I don't have this beast in my lab, so, I would like to know if it is possible to use two 32V zener diodes in series, back-to-back, to obtain approximately the same effect, a question that may be interesting for its own.

Here is a somewhat simplified schematic (snubber and some filtering cap not included)


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2 answers


No, two zener diodes are not equivalent to a DIAC.

While they both have a specific breakdown voltage, the DIAC is a special 3-layer device that exhibits a negative resistance characteristic once it breaks down:

DIAC characteristic curve

In contrast, the two zener diodes simply stop conducting once the voltage is reduced below the breakdown level.

In a typical phase-control dimmer, it is the charge on the timing capacitor that is used to trigger the TRIAC. The negative resistance characteristic of the DIAC is what allows this to happen. The back-to-back zener diodes would not pass enough current to trigger the TRIAC.

You can use a couple of SCR-connected transistors (or an actual SCR) to simulate the negative-resistance characteristic of a DIAC, but only in one direction. Each of these circuits only works correctly when the upper terminal is positive with respect to the lower. But you can work around that limitation by "wrapping" either circuit in a bridge rectifier.

alternate circuits


That's exactly what I suspected. Thank you. An almost rhetorical question: do you know some ersatz for the diac? coquelicot 23 days ago

See edit above. Dave Tweed 22 days ago

Nice. Perhaps even more clever: in your second schematic, two zener diodes back-to-back in place of the single zener, and a triac in place of SCR1? coquelicot 22 days ago

Oh, and of course with one more resistor at the top, to make the schematic entirely symmetric. coquelicot 22 days ago


It's not clear what you are really trying to do, but if you're trying to replace the left circuit with the right one, that's a bad idea:

The left circuit switches to appear to pass partial voltage. That means it doesn't dissipate much power. Power into something is the voltage across it times the current thru it. A fully off switch doesn't dissipate power because the current thru it is 0. A fully on switch doesn't dissipate power because the voltage across it is 0.

The second circuit always "eats up" a fixed voltage. When in series with a load, it will dissipate the 32 V or so it drops times the load current. That could be quite significant power.

Step back and show the original circuit, and explain what you are trying to accomplish.

Response to schematic

Now that you have posted a schematic, we can see that you want to replace the part that triggers the pass element, not the pass element itself.

Two back to back Zener diodes are not the same as the diac. Both can be arranged to have a threshold below which they won't conduct. However, the diac has the special property that once conduction is triggered, it the voltage drop becomes much lower and the device stays on until the current falls below some threshold.

There are some other devices that exhibit this kind of foldback behavior, but probably not appropriate for your case. Neon bulbs have this property, but usually require about 80 V to trigger for the common type. Spark gaps have this property too, but are much higher voltage devices. It would be difficult to make a spark gap reliably trigger at only 32 V.

You could try another triac with the gate being driven from a resistor divider between the anode and cathode. Or use a Zener diode between anode and gate. This will be more finnicky than a diac, but could possibly be made to work when that's all you've got. You can think of a diac as a triac that self-triggers at a pre-determined voltage threshold.


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