3
\$\begingroup\$

Not after a full breakdown of how this specific light dimmer works, but rather how a device in series can control a load.

A light dimmer has the following wiring diagram. As you can see, the light dimmer is in series with the load.

enter image description here

From my research, most household dimmers modify the AC waveform to reduce the RMS voltage, hence lower the light, rather than using a rheostat.

enter image description here

Most of the information suggests the switching is done by a TRIAC, but in here there seems to be a power MOSFET (IRF840A) with a heatsink. There seems to be a 555 timer and a 324 op amp, which I'm guessing does the modulating.

Of course a definite answer of how this device works cannot be observed without reverse engineering it, however in general I don't understand how a device in series can operate when it's purpose is to control the load. The circuitry needs current to operate, but it's purpose is to (partially) kill the power to the load, and thereby killing it's own power (since the device is in series).

This is my interpretation of how a device could modulate it, but it doesn't make much sense to me.

schematic

simulate this circuit – Schematic created using CircuitLab

|improve this question
\$\endgroup\$
  • 1
    \$\begingroup\$ I believe the actual question being asked here is, "how does the control circuitry get power without a connection to neutral?" \$\endgroup\$ – gbarry Jan 5 '15 at 18:04
6
\$\begingroup\$

There are even simpler in-series circuits that can achieve the same: diac+triac (for bipolar waveforms) or just a bridge-rectifier+scr (thyristor) [this will only produce unipolar waveforms], or... if don't care about adjustment, just a diode! I'm honestly not sure what is that you don't understand. Reduce the problem to the simplest idea: a diode in series with a load will reduce power on AC because it cuts out half the waveform.

EDIT: Ok, since based on your comment below you're confused about the adjustability part, it helps to look at a simple rectifier+scr circuit. There are dozens on the net, but the one below seems the simplest (courtesy of Kuphaldt's Lessons in Electronic Circuits), so most useful for educational purposes:

enter image description here

Yes, the dimmer does steal some power for its control function/sub-circuit (which in this case is simply the pot and diode on the right side of the figure), but relatively little.

EDIT2: As to your theory that it must "kill its own power" (and thus not work)... again ask yourself: is a diode "killing its own power"? Maybe so, but should you infer from that that it only works for exactly one half-period and then stays off forever? No, because it reacts to change in potential at its inputs caused by the ever-changing AC waveform... which keeps passing through the load.

EDIT3: As for your MOSFET+opamp circuit, I found a good article on EDN explaining how it works: "Use a self-powered op amp to create a low-leakage rectifier". If you understand the basics from the previous simpler circuit, you should have no trouble following that.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Maybe active wasn't the correct term. The device allows variable control of the light, so the circuitry to do this requires power, yes? What I don't understand is how the circuitry get's power, when it's job is essentially trying to (partially) kill the power to the load, thereby killing it's own power.. \$\endgroup\$ – tgun926 Jan 5 '15 at 9:58
5
\$\begingroup\$

The trick is that you can leak a small amount of current through a lightbulb without turning it on. If you probe the light socket with a multimeter (CAUTION) you can observe this: http://www.newton.dep.anl.gov/askasci/eng99/eng99750.htm

The control circuitry will require something like a miliwatt. There will be a capacitor-based rectifier in there; probably the large electrolytic capacitor will have 5-10V across it in operation. The light is never completely off; the current to power the dimmer will always flow through the light.

What's the large black component left-foreground in the picture?

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Yeah I just read the label and it says minimum 10VA. That's a fuse I believe. \$\endgroup\$ – tgun926 Jan 5 '15 at 11:05
  • \$\begingroup\$ I've upvoted you because this is something that I should have perhaps explained better: "killing power" is not an absolute off;even a diode has some extremely small reverse current, IS in the Wikipedia figure. The control circuit in these in-series dimmers is effectively in parallel with the "electronic switch" [MOSFET, scr, triac etc.] and both are in-series with load. So the control circuit may be always powered. It's just that its quiescent current is so low that it is negligible to the load (won't light a lamp etc.) \$\endgroup\$ – Fizz Jan 5 '15 at 11:25
  • \$\begingroup\$ Another trick is that once triggered, the triac will keep conducting for the duration of the half-cycle without any input needed. \$\endgroup\$ – gbarry Jan 5 '15 at 18:02
  • \$\begingroup\$ Large black thing looks like a capacitor, for noise suppression, or possibly a reactive voltage dropper as part of the power supply. \$\endgroup\$ – gbarry Jan 5 '15 at 18:03

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.