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I've opened up a Lutron dinner, and I'm trying to understand how the microprocessor gets its DC voltage. With my limited knowledge I'm not seeing an AC/DC converter. I've attached pictures, can someone explain how this works?

front back

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    \$\begingroup\$ What's a Lutron dinner? How does it taste? \$\endgroup\$ – Curd Jul 9 '16 at 20:21
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    \$\begingroup\$ I'm not seeing an AC/DC converter You'd need at least one diode, I see at least two. \$\endgroup\$ – Bimpelrekkie Jul 9 '16 at 20:26
  • \$\begingroup\$ Also I doubt it's a dimmer, I see no TRIAC or such. Looks like a switch to me. With a PIR movement detector. And don't assume there's a microcontroller there. For a PIR switch it's not needed, a few opamps or a dedicated chip can do the job. It seems to have a simple resistive voltage dropper, the 3 light blue resistors, elco on one side, Z2 to limit the voltage. Not rocket science ;-) \$\endgroup\$ – Bimpelrekkie Jul 9 '16 at 20:29
  • \$\begingroup\$ @FakeMoustache. Normally these have a reactive power supply using a Mylar capacitor of about .68uF. The rest looks much like you mentioned , but maybe using a relay instead of a triac. The two buttons indicate this dimmer has user controls as well. \$\endgroup\$ – Sparky256 Jul 9 '16 at 20:36
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    \$\begingroup\$ Althoug you accepted AndyW's answer I think he's wrong about how the low voltage is made. I am quite sure that Bruce's answer is actually the correct one. So reverse engineer the schematic and see who is right. \$\endgroup\$ – Bimpelrekkie Jul 10 '16 at 10:22
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There is undoubtedly an "AC/DC" convertor on the board, since the TI micro needs 3.3V DC nominally to operate. However, you're probably expecting to see an inductor or two, and a switching power supply control IC, which there isn't. That's because the circuitry consumes so little power, it's not necessary to go this route. Instead, I think the AC is directly rectified (D3, D4), and then probably "chopped" by a series pass transistor (Q8?) operating at a fixed duty frequency/duty cycle directly into the large electrolytic filter cap, with a paralleled zener diode for regulation (Z2?).

Of course, this is merely a guess, since we don't have anything approaching a circuit schematic to reference.

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  • \$\begingroup\$ This was a very helpful answer, thanks! I was expecting to see a switching power supply, but you've explained that the circuitry doesn't consume much power, that makes sense. The two diodes D3 and D4 are SM56 models, which shows them to be rated w/breakdown voltage of 61.6. I suspect that's because there is one for each leg (120v coverage). [0] Now I know where to head, I'll keep reading on chopped rectification and break out the oscilloscope. Any other thoughts to leave me with? [0] pdf1.alldatasheet.com/datasheet-pdf/view/275277/MIC/SM56.html \$\endgroup\$ – beeudoublez Jul 9 '16 at 21:24
  • \$\begingroup\$ Be very careful with the scope. unless the scope inputs are the floating input type, connecting the ground probe to places in this circuit will short circuit power to ground. Remember neutral and ground are connected together. see youtube.com/watch?v=xaELqAo4kkQ \$\endgroup\$ – Spoon Jul 10 '16 at 6:48
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The AC/DC convertor is a rather crude circuit with dropper resistors, rectifier diodes and a Zener shunt regulator. After this there should be a regulator IC which supplies the MCU with a more stable voltage.

Due to the high voltage drop it will only be capable of supplying a very low average current, but the large filter capacitor may hold enough charge to pulse a latching relay on and off.

Your images are too fuzzy to trace reliably, however I think the circuit may look something like this:-

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the 'ground' symbol in this diagram is just to satisfy the simulator. In reality the entire circuit should be considered live!

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  • \$\begingroup\$ My guess is also that this is what is used instead of what AndyW says in his answer. \$\endgroup\$ – Bimpelrekkie Jul 10 '16 at 10:18

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