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Forgive the simplicity of the question, since I'm not a professional, just a tinker with a soldering iron.

Just dancing with the idea in my head, I believe that if I rectify a current that's been phase cut (ie through a dimmer), it will just produce added noise in the output. Is my analysis correct? How would adding a capacitor to filter the output of the rectifier be affected by the modulated current?

For anyone curious, I bought several 50 watt LEDs on ebay along with a rectifier for an LED light rope. The plan is to link the LEDs in series on 120v instead of spending $50 on a 32v power supply. This is totally just for the fun of playing with it. Adding a dimmer was a thought experiment because I like thinking about things. I understand that hooking this fire hazard up to a dimmer probably won't work, but I'm interested in why it won't work. I suspect it's because the capacitor in the rectifier will fail to filter the modulated output once the current being drawn drops below a particular threshold. I'm unclear on exactly why that is or how it would affect the output from the filter.

:Edit:
A circuit diagram drawn by someone who doesn't know how to draw circuit diagrams, by request. See comments.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ You need to be clearer I think. There is not a capacitor in a rectifier. A rectifier is a diode or a bridge of diodes - no capacitors. Maybe you can show a circuit? \$\endgroup\$ – Andy aka Mar 20 '15 at 22:17
  • \$\begingroup\$ @Andyaka My understanding is that, in practice, most rectifiers smooth the output using a capacitor. I never understood circuit diagrams - the only reason I didn't go into EE professionally :( I will see what I can do to clear it up. \$\endgroup\$ – Thomas Mar 20 '15 at 22:19
  • \$\begingroup\$ You are talking about a bridge rectifier AND a smoothing capacitor - neither are integral parts of each other. Please try and link to a circuit diagram. \$\endgroup\$ – Andy aka Mar 20 '15 at 22:22
  • \$\begingroup\$ Put the capacitor before the bridge and it stands a chance of working. At the moment it will just attain a peak dc voltage and prevent the diodes from conducting current into the LEDs. \$\endgroup\$ – Andy aka Mar 20 '15 at 23:01
  • \$\begingroup\$ @Andyaka I can see that you've given some very good and easy to understand answers to other questions on this site. So I can only guess that you don't actually need a diagram yourself, and that this is some kind of educational exercise. If so, you might tell me. \$\endgroup\$ – Thomas Mar 20 '15 at 23:11
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If the cap is big enough, then its voltage will be mostly the peak of the input voltage with a little ripple.

Simplistic dimmers modulate the RMS output by keeping the output off a variable delay after each zero crossing. You can think of the full output as having 0 delay, and full off as having ½ cycle delay (8.3 ms for 60 Hz).

Since the capacitor voltage rides the peaks, there will be little change in this voltage during the first half of the delay range. Since the peak voltage occurs half way thru the possible delay range, the whole first half of the delay range will have the same peak voltage, even though the RMS value coming out of the dimmer is half at half way thru the delay interval. In the second half of the delay interval, the capacitor voltage will drop since the peak voltage coming out of the dimmer also drops.

These kinds of dimmers are rather frowned upon today due to the high frequency components they create in the current. They were also intended for loads like incandescent bulbs that respond to the RMS voltage, not the peak. They are poorly matched to modern electronic loads.

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I do not know how you feel about using a transformer. Hooking the output of your dimmer to the 110 VAC input primary and selecting a secondary output voltage that could be rectified and filtered to run your LEDs. I have used similar circuits before. The transformer smooths out the voltage, and the output voltage at the secondary is relative to the input voltage from the dimmer. Full to minimum Voltage. ------------ After seeing your comment my thought is, that their possibly is not enough load on the circuit when going above the dimmers threshold. Diodes of any type are can act strange with improper loading. 30 years as an electronics technician. I don't know if your know how a dimmer works. It uses a triac and a circuit with a potentiometer to control when each of 2 SCRs (the 2 back to back SCRs are contained in one device) will turn on each half of the phase of the 60Hz cycle. Without a proper continuous load the SCR could keep conducting, therefore never shutting off. The SCR only should turn on after reaching a predetermine voltage level, set by the potentiometer. Normally crossing over the Zero voltage level will shut off the SCRs. That is why a transformer would work the winding are a continuous load on the triac. Another possible solution is putting a Resistor in parallel with the Series of LEDs. Noise is an entirely different issue. When a diode crosses over it's threshold it can produce a noise pulse consisting of many harmonics (ELECTRICAL NOISE) an SCR is basically a control diode. Look up SCR and TRIAC on the internet.

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  • \$\begingroup\$ I'm less concerned with finding a way to dim the LEDs. In this case, the minimum voltage is too high for dimming to be effective no matter how we do it. I'm just trying to understand what happens to phase cut current as it passes through a rectifier and filter capacitor. \$\endgroup\$ – Thomas Mar 21 '15 at 16:43
  • \$\begingroup\$ So, if I'm understanding Wikipedia and applying it correctly, lowering the effective output voltage using a dimmer reduces the current draw of the LEDs below the holding current for the SCRs, forcing them into an OFF state, effectively switching off the circuit. Yes? Or am I way off? \$\endgroup\$ – Thomas Mar 23 '15 at 18:08
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    \$\begingroup\$ SCR's do have a minimum holding current. A Led begins conduction at a voltage of 1.5 to 2 Volts, but LED's current draw is not linear, any limiting resistor in series will continue to provide voltage to the LED until it reach it minimum turn on voltage of about 2 Volts.The LED will maintain voltage draw near the full cycle of 120 * 1.41 = about 170 Volt peak. If you were running on a much lower voltage nearer to it's operation Voltage. The Scr would shutoff when the LED reached it minimum operating voltage. \$\endgroup\$ – user66377 Mar 26 '15 at 22:12

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