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Williams GI Circuit 6.3VAC converted to DC - math lesson and question!

By KeithinMI

9 years ago


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    #1 9 years ago
    OP
     
    TMI

    KeithinMI

    Inactive member

    US Farmington Hills, MI

    11y 43,400 518

    ok, so I am working on some mods that I want to light through the GI.

    My circuit that I have been using is:

    GI Voltage Source--->1n4006 diode--->current limiting resistor----->LED (or 2 in parallel)

    Here's my quandary -- what I have above is basically a half-wave rectifier circuit. There are obviously 3 voltage drops, the 1n4006, the CLR, and the LEDs. I have some flicker LEDS that are Vfwd 2.0V with 30ma of forward current. In that. The 1n4006 measures 1.7v Vfwd with my meter.

    The LED calculators all presume DC voltage at the source. Using one, I get about a 100 ohm current-limiting resistor. Problem is that the flicker LED's don't flicker.

    I get that I'm only getting basically a 30Hz pulsed DC.

    So here's where I need some confirmation. According to the formulas for a half-wave rectifier:

    Vrms=Vpeak/2
    Vdc=Vpeak/π

    so, substituting, you get Vdc=(Vrms*2)/π.

    **(note that is NOT "N", it's PI above in both equations)**

    that's 12.6 (6.3VAC*2)/π or 4.01V. That is my actual DC voltage in the half-wave rectifier circuit, right?

    If that's correct then my VFwd is about the same as the source DC voltage. I know diodes need a current-limiting resistor but you can't put one in without dropping the source voltage down.

    Anyone have anything they can chime in on this?

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    #2 9 years ago
     
    DAV

    wayout440

    Pinside member

    US Berea, OH

    11y 113K 11,264 3 12

    Using your math, I get 68 ohms as the value for your current limiting resistor, not 100 ohms. This is for only one LED in a circuit. If you want to use 2 (or more) in series, your combined forward voltage can not be equal or more than the supply volts (Vfwd 2.0V*2= 4VDC, where your supply is roughly 4VDC so it's not going to work unless you up the supply voltage). Two LEDs in parallel would require a 39 ohm current limiting resistor, the more LEDs in parallel you would add, the lower the value of the CLR would become (and the higher it's wattage). I used the calculator at http://ledcalc.com/#calc

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    #3 9 years ago
     
    SBM

    steve1515

    Pinside member

    10y 36,400 393

    Something doesn't seem right in your equations. What is "n"?

    To get Vpeak from VACrms, you can use the equation Vpk = Vrms * Sqrt(2)

    So, if you rectify 6.3VACrms, than you get VDC with 8.9V peaks. Then you'd have to subtract 1VDC (for the 1N4006, depends on current, though) for the diode drop.

    So, you should use 7.9V for your LED resistor calculations.

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    #4 9 years ago
    OP
     
    TMI

    KeithinMI

    Inactive member

    US Farmington Hills, MI

    11y 43,400 518

    that's not "n" it's PI. Apparently the character set on the website doesn't display it so well.

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    #5 9 years ago
     
    HLT

    herg

    LED OCD

    13y 110K 5,551

    OP
    Quoted from KeithinMI:

    I get that I'm only getting basically a 30Hz pulsed DC.

    You're actually getting a 60Hz half wave that's not really DC. To make DC out of it, you'd need a smoothing cap.

    If you're just trying to light a standard LED, you also don't need the 1N4006. The LED is a diode itself, and it would act to block reverse current.

    Depending on how the "flicker LED" works, you may need a full-wave rectifier and smoothing cap to provide something closer to DC to it. Have you tested the flicker LED with a DC source to see how it works?

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    #6 9 years ago
    OP
     
    TMI

    KeithinMI

    Inactive member

    US Farmington Hills, MI

    11y 43,400 518

    Quoted from steve1515:

    Something doesn't seem right in your equations. What is "n"?
    To get Vpeak from VACrms, you can use the equation Vpk = Vrms * Sqrt(2)
    So, if you rectify 6.3VACrms, than you get VDC with 8.9V peaks. Then you'd have to subtract 1VDC for the diode drop.
    So, you should use 7.9V for your LED resistor calculations.

    I think your formulae are for a full-wave rectifier. The circuit that happens with the 2 diodes is actually a half wave rectifier.

    Would putting a capacitor across the circuit help at all?

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    #7 9 years ago
     
    SBM

    steve1515

    Pinside member

    10y 36,400 393

    herg is correct, I should have stated that you'll get a DC square wave with the half wave rectifier. You can add a cap or use a full wave rectifier to get actual DC.

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    #8 9 years ago
    OP
     
    TMI

    KeithinMI

    Inactive member

    US Farmington Hills, MI

    11y 43,400 518

    Quoted from herg:

    You're actually getting a 60Hz half wave that's not really DC. To make DC out of it, you'd need a smoothing cap.
    If you're just trying to light a standard LED, you also don't need the 1N4006. The LED is a diode itself, and it would act to block reverse current.
    Depending on how the "flicker LED" works, you may need a full-wave rectifier and smoothing cap to provide something closer to DC to it. Have you tested the flicker LED with a DC source to see how it works?

    yes. I hook up 1 (or more in parallel) with a CLR to a 5v DC wall-wart power supply and they flicker beautifully

    the reason for the 1n4006 would be just to insure that if there's a large PIV spike that the led is protected since PIV ratings for LEDS are a lot less than 4006 which I think is 600v PIV

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    #9 9 years ago
     
    SBM

    steve1515

    Pinside member

    10y 36,400 393

    OP
    Quoted from KeithinMI:

    I think your formulae are for a full-wave rectifier. The circuit that happens with the 2 diodes is actually a half wave rectifier.

    Would putting a capacitor across the circuit help at all?

    The formula is for AC in general. A half wave rectifier uses one diode and a full wave rectifier uses 2 diodes. If you use the full wave rectifier, then you subtract 2 diode drops.

    Yes, a smoothing cap would help a little to get you closer to DC. I don't think you will be able to get nice clean DC without a full wave rectifier though.

    Do you have a scope? It's really the only way to see what you are actually getting on the output of you circuit.

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    #10 9 years ago
     
    SBM

    steve1515

    Pinside member

    10y 36,400 393

    Quoted from steve1515:

    The formula is for AC in general. A half wave rectifier uses one diode and a full wave rectifier uses 2 diodes. If you use the full wave rectifier, then you subtract 2 diode drops.

    Maybe I'll try this out this weekend with my scope just to verify. If I have time that is.

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    #11 9 years ago
    OP
     
    TMI

    KeithinMI

    Inactive member

    US Farmington Hills, MI

    11y 43,400 518

    Quoted from steve1515:

    herg is correct, I should have stated that you'll get a DC square wave with the half wave rectifier. You can add a cap or use a full wave rectifier to get actual DC.

    I'm presuming a small value electrolytic would be fine? like a 10V 50uF?

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    #12 9 years ago
     
    SBM

    steve1515

    Pinside member

    10y 36,400 393

    I'd put in 20V just to be sure it will last. You can try different values to see what works.

    Also, just looking around some more, maybe you are correct about the half wave equation. Hmm... I'll have to look into it some more.

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    #13 9 years ago
    OP
     
    TMI

    KeithinMI

    Inactive member

    US Farmington Hills, MI

    11y 43,400 518

    According to this:

    http://www.tutorvista.com/physics/half-wave-rectifier-calculation

    the formula is right for half-wave rectifiers. Full wave rectifiers use this:

    Vdc=Vav=2Vpeak/π (pi)
    Vrms=Vpeak/Sqrt(2)

    Source: http://en.wikipedia.org/wiki/Rectifier

    btw, you can get PI on a pc by turning on num lock on the keypad, holding the left-alt key and typing 227 on the keypad.
    (or you can just go get Pie at a grocery store which tastes a lot better)

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    #14 9 years ago
     
    SBM

    steve1515

    Pinside member

    10y 36,400 393

    It's been a long time for me too.

    So, maybe it's just that the LED needs real DC to work correctly. I'm not sure how the flicker LEDs work internally.

    A smoothing cap should fix your problem.

    If you want to do the math (tau = RC) you can calculate the cap size you need. See here: http://my.ece.ucsb.edu/York/Bobsclass/2B/Extras/Half-Wave%20Rectifier.pdf (ignore most of the complex math stuff)

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    #15 9 years ago
     
    DAV

    wayout440

    Pinside member

    US Berea, OH

    11y 113K 11,264 3 12

    Quoted from steve1515:

    It's been a long time for me too.
    So, maybe it's just that the LED needs real DC to work correctly. I'm not sure how the flicker LEDs work internally.
    A smoothing cap should fix your problem.
    If you want to do the math (tau = RC) you can calculate the cap size you need. See here: http://my.ece.ucsb.edu/York/Bobsclass/2B/Extras/Half-Wave%20Rectifier.pdf (ignore most of the complex math stuff)

    Ah, yes. The LED calculators assume a straight DC (filtered, rectified). Best to get something at least close, a smoothing cap will help. Then you could a obtain fairly useful DC supply measurement with a DMM to base your LED calculations on.

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    #16 9 years ago
     
    JON

    Schwaggs

    Gameroom Mods

    10y 105K 4,760 10

    Make sure you have the LED and 1N4006 diodes anode to cathode (not cathode to cathode or anode to anode).

    If you need pure DC, you can try the cap but you may need a bridge rectifier.

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