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Calculating peak voltage, and selecting proper capacitance

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I'm working on a power filter for an LED lamp; the lamp is this bad boy: https://www.amazon.com/gp/product/B00LN3Q37C/
I like it because it can run on 10-30VDC, which makes my life easier since I'm not hooking it up to your run-of-the-mill 12VDC automotive system.

The power supply I have handy is ~16VAC (I measured about 15.5V AC no load, I figure I should round up to give myself some margin of error). The first order of business is to slap in a full-wave bridge rectifier, so I'm not losing half the juice. That's no problem, I have one rated 1KV/35A that shouldn't even need heatsinking :)

After that though, I want to filter out the rectified sine ripple, and for that I need caps. If I recall correctly, the formula for calculating RMS voltage is
[latex]V_{rms} = \frac{V_{pk}}{\sqrt{2}}[/latex] which means to peak voltage should be
[latex]V_{pk} = V_{rms} \times \sqrt{2}[/latex]that is,[latex]V_{pk} = 16V \times \sqrt{2} \approx 23V[/latex] correct? So I should be fine and dandy with 50V capacitors? This value also makes me happy because it's well below the 30V rated max input.

The next order of business is figuring out capacitance... I don't even know where to begin on this. On past projects I just grab a decent-sized one and hope for the best, but is there any way to determine what will give me the best filtering without overkill? Instinct tells me a 2200uF cap might do the trick here... and since it's usually easiest/cheapest to get them in packs of 5 or 10, would it hurt to just throw two in parallel and call it a day (I'm pretty sure 4400uF is overkill)?

Thanks!
 
After that though, I want to filter out the rectified sine ripple, and for that I need caps. If I recall correctly, the formula for calculating RMS voltage is
which means to peak voltage should be
that is,

correct? So I should be fine and dandy with 50V capacitors? This value also makes me happy because it's well below the 30V rated max input.
All correct.

The next order of business is figuring out capacitance..
Instinct tells me a 2200uF cap might do the trick here...
It all depends on how much ripple you can live with, the current and the ripple frequency.
I guess the ripple frequency will be 120hz if your supply is 60hz.
The current is a bit of an unknown, but the advert says it is a 12Watt lamp, so lets guess at around 1 amp.
Ripple voltage ???? It is up to you.

All in all, 2200uF is a reasonable starting point, you will probably have about 1 volt ripple. Try it and see.

JimB
 
Your transformer has no current rating. Its voltage with no load will be higher (maybe MUCH higher) with no load. You need to know its loaded voltage. The full wave bridge rectifier will drop 2V.
The LED light has no manufacturer's name so it has no detailed datasheet. XCSOURCE never heard of it. Is it 12W when the voltage is 30V, 6W when the voltage is 15V and only 4W when the voltage is 10V?
How many minutes will the LED light last when its voltage is more than in a 12V vehicle?
How will you cool it when it is not moving fast?
 
JimB thanks for the pointers, I've got a pack of 50V 2200μF caps en route so if I don't like the behavior of the lamp @ 2200, I can always slap another in parallel and that should be more than enough.

audioguru this is actually for a snowblower, and is powered off the blower's alternator, so it'll have different characteristics from a transformer, but yes I imagine the voltage under load will sag some. Also excellent point about the rectifier drop, I completely forgot about that -- though it theoretically works in my benefit, putting the filtered voltage closer to 20V (and buying me a little more headroom on cap voltage rating as well as max input voltage on the lamp).

I'm not worried about the temperature too much, because it's going to be operating in below-freezing temperatures anyways (it's currently 20°F/-7°C out right now); also, if you look at the back of it, it looks like it's got pretty good built-in heatsinking -- I'd imagine enough to keep the LEDs decently cool in cold weather:
71sHdazfhvL._SL300_.jpg
 
In most unregulated supplies you want a ripple voltage of about 10% of the DC. In your case 1.5V. The rule of thumb is 10000 uF for 1A 1V ripple. So in your current of 1A and 1.5V ripple you need about 6600 uF .
 
"Capacitor Sizing - The following equations provide the approximate capacitance required to achieve a given ripple percentage. These equations are generally good for ripple percentages less than 10 and are derived empirically. In general 3 percent ripple at the maximum load is reasonable. Calculate the required capacitance for nominal 3 percent ripple and then round up or down to the nearest standard size. Note that designing for 1 percent ripple takes three times the capacitance for only a marginal improvement –although there may be situations where that is important. Design always takes consideration of many variables and there is rarely such a thing as a singular solution.

For full wave rectification

C = 8.6 / F * R * %ripple.

C in Farads
F = ripple frequency
R = Effective Load Resistance (Vo / Io) "

Does this seem reasonable?
 
Honduras for the ripple frequency in that calculation, would I use double the AC output frequency (i.e. if my meter measures 40Hz on the AC output, I'd use 80Hz)? Since full wave rectification inverts the negative peak, the Δt from +Vpeak to +Vpeak is effectively halved right?
 
I think your application is a little different than everyone thought it was. We all thought the LED was the only thing out there when in fact it is everything on the snow blower. I don't know what that might be besides the ignition stuff. I would think it was already rectified to charge the battery. Or is there no battery?
 
Nope, no battery; it has an electric start that plugs into the wall, as well as the usual recoil (pull) starter. I measured at the headlamp connector, and it's definitely AC (when measuring in AC I get ~15.5V, when measuring in DC I get meaningless bouncy numbers). I guess they figured (back in 1996 when it was built) a regular incandescent bulb doesn't need rectified output, and besides the ignition coil I think the lamp is the only load.
 
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Honduras for the ripple frequency in that calculation, would I use double the AC output frequency (i.e. if my meter measures 40Hz on the AC output, I'd use 80Hz)? Since full wave rectification inverts the negative peak, the Δt from +Vpeak to +Vpeak is effectively halved right?

That's the way I understand it. :)
 
Snow blowers like this generally have a simple alternator where a flywheel with fixed magnets pass over coils which are stationary. The output AC frequency will be a function of the number of poles (coils) and the engine speed so the frequency really isn't a fixed value. This works fine on the AC only alternators for running a basic headlight which were originally incandescent bulbs and didn't care about the frequency. I would just put together a full wave bridge as planned then hang some filter caps out there. I would start with around 2200 uF and see what I get, then add another 2200 uF if my LEDs don't like their power. Snow blowers that actually use DC just come off the alternator with a rectifier pack which also tends to battery charging on units that have a battery. I wouldn't over think this. Also, it would be nice to know what the alternator output is, they can run from 12 watts right up to 60 watts.

Ron
 
Reloadron without a manual/specifications, is there a way to determine the max output capability? As best as I can tell, the original lamp was a 35W halogen unit, so it should have enough juice to run my LED setup, at least.
 
Reloadron without a manual/specifications, is there a way to determine the max output capability? As best as I can tell, the original lamp was a 35W halogen unit, so it should have enough juice to run my LED setup, at least.

That would be my thinking. Short of applying various increasing loads to it and not having any documentation or data sheets I wouldn't worry about it much. On the "bright" side my neighbor does snow plowing and added I think 3 or 4 of those LED headlamps above his blade. Wow! Those things really light things up!

Ron
 
That would be my thinking. Short of applying various increasing loads to it and not having any documentation or data sheets I wouldn't worry about it much. On the "bright" side my neighbor does snow plowing and added I think 3 or 4 of those LED headlamps above his blade. Wow! Those things really light things up!

Ron

Try surplus aircraft landing lights some time. :)
 
Try surplus aircraft landing lights some time. :)

Actually I once saw that done with a test vehicle at OSU (Ohio State University). Talk about some bright lights! :)

Ron
 
One night a Jeep passed me in Denver. Apparently he had trouble getting around. I flashed my headlights and he lit me up with one of those that he had mounted on the rear of the Jeep. Not a nice surprise. :)
 
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