# Voltage too high



## strtch5881 (Oct 6, 2018)

Hi all. 
i have added an led light and hand warmers to my machine, but I keep blowing the led lamp. I have followed the tutorial on this site and checked the wiring 4 times now. Testing shows 18.6 vac to the rectifier, but, 45.3 vdc coming out of the rectifier. I have tried 2 different full wave rectifiers from different companies with the same result. Any thoughts?????


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## strtch5881 (Oct 6, 2018)

I forgot to say Tecumseh HMSK100 on a 1996 Yardman 1030.


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## bigredmf (Jan 16, 2018)

You are gonna need a higher voltage bulb or perhaps lose the rectifier 
The led conversion post has gotten to large to research effectively.

Red


Sent from my iPhone using Tapatalk


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## strtch5881 (Oct 6, 2018)

bigredmf said:


> You are gonna need a higher voltage bulb or perhaps lose the rectifier
> The led conversion post has gotten to large to research effectively.
> 
> Red
> ...


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## rslifkin (Mar 5, 2018)

As another option, you could run the output of the rectifier through a voltage regulator to limit the voltage to something the LED will be happy with.


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## Greg13 (Nov 25, 2018)

Are you sure you are measuring it correctly? It should be like 45v AC to the rectifier and 14-16v DC


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## strtch5881 (Oct 6, 2018)

Greg13 said:


> Are you sure you are measuring it correctly? It should be like 45v AC to the rectifier and 14-16v DC


I am reading it correctly, and am very confused.


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## tuffnell (Dec 1, 2011)

Sounds like your rectifiers are voltage doublers as well.
With 18.6 VAC, it gives me the impression your voltage measuring device is reading peak voltage rather than RMS. If it is a peak reading meter than the RMS value would be 13.2 VAC RMS. Most 

If the meter is reading 18.6 VAC RMS, than with a full wave rectifier and capacitor filter, max voltage would be 26.3 VDC.
Something just dosen't add up.


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## drmerdp (Feb 9, 2014)

tuffnell said:


> Sounds like your rectifiers are voltage doublers as well.
> With 18.6 VAC, it gives me the impression your voltage measuring device is reading peak voltage rather than RMS. If it is a peak reading meter than the RMS value would be 13.2 VAC RMS. Most
> 
> If the meter is reading 18.6 VAC RMS, than with a full wave rectifier and capacitor filter, max voltage would be 26.3 VDC.
> Something just dosen't add up.


Agreed. 

Is the OP using a capacitor(s). Sounds like you might not be. Capacitors are great for smoothening the current, reducing ripple, and suppressing spikes.


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## JayzAuto1 (Jul 25, 2016)

strtch5881 said:


> Hi all.
> i have added an led light and hand warmers to my machine, but I keep blowing the led lamp. I have followed the tutorial on this site and checked the wiring 4 times now. Testing shows 18.6 vac to the rectifier, but, 45.3 vdc coming out of the rectifier. I have tried 2 different full wave rectifiers from different companies with the same result. Any thoughts?????


Do you have a pic or model numbers of the rectifiers you have used??? Almost sounds as if it's wired incorrectly somehow. What are you using for a ground?? Is it the engine block?? Using factory connectors?? Can you supply a wiring diagram of how you presently have it wired?? 

What are you using for an LED Bulb??

Those are wacky numbers,,,,,Can't say I've ever seen results like that.


GLuck, Jay


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## strtch5881 (Oct 6, 2018)

I will try to explain the best that I can. KBPC5010 rectifier, 2- 2200uf 50v radial capacitors and a single 880 1200lmn fog light bulb to replace the stock halogen bulb.
Stator wire to the ac+. The other ac pin to ground. 
dc+ to heater switch and light switch. dc- to ground.
Capacitors between - and + light wires before switch.
Used factory ground next to light. The ground is on a metal plate mounted between the handlebars.
Vom meter used is an Innova auto ranging meter.
I have followed the diagrams on this site, and, double checked my wiring of the rectifier with various other sites on the web.


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## SayItAintSnow (Dec 15, 2017)

Strtch',


*A couple of things.....*


When you say you are getting 45 volts DC, are you taking that measurement under a "no-load" condition, or are your heaters and lights connected into the circuit at that point?


The purpose of adding a capacitor, is to reduce the ripple in the rectified AC, to create more of a "saw-tooth" looking voltage. (Still not pure dc, but closer than what comes out of the bridge...lol). To me, 2200uf seems unnecessarily high, and you are using two of them. (Much over 470uf is not likely to affect the ripple to any degree better than say, 470uf). But are these caps wired in series or parallel across the dc output?


You also mentioned that you grounded the negative terminal of the DC out from the bridge. I'm not sure if that's the best approach. In a conversion that I did last year, I kept the negative isolated on its own wire, and it all works very well.


Since you are powering both hand grip heaters and a light, that means that they should be separately fused, and I'm guessing that's not the case. The right fuse just ahead of the light would have protected the led from burning out, allowing you to troubleshoot the problem without further damage.
.
.


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## strtch5881 (Oct 6, 2018)

First, at this time, I only have the light fused.
The capacitors are wired in parallel.
The testing is done no load. I don't want to burn out another bulb. I used a 5 amp fuse and it didn't blow, but the bulb did. Perhaps the fuse is too strong?


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## LDRider (Jan 24, 2018)

I believe you are getting erroneous readings because you are measuring both AC an DC but probably do not actually have a pure form of either, and certainly not DC. 

The AC coming from the alternator(s) should be a true sine wave, which is 'correct' AC (as opposed to chopped AC, stepped AC, etc. such as many inverters put out). But when you rectify AC you do not get true DC but rather pulsed, sine wave AC which will confuse a simple voltage meter set to DC (VOM). As you say, I cannot see any way that diodes (a rectifier is just four diodes) increasing voltage. 

That said, you do have to be careful using diodes with capacitors and AC because you can make a voltage ladder or voltage multiplier by mistake and that would indeed increase the voltage. This is the method used to increase the voltage for such things as CRT's in older TV's, oscilloscopes, etc. That said, do not think you are actually making one if only using capacitors after the rectifier exclusively.

The best thing would be to view the output(s) with an oscilloscope if at all possible, and then you would see the actual voltage curve(s) and possibly find a spike or something else odd. ??

Best of luck with this but at least it is an interesting problem (yeah, I know- that does not help much: sorry).



strtch5881 said:


> I am reading it correctly, and am very confused.


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## bkwudz (Jan 14, 2013)

that bulb probably have some internal rectifier, did you try it wired directly to the stock wiring?


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## strtch5881 (Oct 6, 2018)

that bulb probably have some internal rectifier, did you try it wired directly to the stock wiring?


No, but I have tried the second bulb in the pack on a 12 volt battery and it worked perfect.


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## bkwudz (Jan 14, 2013)

Im running that type bulb off the stock wiring in the stock housing on my Ariens, the extra LED flood lights run through the rectifier, capacitor setup


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## strtch5881 (Oct 6, 2018)

Interesting, do I chance it and blow another bulb?


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## tadawson (Jan 3, 2018)

Reference this site for schematics . . . https://www.electronics-tutorials.ws/diode/diode_6.html

A full wave rectifier wired as shown (and sounds like it is as the OP has his wired, with the possible exception of tying AC and DC grounds . . ) cannot increase voltage. Voltage multipliers are a radically different circuit. Voltage drop across silicon diodes is also a constant at .7v per diode - again, that's physics . . . 

So, the output voltage *will* be .637 times the peak voltage of the input. (Note that the meter reads *RMS* voltage, not peak . . . ). For a pure sine wave, the peak value will be 1.414 (the square root of 2) times the RMS reading, so assuming that the OP's meter isn't a wildly inaccurate $5 overseas pile, the peak voltage should be 26.3, and the output of the rectifier then 16.75 . . . (not considering diode drop) which is more than a 12V light wants to see. Were your really putting 45v into the lights, I suspect that they would go off like flash bulbs, but 16 and change will take them out over a short period of time. 

A half wave rectifier (single diode) would give you 8.3 volts or so DC output . . . so looks like the alternator output on this machine splits the difference and misses the 12v nominal (13.8?) on either side. *Or* the incandescent bulb had put enough load on it to pull it down to what was good for that bulb, and the LED does not draw enough . . . 

https://www.electronics-tutorials.ws/diode/diode_5.html

So, either add a load, or perhaps a dropping resistor or regulator to the LED.

And conclude that your meter is on crack regarding the DC value you see. The physics and engineering clearly define that output as impossible from a bridge (no matter what caps you use) no matter how you wire it.


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## tadawson (Jan 3, 2018)

bkwudz said:


> that bulb probably have some internal rectifier, did you try it wired directly to the stock wiring?



Even if it did, it would be irrelevant. DC going through a second rectifier will just see a voltage drop due to the diodes (.7 single wave, 1.4 full wave) and will go through with no issues. That's basically how devices can be made to not be voltage sensitive - put a full wave bridge on the input of a DC device, and you can then plug in any polarity without risk.


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## strtch5881 (Oct 6, 2018)

Wow!!!! I can offer a good blueberry pancake recipe.


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## drmerdp (Feb 9, 2014)

I have done a dozen led conversions using a full wave rectifier and 2 2200uf capacitors. I’ve never had a problem, but I have seen high voltage when unloaded. But adding load brough everything back into spec. 

If you are afraid of hooking up the LED bulb Reinstall the hologen bulb and test the voltage. You might be in a position where the single led replacement bulb is too low wattage and you need additional wattage to keep it in the 9-32v window that most LEDs need.


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## SayItAintSnow (Dec 15, 2017)

strtch5881 said:


> _First, at this time, I only have the light fused.
> The capacitors are wired in parallel.
> The testing is done no load. I don't want to burn out another bulb. I used a 5 amp fuse and it didn't blow, but the bulb did. Perhaps the fuse is too strong?_



I hear ya' man, I wouldn't want to risk another led lamp until I knew what was going on either....:wink2:


As to your question about the fuse....well, obviously the fuse is rated too high for the amount of current you have passed through that circuit, or else the fuse would have let go, and protected the light. But that's a side problem that's easily corrected by substituting a lower amp fuse, once the initial problem is resolved.


I went back, and I can't see anywhere where you mentioned the rated wattage for the light. Can you tell me what that is? Then we can calculate the best size fuse to use.


The reason I asked about the caps being in series or parallel, is that, as mentioned previously, I think that the caps are higher than they need to be, and you have wired them in parallel. Keep in mind, with caps it is the _*opposite*_ of resistance. i.e. you wire two equal resistors in parallel and you cut the overall resistance in half, but put two equal caps in parallel and you _*double*_ the overall capacitance. So think about the charge you are storing up in those caps with an open circuit, because you have no load on there. This could be giving you a false voltage reading.


So, let's find something cheaper to place in the circuit to test the voltage with a load, that is draws comparable watts to your led. If you can do that, I'll bet you'll find that the real voltage (under load) is not that high.


.


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## strtch5881 (Oct 6, 2018)

Easier to take a pic instead of typing it all.


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## RedOctobyr (Mar 2, 2014)

It's not a lot of watts. But if you chose to run 2 bulbs, connected in series (not parallel), each bulb would see half the voltage. The bulbs can handle up to 30V, so even if you really do have 40V+ somehow, this would still bring it into an acceptable range.


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## strtch5881 (Oct 6, 2018)

I figured that I would replace the halogen with a good led so I would have enough juice for the handwarmers. By the time I'm done, I probably could have bought a pair of heated gloves.


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## SayItAintSnow (Dec 15, 2017)

strtch5881 said:


> _...... By the time I'm done, I probably could have bought a pair of heated gloves._





:grin: lol....


Don't give up yet Strtch', there's a solution here.....
So 5 watts each....so how many did you hook up at a time? Just one.... or two at a time?


Red' seemed to imply that you had two hooked up in series. Is that true? Two would be no problem, but you should hook them up in *parallel*, NOT series with your DC source. When you place multiple loads in series you are drawing the same amount of amps from the supply as you would if they are in parallel, *but* the total of all the loads is drawn through each individual component of the load. That can be a problem for a device not capable of handling that much current.


_So here's something to try:_ Go to your nearest big box store and buy one or two of those little incandescent bulbs they use for night lights or sewing machines. You can usually find them around 5 or 6 watts, and a package of two is usually about a buck and a half. Yes, they are meant for AC, but they are simply a resistive load, and won't care if you feed them DC.



Tape up a couple of wires to the socket on one, (better to solder them if you can) and use that for your load, only for testing purposes. With that hooked up and the blower running full throtle, now measure your voltage to the bulb, and I'll bet you it'll be more like 12-18vdc


Let us know what you found out. :thumbsup:
.
.


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## bkwudz (Jan 14, 2013)

tadawson said:


> Even if it did, it would be irrelevant. DC going through a second rectifier will just see a voltage drop due to the diodes (.7 single wave, 1.4 full wave) and will go through with no issues. That's basically how devices can be made to not be voltage sensitive - put a full wave bridge on the input of a DC device, and you can then plug in any polarity without risk.


Your speaking another language to me...LOL, dont see how its irrelevant though, if he’s made a mistake in the wiring, and finds the light will work with the stock wiring.


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## strtch5881 (Oct 6, 2018)

My plan was to replace the single halogen with a single led. It has a single bulb headlight.


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## tadawson (Jan 3, 2018)

SayItAintSnow said:


> I hear ya' man, I wouldn't want to risk another led lamp until I knew what was going on either....:wink2:
> 
> 
> As to your question about the fuse....well, obviously the fuse is rated too high for the amount of current you have passed through that circuit, or else the fuse would have let go, and protected the light. But that's a side problem that's easily corrected by substituting a lower amp fuse, once the initial problem is resolved.
> ...



Fuses protect against overcurrent - this failed due to overvoltage. Likely, a fuse change would have accomplished nothing. 



Power is current times voltage, so if the voltage is high, and the current the same, you put more power into the load, but the fuse sees exactly the same thing . . .


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## strtch5881 (Oct 6, 2018)

:dizzy: I feel like I walked on the set of Big Bang Theory. 
I just want the candle to light when I put the match to it.


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## RedOctobyr (Mar 2, 2014)

SayItAintSnow said:


> Red' seemed to imply that you had two hooked up in series. Is that true? Two would be no problem, but you should hook them up in *parallel*, NOT series with your DC source. When you place multiple loads in series you are drawing the same amount of amps from the supply as you would if they are in parallel, *but* the total of all the loads is drawn through each individual component of the load. That can be a problem for a device not capable of handling that much current.


I wasn't implying he was using 2 lights in series. I was suggesting that as an option to make a high voltage more moderate, cutting it in half (as far as each light is concerned), by using 2 lights in series. 

In this scenario, if the voltage remains too-high, connecting 2 lights in parallel might just blow 2 of them. Assuming that the extra load doesn't bring the voltage down. 

That's an interesting point about flowing too much current through lights connected in series. For a simple resistive load, like incandescent bulbs, adding more bulbs in series would increase the resistance of the circuit, and reduce the current, rather than increasing it. 

But these are not a simple resistive load. Going by the specs shown in post #24, if we assume these lights put out a constant 5W, using any of their input voltage range (9V-30V), then that would be different. If you ran a single 5W light on a 20V source, it would draw 0.25A. But connect two 5W lights in series, on that same 20V source, and each light gets only 10V. Meaning it has to draw 0.5A, to output the 5W. But there are 2 lights, so the total current draw actually becomes 1.0A, all of which has to flow through each light, as you said. 

For comparison, if wiring them in parallel, each light would still get 20V, so each would draw 0.25A, for a total, in parallel, of 0.5A. But each light would still only see 0.25A of current flowing through it. The downside here is that, if your voltage really is too high, then each light still gets the full voltage, and could be damaged. 

I do like the idea of using some cheap little incandescent lights for testing purposes. As another possible source, something like car turn-signal bulbs would be 12V, or maybe car interior light bulbs, etc. Something that will draw around 5W, like your LED, would be nice, as it would approximate the load you'll be putting on the alternator.


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## SayItAintSnow (Dec 15, 2017)

tadawson said:


> _Fuses protect against overcurrent - this failed due to overvoltage. Likely, a fuse change would have accomplished nothing.
> 
> Power is current times voltage, so if the voltage is high, and the current the same, you put more power into the load, but the fuse sees exactly the same thing . . _.





Sorry, but that's mostly incorrect...

While it's true that fuses protect against over-current, your assumption was that if you increase the voltage to a given load, the current can stay the same. That's not true. 

_Ohm's law disagrees._

To illustrate this, let's use an example.












Let's say you have this simple circuit where the resistance load is 4 ohms.

If the voltage being supplied is 12 volts, the current will be 3 amps.

Ohm's law: I=E/R (or current= voltage/resistance)

In this case E/R = 12/4 = 3amps


Now, let's increase the voltage to 40 volts. The resistance of the load stays the same of course, but does the current stay the same? Heck no....












I=E/R E/R is now 40/4 = 10 amps


I'm thinking that the O/P issue with blowing out the led is a factor of two things:
1) He's using an led module that is particularly sensitive to even a very fast over voltage/high current surge
2) The capacitors, while they will certainly work to smooth out the wave form, are oversized for the job and can store a very large charge which when dissipated into the led load causes it to fry. This would be a problem especially if a switch is connected directly to the led. The caps would build up a charge, when the switch is turned on, and the caps discharge into the led.....ZAP!


Further suggestions for the O/P to try are in the next post......
.
.


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## SayItAintSnow (Dec 15, 2017)

If the O/P will try a couple of things, we can prove this is what's happening.....:wink2:
1) place the 5w incandescent bulb into the circuit. Measure the voltage. If the voltage is within an acceptable range, go to step 2.
2) Remove the caps from the circuit. Their only purpose is to smooth out the pulsating DC waveform into a sawtooth waveform, so the led should still work, but you might notice a pulsation.

If all that works, you can either
1) leave out the caps completely, if the lighting is acceptable with minimal pulsation
2) use a different, less sensitive led
3) use a single, smaller cap like 470uf
4) leave the caps in, and wire in a "bleeder resistor" across the plus and minus of the dc supply, to dissipate the excessive charge in the caps (This can be tricky to calculate what value of ohms is the best compromise....)
.
.


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## RedOctobyr (Mar 2, 2014)

SayItAintSnow said:


> 2) The capacitors, while they will certainly work to smooth out the wave form, are oversized for the job and can store a very large charge which when dissipated into the led load causes it to fry. This would be a problem especially if a switch is connected directly to the led. The caps would build up a charge, when the switch is turned on, and the caps discharge into the led.....ZAP!


I've used capacitors to help smooth out voltage dips from a power supply for a radio-controlled aircraft. It took the 25V from the main battery, and dropped it to 6V, to run the control electronics. Any momentary dip too low, like from a sudden servo load, could cause an expensive crash. The capacitors I added on the power supply's output helped avoid critical voltage dips. 

I've seen capacitors cause arcing at connectors from the main battery, when large capacitors on the speed controller would suddenly charge in an instant as the battery was connected. 

But how would capacitors, through a switch, blow the LED? Wouldn't the capacitors just charge up to their normal voltage? They'd be *able* to provide a large current, but they wouldn't force too much current, would they? Or do you mean that, without the load from the light to bring the voltage down (that is, the light switch is off), the alternator might charge the capacitors to a high voltage, like maybe 40V? In that case, I can see how you'd cause problems by suddenly dumping that 40V into a light rated for 30V. 

strtch5881, I hope you aren't regretting asking the question  I'm finding the discussion interesting, and educational. And I think your problem should be solvable.


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## strtch5881 (Oct 6, 2018)

I find it quite entertaining, although beyond me. The only thing I learned from a book is, duck when it's thrown at me. I learn from bits and pieces as I need them. Besides, while you guys are having at it, a chainsaw showed up for me to tear down. So, go nuts.
I will get back to the problem at hand tomorrow afternoon.


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## SayItAintSnow (Dec 15, 2017)

[COLOR=Red said:


> _Or do you mean that, without the load from the light to bring the voltage down (that is, the light switch is off), the alternator might charge the capacitors to a high voltage, like maybe 40V? In that case, I can see how you'd cause problems by suddenly dumping that 40V into a light rated for 30V.
> 
> strtch5881, I hope you aren't regretting asking the question  I'm finding the discussion interesting, and educational. And I think your problem should be solvable_[/COLOR].





Exactly.....Some led lamp assemblies may be robust enough to handle that momentary dump from the caps, but maybe the ones he's been using aren't. They're not rated for very much power in the first place. (5 watts?)
But he's got two 2200uf caps in parallel, so that's well over 4000 uf total, which will hold quite a charge.



If it were me, I'd put the resistive load on there....could be an automobile lamp like you suggest, so long as we're close on watts, measure the voltage then. If the dc voltage is o.k., take the caps out of the circuit temporarily and hook up the led to see how it works. It may be good to go right there, if there's no flickering problem. I don't know what the typical frequency of the ac output of his snowblower would be, but perhaps it would be fast enough after running through the bridge that it might not be objectionable.



*As for Strch possibly having regrets......at this point he's probably out at the store buying a rechargeable 12v battery, and decided to scrap this whole idea!*:devil:


BTW Red', I agree completely with your post prior to the one quoted above, especially the bit about these not being simple resistive loads. I've just been sticking with the fundamentals, because we can't be at Strch's machine to see exactly what's happening, like with a scope.


I hope he doesn't just give up on this, but he's probably hiding from us now.......:devil:
.
.


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## Mike C. (Jan 28, 2016)

Is it possible that the higher than expected DC output voltage is a result of high peak voltage being attained due to the alternator's output being relatively high frequency AC and the large-value caps?


The frequency of my HM80's alternator at full throttle is about 360hz.4400uf of capacitance seems overkill,to me.


Could a bleeder resistor be used to discharge the caps when the engine is shutdown and to keep the output voltage to something reasonable?


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## SayItAintSnow (Dec 15, 2017)

Mike C. said:


> _Is it possible that the higher than expected DC output voltage is a result of high peak voltage being attained due to the alternator's output being relatively high frequency AC and the large-value caps?
> 
> 
> The frequency of my HM80's alternator at full throttle is about 360hz.4400uf of capacitance seems overkill,to me.
> ...



Mike,


Yup...that is what I have been suggesting. You don't need 4400 uf to act as a smoothing capacitor. You're right. It's overkill, and can lead to potential problems. Many have probably set their leds up this way and gotten away with it because the particular leds they used were not that sensitive, or they simply have the leds connected all the time, imposing a load immediately on the supply, therefore allowing the caps to immediately start discharging into the load, rather than just building up. But if you are using a switch, it would be important to wire the switch in _"ahead"_ of the caps, so when you don't need the light, the caps aren't charging.


That's interesting about the frequency....didn't know it was that high. So yes, that's a good point.


As for the idea of adding a bleeder, it's a good idea to act as an additional safety feature to help preserve the leds, and prevent them from receiving an unwanted spike. High voltage supplies use them to discharge potentially dangerous residual charges in electronic equipment. But the problem with bleeder resistors is always determining the best compromise between a resistance low enough to quickly discharge the caps when there's no other load, vs. a resistance that's so low that it robs the system of power all the time, turning it into wasted heat.



Another fairly uncomplicated solution to protecting the leds would be to install a Zener diode rated for a constant 15vdc or so, just ahead of the led on the positive feed. The zener will clip any voltage over 15v and represents a relatively cheap form of voltage regulation.
.
.


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## strtch5881 (Oct 6, 2018)

Ok fellas, I don't give up that easy. My first attempt was without the caps. I blew out the first 2 bulbs. That's why I bought the caps. I had hoped they would take care of any spikes. I don't understand the fundamentals of electronics and it's beyond me. That does'nt stop me from soldering a bunch of stuff together though. The Zener diode has my attention and I will look into that further.


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## bkwudz (Jan 14, 2013)

i still have to ask if you tired the LED bulb with the stock wiring? No rectifier, no caps. My areins is running an 881 LEd bulb in the stock housing, stock plug and wiring. I tired 4 different bulbs( different wattage's) and they all work, 

I run 2 bigger LEDs off the rectifier and caps setup, wired separately from the stock light.


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## bkwudz (Jan 14, 2013)

if it still blows with the stock wiring, then i suspect you have a alternator problem. If it works fine, then i suspect its your wiring, but at that point if it works, who cares


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## strtch5881 (Oct 6, 2018)

I just remembered my first experience with caps. About 35 years ago I took apart a 110 pocket camera to see why it didn't work. It was junk, so I went on a mission of discovery. I was playing with the cap for the flash, slowly discharging it. My wife was excited about my discovery, but not the same way that I was.


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## classiccat (Mar 1, 2014)

The no-load high DC voltage is real.

Ditch the switch if you haven't already; wire directly from the bridge to LED lamp.


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## strtch5881 (Oct 6, 2018)

bkwudz
No, I did not try with just the factory wiring. But I wonder, does your Ariens have ac or dc at the pigtail?


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## bkwudz (Jan 14, 2013)

​


strtch5881 said:


> bkwudz
> No, I did not try with just the factory wiring. But I wonder, does your Ariens have ac or dc at the pigtail?


AC, thats why i have to run the Rectifier caps setup for the bigger LED lights. But the 881 bulb works in the stock location, with no modifications. It has some type of rectifier built in.

i would stop wasting time trying to find a problem that might not matter.


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## strtch5881 (Oct 6, 2018)

Well, why not. I'll give that a whirl this afternoon.


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## ELaw (Feb 4, 2015)

strtch5881 said:


> :dizzy: I feel like I walked on the set of Big Bang Theory.


Hi, my name is Sheldon and I'm here to help! :devil:

I think what you may be experiencing is a RMS-vs.-peak issue, combined with a weirdly designed alternator.

The alternator will only produce output when a magnet is moving past it. If your flywheel has only one magnet as many do, that means it's producing zero output most of the time. When you put an AC voltmeter on it, the meter averages that zero-most-of-the-time output and gives a low reading.

On the other hand, your bridge rectifier-and-capacitor combination captures and stores the *peak* output from the alternator which is much higher. And specifically, more than your LED light can handle.

As others have suggested, one way to deal with this would be a zener diode which basically is a device that will draw current to keep the voltage below a specified value. That will definitely work, but the zener could consume a lot of power doing its job, so you'd need one with a fairly high power rating. I would not use one rated less than 5 watts.

For a better but harder-to-implement solution, you could use this device called a "buck regulator" (or buck converter): Click me


That particular unit will take any input voltage from 6.5 to 60 VDC and reduce it to a settable output between 1.25 and 30 VDC. The only downside of that device is you'd have to find a way to protect it from moisture.


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## Lottstodo (Feb 16, 2018)

Sorry not a repair tip but since I have a lot of people asking if I can install leds on their machine I normally tell them that I wont just because of wiring/ mods and possibilities of something that may go wrong, So I just dont. I would like to know if anyone has tried one of these on a snowblower. As I would be more inclined to do a plug and play for customers, and as well for myself. Any thoughts??


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## SayItAintSnow (Dec 15, 2017)

classiccat said:


> _The no-load high DC voltage is real.
> 
> Ditch the switch if you haven't already; wire directly from the bridge to LED lamp._





Exactly right 'Cat.....at least for the circuits that have been routinely discussed on this forum. 



_The higher voltage *is* real_. 

I like the "ditch the switch" idea. That way there is always a load on the circuit. It's also a cost free, simple solution.


However, for those that have their heart set on having a switch, just make sure you wire the switch in right after the plus output of the bridge, and BEFORE the smoothing capacitor. That way, the cap is not connected when the led load is not connected, and no charge will build up.


*The last component that should be just ahead of the led is a properly sized fuse.*


I'm not sure how it came to be, insofar as the suggestion of using two caps across the dc output. Why two? Why not one? As far as the circuit is concerned all it "sees" is a single capacitance of over 4000uf.
As mentioned previously, a single cap at 470uf (a commonly available size) would do an adequate job of smoothing the ripple, and not build up such a big charge. Using two caps does not buy you "redundancy" in case one of them blows out. If the single smoothing cap blows out (highly unlikely) the worst that happens is that the DC voltage has more of a ripple in it, and your led might produce a noticeable flicker.
.
.


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## tadawson (Jan 3, 2018)

SayItAintSnow said:


> Sorry, but that's mostly incorrect...
> 
> While it's true that fuses protect against over-current, your assumption was that if you increase the voltage to a given load, the current can stay the same. That's not true.
> 
> ...



Read it again - I'm dead on correct. What I said was that for a given current, power delivered varies due to voltage as per P=IV . . . and as such, if the voltage is too high, the fuse may well not have made a difference. Yes, some things *CAN and WILL* draw more if the voltage is higher, but that was not the point at all . . . Regulated devices that can handle multiple voltages *WILL* draw the same *POWER* (not current) across multiple voltages . . . right until the voltage exceeds the circuit design and they fail . . . due to (you guessed it!) overvoltage and not overcurrent . . . Many LED lights are *NOT* pure resistive loads, but have power supplies in them that make the pure V=IR relationship pretty much invalid . . . thus my point. 



If you really want to protect against overvoltage, you need to put something like an MOV or other crowbar circuit just behind the fuse to case it to blow on overvoltage as well . . . 


In any case, a fuse *still* likely would have done very little to help here.


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## strtch5881 (Oct 6, 2018)

Well, I wired the light direct to ac like original, put the led bulb in, started her up. The bulb lit up, started to smoke and went out, all in 15 seconds, just like the other 3. The warmers never worked either, so I cut all the wiring out, soldered and heat shrunk the ac wire like original, plugged the halogen back in. Worked like always. Now I have light, and the even if the hand warmers don't work, they are wrapped around the cold steel. That'll help. Just to let you all know, even though I don't know electronics, I do know wiring, as I troubleshoot auto and small engine electrical systems for friends. I used to help a buddy wire street rods also. So if you fellas want to keep going with this thread, have fun with it. I'm off to port some chainsaws. Thank you to everyone for the help and the entertainment.


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## bkwudz (Jan 14, 2013)

Bummer, so maybe i miss understood what you had done, when I suggested plugging the LED bulb into the stock wiring, i figured you still had that option. But it sounds like you wired the rectifier and caps TO the stock plug going off what you said above. So now it sounds like you have it back to stock. Shoot me your address and ill send you a couple bulbs you can try to see if they work, NOW that you are back to stock wiring. I have bulbs I’m not using, as i got a few different ones to see if they all worked, they all did. They are 881’s so the plug orientation might be different, but at least you can see if they work, if they do, you can feel more comfortable getting another correct bulb


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## strtch5881 (Oct 6, 2018)

The last bulb was just on the ac wire from the stator. I had everything but the switch disconnected. Thanks anyway.


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## underp2 (Feb 1, 2018)

I've got similar LED burnout issues...
The machine is a yard king signature 9hp 29". I believe it's made by Murray, same as a Craftsman, Noma, etc.
My issue is that the chute deflector broke, the part was unavailable, so I had to use a similar part & make it work. Then that broke. Removed it and converted the chute control to DC servo cylinder. But the incandescent bulb (very dim, by the way) had such high power draw that it took the servo about 30 seconds to cycle. Too slow. 
I had to install a bridge to get DC for the servo. Then added a momentary MOM-OFF-MOM switch. I added a relay so when the servo switch was pressed, the light would turn off. This did not function because the power draw of the incandescent lamp was so high the relay coil could not energize. 
I put an LED bulb in place of the factory incandescent, hoping with the low power consumption of the LED, the relay would function. 
VICTORY!
The deflector now works as intended, relay functions to momentarily turn off the light while the servo is moving. Total deflector cycle time is less than 6 seconds. This is acceptable to me, as usually I make a chute adjustment and don't move the chute deflector all the way up or all the way down. 
The bad news is that the LED didnt last more than a day or 2. The first LED was a $9.99 per pair from Amazon. Dead. Tried AStar (I think that was the brand). I guess for $30 a pair I thought maybe it would have something to protect it. Nope. Burnt it up, too. I would like to correct this to make the LED work without burning up. It was so much brighter than the incandescent. 
Also, I did wire in a cap, but I can't remember the Farad rating. Maybe 1000 uf? 
I have a fluke RMS meter, and at one point had all the AC & DC voltages(at idle, max RPM, light on, light off, deflector moving, etc). I'll try to find my records of these voltages, but if necessary, I could get them again. If I remember correctly the max RPM voltage was high (maybe 40 is volts)

Also not sure if this belongs in this thread or if I should start a new one for LED light and DC servo chute control.

Thanks for your help!


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## RedOctobyr (Mar 2, 2014)

I think a new thread might be good, since there is overlap, certainly, but you also have some other stuff going on. 

I know it's not the point of your question, but it makes me wonder. Could you use some sort of maybe DPDT switch? Ideally a switch that, in each momentary position, could have one set of contacts closed, and another set open. So that you wouldn't need a relay. As you held the switch, to activate the servo, it would automatically shut off the light. It seems like there should be a way to do this with just a switch, and maybe a bit of wiring? I guess this is exactly what your relay was accomplishing (appropriately selecting NO vs NC), but the voltage was a problem. Unless you could just get a relay with a voltage range that extended lower. 

I wonder if a giant capacitor could slowly charge while running, and help run the servo? Giving you some extra, temporary, current.


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## Galotem23 (Jan 7, 2021)

Building a zener diode voltage regulator it really depens of zener quality, generated heat and radiation


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