High powered LED work light

[heychris]

All,

I'd appreciate an opinion. I have a harbor freight 45 LED work light. It is sadly not very bright. Instead of throwing it out, I'm thinking of installing brighter LED's.

I can get 110k mcd LED's on ebay. They aren't cheap, but they are bright.

I took apart the circuit inside to see how they wired it. It seems they used a bridge rectifier that's good for 1 amp. The bright LED's draw 100ma each. I figure I could put in a bigger bridge rectifier and adjust the limiting resistor accordingly.

The question that I have is around the AC input part of the circuit. The AC input has a .68uF capacitor in parallel with a 1M ohm resistor. This pair is in series with on leg of the AC input.

Is this for smoothing power spikes? Do I need to put in a higher wattage resistor? The current one appears to be 1 watt.

Thanks for the help

Chris

 

[MikeMl]

They are using the reactance of the capacitor as an (almost) lossless means of reducing the line voltage before it goes into the bridge.

 

[Diver300]

To increase the current you would need to increase the capacitor. With 120 V, 60 Hz power there will be about 30 mA with that capacitor.

You need to limit the current with all LEDs. They don't "draw" current like bulbs do. If you change the LEDs and make no other changes, the circuit will still limit the current to 30 mA

However, LEDs vary a lot in how efficient they are. If you are changing the LEDs, you could try them with the existing capacitor, and they may be much brighter at about 30 mA than the existing ones.

The current rating of the LEDs is their maximum, and they will generate a lot more heat at 100 mA than they will at 30 mA, so you might get a problem if you do increase the current a lot. I suggest that you get another couple of 0.68 uH capacitors (rated to at least 1.5 times the mains voltage) and put them one at a time in parallel with the existing one.

The 1 M ohm is there to discharge the capacitor when you unplug. You don't need to change that. The rectifier does not need changing as just about all rectifiers are rated at 1 A or more.

 

[Boncuk]

Hi,

you'll need quite a lot of 30mA LEDs for a good work light.

Here is the data sheet of a HUEY JANN 20W LED (HPR20D-19KXX).

For a work light I recommend to use warm white (3,330K) or natural white (4,000K).

The radiation angle is 120 degrees and the light output is as much as that of a 100W light bulb.

Boncuk

 

[heychris]

Can you explain or do you know of any resource that give the equations for current limiting an AC side of a Bridge Rectifier? I am used to seeing some smoothing of the circuit on the DC side. I am not familar with using a Capacitor to limit the current on the AC side.

Thanks.

 

[heychris]

those are some serious LED's.. The board I am modifying currently is layed out for 5mm LED. The brightest 5mm packaged LED I could find are these.

**broken link removed**

Hi,

you'll need quite a lot of 30mA LEDs for a good work light.

Here is the data sheet of a HUEY JANN 20W LED (HPR20D-19KXX).

For a work light I recommend to use warm white (3,330K) or natural white (4,000K).

The radiation angle is 120 degrees and the light output is as much as that of a 100W light bulb.

Boncuk

 

[heychris]

I measured the voltage before and after the capacitor/ 1M resistor. The voltage is the same 118.7 V aC

They are using the reactance of the capacitor as an (almost) lossless means of reducing the line voltage before it goes into the bridge.

 

[heychris]

Do you know of a good source for these? Since they aren't a 5mm packag, I'd have to rebuild the front plate. I could do that out of aluminum, which would act as a heat sink.

Hi,

you'll need quite a lot of 30mA LEDs for a good work light.

Here is the data sheet of a HUEY JANN 20W LED (HPR20D-19KXX).

For a work light I recommend to use warm white (3,330K) or natural white (4,000K).

The radiation angle is 120 degrees and the light output is as much as that of a 100W light bulb.

Boncuk

 

[audioguru]

Cheap Chinese LEDs with a very high intensity of 110mcd are simply focussed into a very narrow bright beam that is useless for anything except as a flashlight. You need wide-angle bright LEDs that are not cheap and maybe are not Chinese.

 

[crutschow]

Can you explain or do you know of any resource that give the equations for current limiting an AC side of a Bridge Rectifier? I am used to seeing some smoothing of the circuit on the DC side. I am not familar with using a Capacitor to limit the current on the AC side.

It uses the capacitive reactance to limit the current. A rough calculation of the LED current would be (Vline - Vled) *(2Pi*f*C) where Vled is the voltage required by the LED array and f is the line frequency.

 

[MikeMl]

I measured the voltage before and after the capacitor/ 1M resistor. The voltage is the same 118.7 V aC

With the LEDs lit (i.e. drawing normal current)?

 

[heychris]

Now I'm getting a bit confused. It seems to me there are several very valid points.

1. The LED I choose. It'd be nice to go with 10W LED's. I have seen testing on some of the 5mm LED's that is promising. I'd link/show pictures if I was able (I am new to this board).

What have in mind for the first go around is on ebay listed as "50 PC 0.5W 5-Chips 5mm 60° High Power White LED 110Kmcd". yes the beam is narrow, but it is sort of a large flash light.

2. AC side of the bridge rectifier. I am still a little puzzled why I need to limit the current on the AC side of the rectifier.

3. The bridge rectifier itself needs to be sized appropriately for the total amperage. 22 LED's at 100ma a piece is 2.2A, so I figure a 10A will b fine.

4. DC side. The LED's have a current limiting resistor that is sized appropriately for the 108V DC input and the 22 LED's in series.


The thing I am most confused about is every rectifier circuit I have seen uses AC input, a bridge rectifier, on the DC side has a capacitor for smoothing (which is NOT in mine) and a current limiting resistor.

It seems what is being said here is that I have a capacitor and resistor for current limiting on the AC as well as a current limiting resisitor for the DC side.

Am I way off?

 

[heychris]

No at that point I had removed the LED's so I was measuring only the AC side of the rectifier with no load on the DC side. With the LED load on it, the rectifier sees 73V instead of 118 Volts. That makes more sense.

If I am using LED's that need a higher DC voltage should I just remove the capacitor? Maybe even move it to the DC side for smoothing? I figure with 22 3.4V LED's, I can use the additional DC voltage. Would I be better off dropping the AC voltage a little? With the total voltage drop of the 22 LED's being 74V, I figure I need at least 94V DC to drive them.

Thanks for the help.

With the LEDs lit (i.e. drawing normal current)?

 

[RCinFLA]

If there are 45 LED in the light they can't all be connected in series as the LED voltage X 45 will exceed 120 volts.

A 0.68 uF ballast cap would be Xc of 3900 ohms at 60 Hz. I cannot see any series/parallel combination of 45 LED's to make that work.

A 15 LED stack will have a voltage drop of 45v to 55 v depending on forward voltage of LED's. Assuming they run about 20 mA that would be three strings of 20 mA, or 60 mA total.

Again without knowing exact LED forward voltage at 20 mA I am assuming 50 v for stack. That would be 120v - 50 v = 70 vac drop across ballast capacitor.

70 vac / 0.060 amps = 1167 ohms of reactance required for ballast capacitor. That would be C = 1/(2*pi*f*Xc) = 2.3 uF capacitor. A 2.2 uF should be about right. Usually they put a relatively low value series resistor to cut down on possible surge current if turn on happens to occur at peak of sinewave.

If there is five strings of 9 series connected LED's then there would be 27v to 32v total forward voltage. For that, the parallel combination would draw 20 mA x 5 strings = 100 mA's. That would require a 120v - 30v = 90 v across ballast cap. Xc = 90v/0.1 amp = 900 ohm ballast cap. That would be 2.95 uF.

If there is nine strings of 5 series connected LED's then there would be 15v to 17.5v total forward voltage. For that, the parallel combination would draw 20 mA x 9 = 180 mA's. That would require a 120v - 16v = 104 v across ballast cap. Xc = 104v/0.18 amps = 578 ohms ballast cap. That would be 4.6 uF.

 

[heychris]

There are two recifier circuits that are identical except for the led current limiting resister. One circuit has 22 LED's and the other has 23. Each string get's it's own bridge rectifier. I apologize for the lack of photos or drawings. I am not allowed to post links or photos.

The LED I found on ebay is 100ma and 3.4V.

The 22 led string has a total of 74.8 V and 100mA
The 23 LED string has a total of 78.2V and 100mA

If I stick to the rule of the total voltage drop being 80% of the DC voltage. I need the rectifiers to output a little less than 100 VDC. Considering that there is .7-2 V drop on a bridge rectifier. I'm talking about a little over a 100VAC. With my wall voltage measured at 118. It almost seems like a Ballast on the AC side isn't necessary. I'm an amateur so I'd appreciate any opinions there.

If I need to ballast say 15 V -> (15V/ .1A) = 150 ohms. C = 1/(2*pi*f*Xc) = 1/(2*3.14*60*150)=1.77e-5 =1.77uF

Am I correct that this means The resistor and capacitor that are between one of the AC inputs and the AC input of the bridge rectifier are 1.77uf and 150 ohms?

Is that a good idea?

 

[RCinFLA]

I think you should see what you can do with what you have. I think you can beef up the current on the stock item without issues.

What you are proposing with ebay LED's is going to have about 15 watts of heat generation. That requires a pretty large heatsink.

 

[heychris]

You've got a good point. If it overheats, I'll just switch to better LED's. The 500W halogen equivalent puts off a ton of heat.

So are you saying I should put the AC input straight into the bridge rectifier and power the LED's off full DC voltage? I might not have been clear about what I'm replacing. 10ma LED's at 2.7V. The AC circuit was dropped to 73V. I could easily use all the AC voltage and wire the LED's accordingly. I could even reuse the capacitor on the AC circuit to smooth the DC circuit.

I think you should see what you can do with what you have. I think you can beef up the current on the stock item without issues.

What you are proposing with ebay LED's is going to have about 15 watts of heat generation. That requires a pretty large heatsink.

 

[crutschow]

2. AC side of the bridge rectifier. I am still a little puzzled why I need to limit the current on the AC side of the rectifier.

Limiting the voltage on the AC side of the rectifier with a capacitor dissipates essentially no power. Limiting on the DC side of the rectifier with resistors dissipates the excess voltage as heat.

 

[RCinFLA]

The 15 watts is the heat generated by the LED's. You can't just magically switch to better LED's. You can switch to lower power LED's.

You need the ballast cap to set the current in the LED's. It has to be on the AC side.

There is a company I know of that makes LED flood lights for boats. Their housing is cast aluminum with heat sink fin ridges for a 35 watt LED flood light.

If you are trying to replicate the light output of a 500 watt halogen bulb with LED's you are not talking about modifying the LED light you bought.

 

[heychris]

When designing a circuit that was battery powered, I've always stuck to the LED's should add up to 80% of the DC voltage. In this case, should I let the rectifier voltage be closer to the total voltage drop?