High powered LED work light

Ahh.. Let the AC capacitor do the heavy lifting then. Are you saying instead of my calculation above where I take the strings with a voltage drop of 74.8 (22leds) and 78.2 (23LEDs), I can shoot for Maybe 81 Volts instead of 100?

If I redid the calculation there 81Volts needed +6 for the bridge rectifer =87 V. 118 volts of the line make 31 volts I could drop at the capacitor.

31V/.1A = 310 ohms. C = 1/(2*pi*f*Xc) = 1/(2*3.14*60*310)=1.77e-5 =8.56 uF

On the AC side of the rectifier I would have a 8.56uf capacitor in parralel with the 1M ohm resistor.

Am I on the right track?

**broken link removed**

Here is a similar diagram. I'm using 120V and C1 is now 8.56 uF and I handle the DC side accordingly..

Except now we are talking about a $5 capacitor... ehh...

All,

I'd appreciate an opinion on my calculations before I order the parts. Anyone?

Much appreciated

Chris

The circuit is fine.

If you fit a 100 ohm resistor in series with the LEDs that will reduce the flickering. Without it, the capacitor will charge to the LED voltage, but as soon as it drops below the LED voltage they all go out. With a resistor, the LEDs will light over a range of voltages on the capacitor, so it can actually store some useful energy.

So I ended up ordering the parts and putting this together.

I used 22 LED's rated at 3.4V and 100ma

**broken link removed**

I went with a 8uF capacitor

By my calculation Xc = (1/c)/2*pi*f = 331 ohms
So the voltage at the bridge rectifier should be 120V-33.1V = 87V
I assume a loss of 7V on the rectifier for a voltage output of 80Vdc
With that feed voltage I calculate I need a resistor of 56 ohms with 100ma it should draw .5W (I used 1W)

First off wow this is MUCH brighter. Look at the original LED vs the new one. And the final outcome.

Of course first attempts rarely work. In my case, the resistor burned up quickly. I measure the voltage drop over the series of LED's and got 65V instead of my intended 75V also the series draws 200ma instead of 100ma.

Does anyone know where I may have gone wrong?

The voltage drop of 65 V isn't too far off. There is quite a range and an average of 3 is OK.

You need to cut the current down. If the LEDs are rated at 100 mA they will not last on 200 mA.

Your calculation isn't quite right. 331 Ω as the impedance is correct for an 8 µF capacitor at 60 Hz. However, as that is a capacitive impedance, voltages on there do not add arithmetically to voltages across resistors or LEDs.

The average rectified current is about 1.11 times the RMS current, so you want about 90 mA RMS

The phases will be at 90º, so the voltage after a drop of 331 Ω * 90 mA = 29.8 V is

sqrt(120² - 29.8²) = 116 V

So really the capacitor isn't doing much at all.

There will only be a drop of about 1.5 V across the rectifier, so that can be ignored.

The exact current calculation is somewhat difficult, as the supply voltage is varying, but assuming an average voltage of 116 V, the resistor needs to drop

116 - 65 = 51 volts.

That needs a resistor of 560 Ω and 5 W, but a 10 W resistor would run cooler.

Thank you.

So that we are clear, R2 from the drawing above is 560 ohm. correct? I don't actually have a c2, but could add one if you think there is any value.

560 Ω for R2 is what I was suggesting.

However, that it really not making the circuit work in the way that it was designed.

R1, R2 and C1 limit the current. The idea is that C1 limits the current, and it does so without getting hot.
R1 is only there to discharge C1 when the whole circuit is disconnected from the mains, for safety. It is large so little current goes though it and adds nearly nothing to the current in the LEDs
R2 is there to limit the C1 charging current when the circuit is connected to the mains. It is small enough to make virtually no difference to the current in the LEDs
C2 is there to smooth the current in the LEDs so that they don't flicker.

You should really make C1 about 2 µF so that the current is about 100 mA. The rest of the components can be as they are in the circuit diagram.

Thanks. I appreciate your help as this is very interesting for me.

My intention was to limit the current with C1. I will change c1 to 2uF and R2 to 100 ohms.

I get a little lost in your math.

Xc = (1/c)/2*pi*f = (1/.000002)/(2*3.14*60) = 1327 ohms
at 90ma ->V=1327*.09 =119V
sqrt(120^2-119^2)=15.5

I don't mean to be a pest, but would you mind pointing the errors in my ignorance?

You seem to have understood my maths perfectly, but you missed where I said "about" for the capacitor value.

The quick calculation is to simply ignore the LED voltage drop. The impedance of the 2 µF capacitor is 1326 ohms as you worked out. That will limit the current to 90 mA.

Where the voltage drop across the capacitor is close to the supply voltage, then you can't calculate the voltage across the LEDs with any accuracy.

As you wrote

sqrt(120^2-119^2)=15.5

but if the current is just marginally less, the voltage across the capacitor falls very slightly. Let's just assume it drops to 115 V

sqrt(120^2-115^2)=34.3

As you can see, the LED voltage has doubled but the voltage on the capacitor has only fallen slightly.

If the voltage drop on the LEDs, resistor and rectifier is 65 V, then the voltage left on the capacitor is
sqrt(120^2-65^2)=100.9 V
That would indicate that maybe an increase of the capacitor to 2.4 µF would be needed.

I don't think this analysis is perfect. Taking the sum of the squares is only fully valid for capacitors and resistors, and the voltage drop across the rectifier, resistor and LEDs is not sinusoidal.

What it all comes down to is that 2 µF will limit the current to around 100 mA DC, and that won't be reduced much by the voltage drop on the LEDs.








With the large voltage drop across the LEDs, I am not entirely sure about the details.

So I finally received my digi-key order (3 weeks later). I ended up going with a 2.2uf Capacitor. The Voltage output to the LED's is 56V and 50mA.

Is there a Spice like simulator that would be good to test this? Or are they simularly unpredictable in output?

Maybe I should try a 3uf? or I could go with 4.0uf and add a resistor.

What do you think?

4uF did the trick. I measure 105mA and 61V. A little low on the voltage, but good enough for me.

As you can see from the photo. You can actually use this light now... And it runs very cool.