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Old 09-21-2011, 02:53 PM   #1
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Need Help! About the LEDs


Hi! I need some help from expert. I have bought a LED light, but the brightness seem like too dim, and I want it brighter. Please checked the attached file "diagram.jpg" I was drawn, it is from the circuit board design (I know how to draw, but not the calculation, and sorry for I drawn the + & - DC wrong and LED position wrong too).

My questions are,
1. How much power output of is the DC?
2. How to power up the LEDs brightness?
3. Is it because of the 18 nos LED in series cause dim?
4. Or can I redesign it in 6 nos x 3 rows parallel?

Thanks for the help.
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Old 09-21-2011, 03:50 PM   #2
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Need Help! About the LEDs


Is that the circuit that came with it or did you plan on making a custom power supply circuit?

I'm quite sure that the LED's are already configured to operate at their maximum brightness or close to it. If you up the voltage or rewire them in groups of fewer in series then you will burn them out.

The reason why there are 18 in series is that the voltage supplied is 18 times the voltage needed to power just one LED. Sometimes is is more efficient to have a higher voltage supply for several lamps in series compared with a lower voltage (capable of more amperes) for the same lamps in parallel.

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Old 09-21-2011, 06:12 PM   #3
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Quote:
Originally Posted by Jasch2011 View Post
Hi! I need some help from expert. I have bought a LED light, but the brightness seem like too dim, and I want it brighter. Please checked the attached file "diagram.jpg" I was drawn, it is from the circuit board design (I know how to draw, but not the calculation, and sorry for I drawn the + & - DC wrong and LED position wrong too).

My questions are,
1. How much power output of is the DC? Not much....without knowing the uF of C1, your not going to get much power through that resistor.
2. How to power up the LEDs brightness? If your asking that question, then you should really not be making your own AC/DC converter.
3. Is it because of the 18 nos LED in series cause dim? Depends on what the current is.
4. Or can I redesign it in 6 nos x 3 rows parallel? Like Allen said, it is cheaper to do in series since you don't have to drop the voltage so much. With 18 LED's (assuming a 3v drop at rated power), your at around 54 volts.

Thanks for the help.
If you really want to learn about driving LED's....at good place to research it is the Candlelight Forum....if you think coin collectors are anal....they have flashlight collectors over there that make coin collectors look like extroverts.

By now you start to have an appreciation as to why LED bulbs are so expensive....and the weakness of using them in home lighting....
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Old 09-21-2011, 06:16 PM   #4
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Need Help! About the LEDs


That circuit isn't quite right. I'm sure they are dim - probably VERY dim. In fact, I bet they hardly light up at all. I would also bet that R3 gets pretty hot. I think the circuit is not intended to be configured that way. The C1/R1 combination acts as a capacitive reactance ballast to limit AC current through the circuit. I can't tell from the drawing what the capacitance of C1 is, so I can't calculate what the reactance is to see if it's right or not. However, C2 and R2 are configured similarly but are on the DC side of the rectifier. Thus, C2 does essentially nothing and the 3.1M resistor R2 is in series with the LEDs. That limits the current to a bit under 1mA max, which is far too low. This configuration also has the potential to over-volt C2, since it is a 100V capacitor in series with a 300VDC source.

I'm going to make some assumptions, and then a recommendation. If my assumptions are incorrect, then my recommendation will also be incorrect. It's up to you to find out. First, I'm assuming the LEDs are white or blue, not red or green. Thus, their forward voltage drop is 3-4V at normal operating current. Check this. Second, I'm assuming these are normal LEDs with an operating current of 20mA, not high power LEDs. Check that too.

If my assumptions are correct, then place C2 and R2 in parallel with the LEDs, in place of R3. Get rid of R3. C1 should be a 0.2uF (approximate, +/- 15%) capacitor rated at least 300VDC. A 0.22uF 500V mylar film or ceramic capacitor would be a good choice. Try that, measuring the current through the LED string. It should be approximately 20mA. Adjust the value of C1 to adjust the current.
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Old 09-21-2011, 06:51 PM   #5
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Quote:
Originally Posted by mpoulton View Post
That circuit isn't quite right. I'm sure they are dim - probably VERY dim. In fact, I bet they hardly light up at all. I would also bet that R3 gets pretty hot. I think the circuit is not intended to be configured that way. The C1/R1 combination acts as a capacitive reactance ballast to limit AC current through the circuit. I can't tell from the drawing what the capacitance of C1 is, so I can't calculate what the reactance is to see if it's right or not. However, C2 and R2 are configured similarly but are on the DC side of the rectifier. Thus, C2 does essentially nothing and the 3.1M resistor R2 is in series with the LEDs. That limits the current to a bit under 1mA max, which is far too low. This configuration also has the potential to over-volt C2, since it is a 100V capacitor in series with a 300VDC source.

I'm going to make some assumptions, and then a recommendation. If my assumptions are incorrect, then my recommendation will also be incorrect. It's up to you to find out. First, I'm assuming the LEDs are white or blue, not red or green. Thus, their forward voltage drop is 3-4V at normal operating current. Check this. Second, I'm assuming these are normal LEDs with an operating current of 20mA, not high power LEDs. Check that too.

If my assumptions are correct, then place C2 and R2 in parallel with the LEDs, in place of R3. Get rid of R3. C1 should be a 0.2uF (approximate, +/- 15%) capacitor rated at least 300VDC. A 0.22uF 500V mylar film or ceramic capacitor would be a good choice. Try that, measuring the current through the LED string. It should be approximately 20mA. Adjust the value of C1 to adjust the current.
^ True electronics guy..........

I concure......on a side note....a cheap AC/DC adapter is to just use a diode with a capacitor. The uF of the cap sets the current to the load....the diode rectifies it...works great for charging NiCad batteries.....until you break the ckt....then you have the full voltage available....hurts if you touch it....

One other note....are you using a diode bridge or making your own? If making your own, use at least 1n4004's....
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Old 09-22-2011, 12:01 AM   #6
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Need Help! About the LEDs


Thanks for all of your replied.

Quote:
Originally Posted by AllanJ View Post
Is that the circuit that came with it or did you plan on making a custom power supply circuit?
and

Quote:
Originally Posted by ddawg16 View Post
One other note....are you using a diode bridge or making your own? *If making your own, use at least 1n4004's....
Actually it is a ready product, I was copying their design Into a diagram and need some tech guy to help me figure out what's wrong with it. And so sorry that I am not a tech guy, but i was learnt the basic of electronic during my high school. After nearly 20 years did not touch on electronic stuffs, is like returned the 90% of my knowledge to the lecturer.


Quote:
Originally Posted by mpoulton View Post
That circuit isn't quite right. *I'm sure they are dim - probably VERY dim. *In fact, I bet they hardly light up at all. *I would also bet that R3 gets pretty hot. *I think the circuit is not intended to be configured that way. *The C1/R1 combination acts as a capacitive reactance ballast to limit AC current through the circuit. *I can't tell from the drawing what the capacitance of C1 is, so I can't calculate what the reactance is to see if it's right or not. *However, C2 and R2 are configured similarly but are on the DC side of the rectifier. *Thus, C2 does essentially nothing and the 3.1M resistor R2 is in series with the LEDs. *That limits the current to a bit under 1mA max, which is far too low. *This configuration also has the potential to over-volt C2, since it is a 100V capacitor in series with a 300VDC source.

I'm going to make some assumptions, and then a recommendation. *If my assumptions are incorrect, then my recommendation will also be incorrect. *It's up to you to find out. *First, I'm assuming the LEDs are white or blue, not red or green. *Thus, their forward voltage drop is 3-4V at normal operating current. *Check this. *Second, I'm assuming these are normal LEDs with an operating current of 20mA, not high power LEDs. *Check that too.

If my assumptions are correct, then place C2 and R2 in parallel with the LEDs, in place of R3. *Get rid of R3. *C1 should be a 0.2uF (approximate, +/- 15%) capacitor rated at least 300VDC. *A 0.22uF 500V mylar film or ceramic capacitor would be a good choice. *Try that, measuring the current through the LED string. *It should be approximately 20mA. *Adjust the value of C1 to adjust the current.
mpoulton, I am so appreciate of your valuable comments.

Can I measure the C1 with a multimeter and it's still welded on the circuit board with power on? And I am just remembered I can measure the power output with a multimeter, but is a bit dangerous to me without a full understood of the whole design. Is that possible only romove certain component to power up the LEDs?

you have guessed few things correctly, it is the normal white LED. If I can get the power output measurement, do you think the C1 uF is able to calculate?

I have to check the circuit board, not sure that whether I can rewire the C2 & R2. Because the board is so small about an inch. I will try to do it when all ready with a correct calculation.
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Old 09-22-2011, 04:20 AM   #7
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So this is a commercial product, and it's not working? Did it ever work? Are you sure you drew the circuit exactly as it's constructed? That doesn't quite make sense.

Anyway, measure the current through the LEDs, and the voltage across the entire string of LEDs while operating. We'll start there. You can't measure C1 in-circuit because it's in parallel with a resistor and that screws up the meter's capacitance reading.
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Old 09-22-2011, 06:26 AM   #8
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If the light was very dim to start with then the lamp was probably defective. Then all bets are off regarding volts and amperes. You would need to know the correct voltages and currents first in order for your measurements to have any meaning to anyone. You should have returned it to the store. It is probably too late to return it now because you opened it up to examine the circuitry more closely.
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Old 09-22-2011, 08:49 AM   #9
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Originally Posted by AllanJ View Post
If the light was very dim to start with then the lamp was probably defective. Then all bets are off regarding volts and amperes. You would need to know the correct voltages and currents first in order for your measurements to have any meaning to anyone. You should have returned it to the store. It is probably too late to return it now because you opened it up to examine the circuitry more closely.
Allan, probably not, just discovered my silly mistake, drew the wrong diagram, and thanks for the replied too.

Quote:
Originally Posted by mpoulton View Post
So this is a commercial product, and it's not working? Did it ever work? Are you sure you drew the circuit exactly as it's constructed? That doesn't quite make sense.

Anyway, measure the current through the LEDs, and the voltage across the entire string of LEDs while operating. We'll start there. You can't measure C1 in-circuit because it's in parallel with a resistor and that screws up the meter's capacitance reading.
You are right, after you threw the questions to me, and I went thru the circuit board again and saw I drew it wrong due to too sleepy and also late midnight at that time. Attached the new diagram, and also measured the output voltage too. With 2 more photos of the circuit board with components.

So, is that possible to increase some power to LEDs? Is my new question. May I know the calculations too, appreciate to share with me. Thanks in advance.
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Old 09-22-2011, 10:05 AM   #10
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Just found an useful info, The C1 has an actual value now. from the Code 400V and 224J = 330pF 5%, am I correct?
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Old 09-22-2011, 04:46 PM   #11
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NO ! it is not 330pf,
It is .22uf.
There are two lists, you read the wrong one,
read right side.

The amount of power available,
is controled mainly by the value of the capacitor.
So reducing the value slightly will increase power output.

These types of supplies are only really made
for small loads.
And because it is NOT isolated from the mains
can be dangerous, so use GREAT care.
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Old 09-22-2011, 07:27 PM   #12
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A 0.22uF capacitor has an impedance of 12k ohms at 60Hz. There is an additional 1.2k resistance in series with the LEDs. The 400k in parallel with the capacitor is almost irrelevant. You have a peak input voltage of about 330V, and a load voltage of 54V. Thus, the current through the load will be approximately 21mA. This is a normal operating current for LEDs. If you're on a 50Hz system, then the capacitor impedance is 14.5K, so the current is 17.5mA.

You really need to measure the LED current. Without knowing that, it's impossible to say whether you can safely increase the current or not. You also need to know the maximum operating current for these LEDs. It may be 20mA, or it could be 30-60mA.

DMX: increasing the capacitance will increase current.
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Old 09-22-2011, 08:04 PM   #13
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My two cents.
The bridge rectifier and C2R2 create a filtered DC power supply. The voltage across the photo diode stack will remain pretty close to 54V, 3 volts per diode X 18.
The 1.2K resistor limits the DC current. Lowering this resistor value will increase the DC current.
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Old 09-23-2011, 01:44 AM   #14
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The 1.2K resistor limits the DC current. Lowering this resistor value will increase the DC current.
It will, but only a little bit. The main impedance in this circuit is C1 (approximately 12k ohms at 60Hz), not the 1.2k resistor. Eliminating the resistor completely will only increase the current by about 10%.
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Old 09-24-2011, 12:22 AM   #15
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The purpose of the 400K resistor is to discharge the cap,
once the power is removed, otherwise the cap remains charged,
and will bite if you touch it after switch off.

The 1.2 k resistor probably limits the current thru the leds,
You can run them with out the resistor,
but you risk burning them out,
by running them flat out.

The main element that controls the amount of power is the cap.

Altering this will alter power out.

But again remember NO ISOLATION means more danger!

BE CAREFUL.

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