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Old 18-05-2007, 09:44 PM
Dennis
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LED circuit advice required - thanks!

Hello

Apart from the delightful day-to-day tasks of building a base for our soon-to-be-delivered water tank, measuring up for and organising a pool cover and generally keeping the Universe operating, I’m also learning how to use Corel Draw and designing an LED based light box for flat field frames.

Now like my good mate in Nambour, I am a little electrically challenged so I pulled out my 2006 Dick Smith Catalogue and used the formulae therein as a basis for the attached circuit diagram and calculations, drawn in Corel Draw.

I would be extremely grateful if some learned person could cast their eyes over the sums to see if my monkey-see, monkey-do approach is okay.

I assume that in calculating the power rating of the 120 Ohm Resistors, which look like they should be ¼ Watt, I should only be using the current flow of each LED (20mA) in the P=IV calculation, not the overall circuit current of I = 6x20mA = 120mA?

I happen to have a regulated mains adapter that has a rating of 6Vdc at 800mA which I assume should be sufficient to power the above circuit?

Any corrections and advice would be most welcome and gratefully received.

Thanks!

Dennis
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Old 19-05-2007, 12:57 AM
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edit - much better way:

Rework your circuit so you only have a single resistor R, so its battery in series with R then LED in parallel with each other but in series with R.

Now to provide your LEDs with 3.6V and 20mA. Since they are in parallel they will all be at the same voltage. And since LEDs are all identical, we can assume the current will split itself equally amongst. So 20mA each for 6 LED is 120mA in total. So now we need to find R such that when there is 2.4V across it (since the LEDs have a voltage drop of 3.6V), it delivers 120mA to the LEDs:

2.4/120mA = 20 Ohms.

Power = I^2R = (120mA)^2 * 20 = 0.288 - you might get away with a 1/4W. If you used 2 x 40Ohm in parallel you definitly can as the power in each resistor is only (60mA)^2*40 = 0.144W.

Oh, you won't actually find a 40Ohm resistor. The closest value is 39Ohms.

Your existing calculations are however, fine.

Cheers,
Steve

Last edited by freespace; 19-05-2007 at 01:31 AM. Reason: Better idea.
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Old 19-05-2007, 01:38 AM
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Quote:
Originally Posted by Dennis View Post
I assume that in calculating the power rating of the 120 Ohm Resistors, which look like they should be ¼ Watt, I should only be using the current flow of each LED (20mA) in the P=IV calculation, not the overall circuit current of I = 6x20mA = 120mA?
That's right Dennis. 120mA. Your 6VDC 800mA plug pack will be fine.

If you're electrically challenged, use this online LED calculator. It'll draw you a picture and give you various details about the parallel circuit.

You can do it like freespace mentioned and it will work but not recommended (sorry freespace). It's unreliable having LEDs directly in parallel with each other - which is why you never see them in circuits like that. LEDs need their own current limiting resistors.
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Old 19-05-2007, 05:58 AM
Dennis
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Thanks for your replies guys, much appreciated.

Wow, Mojo – how cool is that LED calculator - thanks!!!!

After plugging in various combinations of source voltage values, I decided to use a spare 24Vdc 1A regulated mains adapter, which gave me the attached series circuit.

A bit of Googling seemed to turn up the advice that it was preferable to connect LEDs in Series rather than Parallel.

There is hope for Ponders and myself yet!

Cheers

Dennis
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  #5  
Old 19-05-2007, 08:40 AM
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[1ponders] (Paul)
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Wow, how cool is that. I even understand some of it.

Now if I can only get my fingers unsoldered from the circuit board.
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Old 19-05-2007, 08:57 AM
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Alternatively, you could feed your leds (all together) from a constant current source, ensuring that they would always be the same voltage (and light output) regardless of current. An LM317 and a single resistor makes a nice constant current regulator, at a cost of less than $2 and the arithmetic isn't hard (current output is 1.2/R1). Depending on your supply voltage you may need a small heatsink as well. FWIW the LM317/single resistor constant current design also makes a good constant current charging source for Nicads. BTW the 6V plugpack you have could have a voltage as high as 9-10 volts unless it's regulated.
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Old 19-05-2007, 10:17 AM
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Quote:
Originally Posted by mojo View Post
You can do it like freespace mentioned and it will work but not recommended (sorry freespace). It's unreliable having LEDs directly in parallel with each other - which is why you never see them in circuits like that. LEDs need their own current limiting resistors.
NP! No one ever told me that :-)

But now I know - the things you learn eh :-)

The series design is less desirable to me because it requires more voltage. The original design of this thread had a resistor for each LED - so they are not directly in parallel - and it should work fine.

And yes, if the 6V isn't regulated I recommend a voltage regulator to drop the voltage down to 5V ( LM715? 785? Forgot :\ I don't think there is a 6V regulator model). Then you can just adjust your equations a bit.
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Old 19-05-2007, 10:55 AM
Dennis
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Thanks again for all the replies, advice and suggestions - mucho appreciated.

Quote:
Originally Posted by acropolite View Post
BTW the 6V plugpack you have could have a voltage as high as 9-10 volts unless it's regulated.
Phil - It is a 6V regulated mains adpater. I was advised once to always use regulated PSU's for my nice, expensive astronomy equipment.

Quote:
Originally Posted by freespace View Post
The series design is less desirable to me because it requires more voltage. The original design of this thread had a resistor for each LED - so they are not directly in parallel - and it should work fine.
Freespace - I already have a 24VDC 1A regulated mains adapter and the series design means I only have to solder 1 resistor into the circuit, so being lazy and not too skilled with a soldering iron, I’ll go that route.

Thanks again guys

Dennis
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Old 19-05-2007, 11:02 AM
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With that hefty plug pack, you could certainly use more LEDs closer together (at less current if more will be too bright) to produce a more even spread over a wider area?
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  #10  
Old 19-05-2007, 01:14 PM
Dennis
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Yes – this is turning into quite an interesting project. I’m using x6 White, water clear LED’s, with a viewing angle of 60° and a light output of 1000 mcd each. I’m not to sure of the overall brightness of the final system, so I’m leaving the design open so I can add more, or remove a couple, to ensure I obtain an even illumination across the field, of the correct intensity, to allow flat field exposures in the range of 1 to 10 seconds.

I won’t know the answers until I build the first 6 LED prototype using a white bucket from Bunnings. Once I have taken some flat field frames, I can then inspect them in one of the imaging applications, to check the evenness of illumination.

What I like about the LEDs are their robustness, low power and low heat characteristics, compared to incandescent bulbs.

Cheers

Dennis
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  #11  
Old 19-05-2007, 03:32 PM
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1. The original circuit is good - With each LED carrying a "current sharing" resistor.

Putting LED's in parallel without the resistors will not ensure that each LED gets the same current because of mismatches in the manufacturing process - see "Thermal Runaway".

A constant current source will not ensure equal current distribution as each LED has a different on resistance. Of course each LED may have its own constant current source.

2. The forward voltage of the LED's was put at 3.6V this is good for White and Blue LEDs but Red LEDs have a forward voltage of 1.6V - What colour LEDs are you using.

Addition: No answer required - White LEd it is!!!
Jerry.

Last edited by 74tuc; 19-05-2007 at 03:49 PM. Reason: Point 2 Modified
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  #12  
Old 19-05-2007, 04:45 PM
Dennis
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Thanks Jerry, I appreciate your helpful comments.

I resorted to visiting the Library today, to read the electrical chapter in a Physics book in order to consolidate some of the knowledge I have picked up as a result of the helpful replies from the IIS community.

I reckon I'm more LED savvy than Ponders right now!

Cheers

Dennis
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Old 19-05-2007, 09:25 PM
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FWIW. Dennis, your first post showing the LEDs in parallel with their own current limiting resistor is fine.
Will work till the cows come home.
Use normal 1/4 Watt resistors in each leg.
Your 6 Volt @ 800mA reg supply will do nicely.
Regards, Laurie..
ps. If you find it necessary to dim them, just wire a 20 Kohm to 50 Kohm pot in the positive lead, (eg; in series), with the LEDs.

Last edited by RAJAH235; 19-05-2007 at 09:31 PM. Reason: extra info...
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  #14  
Old 20-05-2007, 05:18 AM
Dennis
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Thanks Laurie – it’s good to know that I can at least follow the simple tech notes in the Dick Smith catalogue, even though I may not fully understand them! Thanks for the advice re the pot – that sounds like a good improvement for the light box design.

Cheers

Dennis
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Old 20-05-2007, 08:11 AM
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Quote:
I reckon I'm more LED savvy than Ponders right now!
That wouldn't take much Dennis
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Old 20-05-2007, 09:07 AM
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20K to 50K in series with each LED?

The LED brightness may range from very dim to off.

This scheme does not lend itself for future expansion as previously mentioned and would be most user unfriendly for adjusting brightness in the field.

To design a brightness control and allow for future expansion will take a re-design along the following lines:

Maintain the original circuit arrangement with the current sharing resistors about 20 ohms (guess - not calculated) then control the whole load with a pulse width modulated (PWM)source (555 timer and small power mosfet). The PWM scheme is suggested for efficiency and because of low operating voltage(6V).

Jerry.
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  #17  
Old 20-05-2007, 10:10 AM
Dennis
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Quote:
Originally Posted by RAJAH235 View Post
ps. If you find it necessary to dim them, just wire a 20 Kohm to 50 Kohm pot in the positive lead, (eg; in series), with the LEDs.
Quote:
Originally Posted by 74tuc View Post
20K to 50K in series with each LED?

The LED brightness may range from very dim to off.

This scheme does not lend itself for future expansion as previously mentioned and would be most user unfriendly for adjusting brightness in the field.

To design a brightness control and allow for future expansion will take a re-design along the following lines:

Maintain the original circuit arrangement with the current sharing resistors about 20 ohms (guess - not calculated) then control the whole load with a pulse width modulated (PWM)source (555 timer and small power mosfet). The PWM scheme is suggested for efficiency and because of low operating voltage(6V).

Jerry.
Possibly showing my ignorance of electrical circuits here, but I assumed that Laurie was suggesting a single potentiometer in the circuit, placed in the wiring segment just after the +ve terminal but before the 1st LED?

Cheers

Dennis
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Old 20-05-2007, 11:37 AM
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Yes correct - my misunderstanding.

But putting pots in series with a load that may cause a relatively large current is that you have to take into account the dissipation of the resistance in circuit at the time. A pots dissipation is rated for the whole pot and so the dissipation in the part limiting the current may be quite high and well out of specification and in some cases will cause the pot to fail. Remember dissipation goes up as the square of the current so a linear drop in resistance will cause the dissipation to rise faster and faster.

The other factor you may want to take into account is how you want the brightness to vary - a linear increase in brightness with the pot rotation. The power dissipated in the load will not be linear with a simple series resistor but power controlled by changing the mark to space ratio (hence duty cycle) in a PWM controller (very efficient and linear) will cause an almost linear change in brightness with pot rotation.

Addition:

If you would like to change brightness using a series resistor it would be preferable to use a multi-position rotary switch (say single pole 5 position) and switch in individual resistors to get 5 brightness levels for example.

Jerry

Last edited by 74tuc; 20-05-2007 at 12:11 PM. Reason: Addition
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  #19  
Old 20-05-2007, 07:02 PM
Dennis
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Circuit diagram updated

Circuit diagram for LED’s in Parallel updated with addition of 20KΩ to 50KΩ potentiometer.

Cheers

Dennis
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  #20  
Old 20-05-2007, 08:17 PM
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Dennis, connect the "wiper" of the pot to one of the ends.


(This is basically what I'm about to do to illuminate the crosshairs in my finderscope - I'm thinking of four LEDs, lined up with each end of a crosshair, mounted in a new dewguard. Yes, four is overkill, but I like soldering. )
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