LED Matrix - 12V

A "new fashioned" televisor, using an Arduino to drive the motor and display.

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Re: LED Matrix - 12V

Postby Steve Anderson » Mon Jun 12, 2017 12:18 am

I don't know how to phrase this nicely, but this is typically how the manufacturers in Asia would do it. You're right, in the majority of cases where the LEDs come from the same batch the forward voltage drop will be close, very close. However it goes against the grain of my thinking and training, conservative I may be, yes, but I would be putting my reputation on the line here, this is to be a long-term display item. For the sake of a few resistors...

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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 12:22 am

Steve Anderson wrote:I don't know how to phrase this nicely, but this is typically how the manufacturers in Asia would do it. You're right, in the majority of cases where the LEDs come from the same batch the forward voltage drop will be close, very close. However it goes against the grain of my thinking and training, conservative I may be, yes, but I'm I would be putting my reputation on the line, this is to be a long-term display item. For the sake of a few resistors...



Understood, and don't worry about offending me. Couldn't I actually measure the voltage drop across sets of three and make sure that I built a matrix where the strings were very closely matched? Also, I think a bit too much focus is on the "display item" thing. It's just a suggestion at this stage, not a goal.
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 12:33 am

With the quick test I did (video, above) I used a 10 ohm resistor and no balancing resistors.
What's the difference between that and (for example) using a 5 ohm resistor and putting a 5 ohm balancing resistor with each string?
Same total resistance for all pathways, right? I'm not against balancing resistors - I just want to understand the need for them, and also want to *include* their resistance in the calculations, not add their resistance "on top of" things. In other words, not keep the 10 ohm resistor I have now and add another 5 ohms of resistance.
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Re: LED Matrix - 12V

Postby Steve Anderson » Mon Jun 12, 2017 12:48 am

Andrew, this is all you need, no more, no less...

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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 12:52 am

Steve Anderson wrote:Andrew, this is all you need, no more, no less...

Steve A.


Thanks for taking the time to do that, Steve. I am still hanging on grimly for a 12V-only solution.
Just a couple of comments - the LEDs are rated at 20 mA and definitely 3.4V. (measured a dozen or so about 3.35V)

Edit: I'm totally understanding what you have, there. With the corrected values, the resistors would be 440 ohms - say 470.
I like this design in that it incorporates the limiting resistor into the string itself, and that also balances. Rather than my currently working array which has balancing resistors independent of an external limiting resistor. I also like that you can add as many "strings" as you want without having to change other stuff, and will function provided there's enough total current.
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 1:00 am

Using that design, but with 12V power and 3.4V forward voltage on each LED, using 20mA....
3xLED = 10.2V. 12V-10.2V = 1.8V. 1.8V/20mA = 90 ohms.
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Re: LED Matrix - 12V

Postby Steve Anderson » Mon Jun 12, 2017 1:14 am

Yes, in theory that would work, though I haven't checked out the math, assuming it's correct, that should be fine...

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I'm off to bed...
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 1:31 am

I gave it a go - works fine. I didn't have 90 ohm resistors, but I had 82 and I thought I'd give those a shot. That puts about 22 mA across the LEDs which I thought should be OK. Edit: measured 3.34V/LED. Three LEDs = 10.02V. 11.4V (input) - 10.02 = 1.38V left. 1.38V/0.082 = 16.8 mA. I could/should have used a 70 ohm or thereabouts.

Here's a couple of pictures.

ledmatrixv2off.jpg
ledmatrixv2off.jpg (122.64 KiB) Viewed 13394 times


ledmatrixv2on.jpg
ledmatrixv2on.jpg (194.51 KiB) Viewed 13394 times


I appreciate all the help, and have learned/understood something today, so it's all been worthwhile :)

Edit: I tried with 5 strips each with a 68 ohm resistor (I only had 5). All OK.
If I wanted to chance 25 mA, I'd go to 56 ohms. Didn't have any handy but tried a 47 ohm on a single strip of 3 (for ~30 mA). LEDs are working but for how long :)
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 1:47 am

In the darkness, 3 LEDs shining brightly. 11.4V input, 47 ohm resistor.
I thought it was a pretty picture and wanted to add to the thread for posterity.

ledmatrix47ohmdark.jpg
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Re: LED Matrix - 12V

Postby Klaas Robers » Mon Jun 12, 2017 1:52 am

Andrew Davie wrote:Firstly, in that original diagram the 6V was calculated ........


This original diagram features a resistor type Digital to Analog converter, with a transistor in a current source configuration. I assumed that this is the (kind of) circuit that you are using in your monitor. Why other would you have linked us to that original diagram? This is logically not a PWM driven circuit.

This current source circuit, so the transistor and the resistor from emitter to ground, has the advantage that variations in the supply votage have no influence on the current through the LEDs and hence no influence on the light output of the LED cluster. On the other hand, it only works in a proper way if the voltage on the collector of the transistor is 6 volt or more above gound level.
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 2:24 am

Klaas Robers wrote:Why other would you have linked us to that original diagram? This is logically not a PWM driven circuit.

This current source circuit, so the transistor and the resistor from emitter to ground, has the advantage that variations in the supply votage have no influence on the current through the LEDs and hence no influence on the light output of the LED cluster. On the other hand, it only works in a proper way if the voltage on the collector of the transistor is 6 volt or more above gound level.


Incompetence, basically. Yeah, sorry about that error. I was only paying attention to the LED matrix itself. I seem to not have done a diagram for the final configuration. I regret the confusion I have caused :( What I have is definitely a PWM circuit. I have balancing resistors on the LED matrix (it's the original one from my 2007 televisor). I also have a limiting resistor on my LED "driver board". The driver board has a TIP122 with a 16 ohm resistor.

So, to clarify... are you saying that with the configuration just described that I am planning (12v supply, Arduino --> TIP122, LED matrix with balancing resistors) then I have to have collector voltage 6 volts above ground for the transistor to work? At the moment it looks like the voltage at the collector will be about 1.4V above ground.

Edit: I have carefully re-read your reply and I think you are saying that the 6V is a requirement for current-source configuration, but my understanding is that what I have been testing is NOT a current source configuration and thus the 6V is not a requirement.
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Re: LED Matrix - 12V

Postby Klaas Robers » Mon Jun 12, 2017 7:26 am

That is true. If you don't use the current source configuration, you don't need the 6 volt at the collector. Then you drive the transistor in saturation, which is about 0.1 volt from collector to emitter (ground). Then you can use the balancing resistor immediately as a current limiting resistor. So:

- 3 LEDs in series 3 x 3.4 volts = 10.2 volts over the 3 LEDs,
- The transistor needs 0.1 volt, so 10.3 volts in total.
- Remains 12 - 10.3 = 1.7 volt for the current limiting resistors,
- which are 85 ohm (1.7 volt / 0.02 ampère) per branch.
So mount a resistor of 82 ohm (standard value) in each branch. Such a branch (3 LEDs and one 82 ohm) should run directly on 12 volts. Each branch will do this in the same way.

The need of balancing resistors is because if one branch gets hotter than other branches, this branch is drawing more current. A warm LED gets a lower "resistance". LEDs and diodes in general have a negative temperature coëfficient. Thus it is drawing even more current.... and so on. The balancing resistors equalize the 10 currents more or less. The higher the resistance of the resistors, the better this works. So a resistor in each branch and no resistor in the combined path is optimal.

Dissipation of the resistors is low: 1.7 V x 0.02 A = 0.034 watt = 34 mW.
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 11:21 am

Klaas Robers wrote:That is true. If you don't use the current source configuration, you don't need the 6 volt at the collector. Then you drive the transistor in saturation, which is about 0.1 volt from collector to emitter (ground). Then you can use the balancing resistor immediately as a current limiting resistor. So:

- 3 LEDs in series 3 x 3.4 volts = 10.2 volts over the 3 LEDs,
- The transistor needs 0.1 volt, so 10.3 volts in total.
- Remains 12 - 10.3 = 1.7 volt for the current limiting resistors,
- which are 85 ohm (1.7 volt / 0.02 ampère) per branch.
So mount a resistor of 82 ohm (standard value) in each branch. Such a branch (3 LEDs and one 82 ohm) should run directly on 12 volts. Each branch will do this in the same way.

The need of balancing resistors is because if one branch gets hotter than other branches, this branch is drawing more current. A warm LED gets a lower "resistance". LEDs and diodes in general have a negative temperature coëfficient. Thus it is drawing even more current.... and so on. The balancing resistors equalize the 10 currents more or less. The higher the resistance of the resistors, the better this works. So a resistor in each branch and no resistor in the combined path is optimal.

Dissipation of the resistors is low: 1.7 V x 0.02 A = 0.034 watt = 34 mW.



Fantastic, thanks! All understood, and the explanation of the warming LEDs and consequent "runaway" problem is clear. Now I understand the reason/advantage of balancing resistors. As with your explanation and Steve's diagram, I have balancing resisotr in each branch. Also useful to me the "extra" voltage consumption by the resistors and the transistor. Many thanks.
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Re: LED Matrix - 12V

Postby Steve Anderson » Mon Jun 12, 2017 4:59 pm

One thing I would like to give a 'heads-up' on... you need to make sure the 12V supply is truly 12V. With these unregulated 'bricks' it may not be. If you re-do Klaas's calculations where the supply is say 13V you end up with a higher current...how much? I'll leave you to do that, you should know how to do that now (I am a part-time teacher). Do it again for 14V, and as a matter of interest, 11V.

For the moment assume the voltage drop across three series-connected LEDs is a constant 10.5V, it isn't, but this points out the trend and is a worst-case scenario.

You'll be surprised at the variation.

Steve A.

You could also do a +/-1V variation the the circuit I posted above, at 18, 19 and 20V. Notice any change?
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Re: LED Matrix - 12V

Postby Andrew Davie » Mon Jun 12, 2017 7:52 pm

Steve Anderson wrote:One thing I would like to give a 'heads-up' on... you need to make sure the 12V supply is truly 12V. With these unregulated 'bricks' it may not be.


Yep. I understand this issue and I'm investigating how best to do this. Something like a 7812 regulator perhaps... but this would require a few volts above 12V as input.

Steve Anderson wrote:If you re-do Klaas's calculations where the supply is say 13V you end up with a higher current...how much? I'll leave you to do that, you should know how to do that now (I am a part-time teacher). Do it again for 14V, and as a matter of interest, 11V.


Easy, as I understand it :)
13 V input. Let's leave the resistor unchanged at 82 ohms. We have 10.2V used by the LEDs, so 13-10.2 = 2.8 V at the base of the LED matrix. 2.8/82 = ~34 mA which will shorten the LED lifespan significantly as they're rated at 20 mA. For 14 V it's 14 - 10.2 = 3.8 V and thus 3.8/82 = 46 mA bye bye LEDs. I used 10.2 instead of your 10.5, but the principle is the same. For 11V we have 11 -10.2 =0.8 and /82 = ~10 mA.

So, it's important that if I design for 12V that the input is actually 12V.

Steve Anderson wrote:You could also do a +/-1V variation the the circuit I posted above, at 18, 19 and 20V. Notice any change?


The difference being that the voltage changes don't affect the current nearly as much. Thus, 18V - 10.2 = 7.8 and /330 = ~24mA. 19V -10.2 = 8.8 and /330 = ~27 mA, and 20V - 10.2 = 9.8 and /330 = ~30 mA. A two volt range gives results from 24 mA to 30 mA whereas at 12V, a two volt range (11-13V) gives from 10 mA to 34 mA. In other words, the higher the voltage input the less severe the variation in the resultant current, for a given resistance.
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