Constructing trouble

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Constructing trouble

Postby DrZarkov » Thu May 03, 2007 6:55 am

Like Andrew I've started at the same time, and me too, I'm not very experienced (if not to say a beginner) with electronics. I can solder something, if you say me what to do, but that's it. But I'm learning, so I started with my simple device and improved it over the weeks.

But now I've started to build a "proper" monitor using the PCBs and the scanning disc from the club shop. Andrews construction diary is a big help for me to avoid mistakes, but I still struggle over the "resistor-thing".

The problem: I've bought 50 ultra bright LEDs, V: 3.6 typ., 4.0 max, mA: 20 (typ.). I would like to use 48 LEDs in an array. I want to avoid the 100 V for the LED-string, mainly because I don't want to have three transformators (12 Volt, "standard" transformator, 17 Volt from an old notebook-transformator, and 100 Volt, how to get???), so I prefer something with the 17 Volt from the notebook-transformator.

What would ou recommend? Making 16 groups of 3 LEDs, in a parallel circuit, that would make 12 Volt, together 960 mA. What kind of resistors do I need then? I do not understand, what influence the resistors in the LED circuit have to the LED-driver circuit. As I understand, I have to change the two resistors parallel to the diodes near the transistor BUT 11? Only them? To what?
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Re: Constructing trouble

Postby Andrew Davie » Thu May 03, 2007 10:32 am

DrZarkov wrote:Like Andrew I've started at the same time, and me too, I'm not very experienced (if not to say a beginner) with electronics. I can solder something, if you say me what to do, but that's it. But I'm learning, so I started with my simple device and improved it over the weeks.

But now I've started to build a "proper" monitor using the PCBs and the scanning disc from the club shop. Andrews construction diary is a big help for me to avoid mistakes, but I still struggle over the "resistor-thing".

The problem: I've bought 50 ultra bright LEDs, V: 3.6 typ., 4.0 max, mA: 20 (typ.). I would like to use 48 LEDs in an array. I want to avoid the 100 V for the LED-string, mainly because I don't want to have three transformators (12 Volt, "standard" transformator, 17 Volt from an old notebook-transformator, and 100 Volt, how to get???), so I prefer something with the 17 Volt from the notebook-transformator.

What would ou recommend? Making 16 groups of 3 LEDs, in a parallel circuit, that would make 12 Volt, together 960 mA. What kind of resistors do I need then? I do not understand, what influence the resistors in the LED circuit have to the LED-driver circuit. As I understand, I have to change the two resistors parallel to the diodes near the transistor BUT 11? Only them? To what?




OK, here's what I understand...

Firstly, what we need to do is decide what current we want to pass through the LEDs. The LEDs will each 'use' their rated voltage, so we don't worry too much about the voltage side of things -- just ensure there is enough to power all the LEDs in any one string (that is, if our LEDs are in groups of 3, then we have 3 x LED voltage minimum).


The important thing to remember is V = I x R. V=Voltage, I=Current, R= Resistance. Now our input voltage is going to be constant (say, 12V) and the voltage that the LEDs use is also going to be constant. What we do when designing the LED array is to change the resistance so that the current flowing through is correct. Too much current and the LEDs blow up -- trust me on this ;)

Firstly consider a single LED. Let's say that it is rated at 3.5V. One thing to know about LEDs is that they require a minimum voltage to be met before they will light up. So if we put 3.0V across our LED, it simply won't light. Similarly, LEDs in series have a total voltage of the sum of all of the LEDs' voltage. With three of our LEDs, our voltage across the LEDs would be 3 x 3.5 = 10.5V. This will be OK with a 12V supply, because there's a bit 'left over' -- but not much.

Back to the single LED -- our 3.5V LED. Let's assume that its rated current is 25mA. What we need to do is choose a resistor such that when a 12V input is provided to the LED, the formula V = I x R will result in I being 25mA (= 0.025A).

Now the real trick is to understand that the LED will 'use' 3.5 of those 12V. SO in reality, we are only needing to deal with the remainder of the voltage (that is, 12 - 3.5 = 8.5V). Plugging that into the formula, we get 8.5 = 0.025 x R, so R = 8.5/0.025 = 340 ohms. A value close to this will be OK (eg: 360) as the current can be 'around' 25mA and still be OK.

OK, so that's how to calculate for a SINGLE LED. We can treat a string of LEDs in a similar fashion. We still have the same current (0.025A) going through the LEDs, but the total voltage across the LEDs is the sum of the voltage for each. SO with three LEDs, at 3.5V each, our total voltage 'used' by the LEDs is 3 x 3.5 = 10.5V. THe voltage remaining is therefore (12-10.5 = 1.5V). And using the V = IR formula, we get 1.5 = 0.025R, so R = 1.5/0.025 = 60 ohms... or thereabouts.

There we have the method for working out the resistor to use for a single string of LEDs in series. Here it becomes obvious that for longer strings you need a higher input voltage, as the voltages sum together and get rather large quickly. This is why Klaas has advocated 60+ volt supplies -- and I'm totally in the same boat as you with regard to using higher voltages at this stage. Though it may be simpler, it is more dangerous (at least to my untrained eye). I'd rather design low-voltage arrays.

And how do arrays work? Mostly the same as per single strings, but there are other considerations, too. When we place LEDs in parallel, the voltages remain the same (ie: the voltage across our array of 8 strings of 3 LEDs will be the same as for a single chain of 3 LEDs) but the current is cumulative. So we have a current of 8 x 25mA (=200mA).

When the LED array is used in a circuit, though, then things get a little different. It is the external circuit that controls the current that is provided to our matrix and the resistors that we put in act as 'balancing' resistors. They simply ensure that all of the strings get an equal amount of the 'power' coming into our circuit. Then it is not so important that our resistor values are correct -- they just have to be the same.

In this case, we can think of the LED matrix as a single unit, and not worry about each individual LED. So what I did was build my matrix (8x3) and tested it in isolation on a 12V supply. I calculated the correct restistor values (as above) for this 12V supply. Then I placed it in-circuit with the NBTV boards, and re-did the resistors later.

Again, using the V=IR formula, we can treat the completed LED matrix as a single entity. In this case, the voltage across the parallel array will be 3 x 3.5 (10.5V). And the total current will be 8 x 25mA (=200mA). Those values can be used to calculate the value for an external resistor that correctly controls current through the matrix (pretty much independant of the resistors associated with each string of 3 LEDs). The external current control relegates the resistors on the LEDs to the role of 'balancing' resistors -- they simply ensure that the LEDs all light up with the same brightness.

So, let's calculate the exernal resistor for the whole matrix. V = IR. Remember, the circuit has 10.5V over the strings, and the total current is 200mA. So, (12-10.5)=0.2A x R, R = 1.5/0.2 = 7.5 ohms. So I would use a 7.5 ohm or thereabouts resistor external to the matrix. Note that on the NBTV circuits, this is the 'R' value mentioned when calculating the resistance of the gamma part of the circuit. It is this R that controls the current flowing into the LED matrix.

Hope this helps!

Cheers
A
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Postby DrZarkov » Thu May 03, 2007 4:28 pm

Thank you very much! Let me try if I learned a little bit:

I have 17 Volt for my array, my LEDs need 3,6 Volt to light up, they have 0,20 mA. I've splitted the array in 16 strings of 3 LEDs (48 LEDs in total), so I need per string 17 Volt - 3,6 V x 3 = 6,2 Volt "too much", which means 6,2/0,020= 310 Ohm per resistor in my array.

For the circuit: the array has 16 strings with 20 mA LEDs, which means 320 mA in total. So I get 6,2 V (as above)/0,32 A = 19,37 Ohm (I think 19 Ohm or better 20 Ohm will fit?).

Last question: Where on the PCB I find the resistor which is to be replaced? I don't want to take the wrong one...
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Postby Viewmaster » Thu May 03, 2007 6:36 pm

One thing I have done to experiment on LEDs is to put in a 1k wire wound
pot (variable resistor) in an LED circuit. If you have a voltmeter and 100m/a meter you can vary the volts across the LED and check on the current flowing thro it too. This will teach you how the current increases as the volts rise and what the volts are across the LED when the current flowing thro' it is the correct milliamps.

Experimentally, I recently put 70 m/a through a rectangular LED ! :shock:
Just experimentally I hasten to add.
The light output was enormous but the current and light dropped after 6 hours. :cry:


Albert.
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Postby Andrew Davie » Thu May 03, 2007 11:11 pm

DrZarkov wrote:I have 17 Volt for my array, my LEDs need 3,6 Volt to light up, they have 0,20 mA. I've splitted the array in 16 strings of 3 LEDs (48 LEDs in total), so I need per string 17 Volt - 3,6 V x 3 = 6,2 Volt "too much", which means 6,2/0,020= 310 Ohm per resistor in my array.

For the circuit: the array has 16 strings with 20 mA LEDs, which means 320 mA in total. So I get 6,2 V (as above)/0,32 A = 19,37 Ohm (I think 19 Ohm or better 20 Ohm will fit?).

Last question: Where on the PCB I find the resistor which is to be replaced? I don't want to take the wrong one...


Yes, the above is how I understand this so far. Remember, I'm learning too, so be sure to read the construction diary entries related to the LED matrix, just in case I get this wrong :)

Now it wouldn't really matter if (say) 25mA ran through the LEDs -- because they will not be 'always on'. In addition, I've had it drummed into me that 10% or thereabouts is good enough. So I'd be quite happy putting 19 Ohm, or even 15 Ohms as the R-value for the gamma circuit (for your configuration). Assuming 15, then that would give 6.2/15 = 0.38A --> /16 = 24mA across each string. This is still a very reasonable value for those LEDs, I would suggest. So the R-value can be 19... without any trouble at all.

Now on the circuit there ISN'T a resistor to replace. R refers to the combined resistance of the three resistors that make the gamma circuit. See the area in the diagram attached. Each resistor value is calculated based on your value of R for the entire gamma circuit (from your calculations above). One resistor uses 3R, one uses 1.6R and the last uses 1.3R. And here, I did not have to change my resistors -- the original values were OK. So I'm not totally sure that this is the correct thing. Pretty sure, though. I would use 19x3 = 57ohms (roughly!), 19x1.6 = 30 ohms (roughly!) and 19x1.3 = 25 ohms (roughly)!!. You will have to give up on German perfectionism -- this is all very wishy-washy, and these resistor values are not needed to be exact. They just control the 'curve' of the gamma correction circuit which in turn modifies the brightness seen by the eye.

Having answered those questions -- my experience is that 24 high brightness LEDs is total overkill. For one thing, you can't pack that many LEDs into an area that is the size of your image. And besides, it is not necessary. I think that as few as 6 LEDs would be sufficient for a bright image. One other thing, packing the LEDs -- I found that if I put them 4-holes apart in a square pattern on the breadboard, then I was able to put an additional one in the middle of the square. It ended up forming a diagonal crosshatch, quite closely spaced (see image). My next LED array will be 18 LEDs (maybe 20) in the format you see here -- 5 columns of 4 LEDs each. In this picture, only the leftmost column, and column 3 have all LEDs inserted. This grid layout comes to slightly larger than the image size on my disc.
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Postby DrZarkov » Fri May 04, 2007 12:08 am

"German correctness" (or "Deutsche Gründlichkeit" ) passed away many years ago... But I will try not to blow up any LEDs ( I did it once in one of my very early NBTV experiments many years ago, I didn't know that a LED can really explode!), and I want to have a good picture. What I saw on the photos from your monitor, it looks very good, much better than my first attempt with the normal 1 Watt audio-amplifier abused for the LEDs.
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Postby Klaas Robers » Sun May 06, 2007 5:26 am

Volker,

I learnt that the calculation of the resistor values of the gamma cicuit is less straight forward than I wrote it. When sending higher currents through the diodes of the gamma circuit, they change their behaviour. The almost 1A that you intend to use is far to large for the suggested diodes. They can't have more than 100mA.

In your case this is 5 strings in parallel. So for 10 LED's in series you need at least 36 volts, plus some volts more for the driver transistor, so sat 40 to 50 volts.

The circuit of the PCB is ligned out for 40 mA, which is two strings (series) of LED's. Each string is then 24 LED's, which is 86 volt. Then you need a voltage of indeed about 100V and a driver transistor that can withstand that voltage.
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Postby DrZarkov » Sun May 06, 2007 8:06 am

Building in another transformator is not an option, it will make the monitor too bulky and too expensive, not to forget that such a transformator is hard to get. So I will make another array of LEDs, with 12 LEDs from the "Andrew-type". That makes it all easier I think.
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Postby Klaas Robers » Mon May 07, 2007 1:17 am

What is the AC voltage of your transformer? There are tricks to get higher voltages in DC from a transformer secundary, called voltage doubling, or tripling, or quadrupling. However it is best that you have a separate secundary winding for it on the transformer.

If you have the Horowitz book, look in chapter 1.28, just before 1.29

The higher the voltage that you depart from, the better it works. I see in the catalogue of Conrad small PCB-type transformers of 2 x 15 volt. When you connect both windings in series you get 30V AC. Then voltage doubling isn't even needed.

When rectifying it positive as well as negative (two single diodes) you get about 40V positive and 40V negative. This can also be seen as 80V positive if you ground the -40V and leave the center tap floating. Remind that you need no more than 40 mA, so just a few watts. A smal transformer, say 5 watts, is large enough.
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Postby Andrew Davie » Mon May 07, 2007 1:24 am

DrZarkov wrote:Building in another transformator is not an option, it will make the monitor too bulky and too expensive, not to forget that such a transformator is hard to get. So I will make another array of LEDs, with 12 LEDs from the "Andrew-type". That makes it all easier I think.


Tonight, being totally sick and tired of trying to get my motor speed circuit to work, I built myself another LED matrix -- this time packed as I showed above. I used 18 LEDs, configured as four rows of 3, and two columns of three crossing those rows (ie: a # pattern). I calculated 34 ohm resistors were needed, only had 27 ohm on hand, so used those. My array is being fed by 20V which comes from 15V AC rectified but not ripple-removed.

The matrix worked very well first time, except for the bottom row which doesn't light (bad LED, or short somewhere). But the results are quite amazing. I have at least twice the picture sharpness/quality. My images are much more detailed... I'm quite amazed!

So I can heartily recommend an "Andrew-type" array -- especially in the configuration described. Who needs those big voltages :)
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Postby Klaas Robers » Mon May 07, 2007 1:47 am

Andrew, haven't you connected one LED in reverse? Then the row won't work as LEDs are diodes and one diode will inhibbit the current to run. You can simply check this with the ohm meter.

I am pleased to see that you found the way of calculation of the resistor values in the gamma circuit. The R value is indeed just a virtual value to find the real values.

However there is still a problem when sending high current through the diodes of the gamma circuit. The 1N4148 can handle no more than 100mA. But you may exchange these diodes by 1N4001 diodes. I guess that they are fast enough to do this trick as well. Please try it and report about it.
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Postby DrZarkov » Mon May 07, 2007 4:53 pm

Andrew, this sounds very good to me! I found in my 48 LED-array a dead LED (very strange, in the row of three two are lightning up, the third is blind, how is that possible? Short circuit?), so I have to re-arrange it. My club circuits will be ready this evening (I finally got the last missing parts...), so I can do first tests, I hope.

We have a new shop in the neighbourhood of my work, www.led-tech.de, so I'm sitting at the source now :D I was looking there for the ultra-bright LEDs, but then I would have the 1 A problem again, one of it has 370 mA, and I would need 6 of it. So maybe I should indeed start thinking about higher voltages...

Update: I've calculated for my new array 120 mA. the original diodes are made for max. 100 mA, can I replace them with 1N4148, will that be enough?
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Postby DrZarkov » Mon May 07, 2007 11:52 pm

@ Andrew: How do you calculated you resistors? You used the same type LEDs like before? If I use "your" formula which with this thread started, I would calculate: 3 LEDs of "your type" in a row need 10.50 V, your power supply hast 20 V, so you calculate 9.5/0,025= 380 Ohm??? :?:

Where is the mistake?
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Postby Andrew Davie » Tue May 08, 2007 12:15 am

DrZarkov wrote:@ Andrew: How do you calculated you resistors? You used the same type LEDs like before? If I use "your" formula which with this thread started, I would calculate: 3 LEDs of "your type" in a row need 10.50 V, your power supply hast 20 V, so you calculate 9.5/0,025= 380 Ohm??? :?:

Where is the mistake?


You're quite right -- each string is about 10.5V, and my calculation of the resistor values was wrong. I suspect I thought I was running of 12V, and I recall thinking I'd give it a bit more current... so assuming that, then I had three LEDs at 3.6V each, with 35mA current --> (12-3x3.6)/.035 = 34 ohms. SO that's almost certainly what I did. The problem is, I'm running off 20V, not 12.

The reason it works, as mentioned earlier, is that the resistors are NOT acting as current limiters... the limiting of the current is done by my gamma circuit resistors. The resistors on the LED matrix are too small (my big mistake), but it doesn't matter -- because as the matrix is used in the NBTV circuit with the gamma resistors limiting current, all is OK. The too-small-resistors simply balance the load evenly across all strings of the matrix.

I would only have a problem if I attempted to run the matrix off 20V directly, instead of through the NBTV circuits.

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Postby DrZarkov » Sat May 12, 2007 6:01 am

I still have a resistor-problem I think. I've testeds my 18 V transformator, it has 18 V. I've tested the LED-array, it is working. Connected to the circuit, I get a faint glowing, by far not enough. At the LEDs something between 10.4 V and 11 V arrive. Unfortunally I haven't such a good equipment like an oscilloscope to test. So now I have to start debugging in a more conventional way...
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