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