I tried a little experiment today. I wired up three strings of three LEDs,
without balancing resistors. I just had the original 56 ohm resistor in series with the array. I wanted to see/understand the effect of the LEDs without balancing resistors. Basically, none visible. They all worked fine and looked equally as bright. It makes me wonder just how necessary these balancing resistors ARE, because the resistance across each string of three LEDs seems to be pretty constant.
There are a few things I'm not "getting", once again. But thanks Klaas and Steve for your patience and assistance.
Klaas Robers wrote:I agree with Steve: use a higher voltage for the LED cluster supply. If you look at your diagram in the link that you gave some posts up, you see that the voltage at the bottom of the LED cluster, also the collector of the video transistor, is quoted as 6 volt as a minimum. Now follow me step by step.
- The voltage over one of your LEDs is 3.5 volt at the max. current of 20 mA.
- If you place (just) 2 LEDs in series, the voltage over the both LEDs will be 7 volt.
- Add 1 volt for the current balancing resistor (47 ohm for 20 mA, thats correct),
- that will give 8 volts for the cluster chain.
- Place that on top of the 6 volts of the collector,
- and you need 14 volts minimum for the LED power supply, not 12 volts.!
Firstly, in that
original diagram the 6V was calculated with the assumption that the LEDs used 2V each, and three in series were 6V. With a 12V supply, then 6V remained, and that's what's written at the base of the LED matrix... "6V 125mA" (because, 5 strings each 25 mA).
In my current case, the LED voltage is measured at 3.36 (let's say 3.4V) and therefore 3x = 10.2V running at 20mA each string. 10 strings = 200 mA. The "remainder" would be 1.8V and for 200mA that's a resistor of 1.8/0.2 = 9 ohms. In previous calculations the balancing resistors were as far as I'm aware *ignored* for the purposes of calculating the limiting resistor value. However, that looks wrong to me. The total resistance in this current case should be 9 ohms. Note also that I'm
not catering for balancing resistors, because at the moment I can't understand why they are required - given the LED strings are pretty much consistent in resistance - in my view possibly at least as consistent as the difference between resistors.
So my first point of not understanding is "place that on top of the 6 volts of the collector". I don't understand this at all. The "collector voltage" in came from the calculation, and in this case it's 1.8V. I'm confused as to how I'm supposed to calculate a collector voltage which isn't anything other than the "remainder" after LED voltage drops (and balancing resistors if present) are taken into account. I had a look at
the TIP122 data sheet, and couldn't really see what I should be looking at/for. I was trying to find something like "what voltage difference is required for operation" but other than Vbe(on)=2.5V as a possible, nothing seemed like a likely candidate.
Currently I'm stuck on this: LED string requires 10.2V @ 20mA. That's less than 12V. So, what's the extra voltage requirements for the transistor and the limiting resistor?
Klaas Robers wrote:But is you use a laptop power supply of 19 volt, abundantly available from old laptops,
- You can place 10 parallel branches of 3 LEDs in series,
- with one balancing resistor in each branch,
- a total current of 200 mA
- 30 LEDs and 10 balancing resistors of 47 ohm.
Not against using a 19V power supply at all. Just want to make sure that we're talking the same language and I fill in the gaps in my understanding. With the above, you are using 3.5V so a string of 3 LEDs is 10.5V and the current is 20mA/string = 200mA. With 19V that leaves 8.5V left over an 8.5/0.2 = 42.5 ohms,
not taking into account any balancing resistor. OK, 47 is a good "next one up" candidate, but I'm still hazy as to how/if the balancing resistor comes into this calculation.
Klaas Robers wrote:The advantage of a lower total current is that the transistor heats up less,
- for 600 mA it dissipates about 3 watt,
- for 300 mA it is about 2 watt,
- for 200 mA it is about 1 watt.
- In all cases the transistor needs some from of heat sink.
With the current array running at 22V or thereabouts it's 8 strings of 5, about 30mA = 240 mA and the transistor is barely warm. It doesn't have a heatsink. I'll keep the above in mind, but not quite understanding why a heat sink will be necessary if it's already cool enough with a display drawing more current.
Very frustrating that I am having trouble with this basic stuff. Appreciate your comments and honestly not trying to be stubborn and just trying to understand