Commonly used LED driver

[tomkonikkara]

I want to connect about 40 or 50 LEDs in series. I think it is better to use an LED driver. Which is the most commonly used LED drivers? Are they available in local electronics shops?

 

[betwixt]

40 to 50 LEDs might require anything from 60 to 300V depending on the LED types, you need to be more specific. What kind of LEDs, what is their forward voltage rating, what current are they running at and most importantly, what supply are you starting off with?

Brian.

 

[tomkonikkara]

Power supply is 230V AC and is rectified using a bridge rectifier. This supply connected to n number of LEDs in series with an LED driver. n may vary from 10 to 40. I need about 25mA constant current through LEDs. Forward voltage of a single LED may vary between 2.1V to 2.8V

 

[betwixt]

The only way to do this is to use a constant current source or constant current sink. As you are using DC, it would be worth looking at a low side sink, possibly using a current mirror or a linear voltage regulator wired as a constant current load. This makes the number of LEDs irrelevant as long as their forward voltage doesn't exceed the supply. As a suggestion, use a 7805 regulator, wire it's output to ground through a 200 Ohm resistor, and connect the input to the low voltage (cathode) end of the LED chain. You will have to connect a capacitor of about 1uF across the regulator input to ground, don't make it much bigger as it's charge current has to pass through the LEDs. At the top (Anode) end of the LEDs, use a series resistor to drop some of the excess voltage.

Brian.

 

[tomkonikkara]

Your idea is pretty good. Today, I just looked over my old text books and found one circuit with some differences. It is something like this:
A 200ohm resistor is connected between OUT and GND pin of 7805. LED series is connected to GND pin of 7805 (anode) and ground of the supply (cathode). DC supply voltage is connected at IN pin and ground.

Is there any problem if I designed a circuit by avoiding 1uF capacitor? (you mentioned something about that).
Surely this idea is most useful for me.

 

[betwixt]

The current in a series circuit is the same wherever you measure it so you can connect the regulator at the 'top' or 'bottom' of the LED chain. In this kind of circuit, the terms top and bottom are relative, it depends how you draw the schematic. You absolutely MUST use the capacitor though. If you leave it out, there is a very high risk of the regulator oscillating and in that condition it will do nothing to regulate the current at all. If you make the capacitor smaller the risk of oscillation increases, if you make it larger, the current spike as you turn the power on gets bigger and as it passes through the LEDs it could cause damage. A value around 1uF should be adequate to keep it in check, you could add a second capacitor between the output and ground as well to further increase reliability but it should work without it. Mount the capacitor(s) as close to the regulator as possible, preferably no more than 3cm away.

Brian.

 

[tomkonikkara]

Thanks.
I need to use two resistors. One is 200 Ω and other is 5KΩ(variable). Since only the current through these is 25mA, is 1W resistors enough? Also I choose capacitor 10uF/16V. Is there any problem?

I've to use 8 such LED series channels. A 230V supply is first bridge rectified and is used to power up these 8 channels. These 8 channels are not directly connected to the bridge rectified output, but connected via 8 TRIACs. Each TRIAC is switched by a control system (something like a microcontroller) so that I can control the 8 channels. Is it feasible?

 

[betwixt]

Using Triacs (SCRs will work exactly the same) isn't a problem but I'm concerned where the 5K variable resistor is being connected - tell me more...
You can probably use 10uF capacitors but remember that at the point of the power turning on, by Triac or the whole thing being powered up, the capacitor will momentarily look like a short circuit so the LED current will not be regulated. The larger the capacitor value, the longer the current surge will last. You should also have a series current limiting resistor connected for protection, to lower the regulator in-out difference and to share power dissipation with the regulator.

Brian.

 

[tomkonikkara]

I've two more doubts about that:
1) I use 230V AC power supply. It is first bridge rectified. So the peak to peak amplitude of o/p will be:
230×sqrt(2)/2≈ 162V
I use 20 green LEDs having 2.1V forward voltage drop @ 25mA. So totla voltage drop across LEDs will be 2.1×20=42V
I design the curcuit such that voltage drop across the regulator circuit (7805) is 10V. So remaining drop is 162-42-10 = 110V.
So I need to put a series resistor of value = 110/25mA = 4.4KΩ/5W
Is my calculation correct? If not can you correct it?

2)If my assumption is previous question is correct, I need to use 5W series resistors in each channel (I use 8 such channels) and it will waste some power. Can I reduce that?

Thanks in advance.

 

[betwixt]

Your calculation is correct but I'm not sure how you are dropping 10V across the 7805. If you use it in constant current mode with a 200 Ohm load, the current will be 25mA but the drop will be 5 (across the resistor) + a voltage dropped between the input and output pins of the 7805. This would be difficult to calculate as it isn't specified by the manufacturers, I would suspect it was closer to 3V than 5V. It makes little difference anyway as the bulk of the drop is in the resistor.

You need to use a resistor rated AT LEAST 5W in each channel. The actual power dissipated is closer to 4W but the resistor needs to have a safety overhead. You would need a wire-wound resistor in a ceramic package to safely handle the power.

There are several ways to reduce the power but each has it's drawbacks. The first is to replace the resistor with a capacitor, one with reactance equal to 4.4K at your mains frequency. I'm not sure which country you are in so I can't tell the AC mains frequency you have. I'll assume 50Hz as this is most likely. The capacitor would have to be 0.36uF and rated at more than 250V AC. There are two dangers in using this method, firstly, the most likely failure mode of the capacitor is to go short circuit (killing all your LEDs!), the second is that there will be other frequencies and interference spikes in the power lines which are likely to pass through with less hinderance. A compromise might be to use a capacitor and a resistor in series, perhaps 0.47uF ( = 3.4K ) and a 1K resistor to make up the original total. The resistor now only dissipates about 1 Watt. If you use this method, please add a fuse in the chain for protection!

The second method would be to start with a lower mains voltage, either using an isolating transformer (ideal solution) or an auto-transformer. This is more costly and will be bigger and heavier but much safer.

A more technically demanding method is to use phase control to turn the Triac on so it only conducts over a small part of the AC cycle. As you already have a bridge rectifier in the circuit, it would be more economical and simpler to use SCRs instead of Triacs. Basically, you make a switch which turns on in the decaying part of the 100 HZ cycles. By altering the timing at which SCR triggers, you can adjust the output voltage to the LEDs without dissipating power during the rest of the time at all. You will need to use a microprocessor to calculate the exact trigger time but you can use one processor to handle all 8 channels at once so it might be an economical solution. The SCR would turn itself off at each mains zero point (100 times per second) so all you have to do is turn it on a short time before the zero point is reached.

Brian.

 

[tomkonikkara]

Thanks lot. I'm going to implement things as you suggested. But this time I'm not using capacitor, but instead I use 5W wire wound resistor. Because I'm not sure about the frequency. I think the frequency of bridge rectifier output is not exactly 100Hz and expect higher Fourier components because of the occurrence of sharp peak in the lower side of the waveform.