Potentiometer and LEDs question ( 2 Views )

no kitty!

Hi!

I'm planning to make a simple circuit, to light up 30-40 Leds. I'll use 12V from PSU, and the 30-40 Leds, along with their resistors, will be connected in parallel. Assuming that each LED works with approximately 25mA, and that the connection is parallel, the total current of my circuit, will be approximately 750 -1000 mA. (30*25mA - 40*25mA).
Now, using my very poor physics knowledge from high school, the total power (wattage?) consumed by the "system", according to the formula P=U*I, will be 9W-12W. Is that correct? Does this amount seem reasonable to you? Do you have any other opinion? I know, that i could use combination of parallel/series to drop the amperage, but that would not be so significant and i don't want to make things more complicated.

Now, i need to use a potentiometer, so that i can control the brightness of the LEDs. Is this feasible? Is there any better alternative? I found those pots on Ebay, but i'm not sure if they are suitable for this case.

Can anyone help me!!!???

(Recep, Jordan)

You're calculations are spot-on. You can easily chop your current usage by 67%, though. Instead of doing this:

12V-----^^^^----|>|----GND

Do this:

12V-----^^^^---|>|---|>|---|>|---GND

Your new resistor values will be R= (12V-3*LEDVoltage)/.025. This will cut your power consumption by a factor of 3.

Now, with regards to the pot...it's not the best way to control LED brightness, but it'll work. You'll want to put a resistor in series with the pot (with the R you calculated above), so that when the pot is at 0, the LEDs are at their brightest, but still have the current-limiting resistance. The size of the pot should be 10-50 times the size of the current-limiting resistor for a single string of LEDs. My brain's a bit fuzzy on this next part (stupid kids), but you'll want to divide that number by the number of strings. So if you have an R of 220 Ohms and 10 strings of LEDs, your pot will be 50*220/10 = 11kohm. 10k is the closest standard pot, so that should work. 5k might also work.

Now....that pot will be dissipating some power. Let's assume 3V leds, so the pot will be taking the last 3V, and 25mA * 10strings = 250mA. So 750mW. (this is just rough, err-on-the-safe-side estimates). Make sure the pot is rated to dissipate that much energy (it probably is).

(seda, Botswana)

I am no electronics expert.
Would that 0.2 watt potentiometer really be up to the task dissipating all the heat? - 30-40 LED's?

You do not want to end up like this :

That happened when a "customer" used a 0.5 watt potentiometer (rheostat) to control a single 80mm fan.

Here is a 25 watt unit which should handle the heat easily. But will the resistance value be adequate?
LIke I said, I am no electronics expert.

(sunay, Virgin Islands, British)

needs more PWM

(büþra, El Salvador)

You'll need 5k or 10k or something in the near vicinity. 25W is pretty bad overkill, but having a higher power rating won't hurt.

(emre, Macao)

Here, this might help!

(sebnem mine , Niue)

Quote:

Originally Posted by Mohonri

Now....that pot will be dissipating some power. Let's assume 3V leds, so the pot will be taking the last 3V, and 25mA * 10strings = 250mA. So 750mW. (this is just rough, err-on-the-safe-side estimates). Make sure the pot is rated to dissipate that much energy (it probably is).

The pot track has to carry 250mA max, so power needed is I^2R, which even for a 100 ohm pot is 6.25W. A standard 5k 0.25W carbon film pot will only carry about 7mA. Like fusewire, you're not concerned with the watts of heat produced by that fraction of an inch, it's the temperature it reaches under a current.

And whilst a 100R 6.25W pot will turn a fan down fine, it won't do much to dim your LEDs.

PWM is the answer, as Whatsisname said earlier.

(mustafa, Cape Verde)

Quote:

Originally Posted by cpemma

The pot track has to carry 250mA max, so power needed is I^2R, which even for a 100 ohm pot is 6.25W.

The problem is this: When the pot is conducting 250mA, it will be dropping very nearly 0 volts (the other, fixed-value resistor will be dropping the voltage). And since P = current * voltage, power will be near 0. At full range, the pot will be at, say 5kOhms. Now, assuming the LEDs are still conducting a very small amount of current (and thus maintaining the voltage), the power is now P= V^2/R = 9V/5000 = 1.8mW. (current will be under 1mA) Somewhere in between, the power dissipated by the pot will be higher, but not all that high.

(sezai, Peru)

Pots are like transistors and various other parts, they have a Safe Operating Area (SOA) with a boundary for the maximum current they'll carry and another boundary for the maximum heat they'll dissipate. Plus a third for maximum voltage, though that doesn't bother us. To avoid damage you've got to be inside all the limits.

Unfortunately suppliers specs generally only quote wattage, but the wattage figure given only applies when the full track is in use, as with a potential divider. Using only a fraction of the track, current is invariably the limiter.

See Vishay's specs, eg their 12.5W rheostat, for their specified current limits.

With fan control, the pot power dissipation is at a maximum when both fan and pot are dropping the same voltage, ie 6V each, and it's much less than the fan's 12V power rating - at 6V the fan current is roughly halved, so a 3W fan is only using about 0.75W at 6V. With LEDs and a fixed resistor plus the pot I'd have to get the calculator going, but the pot still has to handle the current.

(emre, Sri Lanka)

Quote:

Originally Posted by cpemma

Pots are like transistors and various other parts, they have a Safe Operating Area (SOA) with a boundary for the maximum current they'll carry and another boundary for the maximum heat they'll dissipate. Plus a third for maximum voltage, though that doesn't bother us. To avoid damage you've got to be inside all the limits.
<snip>
With fan control, the pot power dissipation is at a maximum when both fan and pot are dropping the same voltage, ie 6V each, and it's much less than the fan's 12V power rating - at 6V the fan current is roughly halved, so a 3W fan is only using about 0.75W at 6V. With LEDs and a fixed resistor plus the pot I'd have to get the calculator going, but the pot still has to handle the current.

You have a good point with the first comment. This sounds like it could turn out to be a very interesting math/physics/EE problem. I might just have to whip something up in a spreadsheet to make this discussion a bit more interesting... :)

You're correct that with a fan, the pot's max power dissipation is when it's dropping half the voltage. Since LEDs drop a fixed amount of voltage for any amount of current (within certain limits, of course), the voltage across the pot + fixed resistor will be constant. In this example, it'll be around 3V total, with 1.5V across the pot. With the pot dropping the same voltage as each of the 220Ohm (or whatever) resistors on each string of LEDs, the current will be halved. So, P = IV = 125mA * 1.5 = 187ish milliwatts.

Somebody remind me why we're not going with PWM? It may be a more complicated circuit to wire up, but it lets us forget about all the math once it's done...

(recep, Saint Vincent and the Grenadines)

Quote:

Originally Posted by Mohonri

Somebody remind me why we're not going with PWM? It may be a more complicated circuit to wire up, but it lets us forget about all the math once it's done...

I'd certainly go PWM, I was just showing with just a series pot you need a whopper wattage when it's the high resistance value needed to get a dim down to faint. :eek:

Even with PWM most of the dimming is at the bottom end of the control pot's travel (that's our eye's response to light levels), may be worth trying a log (audio) taper one to get a better effect. :confused:

(aykut, Bahrain)

We really *should* make a sticky for all things LED-related...

(MUHSÝN, Congo, the Democratic Republic of the)

Potentiometer and LEDs question ( 2 Views )

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