Hobby Issues

[DonRotolo]

In addition to what was said before, some other considerations:

1. Be sure to consider the power dropped by the resistor. As an example, because P=I*V, 40 mA at 18 volts = 0.72 watts - use a 1 watt or larger resistor or it'll overheat quickly.

2. For LED dropping resistors, close enough is fine. If your calculations call for 310 Ohms, anything from 270 (a bit broght) to 390) a bit dimmer) is fine.

3. You must always have a resistor in series with LEDs, or they will go into hard conduction and overcurrent very rapidly, letting out a tiny pink glow and a small pop as they fail forever. Start too large and move smaller.

4. LEDs are diodes, too, so running two in opposite polarity, but in series, will mean no current flows.

The short answer to what you did wrong is, your 110 Ohm resistor is too small, Al gives good values.

Don

[EricVanWyk]

Originally Posted by Don Rotolo

4. LEDs are diodes, too, so running two in opposite polarity, but in series, will mean no current flows.

Just to be clear, I'm asking for anti-parallel diodes, not series.

[Vikesrock]

Originally Posted by Don Rotolo

In addition to what was said before, some other considerations:

1. Be sure to consider the power dropped by the resistor. As an example, because P=I*V, 40 mA at 18 volts = 0.72 watts - use a 1 watt or larger resistor or it'll overheat quickly.

Shouldn't power calculations be done with RMS voltage and current?

P=IV=V^2/R

For 270 ohms:
P = 6.5^2/270 = .156 W

For 390 ohms:
P = 6.5^2/390 = .108 W

1/4 W resistors should be fine as far as I can tell. I am still learning this stuff though (Sophomore EE major) so please feel free to correct anything here not quite right.

[Chief Pride]

wow, i am really grateful to all of your replies... i wasnt expecting to get so many...

here are the exact LED's i am using: http://ledshoppe.com/Product/led/LE1015.htm

Emitted Colour : OCEAN BLUE
Size (mm) : 5mm T1 3/4
Lens Colour : Water Clear
Peak Wave Length (nm) : 465 ~ 470
Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 6000
Life Rating : 100,000 Hours
Viewing Angle : ±10°
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
Lead Soldering Temperature : 240°C (<5Sec)
Operating Temperature Range : -25°C ~ +85°C
Preservative Temperature Range : -30°C ~ +100°C

i used this to figure out what resistor i should use: http://home.cogeco.ca/~rpaisley4/LEDcalc.html

the data i entered into that calculator is as follows:

Supply Voltage: 6.5
Voltage Drop Across LED: 3.2
Desired LED Current: 30

this is where i recieved my 110 ohm resistor... if i understand AL, he is saying that i calculated the voltage as if it were DC, and didnt take into account the range of AC, so the real current is 18.32, and from a couple people i apparently misunderstood my LED data, and the actual drop across the led is 3.7, the actual resistor needed is a 487ohm and 0.4386 watts...

now, i still have a few questions
how do i know a resistors wattage?
and, why wouldent i be able to use LED's in parallel? according to this: http://led.linear1.org/led.wiz i should be able to...

[Al Skierkiewicz]

James,
The value I calaculated was based on 40 ma, just a guess on my part as max current through your LED. The actual recommended is the 30 ma which is the value you used. The closest standard value resistor is either 470 or 560. The 470 should be just fine. Power=I^^2 * R or V^^2/R or V * I. Since you have already calculated the voltage drop for the resistor, 14.62 and the current is 30 ma, then the power in the resistor is 14.62 * .03=0.4386 watts. A half watt resistor should then be fine unless you severely limit the airflow around the resistor. A 1 watt resistor would work fine as well. If you are unable to find 1 watt resistors, you can use two 910 ohm, 1/2 watt resistors in parallel, to form a 1 watt, 455 ohm resistor. Resistor wattage is more often known by it's size or the package markings. Less frequently it might be printed on the resistor body. I believe Radio Shack has a few 1 watt resistor choices, but many half and quarter watt choices.
To not confuse terms, the 18.32 volts is the Peak voltage developed by the transformer which is the highest voltage available for the LED during the peak of the sine wave. Real is a term used in complex AC circuits in which there exists reactance (capacitors and inductors) that affect calculations using vector analysis.

[DonRotolo]

Originally Posted by Vikesrock

Shouldn't power calculations be done with RMS voltage and current?

Yes, that is correct, because the heat generated is not constant but pulsed, and the important factor is the average power not the peak.

RMS is a mathematical construct used with sinusoidal signals to account for the 'real world' effects of such non-constant voltages. Using RMS in this calculation is good engineering practice.

On the other hand, I doubt the meter is really measuring RMS, more likely just 'close enough'. And that's OK.

Also, in the LED specs, it states the max reverse voltage is 5~6 volts? I'd pick the lower (5) and make sure you manage that. Put a regular 1N400x diode in series to half-wave rectify it (turning it into pulsed DC) and don't subject the LED to any reverse voltage.

Don