Need Resistor value

[jack0987]

Hi:

I have two identical LED's.

One is an indicator LED for the 5VDC outout. I am using a 470 ohm resistor for it.

Now, my problem is, what resistor value should I use on the other indicator LED for the 24VDC output to make them about the same brightness?

Thanks.

 

[DemonicSandwich]

Do you know their forward voltage or current? If not, what color are they?

Assuming it's a Red LED, it will have a Vf of about 1.8-2.0v then a 470Ω is giving it 3.2v / 470Ω = 6.8mA.

For the same brightness then 24v - 1.8v = 22.2, 22.2 / 0.0068A = ~3264Ω.

I think the nearest standard value is 3,300Ω.

 

[jack0987]

Thanks Demonic:

Yes, They are red. There are a lot of numbers on the package I do not understand.
The part no. is B4304H1. It also says 2Vf @ 20mA.

I made a quick test and I think the 33K may be a little low. The LED seems a bit brighter.

 

[DemonicSandwich]

No, 3,300, as in 3.3K.
You say the package says Vf = 2v at 20mA.

Without changing the first LED, then you can assume it's current is 3v / 470Ω = 6.38mA.
Then the second LED would be 22v / 0.00638A = ~3448Ω. So not much change.
LEDs can vary in brightness greatly when driven with low current.

You could try a larger resistor since reducing the current won't hurt it any.

 

[jack0987]

Sorry. I meant to say I tried a 3.3K resistor.

 

[jack0987]

Just one more quick question.
Does it matter if the resistor comes before or after the LED?
In designs I've seen it is usally after.

Thanks, again. I guess I'm in the ballpark and the value will do for now.
Thanks again.

 

[DemonicSandwich]

As long as it is in series with the LED, it won't matter on what side it's on.

 

[jack0987]

Great. Thanks. That's it for now.

I'd like to do a LED for the 115VAC input as well at some time. I am using a neon indicator now. If you would not mind, please point me in the right direction when you have a chance.

 

[#12]

8.2k, 3 W

Power lead, fuse, a diode passing positive to an 8.2k resistor 3 watts, then the LED, then ground.
also...on the positive end of the first diode, a diode passing negative to ground.
The second diode is necessary because the tiny capacitance of the first diode will allow some of the wrong polarity voltage to get to the LED. Use diodes of sufficient voltage rating. 200V is the absolute minimum and 1000 volt diodes only cost 1 cent more.

 

[KeepItSimpleStupid]

Male the resistor a metal-oxide type in this appliaction.

 

[DemonicSandwich]

Using only a simple resistor and diode won't work well as the resistor will generate quite a lot of heat.

lets say for example, you're using 120VAC mains, that same 2v Red LED, a silicon rectifier diode, and a resistor.
And wanting to drive it with ~10mA.
Then 120V - 2v(Led) - 0.6v(Diode) = 117.4v across the resistor.
117.4v / 0.010A = 11,740Ω.
117.4v * 0.010A = 1.174 Watts.
The resistor will generate more than a watt of heat....just to power a small indicator LED.

A better solution is to use a capacitor's reactance to limit the current.
Some topics like this or this could provide some insight.

In my opinion, if the neon indicator is still fully functional, just keep it as it is since it only requires a single resistor to function properly and unlike the LED, it can work on AC and DC current.

Neon lamps even come in different colors like LEDs do...just not as many.

 

[jack0987]

How about something like this?

**broken link removed**

Will look at your links as well. thanks.

 

[audioguru]

Now you show an LED that works from half-wave rectified 50Hz or 60hz mains. It will be seen as flickering all the time.
If you use two back-to-back LEDs or a full-wave rectifier then the flickering will be a double the frequency that almost cannot be seen to flicker.

 

[jack0987]

Hmmm. I was thinking about the flicker.

How about this?

**broken link removed**

 

[#12]

Where's the guy who always chimes in with, "The capacitor method is dangerous"?
(There's always more than one way to skin a cat.)
and there is no need for a resistor with the capacitor method. The impedance of the capacitor is what limits the current.

 

[Diver300]

You do need a resistor. Well, it's a good idea to have a resistor, because of the turn-on surge as the capacitors charge when you turn on. Unless you are lucky and turn on when the mains voltage is near zero, there is a huge current. A 470 R resistor, 0.5 W, will limit that to 1/4 A or so.

A resistor of 100 R or so in series with the LED will reduce the flicker even more, as the capacitor with DC on it will then charge to more than the LED forward voltage, and slowly discharge down towards the forward voltage. The LED will be on all the time that the discharge is happening, and the next mains cycle is only 8 ms later, so the LED stays on, even if the current isn't constant.

A 1 Mohm resistor in parallel with the 0.47 uF capacitor will stop it staying charged and zapping you when you have disconnected it.

 

[jack0987]

Everyones response has been great. Thanks.

Not sure how to connect the parallel 1M resistor. please advise.

 

[Diver300]

The 1 Mohm resistor should be directly in parallel with the 0.47 uF capacitor. It will only generate a tiny amount of heat, but it will discharge the capacitor from 150 V (the peak of a 110 V sinewave) to 10 V in just over a second.

 

[jack0987]

If I have understood correctly, something like this:

**broken link removed**


Not sure I have all the values and the plus and minus signs going the right way.

 

[audioguru]

If I have understood correctly, something like this:
Not sure I have all the values and the plus and minus signs going the right way.

Yes the 1M resistor is in the wrong spot and the polarity of the bridge rectifier is backwards.

Also the 100 ohm resistor is wrongly in parallel with the LED instead of in series.