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building a circuit to power an l.e.d.

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36eliot36

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Hi~ Thank u for any assistance...
I'm wanting to build a circuit to power (1) 3.8 volt, 20ma, l.e.d. & I've never built a circuit designed to provide a steady, adjustable power supply. I have immediate access to any specific component needed to build this circuit but I do not know exactly what I need to utilize or in what order the correct components must be organized. I want this circuit to originate from a standard 110volt a/c plug & I don't want the l.e.d. to fluctuate in brightness.
Can someone inform me which components I need & in what order I need to layout this circuit, please?
 
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Grab a USB charger that supplies 5 V at least > 20 mA. That will be nearly all of them. Cut off the end.

You now have a 5V power supply,

The resistor you need is calculated by the formula R < (5-3.8)/20e-3. This implies a resistor that's less than 60 ohms.
You can check the wattage needed by P = I^2*(R) and you yet 0.024 W; I'd use one of these https://www.digikey.com/product-detail/en/ERG-1SJ560/P56W-1BK-ND/35855 The resistor will also act as a fuse.

You can go this route: https://www.digikey.com/product-detail/en/ERG-1SJ560/P56W-1BK-ND/35855

and since the connector is 5.5/2.1 Digikey doesn't have a Jack, but your local radio Shack will.

With one resistor, you can only get the polarity wrong. If you use a 5V supply, if it doesn't work reverse the connections.

Your LED is marked somehow with the polarity. A pic would help.

Your circuit is a +power supply, resistor, Led (observe polarity) and back to - of power supply.

EDIT: Corrected per Reloadron. Not thinking straight today.
 
Last edited:
KISS
Grab a USB charger that supplies 5 V at least < 20 mA. That will be nearly all of them. Cut off the end.

Shouldn't that read > 20 mA? Greater than 20 mA rather than Less Than?

Ron
 
Grab a USB charger that supplies 5 V at least > 20 mA. That will be nearly all of them. Cut off the end.

You now have a 5V power supply,

The resistor you need is calculated by the formula R < (5-3.8)/20e-3. This implies a resistor that's less than 60 ohms.
You can check the wattage needed by P = I^2*(R) and you yet 0.024 W; I'd use one of these https://www.digikey.com/product-detail/en/ERG-1SJ560/P56W-1BK-ND/35855 The resistor will also act as a fuse.

You can go this route: https://www.digikey.com/product-detail/en/ERG-1SJ560/P56W-1BK-ND/35855

and since the connector is 5.5/2.1 Digikey doesn't have a Jack, but your local radio Shack will.

With one resistor, you can only get the polarity wrong. If you use a 5V supply, if it doesn't work reverse the connections.

Your LED is marked somehow with the polarity. A pic would help.

Your circuit is a +power supply, resistor, Led (observe polarity) and back to - of power supply.

EDIT: Corrected per Reloadron. Not thinking straight today.
Grab a USB charger that supplies 5 V at least > 20 mA. That will be nearly all of them. Cut off the end.

You now have a 5V power supply,

The resistor you need is calculated by the formula R < (5-3.8)/20e-3. This implies a resistor that's less than 60 ohms.
You can check the wattage needed by P = I^2*(R) and you yet 0.024 W; I'd use one of these https://www.digikey.com/product-detail/en/ERG-1SJ560/P56W-1BK-ND/35855 The resistor will also act as a fuse.

You can go this route: https://www.digikey.com/product-detail/en/ERG-1SJ560/P56W-1BK-ND/35855

and since the connector is 5.5/2.1 Digikey doesn't have a Jack, but your local radio Shack will.

With one resistor, you can only get the polarity wrong. If you use a 5V supply, if it doesn't work reverse the connections.

Your LED is marked somehow with the polarity. A pic would help.

Your circuit is a +power supply, resistor, Led (observe polarity) and back to - of power supply.

EDIT: Corrected per Reloadron. Not thinking straight
 
I agree, use a usb/phone charger. I keep a few extra's around just for that and similar purpose's.
 
Hi,

Also, check that resistor value. 60 Ohms sounds kinda low.
 
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