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wall wart and leds

I am trying hookup a 10mm superbright led to a wall wart that is rated output 9v, 300ma. I used a 470 ohm resistor for this connection but .the wire feel a little warm to me. When working with wall warts how does one compute the resistors need for the wall wart strength. Do we go solely by the output info or what? thanks in advance for your help. BTW if I add more leds later is there a simple design for a brightness adjustment...i'm trying to make a sub stage microscope light source. While one bulb works well on the lower power I might need to add more in order to view the higher magnifications and I'd like to be able to adjust the led brightness.

Reply to
medusa
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You compute the resistor needed just like you do any other time. It seems that you did that. With wall warts the specified outputs may only be true when powering the intended device. I would suggest that you measure the output voltage in actual operation with your circuit. You didn't say if it was an AC or DC wall wart. If its an AC wall wart this could be why your wires feel warm. An LED is not a bulb. Tom

Reply to
Tom Biasi

The wall wart says input...120vac 60hz 6 w, output 9vdc 300ma. I used the 470ohms not because I computed it but elsewhere it said for a 12 volt power source a 470 resistor would be adequate.I got the impression there would be no sensation of heat anywhere.

Reply to
medusa

The circuit below regulates the current delivered to the LED, regardless of the voltage provided by your 9V wall wart. Connecting R between the output pin and the adjust pin sets the current that will be available to the load, which is connected to the adjust pin along with the resistor.

+----in[LM317]out---+ | adj | ------- | | [R] | Wall +|---+ | | | 9V | +---------+ | Wart -|---+ | ------- | [LED] | | +---------+

Compute R as follows: R = 1.25/Iled You are looking for Iled ~20 mA with your 470 ohm resistor in your original circuit, and that 20 mA computes to 62.5 ohms in this circuit (1.25/.02). A standard 62 ohm resistor would be fine. You could also go a little lower if you want, say a 56 or a 51 ohm resistor. For example, a standard 51 ohm resistor would yield ~24.5 mA You can add a 1 or 2 watt pot in series with the 51 ohm resistor to adjust the brightness. You can add an LED or two in series with the LED in the diagram.

This is not the only way, but it is simple, requiring only the LM317 IC and the resistor for the basic circuit.

Ed

Reply to
ehsjr

^ 1/2 was what I meant to say (not that 2 is wrong)

Reply to
ehsjr

A 9V 300 mA DC wallwart will likely produce 12 volts when loaded with a light load.

What wires got hot?

If the LED is a white one, then it probably has a voltage drop of

3.2-3.5 volts at a usual amount of current. With 12 volts supply, that leaves a little less than or around 9 volts across the 470ohm resistor. That means current a little less than 20 milliamps flowing through the resistor and the LED.

This means about 70 milliwatts being dissipated in the LED and a little less than .18 watt being dissipated in the resistor. The resistor will probably feel a little warm.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

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