Electronic Circuits and Projects Forum

You can get a lot more power out of that power supply by using the LEDs in series, but you would have to use different resistors, so that might be out.

Basically the 1.1A is the maximum amps the power supply is rated for. Going through the math I estimate each LED is pulling 16.4ma, which is a good number. Every LED/resistor you add is another 16.4. If you have 67 such pairs you would be pulling 1.1 amp, that is the max you could have. You really need to use a bit less, but I suspect this isn't going to be a problem.

Originally Posted by alockrem

Most power supplies I found in my junk drawer are 5v, but 1 of them is has an output of 9v DC, which I believe is what I'm looking for. What I don't understand is the amps, which it says are 1.1 amps. What does that mean to me? Since it's a 9v power supply will it use the same resistors as a 9v battery or am I missing a part of this equation?

If your 9V is typical it means that when suppling a current of 1.1 amps the voltage will be 9V, and if the current supplied(also called a load) is less than the 1.1 amps the voltage may be higher than the 9 volts.

A typical LED as seen in the attachment has a maximum current of 20mA or 0.02 amps. So 1 amp of current would power 50 of the LED's

It has been my expericence that even thou the LED is rated at a maximum current they are clearly visible a less current. I usually operate mine around 10mA or 0.01 amps.

If you are going to use the 9V battery for power you can connect LED in series, but still use a resistor to limit the current of the string. The color of the LED will determine the voltage of the LED. White for example are about 3V. To calculate the resistor needed add up the LED voltages, subtract it from the supply voltage, then divide the difference voltage by the current.

Example: 1 white LED, 9V battery LED current 0.01
9V-3V=6V 6V/0.01= 300 ohms. Since 300 ohms is not a standard value, use either a 330Ω or a 270 resistor.

You didnot say what color the LED's are for sure. I assumed you are using white ones.
The 330Ω resistor may be to small of a value if you use the 9V power supply as the voltage may be some what higher than the indicated 9V.

I would try putting 2 LED's in series with the resistor and see how bright the LEDs on the 9V power supply. See attachment.

I have a bunch of 300Ω, they are one of the standard sizes. My local source (Tanner's Electronics in Carrolton TX) has them for 3¢ each.

I'll attach a list of 5% ¼W resistors that can be had.

If you like what you already have I'd stick with it.

It is nice having a local outlet with resistors, especially at reasonable costs. I would be interested in any mail order equivalents.

I made a project of kitting them a while back.

Hi alockrem,

you might reconsider the arrangement (and use) of current limiting resistors and the number of LEDs connected.

Please refer to the attached schematic and the calculations.

You might connect all LEDs with an individual current limiting resistor, dissipating lots of power in the resistors.

Connecting two LEDs in series does not only save 15 resistors (for 30 LEDs), but also reduces the total power loss by 2.7W (20mA LED current) respectively 1.35W (10mA LED current).

It doesn't matter much if you happen to have an own nuke power plant in the back garden, but even using that will save a lot of energy, thus taking care of less global warming.

Boncuk