Newbie Missing Something

[AirforceMook]

Hello,

I'm trying to understand the concept of amperage. I'm trying to design a simple circuit. [see attachment]

I'm using the applet here: http://www.falstad.com/circuit/
I'm using this calculator here: http://led.linear1.org/led.wiz

My LEDs are 3.5v / 20mA each. I'm trying to connect three in series as displayed in the attachment. Power supply is 12V, 1A max.

What's confusing me is that the calculator suggests I use an 82ohm resistor, but the applet is telling me that I will then get over 32mA through those three LEDs, which I thought was more than they're rated (20mA, right?).

I have to use a 140ohm resistor in the application to make it come out about right (~19.8mA).

If I try to do the calculations myself, I think I need a 75ohm resistor.
(12v - 3.5*3 = 1.5... 1.5 / 0.2 = 75ohms)

I'm definitely missing something here. Any help is appreciated!

-Jason

 

[(*steve*)]

You calculate 75 ohms.

Resistors can not be purchased in just any resistance. The manufacturers make them in a series of "preferred values" that have a mathematical relationship which depends on their tolerance (essentially accuracy).

The nearest preferred value resistor to 75 ohms in is 82 ohms. (It's not quite as simple as this, but that's the general idea).

The applet may be using a model that has a lower Vf than your LEDs.

Try measuring the current you get. As long as you're close to 20mA, you'll be fine. Even 30mA isn't going to kill anything immediately.

See https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html

 

[AirforceMook]

"Even 30mA isn't going to kill anything immediately."

The problem is that this would be used in a lighting application. Wouldn't the LEDs burn out fairly quickly at more than 50% additional amperage?

Thanks for the reply!

 

[(*steve*)]

Yes, they might last only 20,000 hours instead of 50,000 hours.

But that is plenty of time to connect your multimeter and check. If the current is too high, insert a larger resistor. I suspect your calculations are going to be pretty good though.