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# Begginer question

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#### rom7

##### New Member
Hi first off please be easy ive just started learning:

Why is this have a series circuit powered by 9v cell connected to 1x 150 ohm resistor and 3 leds in that order.

I want the forward voltages to be around 2 v across each led(safe Vf limit) but atm they are 6v,4v,2v.The currents(20ma) fine its just the Vf. The point is how do you design a series cricuit like this and not exceed the Vf safe limit for the first two leds?

You use Ohms law. First off, you look at the spec sheet for your LEDs. They will have a typical forward voltage drop @ a specified current (VFtyp). For a garden variety RED LED, that might be ~1.8V@20mA. It will be different for different color LEDs, and different type of LEDs

Three LEDs in series will drop three times as much as one would. So in this example, the forward drop of three RED LEDs would be ~5.4V. If you are starting with a 9V battery, you must have a series resistor between the battery and the LEDS to drop 9V to 5.4V, meaning that the drop across the resistor is 3.6V. (This is Kirchoff's voltage rule)

Also from the LED spec sheet, it specifies that the LEDs are designed to operate at some specified current, typically about 20mA. Note that this is a current that makes them reach their specified brightness; you can usually operated them at a much lower current and they will still be bright enough, thereby prolonging the life of the battery. Note that the forward drop of an LED doesn't change much if you reduce the current.

Finally, you now know that you need a resistor which drops 3.6V @ 20mA (or what ever current you want to operate the LED at). R=E/I = 3.6/0.02 = 180Ω.

One more thing: The power rating of the resistor needs to be checked. P=IE= 3.6*0.02 = 0.072W = 72mW, which means a 1/4W resistor is just fine...

Forgot to add: If you measure the drop across one LED, you should see about 1.8V, across two the drop would be 3.6V and across all three, 5.4V. Another example of Kirchoff's rule.

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you can usually operated them at a much lower current and they will still be bright enough, thereby prolonging the life of the battery

Whoa, the resistor will still diapate the same current...

An LED sets its own voltage. The voltage is not exact, it is a range from 1.7V to 2.4V for a red LED. One might be 1.7V, the next one might be 2.0V and a third LED might be 2.4V. If all three are 1.7V and you have a 150 ohm current-limiting resistor and a new 9V battery then their current is 26mA. If all three LEDs are 2.4V then their current will be only 12mA and maybe the 150 ohm current-limiting resistor should be reduced to 75 ohms.

The safe limit that you need to design for is its safe current. the resistor limits the current, not the voltage.

Whoa, the resistor will still diapate the same current...

First, resistors dissipate power; not current. Second, If you would like to operate the LEDs at 5mA instead of 20mA, then the resistor value would be R=E/I= (9-5.4)/0.005 = 720Ω, and the resistor would disspate P=i*E = 3.6*0.005 = 18mW.

First, resistors dissipate power; not current. Second, If you would like to operate the LEDs at 5mA instead of 20mA, then the resistor value would be R=E/I= (9-5.4)/0.005 = 720Ω, and the resistor would disspate P=i*E = 3.6*0.005 = 18mW.

You will dissapate the same power running your led's at 5mA instead of 20mA because you need a different resistor, so your drain on the battery will be the same...

You will dissapate the same power running your led's at 5mA instead of 20mA because you need a different resistor, so your drain on the battery will be the same...
WRONG!
5ma is not the same as 20mA.

Yes I was mistaken
For 12v @ 20mA: the resistor dissipate 224 mW
the diodes dissipate 40 mW
total power dissipated by the array is 264 mW
the array draws current of 20 mA from the source.

And 12v @ 5mA: the resistor dissipate 55 mW
the diodes dissipate 10 mW
total power dissipated by the array is 65 mW
the array draws current of 5 mA from the source.

Just had to clear that up...

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