# Electric Calculations

Discussion in 'General Electronics Discussion' started by Mr. Oddly Fox, Jul 13, 2015.

1. ### Mr. Oddly Fox

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Jul 10, 2015
I have the formulas handy, but I can not quite get the numbers I require.
(it's been a while since I've actively used all of the formulas)
I have a 5V/2.5A AC adapter. I need to step it down to the area of 3.7V.
I expect that each of the 0.5W LEDs on the circuit to take around 120mA of current.
How many ohms should the resistor be? How much current will I be reduced to if I use that resistor?

2. ### Martaine2005

3,331
910
May 12, 2015
Hi mr oddly,
It could crudely be done by putting two diodes in series. .6v Vf on each.

Gryd3 likes this.
3. ### Gryd3

4,098
875
Jun 25, 2014
It's most certainly AC input, but you need to ensure you have the proper output type as well.
Why do you need the 3.7V ? Is this for the LEDs or for a microcontroller?

You can use a voltage regulator to give you a nice constant 3.7V, otherwise if it's for the LEDs you just need to do a little math.

Additionally... the color of the LED makes a difference (Or better yet, the rated 'forward voltage' on the spec sheet for the LED)
This value will be anywhere from 1.8 to 3.3v usually.

So, time for the math!
Voltage = Source voltage - (LED Forward Voltage * Number of LEDs in series)
Resistance = Voltage / Current

5,165
1,087
Dec 18, 2013
Hello
Please supply the details of the PSU.
Thanks

5. ### Mr. Oddly Fox

19
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Jul 10, 2015
The forward voltage of the LED type is 3.6V. I was aiming for that area of 3.2-3.7V.

The AC adapter specs were listed.
5V/2.5A DC. I'm not sure if that's too much for a 3.6V device.

5,165
1,087
Dec 18, 2013
Ah the reason I asked is because you said AC which makes a difference. Now we know it's DC then that's easier.

7. ### davennModerator

13,802
1,941
Sep 5, 2009
hi there
welcome to EP

have a read of our how to drive LEDs resource, it should answer all your questions ......

Dave

8. ### BobK

7,682
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Jan 5, 2010
We still need the current of the LED at 3.6V in order to calculate the resistor.

Bob

9. ### Martaine2005

3,331
910
May 12, 2015
Agreed, and the amount of leds too.

Martin

5,165
1,087
Dec 18, 2013
You have a couple of options.

1) Use a single resistor for each LED. You need to check what the forward voltage is at 120 mA if it's on the datasheet. So we will say it's 3.6 Volts for the minute. So (5 Volts - 3.6 Volts) / 120 mA = 11.6R (Use 12R) wattage is I^2*R = (0.12^2)*12 = 172.8 mW so go for 1 Watt resistors.

2) Design a simple constant current circuit for each LED. This simplifies the calculations because you don't need to worry about the forward drop of the LED because you have set the current to 120 mA.

Thanks

Martaine2005 likes this.
11. ### AnalogKid

2,482
715
Jun 10, 2015
LED's are current-mode devices. Putting two diodes in series with a 5 V output will not provide any current limiting, and could damage the LEDs. The best way to size a series resistor is to determine what current you want through the LED, look up the corresponding forward voltage on the datasheet, subtract that from 5 V, and kick in Ohm's Law. LED forward voltage is not a rigidly controlled parameter, so the actual forward voltage could be off by 10%.

For most high-brightness LEDs, the brightness is specified as a function of current, not voltage. For the most consistently repeatable brightness, drive the LED with a constant current source designed for the forward current on the data sheet (or less) rather than a resistor.

ak

12. ### Martaine2005

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May 12, 2015
totally agree with you. I wasn't commenting on leds, just a crude way to drop .6v 0r .7v . From 5v to 3.7v. No leds were harmed in this conversation.

Martin

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13. ### Mr. Oddly Fox

19
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Jul 10, 2015
Thank you all.