LED questions

[jeeep]

Three questions:
-If I calculate that I will need a 120 ohm resistor will it matter if it is 1/4 or 1/8 watts?

-If I have 5 LED's in series, each consuming 20mA's, is the total consumption of mA's 100mA's or 20mA's?

-How do you determine the amount of mA's a LED will use in one hour?

Many thanks!

[Bigglez]

Greetings (No Name Supplied),

If I calculate that I will need a 120 ohm resistor will it matter if it is 1/4 or 1/8 watts?

Depends upon how much voltage is present across the resistor.
P = V^2/R, (P*R) = V^2, sq-rt(P*R) = V
For 1/8W 120 ohms, V = 7.5V
For 1/4W 120 ohms, V = 15V

If I have 5 LED's in series, each consuming 20mA's, is the total consumption of mA's 100mA's or 20mA's?

Answer: 20mA. Kirchoff's law; current in a series circuit

How do you determine the amount of mA's a LED will use in one hour?

Energy is measured in joules,not milliamps. Are you asking for the energy consummed in one hour? The current should be the same at the end of one hour compared with the current at the beginning of that hour.

Comments Welcome!

[jeeep]

Depends upon how much voltage is present across the resistor.
P = V^2/R, (P*R) = V^2, sq-rt(P*R) = V
For 1/8W 120 ohms, V = 7.5V
For 1/4W 120 ohms, V = 15V

12Vdc supply voltage, 2Vdc led forward voltage, 20mA of forward current, and 5 leds. My apologies, but what does P stand for? Watts? And what is V telling me here?

Answer: 20mA. Kirchoff's law; current in a series circuit

Thank you!

Energy is measured in joules,not milliamps. Are you asking for the energy consummed in one hour? The current should be the same at the end of one hour compared with the current at the beginning of that hour.

Yes I want to determine what will be the best battery to use,i.e. amp/hours. So if I had a 2 amp/hour battery and my leds are only pulling .02mA that would be 100 hours of use? If this is correct how do I proceed to figure out how much power my leds are using?

Thanks!

[Bigglez]

Greetings (No Name Supplied),

12Vdc supply voltage, 2Vdc led forward voltage, 20mA of forward current, and 5 leds. My apologies, but what does P stand for? Watts? And what is V telling me here?

Five LEDs at 2V each in series is 10V. You'll only
have 2V across the resistor. If R = 120 ohms then
I = 2/120 = 16mA, not 20mA.

P = Power (watts). V in this case is the largest voltage
across a 120 ohm resistor to dissipate 1/8W (7.5V),
or dissipate 1/4W (15V), per your OP.

Comments Welcome!

[Bigglez]

Greetings (No Name Supplied),

So if I had a 2 amp/hour battery and my leds are only pulling .02mA that would be 100 hours of use?

In theory, but in practice the battery will self-disharge
and shorten the calculated run time.

If this is correct how do I proceed to figure out how much power my leds are using?

Electrical power is measured in watts, Electrical energy
is measured in Watt-hours.

Energy is measured in joules (as noted before).
Joules = watt-seconds.
1 watt-hour = 3600 J
1 kilowatt-hour = 3.6 ×106 J (or 3.6 MJ)
1 joule = 1 newton-metre = 1 watt-second

Electrical power is voltage * current. For your
LED this is 2V * 20mA = 40mW

For Five LEDs this is 200mW

For the circuit running from 12V at 20mA = 240mW
(40mW loss in the resistor).

After one hour the circuit will use 0.24 * 3600 = 864J

854/3600 = 0.24 Watt-hours

Comments Welcome!

[bodgy]

Just in case you're wondering Jeep, the 3600 that Bigglez mentions is 3600 seconds which of course equals 1 hour.

A 100 ohm resistor will give you your 20mA, but I'd suggest staying with the 120 ohm and see if the LED's are bright enough for your purpose.

If your resistor has been manufactured with a 5% tolerance then it could be any value between 114 - 126 ohms, if 1% which tend to be the same price these days (in Europe anyhow) then the value will vary by +/- approx 1 ohm


Colin

[jeeep]

Ok so just to be sure.
LED Module 1 pulls 42.2mA's
LED Module 2 pulls 60mA's

There are 6 Module 1's; 42.2mA * 6 = 253.2mA
There are 6 Module 2's; 60mA * 6 = 360mA

Grand Total mA's being pulled: 613.2mA's

I have a rechargeable battery that is 12Vdc 2000mAH.
So: 2000mAH / 613.2mA = 3.26 Hours of run time. Correct?

[bodgy]

The first part of your statement is correct, though R2 should be 200 ohms.

The second part depends how the other modules are connected.

Your sum is correct if you are connecting the modules in parallel with each other. In other words just extend your schematic to the right by adding on copies of what you already have.

Your complete unit is going to get warm with the total current specified.

In real life you'll probably get close to 3 hours from the battery, because as the battery discharges it's internal resistance will get higher and the voltage in the battery will drop. This will depend a little on the type of rechargeable battery you intend to use - make sure you don't drop the voltage below it's cut off point else your battery will never recharge properly again.

Colin

[jeeep]

Thank you!