 # Question about LEDs and resistors 1/4, 1/5 ...

hello guys,
my question here is that by calculation we find that a 5mm LED with 0.025 or 0.03 Ah current consumption should have a 1/4 resistor at 5V right ?

P = VI
with V = 5V and I = 0.025 so P = 0.125W < 0.25W

but if V = 9V and I = 0.03 so P = 0.27W > 0.25W the resistor will get hot

Sure that more voltage increases the power, and you need to adjust the resistor accordingly, but don't forget to subtract the Voltage drop across the LED ( around 2V for a red LED, but have a look at the datasheet) .

P(resistor) = (V-Vled) x I

Edit : anyway, be cautious, if you find 0.21W , do not use a 1/4W resistor, it will get too hot and won't last very long . And.... 0.03A is a lot for a standard LED

alnath: P(resistor) = (V-Vled) x I

i didn't really get this one :(

shouldn't the P resistor = the Voltage that is coming to the resistor directly from the source of 5V and the drop will happen for the LED ?

shouldn't the P resistor = the Voltage that is coming to the resistor directly from the source of 5V and the drop will happen for the LED ?

No. A resistor defines the amount of current that will flow with a given voltage ACROSS it. If you connect it to 5V it can not have more than 5V across it and only then if it is connected between 5V and ground. You have an LED that "takes" some of that voltage so what is left 5V - LED voltage is what is across the resistor.

Grumpy_Mike: No. A resistor defines the amount of current that will flow with a given voltage ACROSS it. If you connect it to 5V it can not have more than 5V across it and only then if it is connected between 5V and ground. You have an LED that "takes" some of that voltage so what is left 5V - LED voltage is what is across the resistor.

but we put the resistor to limit the current and make a voltage drop for the LED not to burn by exceeding it's max forward voltage so why do i have to subtract the LED voltage that's what i don't understand :(

the resistor will get the 5V again after the cycle is done

You're not really using the right equation. The LED will have some voltage across it, Vf, when the 30mA of current is flowing. so (5V - Vf) = voltage across the resistor, Vr. Now knowing Vr, you can calculate R for the current using Ohms law: V = I*R Rearranging, V/I = R So (5V - Vf)/I = R If the LED has 3V across when current flows and you want 30mA: (5V - 3V)/.03A = 67 ohms. Use a 68 ohm standard resistor. Now, P=IV can come into play. You have I, 30mA and you know Vr, so you can calculate the power dissipated in the resistor. Taking Vr=IR, and subbing that in for V above, P = I*IR, so .03A *.03A * 68 ohm = 62mW, so a 1/4W rated resistor would be fine.

Now do the same for your actual LED's Vf, and double check the 30mA - that sounds like the absolute max for a typical LED, with 20mA typical for continuous on current.

ok thanks so let's say i have a 5V battery, so 5V = (V1 + V2 + V3 + ... ) with Vi the voltage of component in the circuit ?

so why do i have to subtract the LED voltage that's what i don't understand

Because the voltage across the LED is not across the resistor. So it has to be subtracted from the 5V to see what is across the resistor.

so let's say i have a 5V battery, so 5V = (V1 + V2 + V3 + ... ) with Vi the voltage of component in the circuit ?

No idea what that means, can you try again please?

Grumpy_Mike: No idea what that means, can you try again please?

i mean the source voltage is the sum of voltages of every component in the circuit ?

i mean the source voltage is the sum of voltages of every component in the circuit ?

Yes, so the total voltage is 5V = V across the resistor + V across the LED. Rearrange the equation ( swap the side swap the sign ) and you get:- V across the resistor = 5V - V across the LED.

ok thanks a lot for the help, i must correct this on my note book :relaxed:

the resistor will get the 5V again after the cycle is done

I think this is a big flaw in your thinking. There is NO cycle, it is all a steady state. Only when current flows through components do you get voltage drops.

Grumpy_Mike: I think this is a big flaw in your thinking. There is NO cycle, it is all a steady state. Only when current flows through components do you get voltage drops.

by cycle i mean the flow, i can't fix my thinking about flow making it look like a flow of water ! :disappointed_relieved:

See the attached image.

Sticking with the water analogy. The LED is a one way check valve. It prevents current going the wrong direction. If you hook an LED up backwards to a battery, it won’t work just like a check value only works in one direction.

Current is like the water flowing through the pipe. The amount of water flowing at one end is the same as through the other end.

Resistors are narrower pipes. The narrower the pipe, the higher the restriction on water flow.

VB - Battery Voltage - 5V
VF - LED Forward Voltage - 3V
VR - Resistor Voltage - 5V - 3V = 2V

CrossRoads provided a pretty good description of the maths.

You’re not really using the right equation.
The LED will have some voltage across it, Vf, when the 30mA of current is flowing.
so (5V - Vf) = voltage across the resistor, Vr.
Now knowing Vr, you can calculate R for the current using Ohms law: V = IR
Rearranging, V/I = R
So (5V - Vf)/I = R
If the LED has 3V across when current flows and you want 30mA:
(5V - 3V)/.03A = 67 ohms. Use a 68 ohm standard resistor.
Now, P=IV can come into play.
You have I, 30mA and you know Vr, so you can calculate the power dissipated in the resistor.
Taking Vr=IR, and subbing that in for V above,
P = I
IR, so .03A *.03A * 68 ohm = 62mW, so a 1/4W rated resistor would be fine.

These calculations are for continuous operation, not pulsed. For pulsed usage, the amount of current can be higher but until you are used to the maths, then it is best to use these values.

There are various pdf’s and web sites that offer an introduction to electrics and electronics.

Hope this helps Hi,

File this image away for reference.
Everybody copy it, use for future reference on the forum. (Just like a certain servo diagram.)

Tom… I love Express and Paint.

TomGeorge:
File this image away for reference.
Everybody copy it, use for future reference on the forum. (Just like a certain servo diagram.)

Done! Not sure I like the direction of the “I” (current) arrow!

Hi, sorry electron flow.

Electrons are repelled away from the NEG terminal and attracted to the POS terminal.

Tom… OK, we are treading on dangerous ground here! :astonished: :astonished:

Paul__B: OK, we are treading on dangerous ground here! :astonished: :astonished:

I don't think so, because it matters not what direction you think of the current flowing just as long as you are consistant. If you think it does matter then it shows up some wrong thinking. Hence the typical beginners mistake thinking it matters which way round the resistor and LED go, resistor to +ve or -ve.

TomGeorge: Hi,

File this image away for reference. Everybody copy it, use for future reference on the forum.

I would like to see an other arrow head on the voltage indicators to show the voltage is measured across those points.