Voltage Drop Question - Electronic Circuits Projects Diagrams Free

shaneshane1 · 27th February 2008, 02:22 AM

I have a simple LED voltage drop question

This is a really stupid question that i cant seem to get my head around

here goes

I have a LED running of 12V

so 12V - 2.1V(voltage drop) = 9.9V

then 9.9V / 0.020(Amps) = 495ohms

so my question is do i NEED to include the voltage drop in the math?

when i put a multimeter on the anode(+) of the LED it shows 2.1V

and when i do the math without the voltage drop included

so 12V / 0.020 = 600ohms

I still get 2.1V on the anode of the LED?

is the current to the LED decreasing if i dont add the voltage to the math?

and also if im reading 2.1V on the LED anode does that mean there is a voltage loss of 9.9V through the resistor?

k7elp60 · 27th February 2008, 03:17 AM

Your calculation of 495 ohms is the correct method. The voltage accross the LED varies a little with different current. I usually go to the next standard value of resistor in this type of calculation so a 510 ohm would be fine. You should calculate the wattage thou. (I^2)R or .02x.02x510=.204 watts.
a 1/4w resistor is fine.

audioguru · 27th February 2008, 03:29 AM

The 600 ohm resistor does not have 12V across it. It has 12V - 2.1V= 9.9V across it. So the current is 9.9V/600 ohms= 16.5mA.

shaneshane1 · 27th February 2008, 06:20 AM

Quote:

Originally Posted by audioguru

The 600 ohm resistor does not have 12V across it. It has 12V - 2.1V= 9.9V across it. So the current is 9.9V/600 ohms= 16.5mA.

Ok thanks!!!

what gives the LED its "votage drop"?

and the voltage on the battery(+) still remains at 12V or close to anyway?

is the led only "using" 2.1V from the 12V supply?

mneary · 27th February 2008, 08:23 AM

The LED "voltage drop" is determined by the physics of the semiconductor junction. Within the range of current you're providing, it doesn't vary much.

The voltage on the battery(+) shouldn't be influenced much by small variations in current.

The LED and resistor "use" power, which is not exactly the same as voltage. Power is current times voltage, so the LED uses 2.1V times 0.02A, and the resistor uses 9.9V times 0.02A.

Hero999 · 29th February 2008, 05:53 PM

The LED forward can vary quite a bit so go for the lowest voltage drop and highest power supply voltage when calculating the current and resistor power dissipation and the highest voltage drop when taking the minimum power supply voltage into account.

For example, powering an LED with a nominal voltage of 3.5V (this can vary from between 3.3V and 3.8V) from three 1.5V cells is a bad idea as the battery voltage will quickly drop to 3V causing the LED to turn off.

The correct way of doing it is to use four 1.5V cells to give 6V and calculate the resistor value assuming the LED's voltage drop will be 3.3V.

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27th February 2008, 02:22 AM	(permalink)
shaneshane1	Voltage Drop Question I have a simple LED voltage drop question This is a really stupid question that i cant seem to get my head around here goes I have a LED running of 12V so 12V - 2.1V(voltage drop) = 9.9V then 9.9V / 0.020(Amps) = 495ohms so my question is do i NEED to include the voltage drop in the math? when i put a multimeter on the anode(+) of the LED it shows 2.1V and when i do the math without the voltage drop included so 12V / 0.020 = 600ohms I still get 2.1V on the anode of the LED? is the current to the LED decreasing if i dont add the voltage to the math? and also if im reading 2.1V on the LED anode does that mean there is a voltage loss of 9.9V through the resistor?

27th February 2008, 03:17 AM	(permalink)
k7elp60	Your calculation of 495 ohms is the correct method. The voltage accross the LED varies a little with different current. I usually go to the next standard value of resistor in this type of calculation so a 510 ohm would be fine. You should calculate the wattage thou. (I^2)R or .02x.02x510=.204 watts. a 1/4w resistor is fine. __________________ The great thing about electronics is unlimited ways to do the job. The only limit is one\'s imagination. I generally think my way is best. Show me a different way. I have an open mind.

27th February 2008, 03:29 AM	(permalink)
audioguru	The 600 ohm resistor does not have 12V across it. It has 12V - 2.1V= 9.9V across it. So the current is 9.9V/600 ohms= 16.5mA. __________________ Uncle $crooge

27th February 2008, 08:23 AM	(permalink)
mneary	The LED "voltage drop" is determined by the physics of the semiconductor junction. Within the range of current you're providing, it doesn't vary much. The voltage on the battery(+) shouldn't be influenced much by small variations in current. The LED and resistor "use" power, which is not exactly the same as voltage. Power is current times voltage, so the LED uses 2.1V times 0.02A, and the resistor uses 9.9V times 0.02A.

29th February 2008, 05:53 PM	(permalink)
Hero999	The LED forward can vary quite a bit so go for the lowest voltage drop and highest power supply voltage when calculating the current and resistor power dissipation and the highest voltage drop when taking the minimum power supply voltage into account. For example, powering an LED with a nominal voltage of 3.5V (this can vary from between 3.3V and 3.8V) from three 1.5V cells is a bad idea as the battery voltage will quickly drop to 3V causing the LED to turn off. The correct way of doing it is to use four 1.5V cells to give 6V and calculate the resistor value assuming the LED's voltage drop will be 3.3V. __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez