## Thursday, October 25, 2018

### Will Someone Remind Me Why the Resistor is Needed Between LED and DC Power?

Is it to reduce power loss?  Shouldn't applying the leads of the LED to a 9VDC battery light it up?

Steve said...

To limit the current.

Richard said...

A resistor in series will act to limit the current through the circuit to no more than V/R. In a circuit with a LED, it acts to prevent the LED from being overloaded or burnt out by excessive current.

Unknown said...

If the LED requires less voltage that what is being applied from the power source then adding a resistor in series is an inexpensive way to adjust this. There will be a voltage drop proportional to the resistance of the two components.

Jim Horn said...

The resistor is to limit the LED current. LEDs are somewhat constant voltage devices, in that a small increase of voltage can cause a large increase of current. To prevent overcurrent failure of the LED, a resistor (or semiconductor acting as a variable resistor) is used.

Supplying 9V to most LEDs will burn them out immediately.

A typical small LED usually will run at 20mA or less. Using Ohm's Law (Voltage = Current times Resistance), we can say that a 9V battery (typically about 7.6V) driving a red LED (about 1.7V) at 20mA needs (7.6 - 1.7)/0.020 or 295 ohms of resistance. That is a very non critical value.

B said...

Once you reach the threshold voltage (more or less 3v) the current rises rapidly (It is all but a short). The resistor is there to keep the current (and thus heat) down to level that will not destroy the LED crystal.

If you use 3 3v leds in series on a 9v Battery, they will last for a while. But even there, you need to use a resistor if the battery will put out more current than the LED's can handle.

Hope this helps. Email me if you need more info or a better explanation.

Papa Bear said...

It limits the amount of current going through the LED which not only sets the brightness but prevents the LED from burning out.

PhaseMargin said...

Shouldn't applying the leads of the LED to a 9VDC battery light it up?

Oh, yeah, it will really light it up. It may well release the smoke from inside the package, which is what you really try to avoid.

A little seriously, it's to limit the current through the diode. To make it easy, think of the diode as something with zero resistance once you hit the on-voltage. If you exceed the current through the diode the (small) resistance inside the diode will cause the diode to overheat and burn out. So you generally choose the resistance to be R>(Vbattery-Von)/Ilimit, and larger than that is only good because LEDs aren't precision components. I generally don't do less than 2x that resistance.

KCSteve said...

As I'm sure others have already said, you have to limit the current so you don't burn out the LED.

Left Coast Conservative said...

To limit the current, avoiding burning out the LED. 0.7V drop across the LED results in too large of current.

rick said...

It drops the voltage to a usable range so you don't over-drive the LED.

StormCchaser said...

Two reasons:

1) To keep the LED from burning out. The LED current will rise *very* rapidly as voltage goes up - exponentially in fact, so you really don't dare leave out the resistor.

2) To minimize the current used by the LED if you care about battery life - trading off better battery life for a somewhat dimmer LED. If that's an issue.

jdege said...

"It may well release the smoke from inside the package"

And we all know that it's the magic blue smoke that makes electronics go. When you let it out, everything stops.

Clayton Cramer said...

Jeff: Glad to report that previous experiments with electronics have never released the smoke.