Conservative. Idaho. Software engineer. Historian. Trying to prevent Idiocracy from becoming a documentary.
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"And we know that all things work together for good to them that love God, to them who are the called according to his purpose." -- Rom. 8:28
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.
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.
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.
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.
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.
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.
To limit the current.
ReplyDeleteA 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.
ReplyDeleteIf 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.
ReplyDeleteThe 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.
ReplyDeleteSupplying 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.
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.
ReplyDeleteIf 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.
It limits the amount of current going through the LED which not only sets the brightness but prevents the LED from burning out.
ReplyDeleteShouldn't applying the leads of the LED to a 9VDC battery light it up?
ReplyDeleteOh, 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.
As I'm sure others have already said, you have to limit the current so you don't burn out the LED.
ReplyDeleteTo limit the current, avoiding burning out the LED. 0.7V drop across the LED results in too large of current.
ReplyDeleteIt drops the voltage to a usable range so you don't over-drive the LED.
ReplyDeleteTwo reasons:
ReplyDelete1) 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.
"It may well release the smoke from inside the package"
ReplyDeleteAnd we all know that it's the magic blue smoke that makes electronics go. When you let it out, everything stops.
Jeff: Glad to report that previous experiments with electronics have never released the smoke.
ReplyDelete