In series with the LED. The value would be whatever it takes to obtain the required voltage and current across the LED. As an example, if you had an LED that required 25ma at 2v, and you wanted to use a 9v battery, you would need a resistance of 280 ohms. (This is (9-2) / 0.025, a simple application of Ohm's law.)
A: Add the proper resistor in series with the LED. What resistor? Simple 10volts source minus the LED source divided by the 20ma current that should be flowing gives you the resistor 10-1.8=8.2/.02=900 ohms. The 1.8 can be anything it is up to the LED voltage drop. LONG LEAD IS POSITIVE SHORT IS NEGATIVE SINCE IT IS A DIODE
current limiter.
The component used to protect a LED from burning up is called a resistor.
In order to determine what size of resistor is required to operate an LED from a 9V battery, first start by knowing the current and voltage required for the LED. That information is available in the LED's specifications. For discussion purposes, lets assume a typical LED at 2.5V and 50mW. The translates to a forward current of 20mA. Build a simple series circuit containing a 9V battery, a resistor of an as yet unknown value, and the LED. By Kirchoff's current law, the current in the LED is the same as the current in the resistor, which is also the same as the current in the battery. This is 20ma. By Kirchoff's voltage law, the voltage across the LED plus the voltage across the resistor equals the voltage across the battery. This is 6.5V. (9 - 2.5) By Ohm's law, resistance is voltage divided by current, so the resistor is 6.5 / 0.02, or 325 Ohms. The nearest standard value to that is 330 Ohms. Cross check the power through the resistor. Power is voltage times current, or 6.5V times 0.02A, or 0.13W. A half watt resistor is more than adequate for this job.
You will have to check the datasheets for both the CMOS gate and the LED, then if the CMOS gate's rated output current is enough to light the LED you will need to do a little arithmetic using Ohm's law to calculate the resistor to put in series with the LED to limit current and avoid damaging the parts.
34Kohms
esistors restrict the flow of electric current, for example a resistor is placed in series with a light-emitting diode (LED) to limit the current passing through the LED.
It doesn't mater.
A: It can be two things one a resistor to limit the current to a safe level and two is a hi resistor placed across the LEDS if placed in series with hi voltage source. That will equalize the leakage current so the string will not fail because of it.
This varies depending on the color. Try driving them with a current source instead. About 5mA to 50mA should work on any LED. Using a voltage source will blow some colors, and not light other colors at all. Of course a current source is likely to be more expensive than a series resistor.
A: That resistor is there to limit the current to the LED it can be any value if the voltage is decreased or increased or no resistor if the voltage across the led is equal to the forward voltage drop.
It does not matter. Kirchoff's Current Law states that the signed sum of the currents entering a node is zero. A consequence of that law is that the current in every part of a series circuit is the same. The only thing that resistor location affects is the potential voltage of the LED terminals with respect to the rest of the circuit. Certainly, if you are driving the LED with high voltage, such as 120VAC, you should consider the resistor location so as to reduce electrocution hazard but, the LED's performance is not affected by resistor location in the circuit.
I bought: reed switch LED Buzzer Button (On/Off) Resistor (if necessary) I connected: <---Button------Reed switch------Resistor (if necessary)------LED------Buzzer---Batteries--->
Typically, a 100 ohm resistor is used to connect a 1.5 volt led to a series 220v ac adapter. Many LEDs can be connected into a string using the resistors.
A: Add the proper resistor in series with the LED. What resistor? Simple 10volts source minus the LED source divided by the 20ma current that should be flowing gives you the resistor 10-1.8=8.2/.02=900 ohms. The 1.8 can be anything it is up to the LED voltage drop. LONG LEAD IS POSITIVE SHORT IS NEGATIVE SINCE IT IS A DIODE
A correctly rated resistor will prevent that the LED is not burnt out if it receives too much power.
Has nothing to do with the intensity of the LED, and all to do with the voltage/amperage of thediode, and the voltage of the system it is supposed to be used with.