Hello guys,
How can I calculate the Resistor value of an IR Emitter, based on datasheet parameters?
Thanks a lot,
Typically a data sheet will tell you the maximum and minimum voltage a part can handle or requires.
You can use that information and the voltage for the circuit you are building to determine an appropriate resistor value.
It can be more complicated.
Perhaps with additional information such as the part in question, the circuit in question, the voltage source, etc. a better answer could be constructed.
Ohm's Law: R = (VSUPPLY - VLED)/ILED
Subtract the IR LED forward voltage (Vf) from the power supply voltage. That gives the voltage that the resistor must drop (Vr). Divide the difference (Vr) by the required LED current (insure that the current is less than the max continuous current from the data sheet). The result is the resistance required. Pick the closest standard value.
That is for a basic IR LED. Like bigred1212 says, your particular LED could be different.
navab1362:
Hello guys,How can I calculate the Resistor value of an IR Emitter, based on datasheet parameters?
Thanks a lot,
It depends on how you plan to use the emitter. If it's part of an IR remote control, then it's pulsed on and off at typically 40 kHz at a 50/50 duty cycle, and the total time to send an IR code sequence is rarely longer than 40 milliseconds (which is actually 20 due to the 50/50 duty cycle).
For example, for my IR remote projects, I use Vishay TSAL6400 high power IR LED's (940 nm). The specs say the max forward current is 100 mA, but for pulsed operation it can handle 1.5 AMPERES!
So, if you want to use the LED as an IR remote emitter and drive it with a BJT (which in this example has a Vce(sat) of 1.2 volts), you would do this calculation:
Forward voltage at 1.0 amp: 2.2 volts
LED pulse current: 1000 mA (1 amp) (being conservative here - not going the full 1.5).
Transistor Vce drop: 1.2 volts
Power supply: 5.0 volts
R=E/I
R=(5-(1.2+2.2))/1
R=1.6 ohms
Closest normal R value: 1.5 ohms
LED current with 1.5 ohms instead of 1.6:
I=E/R
I=1.6 / 1.5
I = 1.07 amperes (OK, close enough)
See what I did? I added up the total voltage drop (BJT Vce(sat) and LED Vf) to get 3.4 volts. Since I am using a 5 volt supply, I have to "resistor away" 5-3.4=1.6 volts. R=E/I, R=1.6/1, R=1.6 ohms.
Now, this is ONLY for a high power pulsed IR led use. Steady state the LED will fry in less than a second. You have to then use the steady state current value from the spec sheet (100 mA in this example).
So, R=3.4/0.1, R=34 ohms, closest standard value is 33 (OK).
Hope this helps.
groundFungus:
Subtract the IR LED forward voltage (Vf) from the power supply voltage. That gives the voltage that the resistor must drop (Vr). Divide the difference (Vr) by the required LED current (insure that the current is less than the max continuous current from the data sheet). The result is the resistance required. Pick the closest standard value.That is for a basic IR LED. Like bigred1212 says, your particular LED could be different.
Thanks a lot for useful information.
krupski:
It depends on how you plan to use the emitter. If it's part of an IR remote control, then it's pulsed on and off at typically 40 kHz at a 50/50 duty cycle, and the total time to send an IR code sequence is rarely longer than 40 milliseconds (which is actually 20 due to the 50/50 duty cycle).For example, for my IR remote projects, I use Vishay TSAL6400 high power IR LED's (940 nm). The specs say the max forward current is 100 mA, but for pulsed operation it can handle 1.5 AMPERES!
So, if you want to use the LED as an IR remote emitter and drive it with a BJT (which in this example has a Vce(sat) of 1.2 volts), you would do this calculation:
Forward voltage at 1.0 amp: 2.2 volts
LED pulse current: 1000 mA (1 amp) (being conservative here - not going the full 1.5).
Transistor Vce drop: 1.2 volts
Power supply: 5.0 voltsR=E/I
R=(5-(1.2+2.2))/1
R=1.6 ohms
Closest normal R value: 1.5 ohms
LED current with 1.5 ohms instead of 1.6:
I=E/R
I=1.6 / 1.5
I = 1.07 amperes (OK, close enough)See what I did? I added up the total voltage drop (BJT Vce(sat) and LED Vf) to get 3.4 volts. Since I am using a 5 volt supply, I have to "resistor away" 5-3.4=1.6 volts. R=E/I, R=1.6/1, R=1.6 ohms.
Now, this is ONLY for a high power pulsed IR led use. Steady state the LED will fry in less than a second. You have to then use the steady state current value from the spec sheet (100 mA in this example).
So, R=3.4/0.1, R=34 ohms, closest standard value is 33 (OK).
Hope this helps.
Thanks for complete info