How Do I Calculate the Resistor Value for an Infrared Transmitter Circuit?

In summary: Basically, you want to calculate the resistance (R) to limit the current on the diode to 1.2A (Ipk). This can be done using the formula R = (Vcc - V_led - Vmosfet) / Ipk, where Vcc is the supply voltage, V_led is the voltage drop across the LED, and Vmosfet is the voltage drop across the MOSFET. Once you have calculated the resistance, you can pulse the signal normally. However, it's always a good idea to double check your calculations and consult a reliable source for confirmation.
  • #1
N468989
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Hi all,

i need some help on this circuit that I am building. i want to get the maximum current for my infrared emitting diode (Kingbright L-53F3C)
datasheet->http://www.soselectronic.com/a_info/resource/pdf/king/L-53F3C.pdf"

the peak forward current permitted is 1.2A (condition: 1/10 DutyCycle, and 0,1ms pulse width. but i don't understand how to apply this..

(the idea is more current equals more distance...)

i have attached my IR transmitter block diagram...

i am also using a N-MOSFET BS170 and the maximum current allowed is 500mA continuous and 1200mA pulsed (not really sure what that means either).

datasheet->http://w3.id.tue.nl/fileadmin/id/objects/E-Atelier/doc/Datasheets/Transistoren/BS170.pdf"

so basically i would help on how do obtain the value of the Resistor R for my signal do be sent more efficiently for my specified frequency (90kHz) and dutycycle (5%) (without burning the IR LED).

any help is much appreciated thanks
 

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  • #2
Wouldn't you find the resistance that gives 1.2A current continious, then make sure that your pulses are 10 mircosec or less with a 10% maximum duty cycle?
 
  • #3
i have calculated the mean current for the IR LED for my signal

DATA:
frequency = 90kHz
Period = 11.11us
Ton = 0.55us
Toff = 10.056us
Dutycycle = 5%

Ipk = 1.2A (peak current)

Im= Ipk * 0.05 <=> I am = 60mA (mean current)

now i must calculate the R to limit the current on the diode

R = (Vcc - V_led - Vmosfet ) / I am <=> R = (5-1.2-2.1) / 60mA = 28.3 Ohm

R gives me 28.3 Ohm, but I am not very sure about this result i might be missing something...

then there is the power dissipation which is 100mW

if someone could help me out with these calculations, i have a book on this but its no very helpfull.
thanks




dlgoff : do mean like this -> R = (Vcc - V_led - Vmosfet ) / Ipk , then i just pulse the signal normally?
 
Last edited:
  • #4
"dlgoff : do mean like this -> R = (Vcc - V_led - Vmosfet ) / Ipk , then i just pulse the signal normally?"

Well that's how I would do it, but I'm no expert here.
 

Related to How Do I Calculate the Resistor Value for an Infrared Transmitter Circuit?

1. What is an infrared transmitter circuit?

An infrared transmitter circuit is an electronic circuit designed to emit infrared light signals for communication purposes. It typically consists of a microcontroller, an infrared LED, and supporting components such as resistors and capacitors.

2. How does an infrared transmitter circuit work?

An infrared transmitter circuit works by converting electrical signals from the microcontroller into infrared light signals using an infrared LED. These signals are then transmitted through the air and can be received by an infrared receiver on the receiving end.

3. What are the main components of an infrared transmitter circuit?

The main components of an infrared transmitter circuit include a microcontroller, an infrared LED, resistors, capacitors, and a power source. Some circuits may also include additional components such as transistors for amplification.

4. What are the applications of an infrared transmitter circuit?

Infrared transmitter circuits are commonly used in remote controls for TVs, DVD players, and other electronic devices. They are also used in wireless data transmission, security systems, and proximity sensors.

5. Are there any safety precautions to consider when working with an infrared transmitter circuit?

Yes, it is important to handle the infrared LED with care as it emits high-intensity light that can be harmful to the eyes. It is also recommended to use a current-limiting resistor to prevent overheating and damage to the LED. Additionally, be sure to follow proper circuit design and precautions to avoid any potential electrical hazards.

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