Introduction: Infrared Transmitter
This article shows you how to make an infra red analogue transmitter.
This is an old circuit. Nowadays laser diodes are used to transmit digital signals via optic fibres.
This circuit can be used to to transmit audio signal via infrared. You will need a receiver to detect the transmitted signal. The signal does not need to be modulated.
Components: NPN BJT power transistor, heat sink, insulated wires, matrix board, 1 kohm resistor - 5, 100 ohm resistor - 3 (depending on the amount transmitters that you use), 100 uF bipolar capacitor, 1 Megohm potentiometer - 2, power source (3 V or 4.5 V - can be implemented with AA/AAA/C/D batteries).
Tools: wire stripper, pliers.
Optional components: solder, 1 mm metal wire, heat transfer paste.
Optional tools: soldering iron, USB oscilloscope.
Step 1: Design the Circuit
Do not increase Rb1 above 1 kohm. Otherwise the transistor will not saturate.
I modeled the infrared transmitter with four diodes. If each diodes has a potential voltage of 0.7 V than the total series voltage will be 2.8 V or about 3 V. This was the voltage drop across my infrared transmitter.
The Ra resistor can be any value from 1 kohm to 1 Megohm.
I found that adding the Rc value to the transistor circuit increased the gain of this amplifier. When the input voltage is very low the transistor is OFF, low biasing current is entering the transistor base with Vce (collector emitter voltage near zero). The Rc resistor increases the transistor Vce voltage when the transistor is OFF. You can try Rc value of 10 kohms or even 100 kohms and see if this will increase the gain because low Rc value (even 1 kohm) creates a loading affect on the transistor output. However, connecting high Rc resistor values is like not using the Rc resistor at all.
However, contrary adding Rc resistor to general purpose transistor LED detectors only reduces the gain and thus was NOT used in those articles:
It is best to assume that each transistor type has its own unique characteristics.
Step 2: Simulations
PSpice simulations show a very high gain and this is why I connected the attenuation potentiometer to input.
High potentiometer values are influencing the high pass filter frequency. However, do not use potentiometers below 1 kohms. Infact better you use at least 10 kohms to avoid possible damage to audio output.
Step 3: Build the Circuit
I used high power resistors. You do not need high power resistors for this circuit. Probably Rd1 and Rd2 need be high power if you raise the supply voltage and use high current infrared diodes.
I specified a 3 V power supply in the circuit design because some infra-red diodes have a maximum forward biasing voltage of only 2 V. That means the maximum diode current will be:
IcMax = (Vs - Vd - VceSat) / Rc
= (3 V - 2 V - 0.25 V) / 100 ohms
= 0.75 V / 100 ohms = 7.5 mA
However, the diodes that I used have a maximum forward biasing voltage of 3 V. This is why I used a 4.5 V supply (not 3 V) and maximum diode current in my circuit current was:
IcMax = (Vs - Vd - VceSat) / Rc
= (4.5 V - 3 V - 0.25 V) / 100 ohms
= 1.25 V / 100 ohms = 12.5 mA
Step 4: Testing
I introduced the potentiometer attenuation because the transistor amplifier had a very high gain, thus saturating the output that is not appropriate for audio signals that require linear amplification and transmission.
I connected the purple channel to one of the infrared transmitter nodes (the second node is connected to power supply).
My signal generator has a maximum output of 15 V peak or 30 V peak to peak. However, for the graphs above I set the signal generator to minimum settings. My USB oscilloscope is showing the wrong scale for the light blue channel. The input signal amplitude was set about 100 mV peak.
My circuit was not tested with infrared receiver. You can make this yourself.