Introduction: Infrared Lamp
This project shows an infrared lamp that turns ON for half a minute after it receives a signal from a TV infrared remote control. You can see the circuit working in the video.
I designed a circuit with BJT transistors after reading this article:
I modified the circuit to drive higher current loads and keep the light ON for a small duration of time.
The IR (infrared) receiver has a maximum range of about 20 meters. However, this range could be a lot smaller outside due to inference from sunlight. I have not tested this IC in the 40-degree summer heat.
However, this circuit can be designed with just one MOSFET:
Yet MOSFETs cost a lot more money. A reliable power MOSFET could be as high as $3 USA. It is best to order a few MOSFETs because it could be very frustrating if you burn one of them and have to wait for weeks until another one comes.
Those links show Instructable articles about the infrared sensor made from transistors:
Components: NPN general-purpose transistors - 5, PNP general-purpose transistors - 5, power transistors - 4, 1 kohm resistor - 1, 100 kohm resistor - 1, 1 Megohm resistor - 1, 100 ohm high power resistors - 10, diodes - 5, 470 uF capacitors - 10, matrix board - 2, Heat sinks TO220 or TO3 - 2, solder, 6 V light bulb or 6 V LED bulb.
Optional components: encasement/box.
Tools: soldering iron.
Optional tools: multimeter, USB oscilloscope.
Step 1: Design the Circuit
I designed the 5 V power supply for the IR receiver TTL voltage. However, nowadays most IR receivers can work at voltages from about 2.5 V to about 9 V or even 20 V. You need to check the specifications/datasheets. This is why my TTL power supply circuit is optional. You should be able to connect IR receiver power supply directly to Cs2 capacitor or make another RC power supply low pass filter circuit by cascading/connecting the Cs1 capacitor and Rs1 resistor to Cs2.
The circuit that I designed is not the most optimum solution because some the transistors are not saturating. I had to use what I had in stock thus applying the voltage following configuration to the Q2 transistor.
You can click on the last two links on the previous page of this article and see for yourself:
Calculate the discharge time constant:
Tdc = (Rb1||Rdc) * Cdc = 470 uF = 156.666666667 seconds
It takes 5-time constants for the capacitor to discharge. However, after about a quarter of a time constant the light bulb should turn OFF. Higher transistor current gains will keep the light ON longer. You can increase the discharge time by connecting another 470 uF capacitor in parallel with Cdc.
Step 2: Simulations
Simulations show that:
1. The IR receiver TTL voltage is about 5 V.
2. The capacitor is discharging slowly.
3. The 6 V light bulb will receive the 300 mA current that it needs to turn ON to full brightness. The light bulb turns OFF after 90 seconds, not 30 seconds shown in the video. This is because of the discrepancy between simulation models and practical transistor current gains.
Step 3: Make the Circuit
I added extra 470 uF capacitors for better power supply noise filtering (this is why I did noted ten 470 uF capacitors in the component list).
I used five normal transistors in parallel and a power transistor to drive the light bulb. If you are using a 6 V LED light bulb then you need to consider the polarity of this component because the LED conducts only in one direction. LED light bulb consumes a lot less current than the traditional incandescent bulb. However, there are bright LED bulbs that consume more current.
You can see the matrix board with the light bulb attached. This matrix board is the 5 V TTL power supply. I used two 100-ohm resistors in parallel then give 50 ohms to reduce the power dissipation for each resistor and ensure that the TTL power supply voltage does not drop too much because of the high power supply resistor values.
Step 4: Encasement and Testing
I used the tomato plastic container to save money from purchasing a box.
Participated in the