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Infrared LED

cheap-circuitsCheap Circuits wrote 09/17/2020 at 07:00 • 3 min read • Like

Many years ago when I was designing an infrared camera slave pointing system the cost of an infrared receiver IC (integrated circuit) was about $7. This IC was TTL (Transistor Transistor Logic) with three pins. Power, Output and Ground. Then the price increased to about $8 which when I actually purchased this IC in late 2002. I also obtained a copy of the datasheet from the electronics store. However, the printing was very low quality and this is why I could not read the pins properly. I must have made a mistake with the pins and partially burned this IC. The IC was receiving the infra red signal well. However, it had a weak output drive because I have partially burned it.

Nowadays the cost is only about $3. This make this ICs appropriate for cheap circuits.

Here is the video of the circuit working:

First I read the following articles about old fashioned transistor amplifier circuits that can be used to amplify signals from infrared sensors:

https://www.instructables.com/id/Transistor-Sensor-Amplifier/

https://www.instructables.com/id/Recycled-Transistor-Amplifier/

I designed the circuit with the ZD1953 IC and a simple PNP BJT (Bipolar Junction Transistor) inverter.

Step 1: Make the Circuit

Two transistors were placed in parallel to improve reliability and increase device life time. However, using a second transistor is not necessary. The circuit can still work well with one transistor. You also might not need a second 1 kohm resistor either. The two 1 kohm resistors are connected in series to reduce the maximum transistor base current.

From top to bottom the pin layout is:

1. Output (connected to 1 kohm resistor)

2. Ground (blue wire)

3. 5 V Power Supply for TTL (red wire)

The PNP transistor collector is connected to four LEDs in parallel with the blue 100 ohm resistor. The voltage potential is:

VoMax = Vled + Vr

= 2 V + 100 ohms * 25 mA = 4.5 V

There are four LEDs. Each LED will receiver about 6.25 mA of current. Those LEDs are made to consume 5 mA.

Suppose we raise the power supply to 6 V:

VoMax = Vled + Vr

= 2 V + 100 ohms * 40 mA = 6 V

Each LED will receive about 10 mA of current.

I placed the circuit in a plastic container because I did not want to pay for the box.



Step 2: Testing

I used my infrared transmitter to keep the LEDs ON long enough so that I could take a photo:

You can see my transmitter sending the signal to the infrared sensor:

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