How do we use an LED as a photodiode? Light detector mounting with LM741 / TL071

Reading time: 2 minute

How does the assembly work?

The photovoltaic effect of a red LED junction is about 2V. It is best to use an operational amplifier TL071 because it has inputs with JFET transistors, so a high input impedance.

The resistor above 10 MOhm ensures the proper polarization of the inverter input (0V) in the dark. Its value can be increased even more, taking into account that The AO has an offset voltage of 10 - 15mV, and an LED offers as photodiode approx. 50mV. 3 resistors of 10 MOhm can be mounted in series, parallel on the diode.

The output of the operational amplifier offers zero volts (0V) with the LED in the dark. Adjust the semi-dimmer so that the corresponding gain and voltage level desired at the output are obtained when the LED is exposed to light.

List of required components

R1 = 3 x 10 Mohm resistors mounted in series;
R2 = 470k;
R3 = 3.3k;
R5 = semi-adjustable 25k;
1 x red LED;
1 x operational amplifier TL071 (or LM741);
2 x terminal contacts for +/- 5V differential supply and for output;
1 x wiring (or bradboard) and connecting wires.

How do we build a 5V dual (differential) power supply?

Many electronics projects require dual power supplies. The electronic diagram below shows a design / diagram of a 5V differential power supply using 7805 and 7905 voltage stabilizer integrated circuits. The LM7805 is an IC with a positive voltage regulator, while the LM7905 is an IC with a negative voltage regulator.

Both electrical circuits are equipped with many built-in features, such as overheating protection, short circuit protection, etc.

How does the electronic circuit work?

The circuit uses a descending transformer 230V AC to 9V. Four 1N4001 diodes are used to rectify AC to DC voltage. It uses a 2200uF / 25V electrolytic capacitor, a diode voltage filter, and other capacitors used for disconnection.

The output of this dual power supply circuit is up to 1.5A. Therefore, you can use this power supply for any project that requires a current below 1.5A.

List of components required for the differential supply part

1 x low voltage transformer (230V-9V / 2A);
D1-D4 = 1N4001 (or 1N4007);
C1, C2, C7, C8 = 2200 uF;
C3, C4 = 10 nF;
C5, C6 = 10 uF;
1 x LM7805 (with radiator);
1 x LM7905 (with radiator);
2 x terminal contacts for supply and output;
1 x wiring (or bradboard) and connecting wires.

Bibliography:

Notebook Circuit - Silicon Chip
http://www.circuitdiagram.org/
Conex Club Magazine - no.7-8 - 2006
https://makezine.com/

 1,897 total hits, 7 hits today

6 comments

  1. I wouldn't complicate myself like that. You can use a normal source and simple mounting with transistors, and the LED that is used as a light sensor connected in the blocking direction (ie with the anode to ground).
    Yellow and green LEDs are more sensitive to light. In principle, you can also take a power transistor and decapsulate it, silicon is photosensitive. In LEDs we usually speak of galium-arsenide.

  2. I played with this effect in the 1970s after my father told me about this. Later I experimented with green, yellow, and amber LEDs. Red by far was best.

  3. The input offset voltage can be positive or negative; it's simply an error term whose magnitude is important because it introduces an offset multiplied by the gain to the output.
    There are op amps with very low input offsets that would be better than the TL071.

Add a comment

Your email address will not be published.

The maximum upload file size: 2 MB. you can upload: image, audio, video, document, spreadsheet, Interactive, text, archive, queues, other. Links to YouTube, Facebook, Twitter and other services inserted in the comment text will be automatically embedded. Drop file here