Photocell Light Switch or Daylight Dusk Till Dawn Sensor Lightswitch, 10a

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WARNING: The circuit is not isolated from the AC mains supply and will be floating at the mains level, which can be fatal for anybody who touches the circuit in powered ON condition, without an insulated enclosure. Circuit Description This last principle is used in the sensitive switch installation you will be presented next. Photocell is placed in a bridge circuit and a comparator is used as detector bridge balanced. The comparator’s output control thyristor through a transistor. In this circuit, protective measures must be taken because it is isolated from the network. Light Switch Circuit Diagram

The gate which is responsible for the sensing can be seen accompanied with the light sensing device LDR wired across its input and the positive via a variable resistor. In other words the circuit can be used like aday activated automatic switch or a darkness activated automatic switch, depending upon the user preference or the specific application need. EDIT - the breadboard circuit you have added looks correct (though it's hard to read..) so go ahead and try it. If it doesn't work let us know. Maybe change the resistor to 2k or larger if you are worried about blowing the LED.Power supply circuit is taken from the bridge rectifier D1…D4 is filtered and stabilized by R1, C1, and D5. The bridge circuit is difficult to identify in the given diagram, but it is made of R2 ….. R4, P1 and photoresist (LDR). IC1 is used in the comparator configuration and power level will be about 1.8 V when potential entry vice versa (negative) exceed that of the entry non vice versa. Resistance R5 conducts a hysteresis of about 1V to prevent thyristor to oscillate at T1 and threshold lighting. Since the pin#3 is fixed at 5 V, means, as long as the pin#2 remains below this reference level, the op amp output remains high, enabling the T1 to remain switched ON, and the SCR/load switched OFF.

Referring to the schematic above, the working of this dual function light activated switch can be understood with the following points:

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R2, R3 being equal in value, the reference voltage is set at the 50% of the zener voltageD5 which is used for stabilizing the rectified 310 VDC to 10 V DC.

This variable resistor is used for setting the triggering point of the gate when the light falling over the LDR reaches the desired specified intensity. To make our circuit operative after dark, a phototransistor is employed, so that when the daylight is void, the LED gets switched ON. You can find lots of applications in which photoelectric detection is employed to turn a circuit on/off. This below shown straightforward circuit is configured like a bistable multivibrator. Each of the gates are formed into inverters by shorting its both the inputs together, so that the input logic level of the gates now get effectively reversed at thie outputs.Once this is implemented, the LDR now gets associated with the positive line, and the R4 end gets connected with the negative line. Some practical advice: if you want to use higher power lamps then diodes D1 …. D4 need to be replaced with other type 1N5404 and thyristor TH1 will be equipped with a heat sink. With these changes, the circuit can control currents up to 3A.

The reason the circuit is usually not the best way to do this is because it relies upon Hfe, which can vary quite widely in a transistor and is subject to temperature changes. This means the base resistor must be chosen according to the particular transistor used. To make the circuit extermely compact one button battery type is preferred here, quite akin to those used in calculators, watches, etc. Its non-inverting input pin#3 is clamped with a fix reference derived from the junction of the resistive divider formed by R2/R3. This is because, during day time the LDR resistancedrops drastically causing the pin#2 potential to drop significantly and below pin#3 potential. This is the circuit you should get. If phototransistor Q2 does not draw current, then Q1's base gets current through R2. The base is at 0.7 V (always 0.7 V higher than the emitter), and the power supply is 3 V, so there's 3 V - 0.7 V = 2.3 V across R2. Then because of Ohm's Law the current through R2 = 2.3 V / 1 kΩ = 2.3 mA. Transistor Q1 will want to increase that 100-fold to get 230 mA collector current. R1 will limit that. If Q1 is on then the LED's cathode will be at around 0 V, and the anode 2 V higher, at 2 V (that's typical for a red LED). So there remains 1 V for resistor R1, and if we want 20 mA through it (and the LED) we apply Ohm's Law again: R = 1 V / 20 mA = 50 Ω. So R1 will make sure that the LED current won't go higher than 20 mA.To give an idea of why the 22k resistor, the BC337 I'm using has a gain of around ~400, and the voltage it will see is 3V - (Vled + collector-emitter drop) -> 3V - (1.8V + 0.7V) = ~0.5V. So 0.5V / 22k = 23uA into the base.