Dark-activated switch needs only three components
Build a dark-activated switch with only a triac, a photocell, and a resistor.
Assume that you have a device that receives its power from the main 120 or 220V-ac line and you need to add a switch between the ac line and the device so that the device works only when it is dark. Although you may think this task would be trivial, it is difficult to find a workable approach because most of the published schematics need 6 to 12V-dc power supplies and relays. Several off-the-shelf dark-activated switches, such as devices from Suns International, are available, but they’re expensive for a consumer product. After looking at products from dozens of Web sites, you may decide to make your own. The solution is simple and inexpensive.
The circuit in Figure 1 employs an internally triggered triac, which Teccor Electronics originally developed. The primary purpose of any triac is bidirectional-ac switching. The Quadrac triac has a built-in triggering device with the threshold-voltage level of approximately 40V. To achieve this level, the circuit uses a voltage divider comprising a photocell and resistor R1. When you light the photocell, its voltage drop is lower than the triggering level of the threshold voltage, and Q1 is locked, so the load disconnects from the ac line. When it becomes dark, the peak voltage amplitude on the photocell increases to 40V, opening Q1 and making the load connect to the power line.
The choice of Q1 depends on the load current and ac-line voltage. This circuit uses the Q4004LT from Littelfuse with a maximum current of 4A rms and a voltage of 400V. You can use any photocell, but this circuit uses an off-the-shelf model and accordingly uses a value of 47 kΩ for R1 to achieve reliable switching. For an inductive load, add a 100Ω resistor in series with a 0.1-µF capacitor between pins 1 and 2 of Q1.
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