Abel Raynus, Armatron International Inc, Malden MA
There are many approaches to the problem of activating a light when it becomes dark, and a recent Design Idea covers this topic (Reference 1). Some approaches require a dc power supply and an electromechanical relay, but a better approach involves feeding the device directly from the ac line, minimizing the number of components (Figure 1).
 |
|
| Figure 1. | The photoresistor activates the TRIAC and the load when darkness falls. |
The heart of the device is a light-sensitive cadmium-sulphide resistor, PR, with a resistance of approximately 200 kΩ in the dark and decreasing to a few kilohms in the light. PR and capacitor C1 form an ac-voltage divider. In daylight, the voltage across PR is too low to generate the required gate-trigger current to turn on bidirectional ac switch Q1, thus keeping the load – usually a lamp – off. When it becomes dark, PR’s resistance rises, resulting in an increase in the TRIAC’s gate current that triggers the TRIAC and lights the lamp.
The circuit uses inexpensive, off-the-shelf components, including the VT90N1 photoresistor; a 0.1-μF, 275V capacitor; and an L2004F61 TRIAC with a load current of 4A rms, a peak blocking voltage of 200V, and a gate-trigger current of 5 mA. The exact specifications of these components are not critical; you could use others instead.
Editor’s note:
Attributes worth mentioning include the fact that the capacitor introduces a phase shift, which places the peak of the gate voltage close to the zero crossing of the load’s sine wave for optimum turn-on timing. Another benefit is thermal hysteresis, which occurs due to the reduction of the required triggering voltage and current as the TRIAC warms up after the initial turn-on.
Reference
- Tran, Chau, “Simple night-light uses a photoresistor to detect dusk,” EDN, Dec 15, 2011, pg 49.
