Halogen lamps are prone to burn-out owing to their low cold current. The rapid heating inside the lamp melts the thin filament and cuts the lamp life short. The circuit described here enhances the life of the halogen lamp by allowing soft turn-on of the lamp.
When switch S1 is pressed, capacitor C1 charges via resistors R1 and R2, causing MOSFET T1 to conduct. It acts like a normal switch as long as the load (bulb B1) does not draw too much current. Resistor R4 limits the current through the load and also monitors the current flowing through MOSFET T1. The voltage drop across resistor R4 increases as the current through MOSFET T1 increases.
If the voltage at R4 reaches 0.65V, transistor T2 conducts and MOSFET T1 stops conducting. As a result, the lamp stops glowing. Capacitor C1 will hold charge when S1 is switched off. The discharge path for that residual charge is through diode D1 and resistor R5. So the soft turn-on functions correctly any time the bulb is turned on.
Assemble the circuit on a general-purpose PCB and enclose in a small plastic box. Use a heat-sink for mounting the MOSFET in order to dissipate the heat. Current-limiting resistor R4 protects the battery in case the output is short-circuited or the lamp draws more current than specified. Connect the bulb through two external wires having high-current-rating gauge.
I think, before the MOSFET fully turn ON, full load current 2.9A will flow through R5. Will that 1K, 1/4W withstand such heavy current….?