Simple fade-in and fade-out switch circuit (1)

The simple dimming/brightening switch circuit is shown in Figure 23. Normally, the switch S is in the open state, the diodes w TI and
VT2 are in the cut-off state, and the inductor E does not light up. When a lamp is needed, the 220V alternating current is rectified by VD1
-VD4 and outputs pulsating direct current. Through the switch
S and the resistor R, the capacitor C is charged, so that the voltage at both ends
gradually rises. Therefore, VT1 can obtain
a basic bias current from small to large through R. VT1 and V i 2
gradually enter the amplified state from the cut-off state, and finally enter the fully
conductive state, so the lamp E gradually dims and becomes brighter,
and finally stabilizes. In the brightest state. When the light is turned off, as long as
S is turned on, the charge stored in C will pass through R. Discharging the emitter junction to VT1 can still maintain conduction of V TI
and VT2. As the discharge continues, the discharge current becomes smaller and smaller, VT1 and VT2 exit saturation, switch to amplification state, and
finally enter cut-off state, so the brightness of UE gradually changes from bright to eye-catching, and finally goes out. The changing times of fading/brightening
are respectively represented by R,, R. Determined by the value of (. VT2 should use high back-voltage and high-power NPN diodes , and there are no special requirements
for other components .