How to switch an 60W bulb using an IC?
Ex:Controlling the bulb using outputs of an IC 7408...
i'm actually using various IC's in my project,
so i want to know how i can do it...
Inputs of IC depends on other parameters...
Don't use an IC, use a relay, if you want to use an IC use a solid state relay, without input requirements your post can't be answered properly we can at best guess.
IC input = 6V ; I'm using 3-4 IC combination...
Outputs of 1st is input to 2nd IC and third ic Output
is the one which gives ON/OFF state...
using this output(3rd IC), i want to switch 60W Bulb
why on earth do you use light bulbs to be switched? multiple white LEDs would've been easier to control, and they don't need relays. Just a suggestion, since you can get your hands on an IC.
why on earth do you use light bulbs to be switched? multiple white LEDs would've been easier to control, and they don't need relays. Just a suggestion, since you can get your hands on an IC.
the 'flyback' (I prefer freewheeling) diode prevents reverse voltage surge from damaging the system, and also to remedy the reluctance of the coil due to residual magnetism.
You NEED it.
LEDs are much more efficient than light bulbs, obviously. You can light a room using those LEDs and a 9V battery (or a low-voltage power supply).
Why not look at the datasheet of a 7408? its output high voltage is only 2.4v to 3.4v, not 5V.
you have the transistoir as an emitter-follower so its output voltage to the relay coil is only 1.7v to 2.7V.
The transistor will be destroyed by the flyback high voltage spike produced by the coil withoput a protection diode across the coil (not across the power supply).
Maybe a demonstration of the flyback high voltage spike is needed. Hold one hand on both coil terminals of a relay coil then apply power. You will feel nothing. Then disconnect the power to the relay coil and you will jump with a high voltage shock to your hand. The driver transistor will be destroyed by the high voltage spike without the protection diode.