Hi. I would like to be able to control mains powered (240v) bulbs using a 5v current. I believe it is possible to do this using relays but am not quite sure which ones. It sounds like a fairly straight forward problem but don;t know where to start. Thanks.
Danny.
Check out this thread and some of the articles linked in it. I think a lot of your questions will be answered. There are even some relays suggestions to get you started.
Thanks. The specs suggested that it would indeed be possible but I am treading lightly so as not to ruin any of my electronics.
Would I need to add a diode in order to stop the "Backlash" which I heave heard about?
I have an arduino microcontroller connected to a transistor to provide the 5V signal which I don't want to fry.
Thank you.
Danny.
Thanks. The specs suggested that it would indeed be possible but I am treading lightly so as not to ruin any of my electronics.
Would I need to add a diode in order to stop the "Backlash" which I heave heard about?
I have an arduino microcontroller connected to a transistor to provide the 5V signal which I don't want to fry.
Thank you.
Danny.
Yes, you'll want to put a diode across the coil of the relay. The "backlash" you are talking about is often called "inductive kickback", but more correctly, it's back-EMF (electromotive force... aka voltage). Current in an inductor (coil) cannot change instantaneously, so when you disconnect power from a coil the magnetic field will begin to collapse, trying to keep the current flowing at the same rate. This creates a large neagative voltage potential, which can be tens of volts, which could be enough to cause damage to your transistor & microcontroller.
Look up at one of my older posts. I linked a thread that is doing about the same thing you are. In there is a link to a SparkFun article, also doing the same thing you are. It even has a nice little schematic.
No, the polarity doens't depend on the circuit. The voltage across an inductor comes from this fundamental differential equation: V = L * di/dt. If your time-rate of change of current through the inductor is negative (going from a finite DC current, to zero), then V is negative (di/dt is negative; times L, its still a negative value). The voltage across the terminals of the coil will most certainly be negative, however, the absolute value of that voltage, relative to the circuit ground, is dependant on the configuration of your circuit.
No, the polarity doens't depend on the circuit. ..... the absolute value of that voltage, relative to the circuit groundsic, is dependant on the configuration of your circuit.
If one end of the coil is connected to a positive supply rail, the other end to a switch and the other side of the switch to 0V; when the switch opens the voltage across the coil will reverse, the negative being connected to the positive rail and the positive to the now open switch. Without a diode, that voltage will be considerably more positive than the positive supply rail.
To call that negative is indicative of muddled thinking. Very muddled indeed.
Methinks I have discovered yet another individual who likes to argue for argument's sake.