Help with timer circuit and relay

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What is the exact part number of the 555? I'm trying to determine if it is a CMOS version or if it is the standard version. The output might need a transistor driver to work with the relay. You will need to place a diode across the relay to protect the preceding circuitry from the inductive 'kick' when the relay is switched off.

What voltage are you applying the timer / relay circuit and what is the resistance of the relay coil?

The coil current of that relay seems to fall well within the output specs of the 555 so it shouldn't be overloading the chip. Have you fitted a diode (1N4148 signal diode) across the relay coil (connected backwards) to protect the 555 from the voltage spike created as the coil is switched off?

It is possible that your circuit is re-triggering itself as the coil is switched off so it's a good idea to also connect another diode (1N4001 rectifier diode)in series with the coil.

You could also drive the relay via a transistor circuit.

It was the two diodes that it needed. Thanks

Note that there is a right way and a wrong way of connecting these two diodes. One diode must be connected directly across the coil, in reverse (so it's off while the relay is activated), and should be able to take the full coil current. It must not have the other diode in the loop.

The other diode (if you need it - probably not) goes to the 555.

The reason for this is that the coil, like any inductance, resists any change in current flow. When it's on, it has current flowing down into the 555. When the 555 turns off, that current *doesn't* stop straight away. Instead, the voltage at the bottom end of the relay coil rises high enough that the current can keep flowing. That is, it rises *above* the supply voltage, and often high enough to push current through the 555 output, even though it's *off*. This continues until the energy that was stored as a magnetic field is dissipated, by which time the output transistor of your 555 may well be toast. If not, well, the high voltage injected into its output terminal has found another way home, possibly disrupting the timing function of the circuit - as you found.

So now you see why the 2nd diode probably isn't needed. When the

If you only use the blocking (2nd) diode, and not the 1st one, or if you put the 2nd one inside the loop, you provide no current path and the voltage will rise high enough to destroy the diode.

Clifford Heath.