Watchdog for asterisk: implementation (Part IV)

Part IV: Revising the circuit to use a transistor to drive the relay

Well, it turns out that I needed to revise the circuit since I'll be using an unregulated power supply---a simple class 2 transformer (like the ones used for charging cell phones or powering computer peripherals). I have one with adjustable voltage (3V, 4.5V, 6V, 7.5V, 9V, and 12V). At nominal 4.5V, the measured open-circuit output is actually about 7.5V, and that is sufficient for my relay (Radio Shack Catalog #275-005) , which is rated at 7V to 9V.

But there were a couple of problems with unregulated supply. Since the voltage will vary with the amount of load, it could affect the accuracy and reliability of the operations of the two ICs on the circuit. Since the LM567 is rated at 9V maximum, I decided to use an LM7805 5V linear regulator. But this gives rise to another problem. Although the output of the 555 timer can drive 200mA, its output voltage is about 1.3V below its Vcc, in this case 5V. Secondly, driving an inductive load (relay) is tricky. The circuit in Part III is actually not quite right, because even with the protective diode, the negative voltage of -0.6V (the kickback voltage of the relay clamped by the diode) on the output of the 555 can actually cause it to malfunction. The application note for the NE555 (Philips AN170) (p.7, Figure 8) recommends that a series diode be connected between its output and the inductive load, but that means another drop of 0.6V, which is far from sufficient for driving the the relay.

The solution is to use an NPN transistor (Q2), with the output of the 555 driving the base (via a resistor [R6] to get the necessary base current). The current to drive the relay will be supplied from the unregulated output of the transformer. The modified circuit is shown below. Note that since Vcc is now 5V, I reduced the value of R5 so that D2 will not be too dim.