27-04-2015, 09:22 PM
Honestly, I wouldn't bother with a 555 timer for such a simple application - but that's because I don't fully trust them 
I've attached something that would probably work. Component count - not including switch or relay - is 8. This is the same as the 555 circuit, but it has the advantage that it's much easier to understand, and you have complete control over how it behaves.
Basically, a pair of transistors form a Schmitt trigger. The Schmitt action isn't strictly required, as the mechanics of the relay will ensure that it is either on or off, but it doesn't hurt... Omitting the 10 Ohm resistor will do away with the Schmitt action and save £0.01 at the same time
At rest, the capacitor charges up (to around 0.6V), turning on the first transistor. This turns off the second.
When pressed, the switch rapidly discharges the capacitor. The first transistor turns off, and the second transistor turns on.
When the switch is released, the capacitor starts to charge, and at some point, the first transistor turns back on, switching off the second one in the process.
Obviously, the relay coil must be matched to the supply voltage (as would be the case in the 555 circuit). The back-EMF diode is definitely recommended. The transistors can be any small-signal NPN devices - the BC108 is a perfectly good bet. Obviously, the second one must be able to handle the current of the relay coil...
The resistors are pretty arbitrary. The 22k needs to supply enough base current to the second transistor - if we assume a Hfe of 200 and a (generous) relay current of 50mA, then the base current is 0.25mA. The 22k supplies approximately double that, which is a decent margin for a one-off. Of course, the 100k and 100u RC combination set the time delay - feel free to experiment. Finally, the 100 Ohm resistor is just to limit the current seen by the switch contacts - depending on the switch, this might not be required (especially if it enjoys the "wetting current") - or perhaps the resistance of a long run of "bellwire" might present enough resistance.
The quiescent current consumed by this (if that matters) is mostly determined by the 22k resistor - about half a milliamp. Much lower than the non-CMOS 555 chip. Fine for a mains powered bell, but a bit high for a battery operated unit. Replacing the second transistor with a Darlington pair would help here. Also, finding the most power-efficient relay will help. For example, the NEC "MR82" types need 17mA, which means the 22k could be increased to 100k with no problems.
If you wanted to drive down the quiescent current still further, you could use a CMOS chip - e.g. the 40106 is good for this sort of thing.
Depending on the bell itself, it might be possible to omit the relay...
Anyway, just a suggestion...
Cheers,
Mark
I've attached something that would probably work. Component count - not including switch or relay - is 8. This is the same as the 555 circuit, but it has the advantage that it's much easier to understand, and you have complete control over how it behaves.
Basically, a pair of transistors form a Schmitt trigger. The Schmitt action isn't strictly required, as the mechanics of the relay will ensure that it is either on or off, but it doesn't hurt... Omitting the 10 Ohm resistor will do away with the Schmitt action and save £0.01 at the same time
At rest, the capacitor charges up (to around 0.6V), turning on the first transistor. This turns off the second.
When pressed, the switch rapidly discharges the capacitor. The first transistor turns off, and the second transistor turns on.
When the switch is released, the capacitor starts to charge, and at some point, the first transistor turns back on, switching off the second one in the process.
Obviously, the relay coil must be matched to the supply voltage (as would be the case in the 555 circuit). The back-EMF diode is definitely recommended. The transistors can be any small-signal NPN devices - the BC108 is a perfectly good bet. Obviously, the second one must be able to handle the current of the relay coil...
The resistors are pretty arbitrary. The 22k needs to supply enough base current to the second transistor - if we assume a Hfe of 200 and a (generous) relay current of 50mA, then the base current is 0.25mA. The 22k supplies approximately double that, which is a decent margin for a one-off. Of course, the 100k and 100u RC combination set the time delay - feel free to experiment. Finally, the 100 Ohm resistor is just to limit the current seen by the switch contacts - depending on the switch, this might not be required (especially if it enjoys the "wetting current") - or perhaps the resistance of a long run of "bellwire" might present enough resistance.
The quiescent current consumed by this (if that matters) is mostly determined by the 22k resistor - about half a milliamp. Much lower than the non-CMOS 555 chip. Fine for a mains powered bell, but a bit high for a battery operated unit. Replacing the second transistor with a Darlington pair would help here. Also, finding the most power-efficient relay will help. For example, the NEC "MR82" types need 17mA, which means the 22k could be increased to 100k with no problems.
If you wanted to drive down the quiescent current still further, you could use a CMOS chip - e.g. the 40106 is good for this sort of thing.
Depending on the bell itself, it might be possible to omit the relay...
Anyway, just a suggestion...
Cheers,
Mark
