voltage drop

[audioguru]

Hi George,
A 555 will switch all relays off at the same time. Its datasheet is here:
**broken link removed**
It will draw 4mA to about 10mA all the time it is connected to the battery.
Since it needs a series diode to protect it from the flyback voltage spike from the relay coils and since it has an output voltage loss of up to 2.25V,
a 9V battery is needed to power its circuit.
Since most 5V relays are guaranteed to operate with a voltage of 3.75, the circuit will still work when the battery voltage drops to about 7V.

 

[ljcox]

how can I use the 555 timer IC to stop the relays from going off one by one.

thanks,

George

Do a search for 555 and you will find many circuits. Many of them won't be suitable for your purpose, but some may.

Note that there is a CMOS version the 7555 which consumes less power.

If you do this and still need help, send me a PM.

Len

 

[BravoKilo]

GeorgeL, don't you have your arrow on the transistor backwards? It should be an NPN, right?Wouldn't you want to use a 555 timer IC to do it? Make pin 3 drive one transistor each for each relay via a 100 ohm resistor.

ljcox, you are right when you see the switch closed. But when it is opened, that is a voltage divider in there.

Miles, I agree that adding a resistor in series does not lessen power used. Ohms law says E=IR and power of course is VA in watts. For a given voltage and a given resistance in a circuit, current I would be V/R. By increasing R,I will fall but the voltage across the series resistances goes up. So, I goes down but E goes up inversely proportionally. Since power if VA, it remains the same. Only switching will reduce power consumption like the scr speed controller for small DC motors where the SCR is oscillating. I would stand corrected if I am wrong.