Ans.
Relay coils have a specified coil resistance. That coil resistance determines how much current flows through the relay coil when a voltage is placed on it. Although relays are specified by voltages, they're
actually current activated devices. A "5V" relay may have coil resistances of 50 to several hundred ohms, depending on the size and physical design.
For Eg: the NEC EA2 relay series has a 5V relay with a
178 ohm coil. What that tells you is that it takes 5v/178=28.1ma to run that coil.
The EA2 will operate (pull in) at voltages of 80% ofnominal, or 4V: that means it will operate on 4v/178 =
22.5ma.
How that relates to the question is this:
- If you hook a 9V battery directly to a 5V EA2 (which is a representative low power relay) the coil current will be 9V /178 = 50.5ma.
A 9V battery is a 160ma-Hr to 320 ma-hr device, so the battery would last 160/50.5= 3 to 6 hours. However, the relay coil would probably
burn out before then.
- if you hook a resistor in series to keep the current down to 28ma, then the coil won't burn out,also the
battery will last 5.6 - 11 hours before being exhausted. You can't pick a resistor to get 5V from a 9V battery; you have to pick a resistor to fit the current that flows in the 5V relay coil when there is 5V across it and to soak up the excess voltage. A 9V battery is usually 8.8 to 9.3V, but sags to as low as 7V when getting low.
-If you had a perfect 9.000V battery, and a 5V relay coil operating on 5V across the coil, you could figure the necessary resistor by dividing the 5V across the relay coil by the coil resistance to get the coil current, then using that current and the difference between 9V and 5V (that is, 4V) and dividing *that* by the relay coil current to get the correct resistor value.
As a hint, it will be 4/5 of the resistance of the relay coil.
Referrence:
http://www.diystompboxes.com/smfforum/index.php?topic=23624.0;wap2
Comments
Post a Comment