Reducing 12vdc to 6vdc on a I C board

You can put a common signal diode on the ground of a 5v voltage regulator to up its output to 5.7v, which might be close enough for the relay operation.

--- With all due respect to your admitted lack of brilliance, your description of what you want to do is very confusing.

You say you've ordered a 4 station remote, whatever that means, and that it has a SPST relay in the circuit. What circuit? And only one? What about the other three stations?

And then you say you want to use that/those relays to control some other relays with 6 volt DC coils, while stating that you'll have

12V coming from the relays that'll be doing the controlling.

Then you state that the driven relay field will be eight relays wide and, ostensibly, that that field will be driven by the 4 station remote and that either one or two 6V relays might be driven at the same time.

Unless you can come up with a better description of your system I'm not interested in helping you.

-- John Fields Professional Circuit Designer

Hi, Charlie. Just find out or measure the resistance R of the 6V coil, and put an equivalent resistance in series with each coil. Half the voltage will be across the coil, half across the resistor.

Be sure the series resistors are sized to handle the power. Wattage will be (6V)^2 / R, or 36/R watts.

Good luck Chris

To run a 6 VDC relay coil from a 12VDC supply, you can use a series resistor with a resistance equal to the coil resistance and a power rating greater than the power spec of the coil.

For example, here's a page that has a relay with a 6VDC coil with a DC resistance of 40 ohms, for an operating current of 150 mA:

Web Data/LY.pdf

And, the power is 6 * .15, which I'm too lazy to do the arighmetic; that's the power rating of the 39-ohm (standard value) resistor you'd use in series with this particular relay coil to run it off of 12V - to do yours, you'll have to look up the coil resistance or operating current, and do the arithmetic for your particular case - in either case, the resistor you want to use to halve the voltage will be equal to the coil resistance - look up "ohm's law" and "voltage divider". :-)

Good Luck! Rich

if you're always wanting to operate the same two relays at onece you could connect them in series, otherwise get some resistors of similar resistance to the relay coils (10% accuracy is sufficient) and rated for enough watts and connect the resistor in series with the relay coil.

the paired resistor-relay will operate fine of 12V

--

Bye.
   Jasen

Hi Charlie, Here is a "picture" - view in fixed font. It uses an LM7805 voltage regulator, two 1N4001 diodes, and two capacitors - a .1 uF and a .22 uF.

------- | o | |_______| /_______/| |LM7805 || | || |_______|/ | | |

+12 ---+-----+ | +---+-----+----> To relays (+) | | | | --- - --- - |.22| |D| | .1| |=| | uF| |i| | uF| |e| --- |o| --- |d| | |d| | |o| | |e| | |i| | |=| | |D| | - | - | | | | Gnd ---+--------+------+-----+----> To relays (-)

Note the banded end of the diodes indicated by an = at the end of each diode. It is important that the diodes be installed that way.

The circuit will provide about 5.7 volts, and your

6 volt relays will work just fine on that.

Mount the LM7805 on a small heatsink. The "package" style drawn is a TO-220 - make sure you get that style package for the 7805 and that the heatsink matches.

The parts are available from Allelectronics:

CAT #7805 1 for $.50 Voltage Regulator CAT 1N4001 15 for $.99 Diode CAT# RMC-104 1 for $.17 .1 uF capacitor CAT# RMF-224 5 for $1.00 .22uf capacitor CAT# HS-134 4 for $1.00 TO-220 Heatsink If you buy from them, do yourself a favor and get CAT# 1N4148 15 for $1.00. Take one of these diodes for each relay and install it across the coil of the relay with the banded end on the (+) side of the coil and the other end on the (-) side. Those relays may already have a diode installed across their coils - if so, the 1N4148 diodes are not needed.

Ed