Re: Which way is better, zener or resistor

On Sat, 18 Aug 2007 16:46:13 -0700, Allen Bong
<allenbsf6502@xxxxxxxxx> wrote:

I have a 12V relay and I wanted to operate it with 18V supply. Could
I use a 6V 1W zener to drop the extra 6V from the supply or is it
better to use a 150 ohm 1/2W resistor. Which will generate more
heat? The relay operates with 20mA current.

Allen


You've already gotten enough responses to your question. The only
questions that remain: do you care how much power is wasted? Can,
will you, does the job warrant it, rewind the relay with smaller gauge
wire?

Relays, as a general rule, will pull in at a lot less voltage than
their ratings - since the rating is usually "must pull in by X volts."

If this is a DC relay - your text implies that - you can do what a lot
of us have done for years. . . .

The relay "pull in" voltage and "hold" (in) voltage are different.
May take 10 volts to pull in and 8 volts to hold in . . .

You choose a resistor that allows the hold voltage to develop across
the coil and put it in series with the relay coil. (calculating the
resistor with ohms law and relay drop and supply voltage).

TaDa!

Now you just need to find a capacitor to bypass the relay that will
allow the pull in current to flow and put it in parallel with the
resistor (observing polarity since it will need to be an electrolytic
for large capacity)

Calculate the resistor using Ohms law - calculate for hold current
(look up the data ***) - or just calculate for 12 volts across the
coil.

Relay just sits there. Suddenly 18 volts is applied and relay current
charges the capacitor (which is discharged so appears like a dead
short) all that current flows into the coil pulling the armature in.
Now in time, the capacitor reaches some equalizing voltage and passes
no more current, but the resistor supplies the hold current/voltage.

Empirical testing? Calculate the resistor based on hold voltage (the
coil and resistor are in the same voltage divider - 12 volts on the
relay 6 volts across the resistor - series circuit)

Try caps until the relay pulls in when 18 volts is applies then double
it.

Resistor? 80 milliamps and 6 volts to drop = 75 ohms (ohm's law) (you
still have the "must pull in" specification for a fudge factor). So
you could use just a 75 ohm resistor in series with the coil to make
it work (it would have to be a 1/2 watt resistor or more to keep it
cool)

That alone will work. No capacitor, no hassle.

Using a capacitor allows you to pull in with a lower calculated
resistor. Let's say it takes 8 volts and 150 ohms to "hold" in.
That's only 53 milliamps to hold. Then the resistor to do that is 113
ohms (or 100 for ease of procurement) and .4 watts (still use a half
watt resistor)

Try a 200 ufd cap with 20 volt rating across the resistor. Whether it
works or not depends on the mass and friction in the armature.

This is only worth doing if you want to save power - like you are
using batteries to supply the 18 volts.

Years ago I applied it into a two way radio system in the mobile units
- yes it saved battery energy - but it also meant the person in the
field had to run his engine to make the transmitter work when the
battery got low. No more dead batteries in the field - and saved a
ton of money - but that wasn't my goal.

Otherwise the field rep would talk on the radio and kill the battery
and couldn't restart the engine - just a unintended benefit or
synergy.
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