Vac. Relay

[Kenneth G. Gordon]

On 18 Apr 2010 at 14:11, mtippin wrote:

> Question: thinking about installing a 335 ohm- 26.5VDC relay in a 922. The
> supply voltage is about 110 +or-dc .What size is the dropping
> resistor??

About 1050 ohms at nearly 7 watts. I would use a 15 or 20 watt resistor 
in the interests of safety.

The calculations go like this:

1) I=E/R, which equals 26.5/335, which equals almost 80 mA (0.08 
amps). This is the current the relay will draw from a 26.5 volt supply 
when it is closed.

2)R=E/I. But in this case, the E is 110 - 26.5 or 83.5 volts, or "about" 84 
volts., Then R=84/.08 which equals 1050 ohms. 

We have to use the voltage that is DROPPED in our calculation: i.e., 
the voltage that we have to get rid of to equal 26.5 volts at the relay.

3) Then, P=EI, equals 84*.08 =  6.72 watts

This means that at full load, the 1050 ohm resistor will be dissipating 
~84 volts at 80 mA or nearly 7 watts.

As I said, I'd use a 20 watter, although you MIGHT get by with a 10 
watter if you didn't hold the key down too long.

Also, are you certain that your 110 volt supply will provide 80 mA in 
addition to its other loads?

One definite advantage to doing it this way rather than by building a 
separate 26.5 volt power supply is that until the relay draws its full 
rated current of 80 mA, the voltage on the relay coil will be 110 Volts.

This method has been used for many years in order to "hot rod" a 
relay, making it close MUCH faster than it would if it were operated at 
its "correct" voltage.

In fact, if your 110 volt supply can provide the necessary current, I 
would definitely do it this way.

In one case, in my job, we used a 500 VDC supply and dropping 
resistors to operate a 3 VDC solenoid-operated valve in order to get it 
to open in microseconds, instead of milliseconds or longer. We built 
this at least 10 years ago, and it is still being used today with the 
original solenoid.

Ken W7EKB

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