# Re: G-5500 Wattage (Power) Requirements

```At 03:10 PM 5/12/03 -0400, RogerKola@aol.com wrote:
>I took this question as a learning challenge...this is what I came up with...
>
>Specifications (off Yaesu Web Site)
>G-5500
>Power Supply Voltage: 117 VAC, 50-60 Hz
>Power Supply Current Consumption: 120 VA
>Rotor Voltage: 24 VAC
>
>Elsewhere.....
>
>"In order for the power protection equipment to work effectively then it
must
>be sized according to the load it must carry. Power protection is normally
>sized in Volt/Amps but can be rated in watts or amps. Most types of
equipment
>will have a power rating on the back of each unit this will be given in
>either Watts or amps, to convert to VA the following formulas can be used.
>
>Volts (V) x Amps (A) = Watts
>W / Power Factor (Pf) = Volt/Amps (VA)."
>
>or
>
>"Watts x 1.4 = volt amps"
>
>or
>
>"Typically, to convert volt amps to watts, multiply the volt amps by 66%."
>
>Only research...I am by no means an expert...
>
>
>Roger
>WA1KAT

Good approach Roger.  If you don't find the power rating on the tag, then
use the AC fuse for determning current.  This will usually be a good amount
over the operating current.  Remember that ther rotator motors often are
capacitor started AC motors so they shift the Pf hard to start the rotator
motor.  Then you want to add allowance for invertor efficiency which for
the cheap square-wave type is less than 80%... my wild guess is at least
300w and maybe 500w inverter will be needed.

If you run an 80% eff inverter don't forget it is using 20% of the max
power continuously, even when the rotator is idle.  So expect the battery
to go down in a time of = amp-hour/inverter idle current (amps)

Good exercise of ohms law in this!  ALL hams know ohms law don't they?   P
= E*I  and E= I*R   =>  P = (I)^2 * R      ....Roger provided the Pf formula.

Have fun!

Ed - AL7EB

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