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Re: [Amps] RE : Plate Impedance

To: on4kj@skynet.be, Xmitters@aol.com, amps@contesting.com
Subject: Re: [Amps] RE : Plate Impedance
From: TexasRF@aol.com
Date: Mon, 10 Apr 2006 20:14:51 EDT
List-post: <mailto:amps@contesting.com>
 
Yes, that should be 3000 divided by 1.8 = 1666.6 ohms.
 
In this case power equals current squared times load resistance which is  
1666.6 watts. This power level is consistent with 3000 watts input and 55.5%  
efficiency. If the load was 5400 ohms we would be generating 5400 watts of 
power  
which is more than the power input; this is of coarse impossible.
 
 
73/k5gw
 
 
In a message dated 4/10/2006 4:43:24 P.M. Central Standard Time,  
on4kj@skynet.be writes:

n  


5400 ohms plate load  
calculating using the K factor  stuff.

3000v/1A/1.8 is equal to 1666 ohms so one of your numbers is  flawed.

3000 X 1.8 = 5400  is there any relation to be aware of  ?

Jos 

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