http://elecraft.85.s1.nabble.com/KPA-1500-Efficiency-As-A-Function-of-Input-Power-tp7644440p7644470.html
Today in broadcast we generally rate transmitters by AC in to RF out. For AM transmitters that figure is in the low 90 percent range and the current FMs run at about 74% AC in to RF out.
> On Aug 17, 2018, at 11:48 AM, Fred Jensen <
[hidden email]> wrote:
>
> Well, only if you've figured out a way to divide by zero. [:-)
>
> The term "amplifier efficiency" must have changed dramatically since I worked in broadcast nearly a lifetime ago. Then, it was the ratio [expressed as a percentage] of the RF power delivered to the 3 1/8" hardline divided by the DC power supplied to the amplifier plate circuit. Power to the filaments was generally excluded by manufacturers seeking to have higher efficiency numbers. One 10 KW FM transmitter I helped build used four 4-1000A's in parallel and the filament power was 600 watts. Efficiency, including filament power was just over 70%.
>
> To measure the efficiency of a KPA1500 [or 500, or KXPA100, or KPA2] one would measure the key down RF output and divide it by the key down DC input power. If it was water-cooled, I suppose one should include the input power to the pump(s), but no one ever did. SS amplifiers like the KPA(1)500 probably run class AB<mumble> or its equivalent and have substantial DC input power even at zero RF input.
>
> RF output power vs RF input power defines the amplifer's transfer function, not its efficiency.
>
> 73,
>
> Fred ["Skip"] K6DGW
> Sparks NV DM09dn
> Washoe County
>
> PS: I really don't recommend trying to get four parallel 4-1000A's to work anything close to "well" at 90.7 MHz. Apparently, it seemed like a good idea at the time. [:-)
>
>> On 8/17/2018 6:24 AM, Charlie T wrote:
>> Think zero output with zero drive = zero efficiency.
>> Apply some drive, read some output and the efficiency goes up from there.
>>
>> 73, Charlie k3ICH
>>
>
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