6500 RF Power

There used to be a lot of discussions on this topic a few years ago... But you should have  13.8 V at the radio, and that the power supply can take the load. Make sure the Powerpole connector sits firmly and make contact. You may have to re-seat it. 
I never had 100 w o/p on my 6500, always a bit on the low side, but no biggie. Usually run with external amp and drive power is a lot lower. But one have to realize 80 watt vs 100 watt is just 1 (one) dB.

This question is raised a lot, both here and for other brand radios as well.  From Flex:

Let me first address how we calibrate our radios at the factory.  The entire process is automated and cannot be bypassed by the factory operator or test technician.  It is not possible for a radio to complete the test process without PA calibration over its linear power range on all all bands.  After the automated calibration and 24 hour burn in we perform a 100% QA using the SmartSDR GUI and run the power up to manually confirm 100W on the center of each band before shipment.

All power calibration is performed with the voltage set to 13.8V under full load at the DC terminals on the back of the radio.  We use a 40 dB precision power attenuator connected to a Mini-Circuits PWR-4GHS Power Sensor to precisely measure the power during both the automated test and the final QA.  The measurement accuracy of this test station is significantly better than virtually all ham measurement equipment on the market including the popular Bird "slug" watt-meters.  

MOSFETs used in HF power amplifiers are effectively variable resistors where the resistance will increase as they heat.  As the resistance goes up with heat, the power will decrease (P = E^2/R).  If the voltage on the input terminals decreases the power will go down proportional to the square of the voltage.  

Not all dummy loads are 50 ohms.  Not all ham dummy loads are still (or were ever) good.  Cables don't have 0 dB loss.  Not all connectors are good.  DC connector crimps and connections are not 0 Ohms.  What is the absolute accuracy of the power meter at the frequency of measurement?

Let's say we have a peak current of 20A and a series resistance of 0.1 Ohm in the DC cable and connector.  P = I^2 * R = 20^2 * 0.1 = 40W  That means that the cable and connector will dissipate 40W before you get to the back of the radio.  The voltage drop at 20A will be 20 * 0.1 = 2V.  It all makes a difference.

Now let's look at the effect of a simple change in impedance of the RF load.  Let's say that the load is 55 Ohms instead of 50 Ohms.  100W is 70.7 Vrms into 50 Ohms.  Since P = E^2/R then P = 70.7^2/55 = 90.88W.  A 5 Ohm (10%) increase in the load drops the power to ~91W.   55 Ohms is a SWR of 1.1:1.  You can experience that SWR on a cable, connectors and ham shack dummy load fairly commonly.  Drop the impedance to 45 Ohms at the same voltage and you get 111W.  Same 1.1:1 SWR!  

Finally, in the 90.88W example the power output difference from 100W is only 0.4 dB, which is totally imperceptible on the other end of the contact.  

73,
Gerald

Great explanation, thanks.