6500 power

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The power meter on my 6500 shows a maximum of 85W. I checked with LS2K4 Wattmeter that is a little bit more accurate than a bird and tha value is correct. Power supply gives 13,53V. Why my Flex doesn't give full power ouput? The specifications says 100W if am not wrong. I read some posts on the same matter but I didn't undestood if this is a general problem, there is a fault on my 6500 or what else. 15W missing on 100W it's a lot...
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Stefano - W2WTZ

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Posted 2 years ago

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Gerald - K5SDR, Official Rep

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Official Response
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