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E-mail: Jack.Smith@cliftonlaboratories.com
 

As part of characterizing my K3's performance, I've measured its DC transmit efficiency, i.e., the percent of DC power that is converted to RF energy:

Efficiency = RF Power / DC Input Power

Although I had an accurate RF wattmeter connected, the efficiency data presented is based upon the K3's commanded output power. My K3's wattmeter, after calibration, is within 5%. This should be kept in mind when comparing the efficiency data with your readings. Also, the data presented on this page does not subtract the "idle" power, i.e., the power consumed in transmit mode with zero watts output. Hence the efficiency includes these idle loads, which represent lower level stages, DSP stages, microprocessors, PIN diode drive, etc. My program collected idle power data, but I've chosen not to present it on this page.

The test equipment is under the control of a program I wrote running under the EZGPIB compiler, about which I've written at EZGPIB and Prologix GPIB Adapter.

The DC power is computed from current and voltage data reported by the HP 6652A DC power supply. The program commands the K3 to go to a particular frequency and set its output power at a specific level and then enter tune mode. The DC power data is captured and saved, along with other parameters to a disk file. These steps are repeated for DC voltages from 11.8 to 15.0 volts, in 0.4 volt increments.

• Efficiency increases as the supply voltage decreases.
• Efficiency is not simply related to frequency

The relationship between supply voltage and efficiency is a function of how the K3 acts to varying supply voltage. For a given output power and frequency, the K3 draws essentially constant current for a given supply voltage.

For example, consider the following data set, for 3900 KHz, 100 watts output power. Although the DC supply voltage changes from 11.8 volts to 15 volts, the current increases by less than 200 mA, or just over 1%.

Why is this? First, the low power stages will generally appear to be a constant current load as they are supplied by DC regulators within the K3. Hence the individual components are supplied with a constant voltage and therefore have a constant current consumption for most purposes. Hence with respect to the supply voltage, the combination of regulator and load are constant current loads. Indeed, the idle current varies by only a trivial degree.

Of the 17 amperes or so of current, nearly 15 are consumed by the K3's low power amplifier (LPA) and high power amplifier (HPA). These transistors are operated at a more or less constant base current drive and hence the LPA and HPA transistors operate along constant current load lines with respect to the average DC current.

If efficiency is the end objective, therefore, the K3 performs best at lower voltages. However, my measurements also confirm that the K3's transmitted intermodulation in SSB mode (or linear data modes such as PSK31) improves with supply voltage. Hence if one is to take advantage of increased efficiency at lower supply voltages, only CW should be used.

It's also the case that efficiency must consider transceiver as part of a complete system that includes the DC power supply. A linear DC power supply will dissipate greater power in its pass transistors as the supply output voltage is set lower. (The power supply's  transformer and rectifier outputs a roughly constant supply voltage. Thus any reduction in DC power in the K3 is almost exactly offset by an increase in loss in the linear power supply's pass transistors.)

If a switching supply is used, the total supply plus K3 efficiency will act differently. Lower supply voltage to the K3 will not increase the switching supply's loss on a one-for-one basis. Rather,  the switching supply loss will increase only modestly, thus providing a net gain in efficiency when computed on AC input power to RF output power basis.

As far as how the K3's efficiency is a function of frequency, I suspect it is related to efficiency of the various ferrite core transformers in the LPA and HPA which may have increased losses at lower frequencies, such as 1.8 MHz. I'm at a loss for an easy answer to why efficiency increases with increasing frequency hower.