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Author Topic: Operating Solid State Linear Amplfiers in High Duty Cycle Modes  (Read 4251 times)
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AB2EZ
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« on: April 16, 2017, 04:12:05 PM »

Solid state linear amplifiers have RF output circuits that provide a fixed (designed in) RF load impedance to the output transistors*.

*In contrast to traditional tube amplifiers that incorporate a pi network with an adjustable loading capacitor.

When operating in a high duty cycle mode… such as JT65 or one of the other digital modes… reducing the RF output power level below the maximum rated RF output power level (by reducing the drive) does not typically reduce the power that will be dissipated by the output transistors. In fact, reducing the RF output level below the maximum rated RF output level will usually increase the power that will be dissipated by the output transistors.

This is because, with the transistors looking into a fixed RF impedance, the total electrical power consumed by the RF output stage is proportional to DC (average) value of the current being drawn by the transistors, but the RF output power is proportional to the square of the RF component of the output current being drawn by the transistors.

For example, with my Elecraft KXPA100 linear amplifier, running key down at full rated output on 20m:
RF output power = 100W
Average drain current = 11.3A
Drain voltage = 13.4V
Total electrical input power to the output stage = 11.3A x 13.4V = 153W
Power dissipated as heat, by the output transistors = 153W - 100W = 53W

If I reduce the RF output power to 30W, by reducing the drive:
RF output power = 32W
Average drain current = 7.2A
Drain voltage = 13.5V
Total electrical input power to the output stage = 7.2A x 13.5V = 97.2W
Power dissipated as heat, by the output transistors = 97.2W - 32W = 65.2W

If I reduce the RF output power to 15.3W, by reducing the drive:
RF output power = 15.3W
Average drain current = 5.1A
Drain voltage = 13.6V
Total electrical input power to the output stage = 5.1A x 13.6V = 69.4
Power dissipated as heat, by the output transistors = 69.4W – 15.1W = 54.1W

Stu





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Stewart ("Stu") Personick. Pictured: (from The New Yorker) "Season's Greetings" looks OK to me. Let's run it by the legal department
KD6VXI
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« Reply #1 on: April 16, 2017, 04:21:57 PM »

Here is where a drain or collector power supply that is modulated by the envelope is good.

The amp always (hopefully) stays in its efficient range.

--Shane
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w4bfs
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« Reply #2 on: April 17, 2017, 03:17:36 PM »

hi Stu...

It would be very interesting to see the distortion figures that attend those power levels and the effects that bias
changes have on those figures as well....It should be possible to get to get good thermal performance at partial  power this way ... of course there is also the possibility of drain voltage changes ...


J J Tom has been advocating the use of a SDR for this type of measurement ...

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Beefus

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to see ourselves as others see us.
It would from many blunders free us.         Robert Burns
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« Reply #3 on: April 17, 2017, 09:10:53 PM »

I have what you want!

I saw Pascal's presentation in SLC.  Nice job!

Who wants pdf??

:-)

http://www.ismrm.org/13/session72.htm
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« Reply #4 on: April 18, 2017, 03:55:29 PM »

I guarantee the old Kenwood TS430-S gets hot making A 25W carrier to drive the 4-1000 stage here. I put a small 12V blower behind it and it stays cool now with that jet of air right into the fan.

That's a bipolar setup in the Kenwood. I wonder how lowering the DC supply would affect it.
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« Reply #5 on: April 19, 2017, 02:58:07 PM »

Hi Stu:

Can you confirm your dissipation numbers with thermal readings (do they both track at various power levels)?

What class is the amp running in?

Interesting!!

Thanks,
Dan

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AB2EZ
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« Reply #6 on: April 20, 2017, 06:35:08 AM »

Dan

The KXPA100 amplifier is a push-pull amplifier running in class AB. The resting (bias) drain current for the pair of transistors is about 1.5A. The operating drain current at full (100W) RF output on 20m is approximately 11A.

The built-in temperature measuring function of the amplifier confirms the higher dissipation at lower RF output power (no surprise). I.e., on key-down... the amplifier temperature increases by a larger amount at reduced RF output power.

The amplifier also performs and displays a calculation of the dissipation...  (probably based on the amplifier's internal measurements of drain current, drain voltage, and RF output power)

A theoretical calculation for an ideal class B amplifier indicates that the efficiency is [pi/4] x 100% x [the peak value of the RF drain voltage swing] / [the DC drain voltage]


Stu
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« Reply #7 on: April 25, 2017, 12:11:08 AM »

Great topic !

I have been looking into building a class H (or class G) high efficiency linear for many years.  I think for this application, the output would have to be a source follower, so the drain voltage presented to the final RF amplifier does not affect the output.  The design presents some challenges, but nothing like envelope elimination and restoration, where the phase relationships are absolutely critical.

Anyway, thoughts on a class H or class G linear amplifier are welcome.

Regards,  Steve
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