Call Search

New to Ham Radio?
My Profile

Friends Remembered
Survey Question

DX Cluster Spots

Ham Exams
Ham Links
List Archives
News Articles
Product Reviews
QSL Managers

Site Info
eHam Help (FAQ)
Support the site
The eHam Team
Advertising Info
Vision Statement

   Home   Help Search  
Pages: [1]   Go Down
Author Topic: Measuring drain voltage wave form  (Read 1635 times)

Posts: 479

« on: August 31, 2012, 01:14:09 PM »

Hello all, I've read somewhere in the past a special probe was required to measure the wave form and voltage on the drain of a MOSFET amp - is this true?  If so, what is this special probe?  How is it made?

The datasheet for a SD2933 says Max Rating for Drain Source Voltage = 125 V:  I'd expect that to be Volts peak to peak?  OR is that RMS?

On an amp like the Elecraft 500 - that uses 2 MOSFETs in Push-Pull (SD2933), what would be the expected voltage (Vp-p?) at 500 watts?

Wouldn't the drain Vp-p depend on the matching of MOSFET drain to transformer impedance?  Close match would give a lower voltage, while a mis-match would give higher voltage at same output watts?

Thank ya'll for any good guidance and knowledge,

73 de Ken H>


Posts: 5917

« Reply #1 on: August 31, 2012, 02:20:24 PM »

A 10:1 passive oscilloscope probe should be fine for measuring the drain voltage with RF applied. The probe will present a 10 pF load.

The SD2933 drain voltage rating is maximum DC. If this is exceeded for even an instant the MOSFET can avalanche.

In a push-pull amp the drain voltage will swing up and down from the DC supply voltage. If 50 volts, the drain may swing from a few volts to close to 100 volts.

Yes the transformers are, I believe, designed so the amp will deliver rated power into a 1.5:1 VSWR. So what you have is an amp optimized for a 75 ohm load. Run it with a higher impedance and it will not make rated power. Run it with a lower impedance and it will make rated power but efficiency will suffer. The drain will not swing to the rails and the drain current is increased.

Posts: 9304


« Reply #2 on: August 31, 2012, 03:47:56 PM »

Adding to that....

Think about what a push-pull amp does. When one side pulls low, the transformer pushed the other side up. Depending on how close the "on" side pulls the drain down, it will push the other side up.

The issue, just like with tube amps, is the Q in the load. If the load stores more energy than it absorbs, the drain voltage can actually swing negative. When it swings negative, it also allows the other side to go higher than double.

This is why U-tube videos showing 99:1 SWR and other things from marketing departments ruffle feathers of anyone who has looked at amplifiers in working conditions with modest or high Q loads. If we dumped into a short without enough Q for ringing, and limited time or current to not overheat the junction, an FET could take a terrible mismatch.

So we watch heat, and we watch or limit excessive voltage, and SWR doesn't matter.

For example, we ran very tiny FET's in meter magnet "chargers" that could add or subtract from magnetic flux to calibrate meter movements. We ran dozens of times the rated current, because we hammered the FET into and out of conduction, and the backpulse was dampened by a diode. This was nearly infinite SWR and many times rated drain current, and we used dozens of machines on the assembly line without FET failures. This was in the early 1980's with garden variety cheap plastic audio and low frequency JFET's. They weren't even power FET's.

You can measure at the drain with a normal scope provided the probe is suitable. Normal peak voltage should be a bit less than twice supply voltage.

Look here to see what happens if you have the wrong load: 


Posts: 479

« Reply #3 on: August 31, 2012, 07:32:17 PM »

Tom, Dave, Thank you both for your reply, they have helped me put together some things I'd read before and understand more.

Tom, I'd read your link before, but it didn't really come together for me until now.... I think anyway.  The voltage on the drains is more a function of the drain voltage and match to the output transformer than it is to the power out - is that correct?

I think I'll move back to my 42 vdc power supply for testing/playing with the MOSFET amp to give the drain a bit more "headroom" in case of mis-match.  Those SD2933s are expensive to blow, even buying the used devices.

Thank you both again, if I have the above correct, then my understanding is much greater than before.

73 de Ken H>


Posts: 9304


« Reply #4 on: September 01, 2012, 05:05:32 AM »


Any device acts as a time-varying resistance. There might be different characteristics of that resistance as voltage and current changes, making it vary with current or voltage, but it is a varying resistance. It doesn't supply anything itself, it either pulls on something or passes something through in series.

A push-pull amp and a single-ended amp, in configurations like we use, "pull" down on a voltage though an impedance at the drain (collector or plate). That impedance is set by whatever couples the device to the load.

If we make that impedance higher, the device can change voltage more. If we make the impedance lower, the device changes voltage less and pulls more current when on.

We either depend on the tank to act like a flywheel and pull it back high smoothly, or we might depend on an opposing device through a "center tapped" transformer or tank in a push-pull stage.

Because of energy stored in the tank, filter, or in some cases out in the load, the device (when off) can be forced much higher than twice supply rail voltage. Instantaneous peak current can also be much higher than the longer term peak of average current we talk about.

For example a "very near-class B" AB2 3-500Z at 800 watts PEP with a 3000 volt supply, with a peak dc anode supply current of 400 mA, will have over 1 ampere of instantaneous peak anode current and swing between maybe 500 volts and 5500 volts when properly loaded. The tank Q pushes the anode up smoothly while the tube is high-resistance or off. If we grossly underload the amp for a given drive level, the anode can actually go negative (below chassis rail) and the positive swing go well past 8000 volts.

Since current makes heat and voltage causes dielectric failures, current is a longer term failure mechanism and voltage can be instant. The big problem with all these FET's is voltage, if the break down back through the gate at 150-250 volts and we let peak voltage get that high through a mismatch and "ringing" in the load, they will instantly fail. It doesn't matter if the market department says they handle 99:1 SWR, or they make a U-tube video showing sparks as they short the load.

73 Tom

Pages: [1]   Go Up
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.11 | SMF © 2006-2009, Simple Machines LLC Valid XHTML 1.0! Valid CSS!