The AM Forum

On the Class E website, the 24-FET schematic shows 300 ohm resistors in series with the gate driver inputs. I could never see the benefit for this since it is a high input impedance device, and the resistor’s value is of little significance. Also, the 8-FET version doesn’t use these, so I’ve never used them.

I was then reading in the 2018 ARRL Handbook in the MOSFETS section of the amplifiers chapter about placing small value resistors (3.3 ohms) in series with the gate leads of FETs themselves when you’re using devices in parallel. This is to reduce the chance of UHF oscillation. I don’t want to copy the text for copyright reasons, but it is on page 17.31.

I know it’s bad to do that in the gate leads of class E transmitters for stability reasons, but is this the reason for using them in the input leads of gate drivers?

Also, is the number of gate drivers in parallel a factor since the KW transmitter design uses them, but not the 400 watt version?

Jon

Fair use doctrine would Trump copyright.

--Shane
KD6VXI

I claim to be NO expert,   but,   Reply #7 in this thread touches on the subject:

http://amfone.net/Amforum/index.php?topic=43643.0

It may come up again later in that Thread.

Termination of the cable is important,   and isolating the driver input from the cable is part of what the 300 Ohm series R might do ...   

FWIW,   Vic

I strongly recommend the resistors in series with the driver input for all designs.  I guess I need to update the schematic.

The reason the resistors work is that the driver has capacitance, and the R-C network formed very effectively stops parasitics that involve the drivers.  I have had this issue during the development of various class E transmitter, and have seen it in other forks' transmitters, so the resistors really are necessary.  I use anything around 300 ohms.

So for the 8 FET leave the 100 ohm resistor to gnd and add a 300 ohm in series?

Yes.  The reason the 8 FET transmitter uses a 100 ohm resistor to ground at the end of the short feedline that runs from the common point to the driver, is so the common point impedance is 50 ohms - and there are 2 drivers per phase, with a 100 ohm termination at the end of each short feedline.  The short feedlines are in parallel at the common point.

With the 24 FET transmitter, there are 6 drivers per phase, so the resistors to ground at the end of the feedlines will be 300 ohms (50 * 6) to make 50 ohms at the common point.

The 300 ohm series resistor goes AFTER the 100 ohm resistor - between the 100 ohm resistor (and the feedline) and the input to the driver.

This may all seem very obvious to anyone who has studied the schematics of the various class E rigs, but I thought the clarification might be helpful.