Warning: Creating default object from empty value in /homepages/11/d132647312/htdocs/Amfone/mkportal/include/SMF/smf_out.php on line 47
input matching to grounded grid amplifier




 
The AM Forum
September 20, 2017, 06:59:57 AM *
Welcome, Guest. Please login or register.

Login with username, password and session length
 
   Home   Help Calendar Links Staff List Gallery Login Register  
Pages: [1]   Go Down
  Print  
Author Topic: input matching to grounded grid amplifier  (Read 777 times)
0 Members and 1 Guest are viewing this topic.
Opcom
Patrick J. / KD5OEI
Contributing
Member
*
Offline Offline

Posts: 6687



WWW
« on: August 13, 2017, 09:40:45 PM »

The job is to use a pi network to match a typical 50 Ohm exciter to the '50 Ohm' cathode of a tube.

I read this:
"(www.somis.org/D-amplifiers4.html)  Most of the tuned input and tuned output circuits in HF amplifiers are pi-networks. There are a number of ways to define the Q of a pi-network. In what follows, Q is defined as the input impedance of the pi-network divided by the reactance of the input, shunt element--typically a capacitor. This definition of Q is the one used by Eimac® in Care and Feeding of Power Grid Tubes.  Eimac® recommends using a pi input network Q of c. 5 for Class AB2 grounded-grid operation. To arrive at a Q of 2, the reactance [X] of the input capacitor, C1, is minus j50 ohm÷2=minus j25 ohm. Using C=1÷[25(2Pi x f)], approximately 220pF of input capacitance is needed for a Q of 2 on the 10m band. In actual practice, however, 220pF may be far from the value that produces a satisfactory SWR with a particular model transceiver and a particular length of coax. It may be possible to find a length of coax that would ameliorate this problem on 10m--but there are eight other bands to contend with below 30MHz. Since band switching different lengths of coax is hardly practicable, it would be useful if the input capacitors were adjustable in a grounded-grid amplifier's tuned input circuit"

Excel came up with these values but the above 'instructions' are incomplete as are most articles.
F KHzL uHC pFQ @50 Ohms
289900.1372202
215780.1703202.2
141280.2704702.1
71140.559102
36801.117002
18682.233001.9
He mentions the pi network, but does his paragraph about this mean that if the input and output Z are both 50 Ohms that the output C is the same as C in the above table? No complete method I could put in excel was given.

The blocking question is: "Since most pi networks are calculated by input C, partial L, other section pf partial L, and output C, then is the calculated value of L, above, the whole inductor in the pi network?

Would it be as well to try one of the 'pi network calculators' for this?
Got to make the cathode tuning circuit before the amp building can continue.
This article http://ham.homelinux.org/schemi-pdf/68hb199.pdf gives values for the 3-1000 which has an input Z of 65-67 ohms, close to 50 Ohms, but does not give the inductances, only turns on a specified slug tuned forms form which may or may not be available, and it was to be run at 65W drive, not 100-400 so those slug tuned forms may not be appropriate.




Logged

Radio Candelstein - Flagship Station of the NRK Radio Network.
DMOD
AC0OB - A Place where Thermionic Emitters Rule!
Contributing
Member
*
Offline Offline

Posts: 1037


« Reply #1 on: August 13, 2017, 10:39:48 PM »

The L should be the total series L in the C-Shunt L-Series C-Shunt Pi circuit.

For a loaded Q of 4.6, here is what my Pi-Net equations in MatLab gave me for 3.5 MHz:

Zin = 50 ohms, Zout = 67 - j1700 (This is taking into account the 27 pF input capacitance)

Q1 = 4.6201


C1 =  420 pF


L = 1.1 uH


C2 (tubegrid) = 370 pF


For a loaded Q of 4.6, here is what my Pi-Net equations in MatLab gave me for 7.2 MHz:

Zin = 50 ohms, Zout = 67 - j1700 (This is taking into account the 27 pF input capacitance)

Q1 = 4.6201


C1 =  200 pF


L = 0.5 uH


C2 (tubegrid) = 180 pF


Phil - AC0OB
Logged

"What kind of Koolaid do they make you drink in the Physics Department?" Charlie Epps to Dr. Larry Fleinhardt, NUMB3RS   Smiley
Opcom
Patrick J. / KD5OEI
Contributing
Member
*
Offline Offline

Posts: 6687



WWW
« Reply #2 on: August 14, 2017, 12:04:27 AM »

Hmm I got 420pF and 1.1uH resonate at 7.404KHz
and
Xc=51.2 Ohms.

For some reason I get Q=0.98, calculated per the somis.org article of Q=R/X. R being the '50 Ohm' input and X being the input C reactance. There are many reasons I'm messed up on this apparently believing articles on the internet may be one of them.

An old DOS program that's worked well before on tube pi outputs gives me this:
Q=5
Zin=50
Zout=67
Input C reactance=10
output c reactance=-11
L reactance=21.5
F=3.5MHz
Input C=4542pF
Inductance=0.979uH
Output C= -3944pF

It does not take into account 'j' either.. or does not mention it.

Where does the -j1700 Ohms come from?

Otherwise it looks like it is oddly close to a decimal point off from your capacitance calculations.

The summary is I don't know how to do this correctly. I think I've been here before looking for the correct math.


Logged

Radio Candelstein - Flagship Station of the NRK Radio Network.
DMOD
AC0OB - A Place where Thermionic Emitters Rule!
Contributing
Member
*
Offline Offline

Posts: 1037


« Reply #3 on: August 14, 2017, 12:23:24 AM »

I am using the Wingfield Pi-Net equations.

I have to input an initial Qin = 10 in order to get a loaded Q near 5.

The -j1700 comes from the tube's input capacitance
Xc = 1/(6.28X27.2e-12 FaradsX3.5e6 Hz).

So the tube's input impedance at 3.5 MHz is 67 - j1700.

At 7.2 MHz it is = 67 -j813 ohms.


Phil - AC0OB

Logged

"What kind of Koolaid do they make you drink in the Physics Department?" Charlie Epps to Dr. Larry Fleinhardt, NUMB3RS   Smiley
W1ITT
Member

Offline Offline

Posts: 220


« Reply #4 on: August 14, 2017, 06:05:43 AM »

What makes this all troublesome is that the input impedance of a grounded grid amplifier varies somewhat with drive power.  I have used an employer's high power network analyzer setup, feeding as much as 150 watts into a directional coupler to overcome adjacent transmitting antennas at shortwave broadcast sites.  It occurred to me that it would be fun to use that setup to quantify the impedance change with some of the popular grounded grid tubes, across a range of drive power levels.  We usually employ the high power network analyzer in a swept configuration, but it could just as easily be used in a CW mode.
The point of this is that, despite good calculations, it's a moving target.  And with a mode like SSB where the drive varies from nothing to 100%  it is even more troublesome.   Be prepared to tweak and optimize at your preferred drive power level.  If the Q of the input network is kept low enough, things will probably work out OK.  Sometimes "what works" wins out over engineering.
Logged
NA3CW
Member

Offline Offline

Posts: 11


« Reply #5 on: August 14, 2017, 12:33:06 PM »

One tip about physical layout:  If at all possible, place at least the output capacitor side of the pi as directly on the tube socket as possible because this is part of the RF current return through the tube.  On the higher bands even a few inches of coax from a pi board to the tube can have major effects of efficiency.  It's probably not news to many but in a now-defunct amplifier company which shall remain nameless, we found the hard way that with a pair of 3CX800A7's in grounded grid, we had to put the 10 meter output capacitor right on the socket.  The rest could be a little farther away and that capacitance just subtracted from the output cap of the other bands.  It made a very noticeable difference in output.

YMMV. This effect may be different for different tubes. 

73,
Chuck
NA3CW
Logged
Dave K6XYZ
Member

Offline Offline

Posts: 80



« Reply #6 on: August 15, 2017, 09:08:19 PM »

Pat.....take a look at the tuned input ckts for the Collins 30S-1...those component values may get you into the ballpark.
You can dwnld a manual from the Collins Collectors site.
Logged
Opcom
Patrick J. / KD5OEI
Contributing
Member
*
Offline Offline

Posts: 6687



WWW
« Reply #7 on: August 20, 2017, 02:54:17 AM »

I am using the Wingfield Pi-Net equations.

I have to input an initial Qin = 10 in order to get a loaded Q near 5.

The -j1700 comes from the tube's input capacitance
Xc = 1/(6.28X27.2e-12 FaradsX3.5e6 Hz).

So the tube's input impedance at 3.5 MHz is 67 - j1700.

At 7.2 MHz it is = 67 -j813 ohms.


Phil - AC0OB



That explains something. Should the input Q be around 10 on all bands possible? I should shoot for that?

There is enough room under the chassis to put the output cap right next to the socket. How about an air variable? it'll probably fit.
Logged

Radio Candelstein - Flagship Station of the NRK Radio Network.
ka1tdq
Member

Offline Offline

Posts: 853



« Reply #8 on: August 20, 2017, 06:42:04 AM »

I'm not the expert on this topic, but I remember an experienced amplifier builder telling me to use an tuned input Q of 2.

I'm in the process of building a small, 20 meter GG amp using a single 3-500 and I'm not going to use a tuned input at all.  The driver is a DX-60 and that'll match to the tube's input. 

Jon
Logged

KD6VXI
Contributing
Member
*
Offline Offline

Posts: 1585

Making Amplitude Modulation GREAT Again!


« Reply #9 on: August 20, 2017, 09:08:15 AM »

A tuned input will give you much better IMD, Jon.

Not using a tuned input is bad Juju.



And yes, a tuned input of 2 to 5 is what Eimac recommends, depending on the data sheet.  2 to 3 is the most common values mentioned.

--Shane
KD6VXI
Logged
ka1tdq
Member

Offline Offline

Posts: 853



« Reply #10 on: August 20, 2017, 11:09:35 AM »

Oh, okaaay... you convinced me. It's just for CW anyway, but I guess I'll go along with that good practices thing and all.

Jon
Logged

Opcom
Patrick J. / KD5OEI
Contributing
Member
*
Offline Offline

Posts: 6687



WWW
« Reply #11 on: August 21, 2017, 01:30:12 AM »

In the Radio Handbook 20th ed 1977 , sec 7.23, it says that if the  tube's average cathode impedance is close to 50, then a simple parallel resonant circuit Q=2 should do. It recommends 20pF per meter and 'make the coil to resonate'. They don't give formulae but OK from the cap the rest can be figured.

This is not the same as the pi network, no doubt the source of my misunderstanding as to what the book and program was addressing and the discrepancy with the results I got vs. the correct pi network figures shown in this topic.

The existing tapped coil allows continuous coverage, so it would be fine to allow that on the input. It would seem easier to do that with fewest knobs using a parallel circuit. Something to think about when thinking about the PI network.
Logged

Radio Candelstein - Flagship Station of the NRK Radio Network.
KD6VXI
Contributing
Member
*
Offline Offline

Posts: 1585

Making Amplitude Modulation GREAT Again!


« Reply #12 on: August 21, 2017, 09:31:19 AM »

The pi will allow for the highest efficiency of input to grid current of any of the networks I've tried.

Yes, there is a second control to have to twiddle, but done correctly your drive power will be lower.

That may or may not be an issue.

And to clarify my prior post, I made a typo.  Yes, the pi out cap at the tube(s) socket is best, but not just for lower imd.  Moreso for efficiency and harmonic rejection.

--Shane
KD6VXI
Logged
Opcom
Patrick J. / KD5OEI
Contributing
Member
*
Offline Offline

Posts: 6687



WWW
« Reply #13 on: August 21, 2017, 10:03:05 PM »

Drive efficiency is somewhat important since my transmitters are mostly 20-30 year old 100W solid state units.  How much worse have you found the parallel circuit to be efficiency wise?
Logged

Radio Candelstein - Flagship Station of the NRK Radio Network.
w8khk
Marietta Georgia
Member

Offline Offline

Posts: 740


You never heard of a Ham Operator?


WWW
« Reply #14 on: August 21, 2017, 11:51:18 PM »

Please disregard.  Posted to incorrect thread.  My apologies.  Late at night!
Logged

Rick / W8KHK  ex WB2HKX, WB4GNR
Global Warming? Actually, it's much worse! On any given day, half the planet faces global warming whereas the other half faces global cooling! The warming half is facing the sun, and the cooling half is opposite the sun
KD6VXI
Contributing
Member
*
Offline Offline

Posts: 1585

Making Amplitude Modulation GREAT Again!


« Reply #15 on: August 22, 2017, 09:28:58 AM »

A 100 watt TX is about the cats meow for that tube.

It took about 15 pct more drive using a parallel resonant circuit instead of a pi circuit.

--Shane
KD6VXI 

Logged
DMOD
AC0OB - A Place where Thermionic Emitters Rule!
Contributing
Member
*
Offline Offline

Posts: 1037


« Reply #16 on: August 22, 2017, 08:20:16 PM »


That explains something. Should the input Q be around 10 on all bands possible? I should shoot for that?

There is enough room under the chassis to put the output cap right next to the socket. How about an air variable? it'll probably fit.

The program has an input Q but also tells you the loaded Q. One has to input a Q of 10 in order to get a loaded Q 4.6, which should be sufficient.

I believe MFJ will sell you a set of pi-net input coils and caps as an assembly.

The caps are fixed but the inductors are tuneable.


Phil - AC0OB 
Logged

"What kind of Koolaid do they make you drink in the Physics Department?" Charlie Epps to Dr. Larry Fleinhardt, NUMB3RS   Smiley
Pages: [1]   Go Up
  Print  
 
Jump to:  

AMfone - Dedicated to Amplitude Modulation on the Amateur Radio Bands
 AMfone © 2001-2015
Powered by SMF 1.1.21 | SMF © 2015, Simple Machines
Page created in 0.058 seconds with 19 queries.