RadioBanter - Losses in PI-filter output

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Losses in PI-filter output

Maybe not quite appropriate for this news-group..

Does anybody know a pointer to a program or the methods
of calculations so I can determine the losses occuring
in a PI-filter output for a high-power tube amplifiers.

I can calculate the required C1, L and C2 values given a
certain anode voltage and current and a certain output
impedance for a loaded Q of 10, 12 or other values, but
what about losses depending on selected loaded-Q values
related to harmonics suppression...

Thanks in advance,
Arie.

Efficiency of a tank circuit or a Pi-tank circuit is given by -

Efficiency = Qi / ( Qi + QL ) * 100 percent,

where Qi is the intrinsic Q of the coil and QL is the loaded Q of the tank
circuit.

PROGRAMS AVAILABLE .

The following calculate L and C values for given load values and/or tube
operating voltages.

Pi_L_NET and Pi_TANK also calculate harmonic suppression.

TRIODE1 and TETRODE1 compute grid drive, tank circuit conditions and
efficiency for Class A, B & C operating conditions and angles.

Download progs in a few seconds from website below and run immediately.
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

"arie" wrote in message
om...
Maybe not quite appropriate for this news-group..

Does anybody know a pointer to a program or the methods
of calculations so I can determine the losses occuring
in a PI-filter output for a high-power tube amplifiers.

I can calculate the required C1, L and C2 values given a
certain anode voltage and current and a certain output
impedance for a loaded Q of 10, 12 or other values, but
what about losses depending on selected loaded-Q values
related to harmonics suppression...

Thanks in advance,
Arie.

Reg, thanks for your fine program,

but.., I get strange values.

When considering the below input-values I get 'strange' results:

- Frequency = 1.8 Mhz
- RF power = 1400 W
- RF peak amplitude = 5400 Volts
- External load res = 50 ohms
- Phase shift = 90 degrees (to get a Q of around 12)
- intrinsic Coil Q = 100

I get:

C1 = 122.5 pF
L = 63.80
C2 = 122.6 pF

Is sounds very strange to me that both C1 and C2 have the same
value whereas the Load resistance at anode is around 10414 ohms
or at least much higher than 50 ohms and an external resistance
of 50 Ohms

What about the Load resistance at anode, I would think this
should be somewhere aound 2800 ohms, considering the Power
output and anode voltage swing.

Arie.

"arie" wrote
- Frequency = 1.8 Mhz
- RF power = 1400 W
- RF peak amplitude = 5400 Volts
- External load res = 50 ohms
- Phase shift = 90 degrees (to get a Q of around 12)
- intrinsic Coil Q = 100

I get:

C1 = 122.5 pF
L = 63.80
C2 = 122.6 pF

Is sounds very strange to me that both C1 and C2 have the same
value whereas the Load resistance at anode is around 10414 ohms
or at least much higher than 50 ohms and an external resistance
of 50 Ohms

What about the Load resistance at anode, I would think this
should be somewhere aound 2800 ohms, considering the Power
output and anode voltage swing.

====================================
Hi Arie,

Circuit-Q is a dependent variable. It depends on nearly all program input
parameters. It's value cannot be entered in the program because you don't
know it. You know only what you would like it to be. It can be calculated
accurately only AFTER parameters such as efficiency have been calculated.
You may not be directly interested in phase shift but by entering it in the
program it does allow you INDIRECTLY to control circuit-Q.

To design an amplifier starting with values of Circuit-Q, DC Plate volts,
Plate current operating angle, etc., there is no alternative to using the
program to produce a multi-dimensional table of results. And even then they
would be approximations.

A phase shift of 90 degrees through a symetrical Pi-Network should set bells
ringing. With all 3 reactance magnitudes being identical the network
behaves as a 1/4-wave transmission line transformer. Terminating impedances
can be anything you like provided Line Zo = Sqrt(Z1*Z2).

You will notice you are not allowed to enter a phase shift as high as 180
degrees. That would correspond to 1/2-wave length of line which matches any
impedance Z only to itself. It then makes itself redundent. The program
would object to being asked to do something silly and would abort into the
operating system. The program user would then blame the long-suffering
author and not himself. ;o)
----
Reg, G4FGQ

A little more on a symetrical C-L-C Pi-network having all reactances of the
same magnitude.

Short-circuit one end and the input impedance at the other end is infinite.

Open circuit one end and the input impedance at the other end is zero.

Which is the same behaviour as a 1/4-wave line.

L-C-L and C-L-C T-networks behave in exactly the same way.

If the L's and C's are interchanged, ie., series C's and shunt-L's, the same
behaviour accurs again. But in these cases there are no corresponding
lengths of transmission line.
---
Reg, G4FGQ

Arie,

Your intrinsic coil Q of 100 seems wimpy. You should be able to get at least
200. You messed up on calculating the anode load impedance - do you really
have a 5400 VDC supply? Or did you mean the peak - peak RF was 5400V? At any
rate,. you ignored the plate saturation voltage, which will be around
300 -400 V. For 2700 VDC supply, and 300Vsat, the anode load should be about
(2700 - 300)^2/(2 x 1400) = 2057 Ohms. You should be able to find tables for
Pi network values. The output capacitor will come out a lot bigger than the
plate capacitor.

Tam/WB2TT
"arie" wrote in message
om...
Reg, thanks for your fine program,

but.., I get strange values.

When considering the below input-values I get 'strange' results:

- Frequency = 1.8 Mhz
- RF power = 1400 W
- RF peak amplitude = 5400 Volts
- External load res = 50 ohms
- Phase shift = 90 degrees (to get a Q of around 12)
- intrinsic Coil Q = 100

I get:

C1 = 122.5 pF
L = 63.80
C2 = 122.6 pF

Is sounds very strange to me that both C1 and C2 have the same
value whereas the Load resistance at anode is around 10414 ohms
or at least much higher than 50 ohms and an external resistance
of 50 Ohms

What about the Load resistance at anode, I would think this
should be somewhere aound 2800 ohms, considering the Power
output and anode voltage swing.

Arie.

Arie,
First, I should not have said you ignored Vsat. I got confused. Anyhow,
using standard tables, the Pi network values for a 2100 Ohm to 50 Ohm
network at 160 meters a

C plate = 427 PF
C load = 2213 PF
L = 18.4 uH

These are rather huge capacitor values, and might explain why the only two
amps that I saw schematics for, that covered 160 m, used either a Pi-L, or
some kind of transformer coupled output.

Tam/WB2TT

Tam, did you miss the programs I recommended?
---
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

"Reg Edwards" wrote in message
...
Tam, did you miss the programs I recommended?
---

Yes, I missed it the first time. Saw it now.

BTW, I might mention that at least one commercially built amplifier uses a
Pi or Pi-L network designed for 200 Ohm output, and follows that with a 4:1
balun inside the amplifier case.

Tam/WB2TT
.................................................. .........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. .........