[R-390] Filters

[Bob Camp]

Hi

Here's a simple lowpass L network cascade that should do the trick:

1) 50 ohm source, one side grounded
2) Series L = 53 uH
3) Shunt C to ground = 2122 pf
4) Series L = 530 uH
5) Shunt C to ground = 212 pf
6) Series L = 5.3 mH
7) Shunt C to ground = 21.2 pf
8) 50 K load resistor shunt to ground

output is from the junction of the final R,L, and C.

Each section does a 10:1 impedance transform and ~ 3:1 voltage transform. Which ever section you stop with gets loaded with 500 or 5,000, or 50,000 ohms to ground (depending on how far you go). My guess is that two sections giving you a 10:1 voltage step up should be adequate. The loaded Q's of all the sections are about 3, so you should try for coils with a Q of at least 30. 

That all assumes that I didn't slip a digit somewhere punching this all in.

Bob

On Feb 19, 2013, at 8:33 PM, Bob Camp <ham at kb8tq.com> wrote:

> Hi
> 
> If you have the RF power, all you need is a transformer / transformation to get the voltage. You could try for a straight transformer. I'd go with some sort of tuned circuit to do the multiplication. A pi network would also be a possible way to go. Cascaded L networks are the classic solution.  What ever it is, low Q is the way to go. You can easily get to high a Q and make it to twitchy to be useful. The problem with any single section approach is that the Q limits the maximum transform you can get.
> 
> Bob
> 
> On Feb 19, 2013, at 8:16 PM, quartz55 <quartz55 at hughes.net> wrote:
> 
>> Yes, didn't think about the 100K vs. 50 ohms.  So, 10VP-P @ 100K would be about 122uW.  That's do-able. But since db are db, 20log10/1(volts)=20db, so yes that would be about right if the filter has a 25db loss. Actually I should probably see about .6VP-P from 10VP-P.  That's easily seen on my 545A or B.
>> 
>> I'm not sure what my R2005 will put out in volts though.  Probably not 10P-P.  Maybe.
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