View Single Post
Old 12th Jun 2021, 6:22 pm   #15
G0HZU_JMR
Dekatron
 
Join Date: Sep 2010
Location: Cheltenham, Gloucestershire, UK.
Posts: 3,077
Default Re: Modifying an inductor?

It looks like you are always using 3.3nF and 8.2nF caps in every version of pi filter and this is restricting what you can do.

Probably the best thing to do is study a basic L match circuit that attempts to match a higher resistance (say 300R) to a lower resistance (say 50R) at 1MHz. The lowpass version of the L match consists of a shunt C and a series L as in the image below.

In this case the shunt C always gets placed across the higher resistance (300R) and then the series L then feeds to the lower resistance (50R). This shunt C series L circuit only has one solution at 1MHz and I've given it in the first image below. The C is 1186.3pF and the L is 17.795uH.

For your application this simple L match doesn't give enough filtering rejection on its own as you can see there is only about 18dB rejection at 3MHz. Matching from 300R to 50R only has a ratio of 6 so the solution doen't produce enough selectivity. Also, in your case it won't produce enough output voltage at the 50R port for your application. You can see the blue trace shows the Vout is less than the red Vin trace.

You could swap the L match around and design for a step up from 300R to 500R and this would give enough voltage at the output. The circuit is given below but this only has a resistance ratio of 500/300 = 1.67. The harmonic filtering for this L match is therefore very poor with just 7.5dB rejection at 3MHz. However, at least it gave out a higher voltage on the blue trace. This is the second image below.
Attached Thumbnails
Click image for larger version

Name:	Lmatch.jpg
Views:	50
Size:	89.4 KB
ID:	235788   Click image for larger version

Name:	Lmatch500.jpg
Views:	54
Size:	88.2 KB
ID:	235789  
__________________
Regards, Jeremy G0HZU
G0HZU_JMR is offline