> Jon wrote.
>
>> I would say if 60 nH is accurate, that could have quite an effect on
>> the
>> overall supressor design. The reason is that the supressor coil
>> itself
>> is in the 60 to 100 nH range. So in effect you have 2 inductors
>> paralled
>> which (for 2 60 nH Ls in parallel with no resistance in either L)
>> would
>> give a total inductance of only 30 nH.
>>
>This 60 nH is the series-inductance (Ls) of the resistor, not the
>parallel-inductance (Lp). You need to calculate the (frequency
>dependent) parallel equivalent first before you start paralleling the
>two inductors.
Gerrard, I stated that specifically when I wrote "no resistance in either
L". I agree with you.
>
>> Yes, there's resistance in the
>> second one (which makes the combination I just did invalid) so you
>> can't
>> do a true parallel combination without doing some parallel/series
>> conversions, but having an indcutance in the resistor as large as the
>> supressor inductor would IMHO, make a big difference.
>>
>Yes, you can do a 'true parallel combination' on any parallel
>combination, but it has to be a parallel combination to start with. Get
>your circuit analysis basics right and everything falls in place.
Yes, I know. I thought this is what I was saying when I said it was
invalid.
OK so, let me present "correct" circuit analysis:
Here is a circuit:
It consists of a 50 Ohm resistor that has 60 nH of inductance. This
resistor is paralled by an ideal inductor which has zero ohms resistance
and 60 nH of inductance (note: IDEAL).
Ok...here goes:
Let's say that the frequency of interest is 100 MHz since that is typical
of where we want a supressor to function and that is what people use
before.
In order to do a schematic in ascii text, I'll use LLLLLLL for an
inductor and RRRRRRR for a resistor.
Here's our circuit:
------------------------LLLLLLL--RRRRRRR--------------
| 60 nH 50 Ohms |
| |
| |
----------LLLLLLL-----------
60 nH
In order to combine the two inductors, we need to do a series to parallel
transformation on the resistor so our circuit will look like:
--------------------------RRRRRRR---------------------
| |
------LLLLLLL------
| |
------LLLLLLL------
So let's do this:
Rp=Rs(1+(Xs^2/Rs^2)) = Rs(1+Q^2)
at 100 MHz, Xs=37.7 Ohms
This gives Q=Xs/Rs = 37.7/50 = 0.754
Rp=50(1+(0.754)^2) = 50(1+.5685)=78.42 Ohms
Xp=Xs(1+(1/Q^2))
Xp=37.7(1+(1/0.5685)) = 37.7(1+1.76)=104.05 Ohms.
So the new circuit is now:
--------------------------RRRRRRR---------------------
| 78.42 Ohms |
| |
------LLLLLLL------
| j104.05 Ohms |
| |
------LLLLLLL------
j37.7 Ohms
We can now do a parallel combination on the inductive reactances to get a
final reactance (Xf) of:
1/Xf = 1/104.05 + 1/37.7 = 0.03613
Xf = 27.67 Ohms
So back to our schematic we have:
--------------------------RRRRRRR---------------------
| 78.42 Ohms |
| |
------LLLLLLL------
27.67 Ohms
So we have reduced the reactance.
Now calculating inductance we get
Lf=Xf/(2piF) = 44 nH
So by having an intrinsic inductance of 60 nH in our resistor we have
reduced the overal inductance in our circuit to 44 nH.
There, is that good enough circuit analysis for you? :-)
73,
Jon
KE9NA
-------------------------------------
Jon Ogden
KE9NA
http://www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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