Hello to everbody can you help me how come C6 to be 10p and c5 13.5 and and l1 to be 190n and L6 240n
i start with 100Meghz, with one circuits 50/XL or 50/XC now become three different thanks
i take it from this tutorial

 

How can we possibly tell you why particular components are particular values unless you tell us what the point of the circuit is to begin with, what the goal is, and what parameters you have control over and which ones are fixed?

 

The components are those values because the person that drew the schematic chose those values. Without knowing what the designer was trying to do, it's impossible to guess the "why". I would assume the values were chosen to provide a tuned oscillator. If this was all done in simulation, it's probably useless because the components and the PCB itself will have real inductances and such that the model does not account for.

 

Hi,

Just for fun I'll take a guess...

"Transform the filter into a highpass, lowpass, and bandpass".
possibly with some other criteria also such as input/output impedances.

This could be checked out by analyzing all three circuits and see what they have in common.
Just a guess so this could be true or totally false

 

How can we possibly tell you why particular components are particular values unless you tell us what the point of the circuit is to begin with, what the goal is, and what parameters you have control over and which ones are fixed?

i put the tutorial

 

Others can download and wade through a 1.2 MB tutorial. I don't have the bandwidth.

 

i put the tutorial

Hi,

Ok i see you added the tutorial, but perhaps you can explain what you want to know in your own words and why you want to know that.

 

i put the tutorial

And did you actually read the text in that tutorial, not just the pretty picutres?

 

Hi,

Ok i see you added the tutorial, but perhaps you can explain what you want to know in your own words and why you want to know that.

-------
this
It is very interesting to check the effects of this modifications on the
filter response in Fig.7
The response show that while the filter with
series inductor and capacitor has a symmetrical
response (blue trace), the two inductor filters has
steeper response at higher frequencies (red
trace), this network is normally referred as low
pass coupling. The two capacitor filter has
steeper response at low frequencies (green
trace), this type of coupling is referred as high
pass coupling.
Fig. 7
Note that in the two coupling capacitor filter, the
value of resonant capacitor C5 has been
decreased in order to compensate the frequency
shift caused by the two coupling capacitors, in
the two inductor filter inductance L1 has been
increased to compensate for the frequency shift
caused by the two inductors.

 

-------
this
It is very interesting to check the effects of this modifications on the
filter response in Fig.7
The response show that while the filter with
series inductor and capacitor has a symmetrical
response (blue trace), the two inductor filters has
steeper response at higher frequencies (red
trace), this network is normally referred as low
pass coupling. The two capacitor filter has
steeper response at low frequencies (green
trace), this type of coupling is referred as high
pass coupling.
Fig. 7
Note that in the two coupling capacitor filter, the
value of resonant capacitor C5 has been
decreased in order to compensate the frequency
shift caused by the two coupling capacitors, in
the two inductor filter inductance L1 has been
increased to compensate for the frequency shift
caused by the two inductors.

That's what you consider "in your own words" or telling us what you want to know?

Seems more to me like a direct copy/paste from the tutorial.

 

That's what you consider "in your own words" or telling us what you want to know?

Seems more to me like a direct copy/paste from the tutorial.

yes that is true, i never work with electronics filter design

 

And did you actually read the text in that tutorial, not just the pretty picutres?

i read but they don come those value til i read twoo circuit was good i understand, but now with three circuit i dont know where they come those new value

 

Hi,

Ok i see you added the tutorial, but perhaps you can explain what you want to know in your own words and why you want to know that.

those
how come C6 to be 10p and c5 13.5 and and l1 to be 190n and L6 240n

 

And did you actually read the text in that tutorial, not just the pretty picutres?

yes i read, first circuit was easy , second circuit was easy, third i dont know
how come C6 to be 10p and c5 13.5 and and l1 to be 190n and L6 240n

 

yes that is true, i never work with electronics filter design

i want to know, how come C6 to be 10p and c5 13.5 and and l1 to be 190n and L6 240n

 

those
how come C6 to be 10p and c5 13.5 and and l1 to be 190n and L6 240n

Hello again,

Well the goal of any filter is to meet the design requirements which are usually at least:
1. Resonant peak at some particular frequency Fo or Fc (many kinds of network applications),
2. Match input and output impedances (networks like this one).
3. Possibly a certain circuit Q.

That means to find out what values you need, you have to be able to analyze the circuit first, then figure out how to get the input impedance to match the required spec, the output to meet that required spec, and the resonant peak to match the required frequency spec.
For this circuit it looks like 50 ohms, 50 ohms, and 100MHz. That's three things you have to accomplish, and the solution will provide you with the required component values. You may have some arbitrary choice too, such as the output capacitor value which you may want to match the input capacitor value. You may also want a certain minimum Q.

It looks like what they did was design three different circuits to match the input and output impedance (did not check this yet though) and the resonant peak point frequency. They then analyzed the circuit Q indirectly to compare sharpness.
Note higher Q's require more exact component values and less change in component values over the life of the equipment too.
In some circuits you can change values also to get a higher Q. I dont know if you can do that with these circuits or not but you could try to determine that.

So can you analyze this circuit? After that you try to solve simultaneously for component values that meet all the specs.

 

Hello to everbody can you help me how come C6 to be 10p and c5 13.5 and and l1 to be 190n and L6 240n
i start with 100Meghz, with one circuits 50/XL or 50/XC now become three different thanks
i take it from this tutorial

View attachment 145319

I'll download your tutorial and see if I can help.

 

I'll download your tutorial and see if I can help.

thnx

 

Hello again,

Well the goal of any filter is to meet the design requirements which are usually at least:
1. Resonant peak at some particular frequency Fo or Fc (many kinds of network applications),
2. Match input and output impedances (networks like this one).
3. Possibly a certain circuit Q.

That means to find out what values you need, you have to be able to analyze the circuit first, then figure out how to get the input impedance to match the required spec, the output to meet that required spec, and the resonant peak to match the required frequency spec.
For this circuit it looks like 50 ohms, 50 ohms, and 100MHz. That's three things you have to accomplish, and the solution will provide you with the required component values. You may have some arbitrary choice too, such as the output capacitor value which you may want to match the input capacitor value. You may also want a certain minimum Q.

It looks like what they did was design three different circuits to match the input and output impedance (did not check this yet though) and the resonant peak point frequency. They then analyzed the circuit Q indirectly to compare sharpness.
Note higher Q's require more exact component values and less change in component values over the life of the equipment too.
In some circuits you can change values also to get a higher Q. I dont know if you can do that with these circuits or not but you could try to determine that.
hello thank for answer
yes it is about higher q, but what do you think you have self to select wich is better of there is a formula
but for this one i understand,
the first input load with 50OHM the second reactane in serie with 50OHM is 100OHM load and here is the resultat, but with three circuits they change

 

Hello again,

Well the goal of any filter is to meet the design requirements which are usually at least:
1. Resonant peak at some particular frequency Fo or Fc (many kinds of network applications),
2. Match input and output impedances (networks like this one).
3. Possibly a certain circuit Q.

That means to find out what values you need, you have to be able to analyze the circuit first, then figure out how to get the input impedance to match the required spec, the output to meet that required spec, and the resonant peak to match the required frequency spec.
For this circuit it looks like 50 ohms, 50 ohms, and 100MHz. That's three things you have to accomplish, and the solution will provide you with the required component values. You may have some arbitrary choice too, such as the output capacitor value which you may want to match the input capacitor value. You may also want a certain minimum Q.

It looks like what they did was design three different circuits to match the input and output impedance (did not check this yet though) and the resonant peak point frequency. They then analyzed the circuit Q indirectly to compare sharpness.
Note higher Q's require more exact component values and less change in component values over the life of the equipment too.
In some circuits you can change values also to get a higher Q. I dont know if you can do that with these circuits or not but you could try to determine that.

So can you analyze this circuit? After that you try to solve simultaneously for component values that meet all the specs.

Yes this is for higher Q