# Manual Calculation of filter

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#### Beau Schwabe

##### Active Member
Well... R4, C1 through C6 and L1 and L4 are the only part that will affect Node6 as V1 changes assuming there is very little resistance within the V1 source.

R4 and C4 are basically a low pass RC filter .... The remaining circuit is a low pass PI filter hybrid ... composed of a "T" filter between L1,L4 and C1, while also composed of a "PI" filter between C3,L1,L4, and C5. Other fundamentals could be found at C3,L1 and C1 as well as C5, L4, and C1. C2 and C6 provide some buffering and DC bias.

Reference:

#### sallehzxc

##### New Member
ahhh issit possible for me then, assuming a AC source of 1V is implemented, to manually calculate the cut-off freq of the circuit?

Since the unknown will be with regards to the cut-off frequency that i wish to attain?

can i group the impedances? and do a voltage analysis to determine at node 6 the effective 3db cutoff which is Vinput/Voutput =2?

is my understanding flawed?

Well... R4, C1 through C6 and L1 and L4 are the only part that will affect Node6 as V1 changes assuming there is very little resistance within the V1 source.

R4 and C4 are basically a low pass RC filter .... The remaining circuit is a low pass PI filter hybrid ... composed of a "T" filter between L1,L4 and C1, while also composed of a "PI" filter between C3,L1,L4, and C5. Other fundamentals could be found at C3,L1 and C1 as well as C5, L4, and C1. C2 and C6 provide some buffering and DC bias.

Reference:

bump!

#### Ratchit

##### Well-Known Member
hi guys,

Is it possible for anybody to shed some light on the manual calculation (how to proceed on calculating the filters) based on the components used?
I have attached the components used.

Thank you for the help!!

View attachment 114789
Are R1, R2, and R3 just hanging in the air so as to pick up spurious signals? Or, are they connected to ground? Where is the output, node 6? If so, document and label it. Does the filter have any load? If so, what is it?

Looks like you can write out 6 node equations and put them through an equation solver. Then graph the node response you are interested in. Need help with that?

Ratch

#### sallehzxc

##### New Member
can i get some help on that?
Are R1, R2, and R3 just hanging in the air so as to pick up spurious signals? Or, are they connected to ground? Where is the output, node 6? If so, document and label it. Does the filter have any load? If so, what is it?

Looks like you can write out 6 node equations and put them through an equation solver. Then graph the node response you are interested in. Need help with that?

Ratch
additionally, issit possible for me to just and calculate an overall value based on the capacitor and inductors used and calculate their cutoff freq using
1/2pi(sqrt L and C).

Thank you so much for your help. The R1, R2,R3 are connected to different parts of the schematic so i assumed since they play no effect in affecting the signal coming through to node6, i can just ignore them since it will be troublesome and heavy on the circuit to input all of them together. BUt, if you want, i can put them all through one circuit and put it here.

Thanks again!

#### Ratchit

##### Well-Known Member
can i get some help on that?

additionally, issit possible for me to just and calculate an overall value based on the capacitor and inductors used and calculate their cutoff freq using
1/2pi(sqrt L and C).
NO, you must use calculate the currents coming/going from/to each node. That means doing a node analysis. You can combine parallel capacitors between each node, however. Use a computer to do the grit work.

Thank you so much for your help. The R1, R2,R3 are connected to different parts of the schematic so i assumed since they play no effect in affecting the signal coming through to node6, i can just ignore them since it will be troublesome and heavy on the circuit to input all of them together. BUt, if you want, i can put them all through one circuit and put it here.

Thanks again!
If those resistors are connected to some low impedance point, then they probably will affect the filter output. You can determine the difference by running the calculations with the resistors first grounded and then open.

Ratch

#### sallehzxc

##### New Member
NO, you must use calculate the currents coming/going from/to each node. That means doing a node analysis. You can combine parallel capacitors between each node, however. Use a computer to do the grit work.

If those resistors are connected to some low impedance point, then they probably will affect the filter output. You can determine the difference by running the calculations with the resistors first grounded and then open.

Ratch

Unfortunately, I am really shoddy in my understanding of what you just said sorry!

Is it possible for you to explain further in detail what you meant?

i have attached the full schematic look on LTSpice for your reference! #### Ratchit

##### Well-Known Member
Now you got me confused. In first schematic, you show R4 going to a capacitor. In the second schematic R4 goes to a coil. I think it is best if you use the Spice program to determine the voltages and currents. The circuit is too complicated for manual calculation unless you can first determine where you can separate the various parts of the circuit where they do not interact with other.

Ratch

#### sallehzxc

##### New Member
Now you got me confused. In first schematic, you show R4 going to a capacitor. In the second schematic R4 goes to a coil. I think it is best if you use the Spice program to determine the voltages and currents. The circuit is too complicated for manual calculation unless you can first determine where you can separate the various parts of the circuit where they do not interact with other.

Ratch
Ahh sorry the schematic at the start is just now at a different node as compared to the overall circuit.

So for what you mentioned, should i do an AC analysis or a transient analysis with a voltage source of 1V?

Following that, how should i proceed though? So sorry for the many questions i am at my wits end with regards to this

Thank you so much again

#### JimB

##### Super Moderator
Where did this circuit originate?
At first sight it looks bizarre.

C7, C8, C9 and C10 are in parallel with the signal source and will present a low reactance as the frequency increases.
At 1kHz the reactance will be about 1600 Ohm, and at 1 MHz about 1.6 Ohm.

Turning out attention to two section "pi" filter L1, L4, C3, C1 and C5.
The values here 2.4nH, 0.7pF and 1.5pF are consistent with a filter which is designed for VHF frequencies of several hundred MHz.

So what is the point of this circuit?
I cannot understand its purpose.

JimB

#### sallehzxc

##### New Member
from what i know, it is to only allow a specific frequency to pass through and nothing else.

i was playing around with analysis of the nodes and i got this formula but again, it looks really weird...

ultimately, the aim is to derive if through calculation, before simulation, we can get the expected frequency we want to not be attenuated.  do you guys have any advice for my implementation and the churning out of the value of s and subsequently freq?

#### JimB

##### Super Moderator
from what i know, it is to only allow a specific frequency to pass through and nothing else.
OK, being quite brutal about this, but, knowing nothing is not a good start for circuit analysis.

As for "allowing a specific frequency to pass through", nonsense!
It is a whole load of low pass filtering and a DC block (C2 and C6).
Where did you get this circuit?
It did not just appear out of thin air, surely?

As I said earlier, it is bizarre, surely not a practical circuit by any stretch of the imagination.

JimB

#### popper

##### New Member
LTSpice is node (and device) limited. Get a freebee PSpice or scilab or mathcad.

#### crutschow

##### Well-Known Member
LTSpice is node (and device) limited.
I know of no such limit.
Why do you say it has?

#### sallehzxc

##### New Member
OK, being quite brutal about this, but, knowing nothing is not a good start for circuit analysis.

As for "allowing a specific frequency to pass through", nonsense!
It is a whole load of low pass filtering and a DC block (C2 and C6).
Where did you get this circuit?
It did not just appear out of thin air, surely?

As I said earlier, it is bizarre, surely not a practical circuit by any stretch of the imagination.

JimB
no no thank you i needed that to further press on and ask more on the application of the circuitry.
Thnk you, I was hoping someone could shed light on this but I'll ask again on the application on the circuit and get back here!

Thanks all!!

#### MrAl

##### Well-Known Member
Hello,

You can approach this in several ways for example:
Combine impedances and use voltage divider formula,
Use Nodal,
Use Norton and Thevenin equivalents.

Keep in mind this will be a somewhat high order solution.

Choose your poison #### MrAl

##### Well-Known Member
Hello again,

Here is a solution found by using Nodal Analysis.
The filter order turns out to be 6th order even though there are 8 storage components and that is because of the way some of them are connected directly together.

I included a redraw of the part of the schematic that was of interest here and relabeled some of the components to be sure any readers would know exactly what components i was using. The original schematic was hard to read so i had to do this, and be aware that some of the components may or may not be the original values.

From the frequency response curve we see this is basically a bandpass, but because the 1 percent amplitude frequency is so much lower than the peak frequency, it can be called a low pass filter just being aware that a frequency of zero (DC) will not be passed. That is also evident from observing the capacitors in series.

Note that the 3db down point will be hard to choose without knowing more about the application. That's because there is a peak just before it dips down and choosing it with reference to the peak would be wrong. A typical way to choose would be to go by the nominal input (if the average gain was 0db) which in this illustration was 1 volt, but note a better way might be to average over the output from maybe 50kHz to somewhere just before the peak and use that average as the reference. Again though, it depends on the application. Last edited:

#### JimB

##### Super Moderator
Interesting MrAl.

My understanding of you graph is that the vertical scale is a simple linear voltage, and the horizontal scale is 0 to 10GHz.
Is that correct?

How did you calculate this, was it done with a simulator program of some kind?
If so, to satisfy my curiosity, could you re-run the program but leave out C23, C22 and C27, also change the value of R11 to 100 Ohm.
I expect that we will then see a reasonable LPF characteristic when the two "pi" sections are terminated with 100 Ohms.

It is a pity that sallehzxc has not returned to tell us the origins of this circuit.

JimB

#### MrAl

##### Well-Known Member
Hi there Jim,

Yes, the vertical scale is simply the output voltage amplitude with 1vac input. You can use RMS or Peak values.

It was calculated using Nodal analysis where we sum the currents into each node and then solve the resulting set of equations simultaneously. This is done in the Laplace domain, and then later the variable 's' is replaced with 'jw' and then the magnitude is calculated by taking the square root of the sum of the imaginary part squared and the real part squared. This is all done in an algebraic program that can handle most of the algebra so all i have to do is present it with the set of equations and go from there.
In the Laplace domain the equation simplifies to 6th order which means some of the components might be redundant.

Two of the caps you mention are in series with the filter, so there is no way we will see DC being passed as in a normal LPF. With a normal LPF the DC offset would also be passed. However, if the application never has to deal with the DC offset or there is none, then in the application it would be considered to be a LPF anyway rather than a BP filter.

I will remove the three caps you talk about and see what happens. Once we remove the two series caps i would expect to see at least some DC being passed because then there is no DC blocking cap to stop it I'll try to get back here later tonight or tomorrow sometime with the new graph.

BTW the equation for Vout is somewhat complicated until we plug in the values then it comes out simpler. It turned out to be first degree in the numerator and 6th degree in the denominator.

[LATER]
Here's the new plot with those three caps removed and R11=100 ohms.
Definitely LP now With 1vdc input now we see 0.5 vdc output.
In case you are interested, here is the transfer function in one form for that setup:
Vout/Vin=1.0e2/(4.2336e-50*s^5+1.2096e-39*s^4+8.256e-29*s^3+1.392e-18*s^2+3.38e-8*s+2.0e2)
and we see now it is down to 5th order.

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