High pass CLC T Filter component values

[LazyTurtle]

I have a multitapped inductor with fixed values. Given this inductor I'm guessing if it is possible to design a switched CLC T filter with specific  centre frequencies, not the ones calculated for the original circuit the multitapped inductor was in.
Would it be possible to calculate the capacitor value for a specific frequency without taking into account the impedance of the line?

Is there a specific equation to obtain the capacitor value for each frequency/inductance value?

I have run LTSpice simulations and obtained specific values for desired frequencies given a inductance value but I can't see how each value of the capacitor relates mathematically to the other ones...

On the other hand, how does not taking into account the impedance value of the line really affect the overall performance/sound?

 

[gyraf]

There is no way around knowing source and load impedances. Look at e.g. http://www.wa4dsy.net/filter/filterdesign.html for dimensions..

Jakob E.

 

[abbey road d enfer]

Is there a specific equation to obtain the capacitor value for each frequency/inductance value?

There are many online calcs dealing with cross-overs. If you know the inductance, you get the capacitors. Note that x-overs assume zero-ohm source impedance, which is OK if you insert the filter after an active stage. However, the response shape varies with the load impedance. The higher the load, the more resonant the response, with a peak at the corner frequency.

I have run LTSpice simulations and obtained specific values for desired frequencies given a inductance value but I can't see how each value of the capacitor relates mathematically to the other ones...

Please re-formulate; I don't understand your question.

On the other hand, how does not taking into account the impedance value of the line really affect the overall performance/sound?

It does, somewhat. There are many actual units in which the response is not exactly what it claims to be. How noticeable it is depends very much on the listener's abilities.
In addition tehre are several effects bopxes that incorporate a deliberately resonant high-pass filter, which actually compensates the absence of low-bass with a boost of low-mid.

 

[LazyTurtle]

Please re-formulate; I don't understand your question.

What I´m trying to explain is that I don´t know which is the transfer function of this filter so that I can obtain capacitor value from inductance value and frequency.

 

[abbey road d enfer]

What I´m trying to explain is that I don´t know which is the transfer function of this filter so that I can obtain capacitor value from inductance value and frequency.

You should target a 3rd-order Butterworth response.

 

[LazyTurtle]

You should target a 3rd-order Butterworth response.

If we take the circuit of the image we can first divide the circuit to be analyzed in two. The first one I have used C1 and L and calculate the transfer function from calculating as a voltage divider.

For the first part I get:

Va/Vin = 1/ (1-(1/(L*C1*s^2))

For the second part I will apply:

Vout = Va*(RL/(RL+1/(s*C2)) --> As C1=C2
Vout/Vin=(1/ (1-(1/(L*C*s^2)))*(RLoad/(RLoad+1/(C*s)) --> Vout/Vin = RLoad*C^2*s^3 /(RLoad*C^2*s^3+L*C*s^2+R*C*s+1)

With this result I will transform s -> jw and divide complex from real parts and check the real value.

 

[abbey road d enfer]

If we take the circuit of the image we can first divide the circuit to be analyzed in two. The first one I have used C1 and L and calculate the transfer function from calculating as a voltage divider.

For the first part I get:

Va/Vin = 1/ (1-(1/(L*C1*s^2))

For the second part I will apply:

Vout = Va*(RL/(RL+1/(s*C2)) --> As C1=C2
Vout/Vin=(1/ (1-(1/(L*C*s^2)))*(RLoad/(RLoad+1/(C*s)) --> Vout/Vin = RLoad*C^2*s^3 /(RLoad*C^2*s^3+L*C*s^2+R*C*s+1)

With this result I will transform s -> jw and divide complex from real parts and check the real value.

I believe you neglect the effects resulting from interaction between both parts.
In a zero-ohm source impedance configuration, C2=3C1. Only when Zs=ZL, C1 and C2 are equal. The filter has 6dB loss then.

 

[LazyTurtle]

I believe you neglect the effects resulting from interaction between both parts.
In a zero-ohm source impedance configuration, C2=3C1. Only when Zs=ZL, C1 and C2 are equal. The filter has 6dB loss then.

Thanks for your answer, but I don't get it. How, if placing two equal value caps, can we have different capacitances between them depending on source impedance?

Could you give some feedback on the correct transfer function for this filter?

Thanks

 

[abbey road d enfer]

Thanks for your answer, but I don't get it. How, if placing two equal value caps, can we have different capacitances between them depending on source impedance?

You cannot isolate components, you need to consider the whole circuit. Your predicament that capacitors are of equal value is correct only for passive filters where the source and load impedances are equal, which also results in 6dB attenuation.

 

[LazyTurtle]

You cannot isolate components, you need to consider the whole circuit. Your predicament that capacitors are of equal value is correct only for passive filters where the source and load impedances are equal, which also results in 6dB attenuation.

Yes, my intention is to used this as a passive CLC filter, sorry I didn´t mention that before.

 

[abbey road d enfer]

Yes, my intention is to used this as a passive CLC filter, sorry I didn´t mention that before.

But are you sure the source impedance is equal to the load impedance?

 

[LazyTurtle]

You cannot isolate components, you need to consider the whole circuit. Your predicament that capacitors are of equal value is correct only for passive filters where the source and load impedances are equal, which also results in 6dB attenuation.

I understand that having different impedances at the input and output will imply in a different behaviour of the circuit.

This might simply be solved by the following way (for example for and input/output impedance of 600Ohm:

- INPUT: Source Impedance (about 150Ohm)--> Add a resistor in series to get the 600Ohm
- OUPUT; Load impedance (about 10KOhm) --> Add a resistor in parallel to load of about 620Ohm to have a load impedance of 600Ohm.

This way I would get the correct values, but the problem here is that inductance values depend in the impedance so, as having an already fixed value multitapped inductor, I cannot select the inductance values.

 

[LazyTurtle]

I understand that having different impedances at the input and output will imply in a different behaviour of the circuit.

This might simply be solved by the following way (for example for and input/output impedance of 600Ohm:

- INPUT: Source Impedance (about 150Ohm)--> Add a resistor in series to get the 600Ohm
- OUPUT; Load impedance (about 10KOhm) --> Add a resistor in parallel to load of about 620Ohm to have a load impedance of 600Ohm.

This way I would get the correct values, but the problem here is that inductance values depend in the impedance so, as having an already fixed value multitapped inductor, I cannot select the inductance values.

This is the reason I came up with this question...having a fixed inductance value multitapped inductor, is it possible to calculate a capacitor value for this T CLC filter for specific frequencies?

 

[abbey road d enfer]

This is the reason I came up with this question...having a fixed inductance value multitapped inductor, is it possible to calculate a capacitor value for this T CLC filter for specific frequencies?

Yes, it's possible. You have to start with the inductance value, then check if you can get a working alignment. For example, let's say you have a 1mH inductor and you want a cut-off frequency of 100Hz.
A calc such as this one https://rf-tools.com/lc-filter/ will calculate capacitors of 1263uF and a characteristic impedance of 1.25 ohm!
So you can see that for 600 ohms, you need 476mH and capacitors of 2.65uF.
If you chose the (almost) zero source impedance option, the values would be 716mH, 1.77uF and 5.3uF. The big advantage is that there would be no in-band loss.

 

[LazyTurtle]

Yes, it's possible. You have to start with the inductance value, then check if you can get a working alignment. For example, let's say you have a 1mH inductor and you want a cut-off frequency of 100Hz.
A calc such as this one https://rf-tools.com/lc-filter/ will calculate capacitors of 1263uF and a characteristic impedance of 1.25 ohm!
So you can see that for 600 ohms, you need 476mH and capacitors of 2.65uF.
If you chose the (almost) zero source impedance option, the values would be 716mH, 1.77uF and 5.3uF. The big advantage is that there would be no in-band loss.

But that calculator will give you a different inductance value from the one I already have, so it will not be possible .
For example, I have a 27mH inductance for tap 1, I would like to get the capacitors values so that, with that inductance, the centre freq for that filter will be one frequency  I choose.

Would that be possible?

 

[LazyTurtle]

For example.
For an inductance of 27mH, if I set 15nF Capacitors I will get a 5KHz  High Pass filter, as you can see in the LTSpice simulation...

If I dramatically change the Source impedance and load imprdance, de frequency of the filter will remain almost the same, so this tells me that impedance will not be so important with centre frequency but with phase issues...

 

[abbey road d enfer]

But that calculator will give you a different inductance value from the one I already have, so it will not be possible .
For example, I have a 27mH inductance for tap 1, I would like to get the capacitors values so that, with that inductance, the centre freq for that filter will be one frequency  I choose.

Would that be possible?

You have to scale the impedance values, until you make it compatible with the induuctance you have.

 

[abbey road d enfer]

For example.
For an inductance of 27mH, if I set 15nF Capacitors I will get a 5KHz  High Pass filter, as you can see in the LTSpice simulation...

If I dramatically change the Source impedance and load imprdance, de frequency of the filter will remain almost the same, so this tells me that impedance will not be so important with centre frequency but with phase issues...

If you choose a termination impedance of 35 ohms, you have a solution for 100Hz, with 45uF caps. I'm not sure you can really use that, but it shows that there is an infinity of solutions.
I checked your simulation; you have an enormous 9dB hump (resonance) at 8kHz.

 

[LazyTurtle]

If you choose a termination impedance of 35 ohms, you have a solution for 100Hz, with 45uF caps. I'm not sure you can really use that, but it shows that there is an infinity of solutions.
I checked your simulation; you have an enormous 9dB hump (resonance) at 8kHz.

Yes, selecting a very low output impedance will dramatically change the filter frequency.

The 8KHz bump will increase with higher load impedances, but as far as the filter frequency remains similar it will be ok for me...

So, maybe a solution to this would be setting constant input and output impedances in the filter no matter whatever you connect to, but I don't really know how to do this without using transformers...

 

[abbey road d enfer]

The 8KHz bump will increase with higher load impedances, but as far as the filter frequency remains similar it will be ok for me...

I gues a large number, including me, would disagree with you.

So, maybe a solution to this would be setting constant input and output impedances in the filter no matter whatever you connect to, but I don't really know how to do this without using transformers...

As you mentioned earlier, you can always use additional resistors to adjust the impedances.