determine impedance of Twin_T circuit

[akbarza]

hi
I want to draw the impedance of the Twin_T circuit, so I run the above schematic in ltspice. I draw v(v1)/i(v1) to show impedance.
Twin_t impedance must show a figure like a bandpass filter.
namely, what I expect is that impedance rises and in f=1/2pi*r*c=25hz takes a pick and then falls.
but after simulation, only a fall is seen.
what is the problem?

 

[KlausST]

Hi,

If I´m not mistaken there is no current through R2 nor C3.
no current --> you may omit them

The remaining
* C1 in series C2 --> resulting in 50nF
* R1 in sereis with R3 --> 60k

So finally you just have 50nF || 60k

***
It can´t be bandpass or any filter at all, becaue a filter needs a signal input (here V1) .. AND it needs an output. But there is no output at all.

Klaus

 

[FvM]

Twin-T acts as bandstop rather than bandpass. Like KlausST I don't understand the purpose of connecting a voltage source between in- and output.

 

[D.A.(Tony)Stewart]

The bridge impedance is fairly flat except as a LPF , but the transfer function is a notch.

If you wanted 50 Hz, I tweaked it. This scale is 20 dB /div.

The Input Impedance drops from -3dB at 10Hz and drops to -14 dB flat attentuation as shown.

your scale was linear 1 to 10 MHz instead of log.

 

[FvM]

Twin_t impedance must show a figure like a bandpass filter.

Besides unsuitable post #1 test circuit, twin-T will never show an impedance similar to a LC filter because it's operation principle is different. The notch characteristic is achieved by cancelation of a lead and lag RC signal path rather than resonance.

The filter output must be unloaded to achieve the intended function, thus impedance measurement between in- and output is inappropriate.

 

[akbarza]

Twin-T acts as bandstop rather than bandpass. Like KlausST I don't understand the purpose of connecting a voltage source between in- and output.

hi
as I said i want impedance vs. frequency for Twin_T circuit

 

[KlausST]

Hi

hi
as I said i want impedance vs. frequency for Twin_T circuit

as already said: A filter has an input and has an output.

Doing a measurement the way you do makes no sense .. and thus gives meaningless results.
For surre you are free to do so ... then, why are you not satisfied with your result?

***
You may give us additional information what´s the idea behind measuring impedance vs frequency?

Klaus

 

[FvM]

hi
as I said i want impedance vs. frequency for Twin_T circuit

If you measure twin-T impedance, it has to be done the way Tony did.

 

[D.A.(Tony)Stewart]

Back in 1975, I used an array of Twin-T's in the audio band tuned to different notes in one octave of the audio scale, with the intent for a quadrapalegic to whistle 2 notes to select and row and column and dial a phone or toggle a light. Mode B scanned slowly and they just had to blow in the mike to select row and column using an LED display.

In order to set the frequency tolerance of my detectors, I needed a BPF where I could vary bandwidth and thus Q without affecting the peak gain much. So I invented my serendipity bandpass filter. This was part of my grad. thesis project.

In this filter as the 1st stage impedance ratio is just a 1st order filter, the output of one T is loaded by the other when combined on the other T node to create the 2nd order resonance effect by having "almost due to tolerance errors", exactly the same amplitude and complementary phase to cancel out. This vector impedance ratio of one T loading the other and visa versa creates the notch. When used with negative feedback, with a high Q , that maximizes the forward gain. In my old case, I needed to adjust the tolerance to match the users precision to hum or whistle a note with the bandwidth of the filter, but without changing the gain.

It looked something like this but in the middle C octave.

 

[BradtheRad]

I surmise that Akbarza might wish to discover whether output amplitude is higher depending on whether we use low Farad values or high Farad values in the caps, and conversely low or High ohm values of the resistors.

Some RC values might drop less voltage and thus have less need for amplification.

I agree Tony's setup (post #4) is correct. The first image inputs the signal correctly and takes output from the correct node.