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I am trying to calculate the input impedance of a multiple feedback low pass filter. What I need is the simplest symbolic expression so that later I fill in the values and get the impedence itself:

enter image description here

I assume that I am on the right track in calculation, the thing is I can not go any further and simplify the equation more. Can someone please help with putting calculated $V_2$ (down in the writings) into equation $(2)$ and then put $V_{in}$ into equation $(1)$ and simplify it so that $i_1$ will be removed?

So the question is how to substitute and simplify to get a clear $Z_{in}$ with no current in the final equation.

Thanks in advance!

enter image description here

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  • $\begingroup$ I know how painful those equations will be ; you will probably have a more successful answer if you post this on physics stack exchange or something. $\endgroup$ – Patrick Da Silva Feb 13 '12 at 7:02
  • $\begingroup$ I agree with Patrick: I think physics.stackexchange.com would yield you better answers. $\endgroup$ – user2093 Feb 13 '12 at 7:16
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    $\begingroup$ Actually from the description it sounds like the question is just asking about some algebraic manipulations, namely how to substitute $V_2$ into $V_\text{in} = V_2 + i_1R_1$ and simplify it. If that's the case it will be off topic on Physics. $\endgroup$ – David Z Feb 13 '12 at 7:28
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    $\begingroup$ Cross-posted to EE.SE; voted to close here. $\endgroup$ – J. M. is a poor mathematician Feb 13 '12 at 8:31
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    $\begingroup$ You waited less than a day, as in "Rome wasn't built in a." $\endgroup$ – Gerry Myerson Feb 13 '12 at 11:50
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I tried answering in electronics.stackexchange.com, but it seems that LaTeX isn't supported there. The derivation seems to be correct.

$v_2=\frac{i_1}{sC_2+\frac{1}{R_3}+\frac{sC_5R_3+1}{sC_5R_3R_4}}=i_1\frac{sC_5R_3R_4}{s^2C_2C_5R_3R_4+sC_5R_4+sC_5R_3+1}$

$Z_{in}=\frac{V_{in}}{i_1}=\frac{v_2+i_1R_1}{i_1}=\frac{sC_5R_3R_4}{s^2C_2C_5R_3R_4+sC_5R_4+sC_5R_3+1}+R_1$

Is this what you are looking for?

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    $\begingroup$ Thanks a lot! I'll check the values and report back if this is correct! $\endgroup$ – Sean87 Feb 13 '12 at 13:06
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    $\begingroup$ Hi Joel, thanks again your answer was correct (Checked it with LTSpice simulation) $\endgroup$ – Sean87 Feb 25 '12 at 22:18

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