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Why is it NOT possible to calculate the transfer function

\$G = Vout/Vin\$

using the voltage divider rule

\$Vout / Vin = Z1 / Z2\$

where

\$Z1 = Xc1 || (R3+R2)\$

is the impedance between Vout and GND (with \$Xc1\$ being the complex capacitor resistance) and where

\$Z2 = R1 + (R2 || (R3 + Xc1))\$

is the impedance between Vin and GND?

I know that the result is wrong and I also know how I could obtain the correct result, but I can't explain myself why this way of calculating it is actually wrong...

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Hm. Because the calculation is completely wrong? R3 is not in series with R2, and C1 is not in parallel to anything, so the calculation of Z1 is wrong. Z2 - I can't even understand the motivation for this calculation. \$\endgroup\$
    – Eugene Sh.
    Jun 4 at 17:35
  • \$\begingroup\$ Use what you know about Rth and Vth to determine the transfer function from Thevenin. \$\endgroup\$ Jun 4 at 17:38
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In voltage-divider mental mode, it looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Above, there are two separate dividers. Starting on the left, \$V_\text{IN}\$ meets its voltage divider that is composed of \$R_1\$ and \$Z_1\$, where \$Z_1\$ is the parallel combination of \$R_2\$ and \$R_3+Z_2\$. This divider yields a voltage, \$V_X\$. \$V_X\$ meets its voltage divider that is composed of \$R_3\$ and \$Z_2\$, where \$Z_2\$ is just \$Z_{C_1}\$. This divider now yields the output voltage, \$V_\text{OUT}\$.

Note: Just a note before seeing the math below. A voltage divider result can be expressed as \$\frac{V_\text{OUT}}{V_\text{IN}}=\frac{R_2}{R_1+R_2}\$. But this is also the same as \$\frac{V_\text{OUT}}{V_\text{IN}}=\frac{1}{1+\frac{R_1}{R_2}}\$. In the newer way of writing, the two parts of the divider appear exactly once in the divider expression. It's a little less cluttered with variables, that way. I'm writing it this way to encourage a more flexible view towards expressions and equations. That's all.

So it goes like this:

$$\begin{align*} \text{Each Stage} \left\{ \begin{array}{rl} V_\text{OUT} = V_X\,\frac{1}{1+\frac{R_3}{Z_2}}&&Z_2 = Z_{C_1}\\\\ V_X = V_\text{IN}\,\frac{1}{1+\frac{R_1}{Z_1}}&&Z_1 = R_2\mid\mid \left(Z_2 + R_3\right) \end{array} \right. \end{align*}$$ $$\therefore \frac{V_\text{OUT}}{V_\text{IN}}=\frac{1}{1+\frac{R_1}{Z_1}}\cdot\frac{1}{1+\frac{R_3}{Z_2}}$$

Can you now compare the method you hold in your mind as correct and compare it with above to see if you can find the flaw in your approach?

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but I can't explain myself why this way of calculating it is actually wrong...

There is no single application of the voltage divider rule which will give you Vout.

But you can apply it twice -- first to find the voltage at the juncture of R1 and R2 and then to find Vout.

Call Vx the voltage where R1 and R2 meet. To determine Vx you have a voltage divider with source Vin, lower leg R2 || (R3+Xc) and upper leg R1.

Now to find Vout you have a voltage divider with source Vx, lower leg Xc and upper leg R3.

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