electronic circuit

[logistic_guy]

The figure shows a circuit that performs a low-pass STC function. Such a circuit is known as a first-order low-pass active filter. Derive the transfer function and show that the dc gain is \(\displaystyle \left(-\frac{R_2}{R_1}\right)\) and the \(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency \(\displaystyle \omega_{\tiny 0} = \frac{1}{CR_2}\). Design the circuit to obtain an input resistance of \(\displaystyle 10 \ \text{k}\Omega\), a dc gain of \(\displaystyle 40 \ \text{dB}\), and a \(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency of \(\displaystyle 1 \ \text{kHz}\). At what frequency does the magnitude of the transfer function reduce to unity?

 

[khansaheb]

The figure shows a circuit that performs a low-pass STC function. Such a circuit is known as a first-order low-pass active filter. Derive the transfer function and show that the dc gain is \(\displaystyle \left(-\frac{R_2}{R_1}\right)\) and the \(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency \(\displaystyle \omega_{\tiny 0} = \frac{1}{CR_2}\). Design the circuit to obtain an input resistance of \(\displaystyle 10 \ \text{k}\Omega\), a dc gain of \(\displaystyle 40 \ \text{dB}\), and a \(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency of \(\displaystyle 1 \ \text{kHz}\). At what frequency does the magnitude of the transfer function reduce to unity?

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[logistic_guy]

The transfer function is defined as:

\(\displaystyle T(s) = \frac{V_{o}}{V_{i}}\)

I will take my node to be at the negative terminal. And by applying KCL, I get:

\(\displaystyle \frac{0 - V_{i}}{R_1} + \frac{0- V_{o}}{R_2 \ || \ \frac{1}{sC}} = 0\)

This gives:

\(\displaystyle \frac{V_{o}}{V_{i}} = -\frac{R_2 \ || \ \frac{1}{sC}}{R_1}\)

Let \(\displaystyle Z_R = R_2\) and \(\displaystyle Z_C = \frac{1}{sC}\).

\(\displaystyle Z = Z_R \ || \ Z_C = \left(\frac{1}{Z_R} + \frac{1}{Z_C}\right)^{-1} = \frac{Z_R Z_C}{Z_R + Z_C}\)

Derive the transfer function

\(\displaystyle T(s) = \frac{V_{o}}{V_{i}} = -\frac{Z}{R_1} = -\frac{1}{R_1}\frac{R_2\frac{1}{sC}}{R_2 + \frac{1}{sC}} = -\frac{1}{R_1}\frac{R_2}{sCR_2 + 1}\)

Or

\(\displaystyle T(j\omega) = \textcolor{blue}{-\frac{R_2/R_1}{j\omega CR_2 + 1}}\)

 

[logistic_guy]

show that the dc gain is \(\displaystyle \left(-\frac{R_2}{R_1}\right)\)

The dc gain happens when \(\displaystyle \omega = 0\).

Then,

\(\displaystyle T(0) = -\frac{R_2/R_1}{j(0) CR_2 + 1} = \textcolor{blue}{-\frac{R_2}{R_1}}\)

 

[logistic_guy]

show that the \(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency \(\displaystyle \omega_{\tiny 0} = \frac{1}{CR_2}\).

\(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency means \(\displaystyle |T(j\omega_0)| = \frac{R_2/R_1}{\sqrt{2}}\)

Then,

\(\displaystyle T(j\omega_0) = -\frac{R_2/R_1}{j\omega_0 CR_2 + 1}\)


\(\displaystyle |T(j\omega_0)| = \frac{R_2/R_1}{\sqrt{\omega_0^2 C^2R^2_2 + 1}} = \frac{R_2/R_1}{\sqrt{2}}\)


\(\displaystyle \frac{1}{\sqrt{\omega_0^2 C^2R^2_2 + 1}} = \frac{1}{\sqrt{2}}\)


\(\displaystyle \sqrt{\omega_0^2 C^2R^2_2 + 1} = \sqrt{2}\)


\(\displaystyle \omega_0^2 C^2R^2_2 + 1 = 2\)


\(\displaystyle \omega_0^2 C^2R^2_2 = 1\)


\(\displaystyle \omega_0^2 = \frac{1}{C^2R^2_2}\)


\(\displaystyle \omega_0 = \frac{1}{CR_2}\)

 

[logistic_guy]

Design the circuit to obtain an input resistance of ..........

This part is a little bit challenging, and we will start it slowly but surely!

They have already given us the first piece of the puzzle.

\(\displaystyle R_1 = \textcolor{blue}{10 \ \text{k}\Omega}\)

We can use the given dc gain to find \(\displaystyle R_2\). The magnitude of the dc gain gives the linear gain which in turn gives us the relation between \(\displaystyle R_1\) and \(\displaystyle R_2\).

\(\displaystyle \left|-\frac{R_2}{R_1}\right| = 10^{\frac{\text{gain in dB}}{20}} = 10^{\frac{40}{20}}\)

\(\displaystyle \frac{R_2}{R_1} = 100\)

\(\displaystyle R_2 = 100 R_1 = 100(10000) = \textcolor{blue}{1 \ \text{M}\Omega}\)

 

[logistic_guy]

Design the circuit to obtain .... and a \(\displaystyle 3\)-\(\displaystyle \text{dB}\) frequency of \(\displaystyle 1 \ \text{kHz}\).

We have already found the \(\displaystyle 3\)-\(\displaystyle \text{dB}\) angular frequency, \(\displaystyle \omega_{\tiny 0}\). That is:

\(\displaystyle \omega_{\tiny 0} = \frac{1}{CR_2}\)

Or

\(\displaystyle 2\pi f_{\tiny 0} = \frac{1}{CR_2}\)

Plug in values.

\(\displaystyle 2\pi(1000) = \frac{1}{C(1000000)}\)

This gives:

\(\displaystyle C = 1.59 \times 10^{-10} = \textcolor{blue}{159 \ \text{pF}}\)

 

[logistic_guy]

At what frequency does the magnitude of the transfer function reduce to unity?

\(\displaystyle |T(j\omega)| = \frac{R_2/R_1}{\sqrt{\omega^2 C^2R^2_2 + 1}} = 1\)


\(\displaystyle \omega^2 C^2R^2_2 + 1 = R^2_2/R^2_1\)


\(\displaystyle \omega^2 = \frac{R^2_2/R^2_1 - 1}{C^2R^2_2}\)


\(\displaystyle \omega = \sqrt{\frac{R^2_2/R^2_1 - 1}{C^2R^2_2}}\)


\(\displaystyle 2\pi f = \sqrt{\frac{R^2_2/R^2_1 - 1}{C^2R^2_2}}\)


\(\displaystyle f = \frac{1}{2\pi}\sqrt{\frac{R^2_2/R^2_1 - 1}{C^2R^2_2}} = \frac{1}{2\pi}\sqrt{\frac{1000000^2/10000^2 - 1}{(159 \times 10^{-12})^2 1000000^2}} = \textcolor{blue}{100092 \ \text{Hz}}\)