Test 2 - Analog Circuits (Electronics and Communication)

For the positive half of $V_i,$ the diode $D_1$ is forward bias, $D_2$ is reverse bias and the zener diode is in breakdown state because state because $V_i > 6.8$

Thus output voltage is $V_0 = 0.7 + 6.8 = 7.5V$

For the negative half of $V_i$, $D_2$ is forward bias thus

Then $V_0 = -0.7 V$

$$CMMR = \dfrac{A_d}{A_c}$$

OR

$$20\ log\ CMMR = 20\ log\ A_d - 20\ log\ A_c$$

OR

$$ = 48-2 = 46 dB$$

Where $A_d \rightarrow$ Differential Voltage Gain
and $A_c \rightarrow$ Common Mode Voltage Gain

$$
\text{This is a Low pass filter, because}\\
\begin{array}{ll}
At\quad \omega = \infty\qquad & \dfrac{V_o}{V_{in}} = 0 \\
and\ at\ \omega = 0 \qquad & \dfrac{V_o}{V_in} = 1
\end{array}
$$

$$I_E = I_s\left(\dfrac{V_{BE}}{enV_r} - 1\right) = 10^{-13}\left(\dfrac{0.7}{e^{1 \times 26 \times 10^{-3}}}-1\right) = 49\ m
A $$

Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e.

$$
\begin{aligned}
I_R = & \quad I_D \\
or \qquad \dfrac{V_i}{R} = & \quad I_s e^{\frac{V_0}{V_r}} \\
or \qquad V_D = &\quad V_T\ ln \dfrac{V_i}{I_sR} \\
\text{For the first condition}\\
V_D = 0 - V_{o1} = V_T \ ln \ \dfrac{2}{I_sR}\\
\text{For the first condition}\\
V_D = 0 - V_{o1} = V_T \ ln \ \dfrac{4}{I_sR}\\
\text{Subtracting the above equation}\\
V_{o1} - V_{o2} = &\quad V_T\ ln\dfrac{4}{I_sR} - V_T\ ln \dfrac{2}{I_sR} \\
or \quad V_{o1} - V_{o2} = &\quad V_T\ ln\dfrac{4}{2} = V_T\ ln 2
\end{aligned}
$$

The $R_2$ decide only the frequecy

The forward resistance of each diode is

$$
\begin{aligned}
r & = \dfrac{V_T}{I_C} \dfrac{25 \ mV}{1 \ mA } = 25\Omega \\
Thus \quad V_{ac} \quad & = V_i \times \left( \dfrac{4(r)}{4(r) + 9900} \right) \\
& = 100 \ mV \ cos(\omega t)0.01 = 1\ cos(\omega t)
\ mV
\end{aligned}
$$

The current flows in the circuit if all the diodes are forward biased. In forward biased there will be 0.7 drop across each diode.
Thus $ I_{DC} = \dfrac{12.7-4(0.7)}{9900} = 1\ mA$

An ideal OPAMP is an ideal voltage controlled voltage source.

This circuit having two diode and capacitor pair in parallel, works as voltage doubler.

The CE configuration has high voltage gain as well as high current gain. It performs basic functions of amplifications. The CB configuration has lowest $R_i$ and highest $R_o$. It is used as last step to match a very low impedance source and to drain a high impedance load.
Thus cascade amplifier is a multistage configuration of CE-CB