Q 1 Attempt any 4 questions:

Q 1 a) With neat circuit explain the working of the comparator circuit **(05 marks)**

**Answer :**

A comparator is an electronic circuit which compares two input signals and produces an output. It compares a signal voltage on one input of an Op-Amp with a known voltage called the reference voltage, on the other input.

Fig(a) shows an Op-Amp as a comparator.

A fixed reference voltage V_{ref} (=1v) is applied to the negative input and a time varying input signal is applied to the positive terminal of the Op-Amps. As the ac signal is applied to the positive terminal, the Op-Amp is considered to be operating the non-inverting comparator.

Now,

(i) When V_{in }< V_{ref }

V_{o} ≅ -V_{sat }(≅ -V_{EE}) because voltage at negative terminal is greater than positive terminal.

(ii) When V_{in }> V_{ref}

V_{o} ≅ +V_{sat }(≅ +V_{cc}) because voltage at positive terminal is greater than negative terminal.

Then V_{o} changes from one saturation level to another whenever V_{in }≅_{ }V_{ref}. At any given time, V_{o} waveform shows whether V_{in }is greater or less than V_{ref}.

The output waveform of the comparator is shown in fig(b) & fig(c)

Fig(b) If V_{ref }is positive

The diodes D_{1} & D_{2} are used to protect the Op-Amp from damage due to excessive input voltage.

Fig(c) If V_{ref} is negative

Q 1 b) Write short note on : Bi FET and Bi MOS differential amplifier circuit **(05 marks)**

**Answer :**

**i) Bi MOSFET Differential amplifier pair-**

Fig (a) shown the basic MOSFET differential pair with MOSFETS Q_{1} & Q_{2} that are matched in all respects. I_{ss} is a constant current source similar to the one used for BJT differential amplifier. The circuit above is common current bias. Differential signals are applied to V_{G1} & V_{G2} (equal amplitude but opposite sign). The differential outputs are obtained at V_{D1} & V_{D2}.

Both the MOSFETs are biased to operate in their saturation region. The current source I_{ss} also uses the dual polarity dc supply.

Hence even with V_{G1} = V_{G2} = 0, it is possible to bias both the MOSFETs in the saturation region is both of them will conduct even when V_{G1} = V_{G2} = 0. The common voltage applied to both Q_{1} & Q_{2} are referred to as common mode, V_{CM}. Common mode inputs usually come from noises and the differential pair should reject V_{CM}.

**(ii) Bi FET Differential Amplifier-**

Bi-FETs are Op-Amps which use FET input stage and BJT for other stages. When high input impedance and high slew rate is desired JFETs are used. In Fig(b), a dual input balanced output differential amplifier using JFETs is shown.

The gain of the circuit is given by-

A_d=\frac{R_c}{r_e} Now,\;\frac1{r_e}=gm\;\&\;R_c=R_D A_d=gmR_DQ 1 c) Design a circuit with OP-AMP, resistors and capacitor that stimulates an inductor of 1H **(05 marks)**

**Answer :**

By using Op-Amps, resistors and capacitors we can replace a passive inductor by an active inductor circuit. This circuit utilizes the properties of Op-Amps to mimic the behaviour of an inductor. The circuit shown in the fig(a) below acts as an active inductor by reversing the operation of a capacitor, thus ancting as a stimulated inductor.

Unlike capacitors, an inductor passes low frequencies more easily than high frequencies and in its ideal form has zero resistance. Therefore, it can easily pass dc without limitation but has infinite impedence at infinite frequency.

The value of the inductor is given by

L = R_{L}R_{1}C_{1} (i)

Let R_{L}=R_{1}=1K ohm

Then eq_{n} (i)

1=1×10^{3}×1×10^{3}×C

C=1μF

Q 1 d) For a regulated dc power supply the output varies from 12V to 11.6V when the load current is varied from 0 to 100 mA which is the maximum value of I_{L}. If the ac line voltage and temperature are constant, calculate the load regulation, % load regulation and output resistance of the power supply. **(05 marks)**

**Answer- **

Given-

V_{NL} = 12 V

V_{FL} = 11.6 V

I_{NL} = 0 mA

I_{FL} = 100 mA

Now,

△V = V_{max} – V_{min} = 0.4 V

△I_{L} = I_{L max} – I_{L min} = 100 mA

Load regulation % = Load regulation × 100 = 3.45%

Output\;resistance,\;R_o=\frac{\triangle V}{I_L}=\frac{0.4}{0.1}=4\OmegaQ 1 e) How can the true RMS value of voltage signal to measured using analog multipliers? **(05 marks)**

**Answer- **

There are different methods to measure the true RMS value of voltage signal by using analog multipliers. One such method is the explicit method. In this method, the input signal is first squared by an analog multiplier. The average value is then taken using an appropriate filter, and the square root is taken using an Op-Amp with a second square in the feedback loop.

By virtual short,

\frac{V_o^2}K=\frac{\left(V_{in}\right)^2AVG}KV_o=\sqrt{\left(V_{in}\right)^2AVG}= RMS Value of V_{in}

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