For the non-inverting amplifiers as shown, find the closed-loop voltage gain.

Option 4 : 11

__Concept of Virtual Ground:__

- The differential input voltage Vid between the noninverting and inverting input terminals is essentially zero.
- This is because even if the output voltage is few volts, due to a large open-loop gain of the op-amp, the difference voltage Vid at the input terminals is almost zero.

Where Vid is differential voltage, Vin1 is noninverting voltage, Vin2 is inverting voltage.

If the output voltage is 10 V and A i.e., the open-loop gain is 104 then,

V out = A Vid

Vid = V out / A

= 10 / 104

= 1 mV.

Hence Vid is very small, for analyzing the circuit assumed to be zero.

Vid = Vin1 - Vin2

(Vin1 - Vin2) = V out / A

= V out / ∞ = 0

__Calculation:__

Circuit diagram:

Two terminals of Op-Amp i.e.; Inverting Terminal and Non-Inverting Terminal are at equipotential.

Apply KCL at node 1V_{pp},

\(\frac{{1{V_{pp}} - 0}}{{10\;k}} + \frac{{1{V_{pp}} - {V_0}}}{{100k}} = 0\)

\(\frac{{10{V_{pp}} + 1{V_{pp}} - {V_0}}}{{100K}} = 0\)

\(11{V_{pp}} - {V_0} = 0\)

Closed-loop gain is given by the ratio of output to the input.

so the Closed-loop voltage gain is given by,

\(\frac{{{V_0}}}{{{V_{pp}}}} = 11\)

Option 2 : 1000 kHz

__Concept:__

The gain of a typical op-amp is inversely proportional to frequency. An op-amp is characterized by its gain-bandwidth product.

For example, an op-amp with a gain-bandwidth product of 1 MHz would have a gain of 5 at 200 kHz, and a gain of 1 at 1 MHz. This low-pass characteristic is introduced deliberately because it tends to stabilize the circuit by introducing a dominant pole. This is known as frequency compensation.

__Calculation:__

gain × bandwidth = 1 MHz

At any frequency, the gain bandwidth product will be equal = 1 MHz

gain = 1

∴ Bandwidth = \(\frac{1~MHz}{1}\)

Bandwidth = 1000 kHz

The circuit shown below uses an ideal OpAmp. Output V0 in volt is ________ (rounded off to one decimal place).

**Concept of Virtual Ground:**

- The
**differential input voltage V**is essentially_{id}between the non-inverting and inverting input terminals**zero.** - This is because even if the output voltage is few volts, due to a large open-loop gain of an op-amp, the difference voltage V
_{id}at the input terminals is almost zero.

Where V_{id }is differential voltage, V_{in1} is non-inverting voltage, V_{in2} is inverting voltage.

If the output voltage is 10 V and A i.e., open-loop gain is 10^{4} then,

V _{out }= A V_{id}

V_{id} = V _{out} / A

= 10 / 10^{4}

= 1 mV.

Hence V_{id} is very small, for analysing the circuit assumed to be zero.

V_{id} = V_{in1} - V_{in2}

(V_{in1} - V_{in2}) = V _{out} / A

= V _{out }/ ∞ = 0

**Calculation: **

Circuit diagram:

Two terminals of Op-Amp i.e._{; }Inverting Terminal and Non-Inverting Terminal is at equipotential.

V_{a} = 50 mV

Apply KCL to node V_{a},

\(\frac{{{V_a} - 0}}{{12}} + \frac{{{V_a} - {V_b}}}{{10}} = 0\)

\(\frac{{10{V_a} - 0 + 12{V_a} - 12{V_b}}}{{120}} = 0\)

\(22{V_a} - 12{V_b} = 0\)

Put the value of Va,

\(22 \times 50 = 12\;{V_b}\)

\(\frac{{22 \times 50}}{{12}} = {V_b}\)

\({V_b} = 91.667\;mV\)

Apply KCL to node V_{b},

\(\frac{{{V_b} - {V_a}}}{{10}} + \frac{{{V_b}}}{1} + \frac{{{V_b} - {V_0}}}{{10}} = 0\)

\(\frac{{{V_b} - {V_a} + 10{V_b} + {V_b} - {V_0}}}{{10}} = 0\)

\(12{V_b} - {V_a} = {V_0}\)

Put the values of Va& Vb, we get

\(12 \times 91.667 - 50 = {V_0} \)

**V _{0} = 1050 mV**

**V _{0 }= 1.050 V**

Option 2 : A bistable multivibrator

- Schmitt trigger is basically bistable multivibrator
- Multivibrator which has both the state stable is called a bistable multivibrator
- Schmitt trigger is to convert any regular or irregular shaped input waveform into a square wave pulse.

For the non-inverting amplifiers as shown, find the closed-loop voltage gain.

Option 4 : 11

__Concept of Virtual Ground:__

- The differential input voltage Vid between the noninverting and inverting input terminals is essentially zero.
- This is because even if the output voltage is few volts, due to a large open-loop gain of the op-amp, the difference voltage Vid at the input terminals is almost zero.

Where Vid is differential voltage, Vin1 is noninverting voltage, Vin2 is inverting voltage.

If the output voltage is 10 V and A i.e., the open-loop gain is 104 then,

V out = A Vid

Vid = V out / A

= 10 / 104

= 1 mV.

Hence Vid is very small, for analyzing the circuit assumed to be zero.

Vid = Vin1 - Vin2

(Vin1 - Vin2) = V out / A

= V out / ∞ = 0

__Calculation:__

Circuit diagram:

Two terminals of Op-Amp i.e.; Inverting Terminal and Non-Inverting Terminal are at equipotential.

Apply KCL at node 1V_{pp},

\(\frac{{1{V_{pp}} - 0}}{{10\;k}} + \frac{{1{V_{pp}} - {V_0}}}{{100k}} = 0\)

\(\frac{{10{V_{pp}} + 1{V_{pp}} - {V_0}}}{{100K}} = 0\)

\(11{V_{pp}} - {V_0} = 0\)

Closed-loop gain is given by the ratio of output to the input.

so the Closed-loop voltage gain is given by,

\(\frac{{{V_0}}}{{{V_{pp}}}} = 11\)

Option 3 : zero

__An op-amp has the following characteristics:__

- Input impedance (Differential or Common-mode) = very high (ideally infinity)
- Output impedance (open loop) = very low (Ideally zero)
- Voltage gain = very high (ideally infinity)
- Common-mode voltage gain = very low (ideally zero), i.e. Vout = 0 (ideally), when both the inputs are at the same voltage, i.e. (zero "offset voltage")
- Output can change instantaneously (Infinite Slew Rate)
- The purpose of bias current is to achieve the ideal behavior in op-amp which is high CMRR, high differential gain, and high input impedance

Characteristics Parameter |
Ideal value |
Practical value |

Voltage Gain (Av) |
∞ |
≈ 106 |

Input Resistance (Ri) |
∞ |
≈ 1 MΩ |

Output Resistance (R0) |
0 |
≈ 10 Ω to 100 Ω |

Bandwidth (B.W) |
∞ |
≈ 1 MHz |

Common mode rejection ratio (CMRR) |
∞ |
≈ 106 or 120 dB |

Slew Rate (S.R) |
∞ |
≈ 80 V/μs |

Find out the type of controller from the circuit diagram given below.

Option 2 : PI-controller

**A schematic diagram for the different controllers is shown below:**

Option 2 : 1000 kHz

__Concept:__

The gain of a typical op-amp is inversely proportional to frequency. An op-amp is characterized by its gain-bandwidth product.

For example, an op-amp with a gain-bandwidth product of 1 MHz would have a gain of 5 at 200 kHz, and a gain of 1 at 1 MHz. This low-pass characteristic is introduced deliberately because it tends to stabilize the circuit by introducing a dominant pole. This is known as frequency compensation.

__Calculation:__

gain × bandwidth = 1 MHz

At any frequency, the gain bandwidth product will be equal = 1 MHz

gain = 1

∴ Bandwidth = \(\frac{1~MHz}{1}\)

Bandwidth = 1000 kHz

Find out the type of controller from the circuit diagram given below.

Option 2 : PI-controller

**A schematic diagram for the different controllers is shown below:**

The circuit shown below uses an ideal OpAmp. Output V0 in volt is ________ (rounded off to one decimal place).

**Concept of Virtual Ground:**

- The
**differential input voltage V**is essentially_{id}between the non-inverting and inverting input terminals**zero.** - This is because even if the output voltage is few volts, due to a large open-loop gain of an op-amp, the difference voltage V
_{id}at the input terminals is almost zero.

Where V_{id }is differential voltage, V_{in1} is non-inverting voltage, V_{in2} is inverting voltage.

If the output voltage is 10 V and A i.e., open-loop gain is 10^{4} then,

V _{out }= A V_{id}

V_{id} = V _{out} / A

= 10 / 10^{4}

= 1 mV.

Hence V_{id} is very small, for analysing the circuit assumed to be zero.

V_{id} = V_{in1} - V_{in2}

(V_{in1} - V_{in2}) = V _{out} / A

= V _{out }/ ∞ = 0

**Calculation: **

Circuit diagram:

Two terminals of Op-Amp i.e._{; }Inverting Terminal and Non-Inverting Terminal is at equipotential.

V_{a} = 50 mV

Apply KCL to node V_{a},

\(\frac{{{V_a} - 0}}{{12}} + \frac{{{V_a} - {V_b}}}{{10}} = 0\)

\(\frac{{10{V_a} - 0 + 12{V_a} - 12{V_b}}}{{120}} = 0\)

\(22{V_a} - 12{V_b} = 0\)

Put the value of Va,

\(22 \times 50 = 12\;{V_b}\)

\(\frac{{22 \times 50}}{{12}} = {V_b}\)

\({V_b} = 91.667\;mV\)

Apply KCL to node V_{b},

\(\frac{{{V_b} - {V_a}}}{{10}} + \frac{{{V_b}}}{1} + \frac{{{V_b} - {V_0}}}{{10}} = 0\)

\(\frac{{{V_b} - {V_a} + 10{V_b} + {V_b} - {V_0}}}{{10}} = 0\)

\(12{V_b} - {V_a} = {V_0}\)

Put the values of Va& Vb, we get

\(12 \times 91.667 - 50 = {V_0} \)

**V _{0} = 1050 mV**

**V _{0 }= 1.050 V**