In the given circuit below, what will be the value of V_{0} ?

Option 2 : 7 V

**Concept:**

- One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
- The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.
- An Op-Amp inverting input (-) is at zero potential (A virtual ground), even though it does not have a galvanic connection to the ground. This is because of feedback due to Rf and the high gain of the Op-Amp.
- The virtual ground is only valid if the Op-Amp gain is much greater than the circuit programmed gain (Rf / R1) in the given figure.

**Calculation:**

Now applying virtual circuit to given circuit:

Let the potential at the non-inverting terminal be V

**Applying KCL at the non-inverting terminal:**

\(\frac{V-V_a}{R} \ + \ \frac{V-V_b}{R} \ + \ \frac{V-V_c}{R}=0\)

\(\frac{V-1}{R} \ + \ \frac{V-2}{R} \ + \ \frac{V-3}{R}=0\)

3V = 6

V = 2

Now due to the virtual ground concept, V will appear across inverting terminal.

**Applying KCL at inverting terminal**

\(\frac{V}{2} \ + \ \frac{V-V_0}{5} =0\)

As V = 2

\(\frac{2}{2} \ + \ \frac{2-V_0}{5} =0\)

2 - V_{0} = -5

**V _{0} = 7 V**

**Hence option (2) is the correct answer.**

In the circuit shown below, the output at B is

Option 1 :

__Concept__:

For an op-amp with feedback and input resistances as shown, the gain when input is at the inverting terminal is given as;

\(\frac{{{V_o}}}{{{V_i}}} = - \frac{{{R_f}}}{{{R_1}}}\)

Since the gain of the above configuration is always negative, it is known as the inverting configuration.

**Analysis:**

Now, if R_{f} _{ }= R_{1}, then the OP-AMP output will be negative to that of OP-AMP input.

Option 1 is the correct choice.

The output Vo of the ideal OpAmp used in the circuit shown below is 5 V. Then the value of resistor RL in (kΩ) ?

Option 3 : 25

**Concept: **

This Opamp is a negative feedback amplifier (negative a terminal of input connected to output) and assumes it is ideal (assume gain = ∞).

In any opamp, if negative feedback is there and Opamp is ideal then we will apply the virtual short concept.

Assume voltages of that node is Va and Vb (From fig. 1) So according to the virtual short concept, node voltage of V_{a} and V_{b} will be equal (Due to high input impedance)

⇒ V_{a} = V_{b }.......(1)

**Calculation:**

V_{o} **= **5 V

Apply KCL at node b;

(V_{b} - 1)/10k + (V_{b} - V_{o})/10k + V_{b}/R_{L}k = 0

⇒ V_{b} = 6R_{L}/(2R_{L} + 10) ----(2)

Apply KCL at node a;

V_{a} /10k + (V_{a} - V_{o})/10k = 0

**⇒ V _{o} = 2V_{a} ** ----(3)

From (1), (2), and (3) we will get the value of R_{L} is,

**R _{L} = 25 kΩ**

Option 2 : -4.4 V

__Concept__:

For an inverting amplifier, as shown above, the voltage gain is given by:

\(A_v=\frac{V_0}{V_{in}}=-\frac{R_f}{R_i}\) ---(1)

**Calculation:**

R_{f} = 22 kΩ

R_{1} = 10 kΩ

V_{in} = 2 V

From equation (1):

\(A_v=\frac{V_0}{V_{in}}=-\frac{22}{10}=-2.2\)

V_{0} = -2.2 × 2 =** -4.4 V**

**Hence option (2) is the correct asnwer.**

Assuming that the Op-amp in the circuit shown is ideal, V_{o} is given by

Option 4 : \(- 3{V_1} + \frac{{11}}{2}{V_2}\)

__Concept__**:**

An ideal op-amp follows the virtual ground concept, i.e. the positive and negative terminals of the Op-amp remains at the same potential:

V_{+} = V_{-}

Also, the Inputs terminals current of an ideal opamp is zero.

__Calculation__**:**

Using KCL at the negative terminal of the Opamp, we can write:

\(\frac{{{V_1} - {V_2}}}{R} = \frac{{{V_2}}}{{2R}} + \frac{{{V_2} - {V_0}}}{{3R}}\)

On solving it, we get:

\({V_{out}} = \frac{{11}}{2}{V_2} - 3{V_1}\)

Consider the circuit with and ideal OPAMP shown in the figure.

Assuming |V_{IN}| << |V_{CC}| and |V_{REF}| << |V_{CC}|, the condition at which V_{OUT} equals to zero is

Option 2 : VIN = VREF

__Concept:__

In the below circuit V_{1} is connected to the ground, so V_{1} = 0.

Thus V_{2} also will be at ground potential.

__Note:__

1. This concept is valid only when negative feedback is applied to op-amp like in inverting amplifiers

2.When op-amp gain has a low value then also this concept is not valid.

**Analysis:**

Using virtual ground concept:

V_{A} = V_{B} = 0 → ground

Applying KVL at node A:

\(\frac{{{V_i} - 0}}{R} + \frac{{ - {V_{Ref}} - 0}}{R} = \frac{{0 - {V_0}}}{{{R_F}}}\)

V_{0} = 0

**VIN = V _{REF}**

A circuit with an ideal OPAMP is shown in the figure. A pulse V_{IN} of 20 ms duration is applied to the input. The capacitors are initially uncharged,

The output voltage V_{OUT} of this circuit t = 0^{+} (in integer) is ______V.

**Method – 1**

Since we are applying at t = 0 a short duration pulse.

At t = 0^{-}; V_{in} = 0; V_{c} (0^{-}) = 0

Since V_{c} (0^{+}) = V_{c} (0^{-}) = 0 V

At t = 0^{+} capacitor will behave as a short circuit.

At t = 0^{+}; V_{in} = 5 V

Since it behaves like an inverting amplifier

\({V_{out}} = \frac{{ - {R_2}}}{{{R_1}}}{V_{in}}\)

R_{2} = 10 kΩ

R_{1} = 0 kΩ

V_{out} = -∞

Since V_{out} < -V_{sat} so V_{out} = -12 V, hence it will work like a capacitor.

**Method – 2**

Using the Laplace transform method

_{CeQ} = 1 + 1 = 2 μF

In s domain –

By virtual ground concept V_{A} = 0

Applying KCL at node A-

\(\frac{{{V_A}\left( s \right) - {V_{out}}\left( s \right)}}{{10}} + \frac{{{V_A}\left( s \right) - {V_{in}}\left( s \right)}}{{\frac{1}{{2s}}}} = 0\)

V_{out} = -10 × 25 V_{in} (s)

\({{V_{in}}\left( s \right) = \frac{5}{s}\left\{ {1 - {e^{ - 20 \times 10 ^{- 3}s}}} \right\}}\\{\begin{array}{*{20}{c}} {}\\ {{V_{in}}\left( t \right) = 5\left\{ {u\left( t \right) - 4\left( {t - 20 \times {{10}^{ - 3}}} \right)} \right\}} \end{array}}\)

\({V_{out}} = \frac{{ - 20s}}{s} \times 5\left\{ {1 - {e^{ - 20 \times 10 ^{- 3}s}}} \right\}\)

Applying initial value theorem in –

\({V_{out}}\left( {{0^ + }} \right) = \mathop {\lim }\limits_{s \to \infty } s\;{V_{out}}\left( s \right)\)

= \(\mathop {\lim }\limits_{s \to \infty } - 100\;s\left\{ {1 - {e^{ - 20 \times 10 - 3s}}} \right\}{e^{ - \infty }} \simeq 0\)

V_{out} (0^{+}) = -∞

Since V_{out} (0^{+}) < -V_{sat} so V_{out} = -V_{sat} = -12 V

Option 2 : 6.24 K ohms

__Concept__:

For an inverting amplifier, as shown above, the voltage gain is given by:

\(A_v=\frac{V_0}{V_{in}}=-\frac{R_f}{R_i}\)

R_{i} = Input resistance

R_{f} = Feedback resistance

__Calculation__:

Given A_{v} = 3.9 and R_{in} = 1.6 kΩ

\(3.9=\frac{R_f}{1.6k}\)

R_{f} = 3.9 × 1.6k

**R _{f} = 6.24 kΩ**

__Important Point__:

The voltage gain of a non-inverting amplifier is given by:

\({A_v} = 1 + \frac{{{R_2}}}{{{R_1}}}\)

In the given circuit below, what will be the value of V_{0} ?

Option 2 : 7 V

**Concept:**

- One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
- The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.
- An Op-Amp inverting input (-) is at zero potential (A virtual ground), even though it does not have a galvanic connection to the ground. This is because of feedback due to Rf and the high gain of the Op-Amp.
- The virtual ground is only valid if the Op-Amp gain is much greater than the circuit programmed gain (Rf / R1) in the given figure.

**Calculation:**

Now applying virtual circuit to given circuit:

Let the potential at the non-inverting terminal be V

**Applying KCL at the non-inverting terminal:**

\(\frac{V-V_a}{R} \ + \ \frac{V-V_b}{R} \ + \ \frac{V-V_c}{R}=0\)

\(\frac{V-1}{R} \ + \ \frac{V-2}{R} \ + \ \frac{V-3}{R}=0\)

3V = 6

V = 2

Now due to the virtual ground concept, V will appear across inverting terminal.

**Applying KCL at inverting terminal**

\(\frac{V}{2} \ + \ \frac{V-V_0}{5} =0\)

As V = 2

\(\frac{2}{2} \ + \ \frac{2-V_0}{5} =0\)

2 - V_{0} = -5

**V _{0} = 7 V**

**Hence option (2) is the correct answer.**

In the circuit shown below, the output at B is

Option 1 :

__Concept__:

For an op-amp with feedback and input resistances as shown, the gain when input is at the inverting terminal is given as;

\(\frac{{{V_o}}}{{{V_i}}} = - \frac{{{R_f}}}{{{R_1}}}\)

Since the gain of the above configuration is always negative, it is known as the inverting configuration.

**Analysis:**

Now, if R_{f} _{ }= R_{1}, then the OP-AMP output will be negative to that of OP-AMP input.

Option 1 is the correct choice.