Option 3 : 20

__Concept__:

Power gain is the ratio of the output power (P0) to the input power (Pin), i.e.

\({A_p} = \frac{{{P_0}}}{{{P_{in}}}}\)

In terms of dB, the power gain is expressed as:

\(P(dB) = 10\log_{10}{\frac{P_{out}}{P_{in}}}\)

**Property: **log_{x} x = 1

__Calculation__:

With A_{P} = 100, we can write:

\(P(dB) = 10\log_{10}{A_P}=10\log_{10}{100}\)

\(P(dB) =10\log_{10}{10^2}\)

\(P(dB) =2\times10\log_{10}{10}\)

Since log_{10}10 = 1, the power gain in dB will be:

\(P(dB) =20 ~dB\)

__Important Note__:

It is the voltage gain or the current gain that is calculated as:

\(A_V=20log(\frac{V_2}{V_1})\)

\(A_I=20log(\frac{I_2}{I_1})\)

Option 2 : 20, 40 and 30 dB

**Concept:**

- A decibel is a
__one-tenth unit__of the Bel (B) which is used to describe a gain of either voltage, current, or power. - It is used to describe the ratio of one magnitude of a power or field quantity to another, on a logarithmic scale, the logarithmic quantity being called a degree of power or a degree of field, respectively.

**Let the voltage gain, current gain and power gain be X**.

Voltage Gain (dB) = 20log(X)

Current Gain (dB) = 20log(X)

Power Gain (dB) = 10log(X)

log(10) = 1

log(X^{n}) = nlog(X)

**Calculation:**

**Given:**

Voltage Gain = 10

Current Gain = 100

**Voltage Gain (dB) = 20log(10) = 20 dB**

**Current Gain (dB) = 20log(100) = 40log(10) = 40 dB**

Power Gain (dB) = 10log(V × I)

Power Gain (dB) = 10log(10 × 100)

**Power Gain (dB) = 10log(10 ^{3}) = 30 dB**

Option 3 : 20

__Concept__:

Power gain is the ratio of the output power (P0) to the input power (Pin), i.e.

\({A_p} = \frac{{{P_0}}}{{{P_{in}}}}\)

In terms of dB, the power gain is expressed as:

\(P(dB) = 10\log_{10}{\frac{P_{out}}{P_{in}}}\)

**Property: **log_{x} x = 1

__Calculation__:

With A_{P} = 100, we can write:

\(P(dB) = 10\log_{10}{A_P}=10\log_{10}{100}\)

\(P(dB) =10\log_{10}{10^2}\)

\(P(dB) =2\times10\log_{10}{10}\)

Since log_{10}10 = 1, the power gain in dB will be:

\(P(dB) =20 ~dB\)

__Important Note__:

It is the voltage gain or the current gain that is calculated as:

\(A_V=20log(\frac{V_2}{V_1})\)

\(A_I=20log(\frac{I_2}{I_1})\)

Option 2 : 20, 40 and 30 dB

**Concept:**

- A decibel is a
__one-tenth unit__of the Bel (B) which is used to describe a gain of either voltage, current, or power. - It is used to describe the ratio of one magnitude of a power or field quantity to another, on a logarithmic scale, the logarithmic quantity being called a degree of power or a degree of field, respectively.

**Let the voltage gain, current gain and power gain be X**.

Voltage Gain (dB) = 20log(X)

Current Gain (dB) = 20log(X)

Power Gain (dB) = 10log(X)

log(10) = 1

log(X^{n}) = nlog(X)

**Calculation:**

**Given:**

Voltage Gain = 10

Current Gain = 100

**Voltage Gain (dB) = 20log(10) = 20 dB**

**Current Gain (dB) = 20log(100) = 40log(10) = 40 dB**

Power Gain (dB) = 10log(V × I)

Power Gain (dB) = 10log(10 × 100)

**Power Gain (dB) = 10log(10 ^{3}) = 30 dB**

Option 3 : 20

__Concept__:

Power gain is the ratio of the output power (P0) to the input power (Pin), i.e.

\({A_p} = \frac{{{P_0}}}{{{P_{in}}}}\)

In terms of dB, the power gain is expressed as:

\(P(dB) = 10\log_{10}{\frac{P_{out}}{P_{in}}}\)

**Property: **log_{x} x = 1

__Calculation__:

With A_{P} = 100, we can write:

\(P(dB) = 10\log_{10}{A_P}=10\log_{10}{100}\)

\(P(dB) =10\log_{10}{10^2}\)

\(P(dB) =2\times10\log_{10}{10}\)

Since log_{10}10 = 1, the power gain in dB will be:

\(P(dB) =20 ~dB\)

__Important Note__:

It is the voltage gain or the current gain that is calculated as:

\(A_V=20log(\frac{V_2}{V_1})\)

\(A_I=20log(\frac{I_2}{I_1})\)

Option 4 : 2 V

**Concept:**

**Differential amplifier:**

- An operational amplifier is internally a differential amplifier with features like high input impedance, low output impedance.
- The differential amplifier configuration is the most widely used building block in analog integrated - circuit design.
**The differential amplifier amplifies the difference between the two input signals.**- In the open-loop differential amplifier shown, input signal V
_{in}_{1}and V_{in2}are applied to the positive (non-inverting) and negative (inverting) input terminals.

- Since the Op-Amp amplifies the differences between the difference between the two input signals, this configuration is called the differential amplifier.
- But, for a practical difference amplifier uses a negative feedback connection to control the voltage gain of the amplifier.

From the circuit, V_{1} = V_{in1} and V_{2} = V_{in2}. And the output voltage V_{0} = A_{d} (V_{in1} - V_{in2})

Where A_{d} is the open-loop gain. Generally, it is a very high value.

__Calculation:__

Given that

Open-loop gain A_{d} = 1000000,

The differential input voltage V_{d} = 2 μV

Then the output voltage **V _{0} = A_{d} V_{d} = 1000000 × 2 × 10^{-6} = 2 V**