Current Mirror – Mathematical Derivation

Consider a basic BJT current mirror consisting of two identical transistors Q1 and Q2.

The base and collector of transistor Q1 are connected together, forcing Q1 to operate in the active region.

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Step 1 : Base-Emitter Voltage Equality

Since both transistors are identical and their bases are connected together:

V_BE1 = V_BE2

For identical transistors operating at the same temperature, equal base-emitter voltage results in equal collector currents.

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Step 2 : Collector Current Equation

For a BJT operating in active region:

I_C = I_S e^(V_BE/V_T)

Where:

• I_S = saturation current
• V_T ≈ 25 mV (thermal voltage at room temperature)

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Step 3 : Equal Collector Currents

Since:

V_BE1 = V_BE2

Therefore:

I_C1 = I_C2

Thus the collector current of Q2 becomes equal to the reference current.

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Step 4 : Reference Current

The reference current is set by resistor R.

I_ref = (V_CC − V_BE) / R

This current flows through transistor Q1.

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Step 5 : Output Current

Since both transistors have equal collector currents:

I_out ≈ I_ref

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Effect of Base Current

Considering base current, the output current becomes:

I_out = I_ref × β / (β + 2)

Where:

• β = transistor current gain

For large β:

I_out ≈ I_ref

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Important GATE Observation

• Ideal current mirror → Iout = Iref
• Practical current mirror slightly lower due to base currents
• Matching of transistors is essential

 

GATE Electrical – Analog Electronics

Page 21 : Current Mirror Circuit

A Current Mirror is an analog circuit used to copy or mirror a reference current into another branch of a circuit.

It is widely used in integrated circuits and operational amplifiers.

                                         

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Basic Idea

The current mirror uses two identical transistors:

• Q1 (Reference transistor)
• Q2 (Output transistor)

If both transistors are matched, the output current becomes approximately equal to the reference current.

Iout ≈ Iref

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Working Principle

The base and collector of the reference transistor (Q1) are connected together. This forces the transistor to operate in the active region.

The same base voltage is applied to the second transistor (Q2).

Since both transistors have identical base-emitter voltage:

VBE1 = VBE2

Therefore the collector currents become equal.

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Reference Current

The reference current is determined by resistor R:

Iref = (VCC − VBE) / R

This current is mirrored to the output transistor.

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Advantages

• Provides constant current
• Useful for biasing circuits
• Simple design
• Widely used in integrated circuits

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Applications

• Operational amplifiers
• Active loads
• Analog IC design
• Current sources

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Important GATE Points

• Current mirror copies reference current.
• Requires matched transistors.
• Output current ≈ reference current.
• Commonly used in differential amplifier circuits.

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Next Page → Widlar Current Source

 

GATE Electrical – Analog Electronics

Page 20 : Differential Amplifier Gain Derivation

The differential amplifier amplifies the difference between two input signals. To understand its operation, we derive the expression for differential gain.

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Basic Differential Amplifier Circuit

The circuit consists of two identical transistors:

• Q1 and Q2
• Collector resistors RC
• Emitter resistor RE
• Two inputs V1 and V2

The output is taken from one of the collectors.

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Differential Input Voltage

The differential input signal is:

Vd = V1 − V2

Each transistor receives half of the differential signal.

Input to each transistor = Vd / 2

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Small Signal Model

For small signal analysis:

• Each transistor has emitter resistance re
• Collector resistor = RC

The emitter resistance is:

re = 25mV / IE

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Output Voltage

The output voltage for one transistor becomes:

Vo = (RC / re) × (Vd / 2)

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Differential Gain

Differential gain is defined as:

Ad = Vo / Vd

Substituting values:

Ad = RC / (2re)

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Important Observations

• Differential gain depends on RC and emitter resistance.
• Higher RC increases gain.
• Lower emitter resistance increases gain.

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Important GATE Points

• Differential amplifier gain = RC / (2re)
• Emitter resistance re = 25mV / IE
• Differential amplifier is the core of Op-Amp circuits
• High CMRR is desirable

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Next Page → Current Mirror Circuit

 

GATE Electrical – Analog Electronics

Page 19 : Differential Amplifier

A Differential Amplifier is a circuit that amplifies the difference between two input signals while rejecting signals common to both inputs.

It is the basic building block of Operational Amplifiers (Op-Amps).

                                                  

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Basic Concept

A differential amplifier has two inputs:

• Input 1 : V₁
• Input 2 : V₂

The output voltage depends on the difference between these two inputs.

Vout = Ad (V1 − V2)

Where:

• Ad = Differential Gain

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Types of Input Signals

1. Differential Mode Input

Two input signals are equal in magnitude but opposite in phase.

Vd = V1 − V2

The amplifier produces maximum output for differential signals.

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2. Common Mode Input

Both inputs receive the same signal.

Vc = (V1 + V2) / 2

Ideally the amplifier should reject common mode signals.

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Common Mode Rejection Ratio (CMRR)

CMRR measures the ability of the amplifier to reject common mode signals.

CMRR = Ad / Ac

Where:

• Ad = Differential gain
• Ac = Common mode gain

In decibels:

CMRR(dB) = 20 log10 (Ad / Ac)

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Applications

• Operational amplifiers
• Instrumentation amplifiers
• Noise rejection circuits
• Signal processing systems

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Important GATE Points

• Differential amplifier amplifies difference between inputs.
• Rejects common signals.
• High CMRR is desirable.
• Foundation of operational amplifiers.

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Next Page → Differential Amplifier Analysis (Gain Derivation)

 

GATE Electrical – Analog Electronics

Page 18 : Miller Effect in Amplifiers

The Miller Effect explains how a small capacitance between the input and output terminals of an amplifier appears as a much larger capacitance at the input.

This effect significantly reduces the high-frequency performance of amplifiers.

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Origin of Miller Effect

In transistor amplifiers, there is a small internal capacitance between:

• Base and Collector in BJT
• Gate and Drain in MOSFET

This capacitance is called the feedback capacitance.

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Miller Capacitance Formula

If a capacitance C exists between input and output, the effective input capacitance becomes:

C_in = C (1 − Av)

Where:

• C = feedback capacitance
• Av = voltage gain of amplifier

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Example

If:

• Feedback capacitance = 2 pF
• Voltage gain Av = −100

Then:

C_in = 2 (1 − (-100)) C_in = 202 pF

Thus a very small capacitor appears very large at the input.

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Effect on Amplifier

• Reduces bandwidth
• Limits high frequency response
• Introduces unwanted feedback

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How to Reduce Miller Effect

• Use Cascade Amplifier
• Use Common Base configuration
• Use Cascode amplifier

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Important GATE Points

• Miller effect increases effective input capacitance.
• It reduces amplifier bandwidth.
• Occurs due to feedback capacitance.
• Cascode amplifier reduces Miller effect.

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Next Page → Differential Amplifier

 

GATE Electrical – Analog Electronics

Page 17 : Bode Plot of Amplifiers

A Bode Plot is a graphical method used to represent the frequency response of an amplifier using a logarithmic frequency scale.

It shows how amplifier gain varies with frequency.

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What is a Bode Plot?

A Bode plot consists of two graphs:

• Magnitude plot (Gain vs Frequency)
• Phase plot (Phase vs Frequency)

For most GATE questions, the magnitude plot is mainly used.

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Gain in Decibels

In Bode plots, amplifier gain is expressed in decibels (dB).

Gain (dB) = 20 log10 (Av)

Where:

• Av = Voltage gain

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Frequency Scale

Frequency is plotted on a logarithmic scale.

Example frequency values:

• 10 Hz
• 100 Hz
• 1 kHz
• 10 kHz
• 100 kHz

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Slope of Bode Plot

The gain decreases with frequency at a certain rate.

Typical slopes are:

• -20 dB / decade
• -40 dB / decade
• -60 dB / decade

Each pole introduces a slope of -20 dB/decade.

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Corner Frequency

The frequency at which gain starts decreasing is called the corner frequency.

At this point gain drops by:

3 dB

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Important GATE Points

• Bode plot uses logarithmic frequency scale.
• Gain is expressed in decibels.
• Each pole contributes -20 dB/decade slope.
• Cut-off frequency corresponds to -3 dB point.

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Next Page → Miller Effect in Amplifiers

 

GATE Electrical – Analog Electronics

Page 17: Frequency Response of Amplifiers

Frequency response describes how the gain of an amplifier changes with the frequency of the input signal. In practical amplifiers, gain is not constant for all frequencies.

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Frequency Response Curve

The frequency response is usually represented by a graph of Voltage Gain (Av) versus Frequency (f).

The amplifier operates in three regions:

• Low Frequency Region
• Mid Frequency Region
• High Frequency Region

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1. Low Frequency Region

At low frequencies, the gain decreases due to the effect of coupling capacitors and bypass capacitors.

Capacitive reactance becomes large, reducing signal transfer.

XC = 1 / (2πfC)

Thus low-frequency gain drops.

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2. Mid Frequency Region

In the mid-frequency range, capacitors behave like short circuits and the amplifier gain becomes approximately constant.

This region provides the maximum and stable gain.

Av ≈ Constant

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3. High Frequency Region

At high frequencies, internal transistor capacitances affect the amplifier operation.

• Base-emitter capacitance
• Base-collector capacitance

These capacitances reduce gain at high frequencies.

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Cut-off Frequencies

Two important frequencies define amplifier bandwidth:

• Lower Cut-off Frequency (fL)
• Upper Cut-off Frequency (fH)

At these frequencies, gain becomes:

Av = 0.707 × Av(max)

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Bandwidth

Bandwidth is the range of frequencies where amplifier gain remains nearly constant.

Bandwidth = fH − fL

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Important GATE Points

• Gain drops at both low and high frequencies.
• Mid-band gain is maximum and constant.
• Cut-off gain = 0.707 × maximum gain.
• Bandwidth = fH − fL.

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Next Page → Bode Plot of Amplifiers