GATE Electrical – Analog Electronics

Page 23 : Wilson Current Mirror

The Wilson Current Mirror is an improved version of the basic current mirror that provides better current accuracy and higher output resistance.

It uses three transistors instead of two and reduces the error caused by base currents.

                                                 

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Why Wilson Current Mirror?

In a simple current mirror, the output current is slightly smaller than the reference current because of base current losses.

The Wilson current mirror compensates for this error.

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Circuit Structure

• Three matched transistors (Q1, Q2, Q3)
• Reference resistor R
• Feedback connection through Q3

The third transistor provides feedback that stabilizes the output current.

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Working Principle

The circuit works by feeding back the collector current through the third transistor. This feedback increases the effective output resistance and improves current accuracy.

As a result:

Iout ≈ Iref

with much smaller error compared to the simple current mirror.

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Advantages

• Higher output resistance
• Improved current accuracy
• Reduced base current error
• Better current stability

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Applications

• Analog integrated circuits
• Operational amplifiers
• Precision current sources
• Bias circuits in IC design

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Important GATE Points

• Wilson current mirror uses three transistors.
• Provides higher output resistance.
• Improves current mirror accuracy.
• Widely used in analog IC design.

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Next Page → Operational Amplifier (Op-Amp) Introduction

 

GATE Electrical – Analog Electronics

Page 22 : Widlar Current Source

                                                   

The Widlar Current Source is a modification of the basic current mirror that allows the generation of very small output currents without using large resistor values.

It is widely used in integrated circuits and operational amplifiers.

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Basic Idea

In a simple current mirror, the output current is nearly equal to the reference current. However, sometimes we require a much smaller current.

The Widlar current source solves this problem by adding an emitter resistor to the output transistor.

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Circuit Structure

• Two matched transistors (Q1 and Q2)
• Reference resistor R
• Emitter resistor RE in output transistor

The resistor in the emitter of Q2 reduces the output current.

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Working Principle

Since the bases of both transistors are connected together:

VBE1 = VBE2 + IE2 RE

Because of this voltage drop across RE, the collector current of Q2 becomes smaller than the reference current.

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Widlar Current Equation

The relationship between reference current and output current is:

Iref / Iout = e^(Iout RE / VT)

Where:

• VT = thermal voltage ≈ 25 mV

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Advantages

• Generates very small currents
• Suitable for IC design
• Requires smaller resistors

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Applications

• Operational amplifier bias circuits
• Integrated circuit current sources
• Analog IC design

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Important GATE Points

• Widlar source produces smaller current than reference current.
• Emitter resistor reduces output current.
• Widely used in integrated circuits.

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Next Page → Wilson Current Mirror

 

Common Emitter Amplifier Gain using h-Parameters

The h-parameter model is widely used to analyze transistor amplifiers. For a Common Emitter (CE) amplifier, four parameters are used:

                                        

• hie → input resistance
• hre → reverse voltage gain
• hfe → forward current gain
• hoe → output admittance

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Step 1 : Input Voltage Equation

From the h-parameter model:

V1 = hie I1 + hre V2

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Step 2 : Output Current Equation

Output current is given by:

I2 = hfe I1 + hoe V2

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Step 3 : Output Voltage

The output voltage across load resistor RL:

V2 = − I2 RL

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Step 4 : Substitute I2

Substituting the current equation:

V2 = − RL ( hfe I1 + hoe V2 )

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Step 5 : Solve for Voltage Gain

After simplification:

Av = V2 / V1

For practical CE amplifier where hre and hoe are very small:

Av ≈ − hfe RL / hie

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Important Result

Voltage Gain (CE amplifier)

Av ≈ − (hfe × RL) / hie

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Important GATE Points

• CE amplifier gain is negative (phase inversion).
• Higher hfe increases gain.
• Larger load resistance increases gain.
• Small hre and hoe usually neglected.

 

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)