BJT - Differential Amplifier (Small Signal Analysis - Differential Gain, Common mode Gain and CMRR)

Key Concepts

• Differential Amplifier
• Differential Gain (Ad)
• Common Mode Input Signal
• Common Mode Rejection Ratio (CMRR)
• Small Signal Analysis
• Virtual Ground
• Single-Ended Input/Output
• Differential Input/Output
• Emitter Resistor
• Input Impedance
• Active Load

Small Signal Analysis of Differential Amplifier

Introduction

The video delves into the small signal analysis of a differential amplifier to understand its behavior in amplifying differential input signals and rejecting common mode input signals. It also derives the expression for the Common Mode Rejection Ratio (CMRR), a key figure of merit.

Small Signal Equivalent Circuit

• Assumptions: DC sources are set to zero (VCC as short circuit, constant current source as open circuit). BJTs are replaced with their small signal models.
• The resulting equivalent circuit shows the transistors replaced by their small-signal models (rπ, gmVπ, etc.).

Virtual Ground at Node P

• Concept: The analysis aims to prove that node P acts as a virtual ground when the differential input signal is small.
• KVL Application: KVL is applied to both loops involving the transistors to relate input voltages (Vin1, Vin2) to Vπ1, Vπ2, and Vp.
  • Vin1 - Vπ1 = Vp
  • Vp = Vin2 - Vπ2
• Differential Input Voltage (Vid): Defined as Vin1 - Vin2. It's assumed Vin1 = Vid/2 and Vin2 = -Vid/2.
• KCL Application: KCL is applied at node P, summing the currents through the transistors' input resistances and dependent current sources.
  • gm1*Vπ1 + gm2*Vπ2 + Vπ1/rπ1 + Vπ2/rπ2 = 0
• Matching Assumption: Assuming perfectly matched transistors, gm1 = gm2 = gm and rπ1 = rπ2 = rπ.
• Derivation: Through simplification and substitution, it's shown that Vp = 0, confirming the virtual ground concept.

Differential Gain (Ad) - Differential Input and Differential Output

• Simplified Circuit: With node P as a virtual ground, the circuit is redrawn, effectively creating two common-emitter amplifier configurations.
• Output Voltages:
  • Vo1 = -gm * Vπ1 * Rc = -gm * (Vid/2) * Rc
  • Vo2 = -gm * Vπ2 * Rc = gm * (Vid/2) * Rc
• Differential Output Voltage (Vod): Vod = Vo1 - Vo2 = -gm * Vid * Rc
• Differential Gain (Ad): Ad = Vod / Vid = -gm * Rc

Differential Gain (Ad) - Single-Ended Input and Differential Output

• Scenario: Input applied only at one end, the other grounded.
• KVL Application: Vin1 - Vπ1 + Vπ2 = 0 or Vin1 = Vπ1 - Vπ2
• Output Voltages:
  • Vo1 = -gm * Vπ1 * Rc
  • Vo2 = -gm * Vπ2 * Rc
• Differential Output Voltage (Vod): Vod = Vo1 - Vo2 = -gm * Rc * (Vπ1 - Vπ2) = -gm * Rc * Vin1
• Differential Gain (Ad): Ad = Vod / Vin1 = -gm * Rc. Since Vid = Vin1, Ad = Vod / Vid = -gm * Rc.
• Conclusion: The differential gain remains the same even with a single-ended input.

Differential Gain (Ad) - Differential Input and Single-Ended Output

• Scenario: Output measured only at one end (e.g., Vo1).
• Differential Gain (Ad): Ad = Vo1 / Vid
• Derivation: Vo1 = -gm * Rc * Vπ1 = -gm * Rc * (Vid/2). Therefore, Ad = -gm * Rc / 2.
• Conclusion: The differential gain is halved when taking a single-ended output. This is because the swing of the other side is not utilized.

Differential Gain with Resistor Biasing

• Concept: The analysis shows that the differential gain expression remains the same even when the differential amplifier is biased using a resistor instead of a constant current source.
• Small Signal Equivalent Circuit: The circuit is modified to reflect resistor biasing.
• Procedure: The procedure to find the differential gain is similar to the constant current source biasing.

Effect of Emitter Resistors (Re)

• Concept: Adding emitter resistors increases the range of the differential input but reduces the differential gain.
• Small Signal Equivalent Circuit: The circuit includes emitter resistors Re.
• Analysis: Each transistor acts as a common-emitter amplifier with an emitter resistor.
• Voltage Gain: The voltage gain for each transistor is −Rc / (1/gm + Re).
• Conclusion: The differential gain is reduced due to the presence of the emitter resistors.

Input Impedance (Zin)

• Method: A voltage source Vx is applied between the two inputs, and the current Ix is calculated. Zin = Vx / Ix.
• KVL Application: Vx = Vπ1 - Vπ2
• Current Relationships: Vπ1 = Ix * rπ1 and Vπ2 = -Ix * rπ2
• Derivation: Vx = Ix * rπ + Ix * rπ = 2 * Ix * rπ. Therefore, Zin = Vx / Ix = 2 * rπ.
• Conclusion: The input impedance of the differential amplifier is 2 * rπ, where rπ = β / gm.

Common Mode Gain (Acm) and CMRR

• Ideal vs. Non-Ideal Current Source: Ideally, the differential amplifier should completely reject common mode signals. However, with a non-ideal current source (finite output impedance Ree), the output voltage responds to changes in the common mode input signal (Vcm).
• Small Signal Equivalent Circuit: The circuit includes the output impedance Ree of the current source.
• Assumptions: Perfectly matched transistors (Vπ1 = Vπ2 = Vπ, gm1 = gm2 = gm, rπ1 = rπ2 = rπ).
• KVL Application: Vcm = Vπ1 + Vp, where Vp is the voltage at the emitter node.
• Emitter Current (Ie): Ie = Vπ / rπ + gm * Vπ = Vπ * (1/rπ + gm)
• Voltage Vp: Vp = 2 * Ie * Ree = 2 * Ree * Vπ * (gm + 1/rπ) ≈ 2 * Ree * Vπ * gm (since gm >> 1/rπ)
• Common Mode Input Voltage (Vcm): Vcm = Vπ + Vp = Vπ * (1 + 2 * gm * Ree)
• Output Voltage (Vo1): Vo1 = -gm * Vπ * Rc
• Common Mode Gain (Acm): Acm = Vo1 / Vcm = -gm * Rc / (1 + 2 * gm * Ree) ≈ -Rc / (1/gm + 2 * Ree)
• Impact of Finite Ree: A finite Ree results in a non-zero common mode gain.

CMRR with Mismatched Collector Resistors

• Scenario: One collector resistor is slightly different (Rc + ΔR).
• Output Voltages:
  • Vo1cm = -gm * Vcm * Rc / (1 + 2 * gm * Ree)
  • Vo2cm = -gm * Vcm * (Rc + ΔR) / (1 + 2 * gm * Ree)
• Differential Output Voltage (Vodcm): Vodcm = Vo1cm - Vo2cm = -gm * Vcm * ΔR / (1 + 2 * gm * Ree)
• Common Mode Gain (Acm): Acm = Vodcm / Vcm = -gm * ΔR / (1 + 2 * gm * Ree) ≈ -ΔR / (2 * Ree)
• Differential Gain (Ad): Ad ≈ gm * Rc
• Common Mode Rejection Ratio (CMRR): CMRR = |Ad / Acm| = (gm * Rc) / (ΔR / (2 * Ree)) = 2 * gm * Ree / (ΔR / Rc)
• Conclusion: High CMRR requires a high Ree and a small ΔR.

Differential to Single-Ended Conversion and Active Load

• Problem: While differential outputs offer good common mode rejection, single-ended outputs are often needed to connect to loads. Simply taking the output from one side halves the differential gain and reduces CMRR due to the finite output impedance of the current source.
• Solution: Active loads are used with differential amplifiers to improve CMRR in single-ended output configurations.

Conclusion

The small signal analysis provides a detailed understanding of the differential amplifier's behavior, including its gain, input impedance, and common mode rejection capabilities. The analysis highlights the importance of matched components and a high output impedance current source for optimal performance. The video concludes by mentioning the use of active loads to improve CMRR in single-ended output configurations.