Topic 3: Class B Push-Pull Amplifiers & The 78.5% Proof
Exam-Focused Concept Note
Core Concepts (Short Note): A Class B amplifier is biased exactly at cutoff (0 V), meaning the transistor only conducts for exactly 180° (one-half) of the input signal cycle. While this zero-bias approach drastically reduces wasted standby power, a single Class B transistor heavily distorts the output because half of the waveform is missing. To solve this, a “Push-Pull” configuration is utilized. It uses two transistors operating on alternating half-cycles—one pushes the positive half of the signal to the load, and the other pulls the negative half—combining them to reproduce a full 360° output. This alternating arrangement significantly boosts the maximum power efficiency to 78.5% while canceling out even-harmonic distortion.
Key Differences: Class A vs. Class B Push-Pull Amplifiers
| Parameter | Class A Transformer-Coupled | Class B Push-Pull |
|---|---|---|
| Operating Cycle | 360° (Constantly ON) | 180° per transistor (Alternating) |
| Maximum Efficiency | 50% | 78.54% |
| No-Signal Power Waste | High (Maximum heat dissipation at 0 input) | Zero (Draws 0 A when there is no signal) |
| Distortion Components | Very Low (but contains all harmonics if overdriven) | Even harmonics are cancelled out completely (leaves mainly 3rd harmonic) |
| Power Supply Draw | Constant average current from the supply | Current draw fluctuates with the signal size |
1. The “Push-Pull” Concept
Core Concepts: A single Class B transistor cannot provide a faithful reproduction of the input signal. The push-pull circuit acts as a team of two complementary halves:
- Why is it called “Push-Pull”? During the positive half-cycle of the input signal, the first transistor is driven into conduction and “pushes” current into the load. During the negative half-cycle, the first transistor turns off and the second transistor turns on, “pulling” current from the load.
- Harmonic Cancellation: The symmetrical nature of the push-pull operation has a massive mathematical advantage: it balances out and completely eliminates all even harmonics in the output, leaving only the odd harmonics (like the third harmonic) as the principal source of distortion.
Exact PYQs to Master:
- Why is the push-pull power amplifier called so? (Asked heavily in: 2020, 2019, 2018, 2017, 2015)
2. Power & Dissipation Formulas for Class B
Core Formulas: In a Class B push-pull amplifier, the current drawn from the supply is a rectified signal.
- DC Input Power: The total power drawn from the supply uses the average current ().
- AC Output Power: The power delivered to the load () using peak voltage () is:
- Power Dissipated by Transistors: The power wasted as heat by both transistors combined is the difference between input and output power: (Note: To find the dissipation of a single transistor, just divide by 2).
3. The 78.5% Maximum Efficiency Proof (Must-Master)
The Core Derivation: This is the single most heavily tested mathematical proof in your syllabus. You must memorize this step-by-step deduction.
- Define Efficiency:
- Substitute the Base Formulas: Simplifying this gives the general efficiency equation:
- Apply the Maximum Condition: Maximum efficiency occurs when the peak output voltage swing reaches the supply voltage limit. Set .
- Final Calculation: Substitute into the equation:
Exact PYQs to Master:
- Show that the maximum efficiency of push pull power amplifier is 78.5%. (Asked in: 2019)
- Deduce the expression for maximum efficiency of push-pull power amplifier. (Asked in: 2018, 2017, 2015)
4. The Class B Design Numerical (PYQ Blueprint)
Core Process: Class B numericals are direct “plug-and-chug” questions. They usually ask for the maximum power conditions. When the prompt asks for “maximum input/output power,” you must instantly assume the maximum voltage swing () and use these specific limit formulas:
- Maximum AC Output Power:
- Maximum DC Input Power:
Exact PYQ to Master (The 2021 Clone):
- For a class B amplifier using a supply of and driving a load of , determine the maximum input power and output power. (Asked in: 2021) (Solution check based on Boylestad Example 12.8: Max Output Power = . Max Input Power = ).