Amplifier calculations form the bedrock of analog circuit design, RF engineering, and audio system integration. Whether you are biasing a discrete BJT stage, configuring the feedback network on an NE5532 op-amp, or chaining RF low-noise amplifiers (LNAs), you must be able to move fluently between linear ratios and logarithmic decibel (dB) scales. This guide strips away the abstract theory and provides the exact formulas, symbol definitions, and step-by-step worked problems you need on the bench.
The Core Amplifier Gain Formulas and Symbol Definitions
Gain is fundamentally a ratio of output to input. We express this in two ways: as a dimensionless linear multiplier (useful for calculating exact voltages) and in decibels (useful for cascading multiple amplifier stages, where multiplication becomes simple addition).
Linear and Logarithmic Equations
Voltage Gain (Linear): Av = Vout / Vin
Voltage Gain (Decibels): Av(dB) = 20 × log10(Vout / Vin)
Power Gain (Linear): Ap = Pout / Pin
Power Gain (Decibels): Ap(dB) = 10 × log10(Pout / Pin)
Symbol Definition Table
| Symbol | Definition | Standard Unit |
|---|---|---|
Av | Linear Voltage Gain (dimensionless ratio) | V/V |
Av(dB) | Voltage Gain expressed in decibels | dB |
Ap | Linear Power Gain (dimensionless ratio) | W/W |
Ap(dB) | Power Gain expressed in decibels | dB |
Vout, Vin | Output and Input Voltage (must be same type: RMS, peak, or p-p) | Volts (V) |
Pout, Pin | Output and Input Power | Watts (W) |
log10 | Base-10 logarithm | N/A |
Assumptions and Realistic Magnitudes
These formulas assume the amplifier is operating in its linear region (the output is not clipping or saturating against the supply rails). For power gain calculations to directly correlate with voltage gain via impedance, the input and output impedances must be matched. If they are not matched, voltage gain and power gain diverge.
What does a realistic answer look like? A standard LM386 audio amplifier IC has a default linear voltage gain (Av) of 20, which translates to roughly 26 dB. A high-performance RF LNA might specify a power gain of 15 dB to 30 dB. If your bench calculation yields a voltage gain of 10,000 (80 dB) for a single-stage audio op-amp, you have likely made a decimal error or the op-amp is acting as an open-loop comparator, not a linear amplifier.
Rearranged Forms for Reverse Engineering
On the bench, you rarely just calculate gain. More often, you know your required output swing and your amplifier's datasheet gain, and you need to find the necessary input drive. Here are the rearranged forms solving for every variable.
Voltage Rearrangements
- Solve for Output Voltage (Linear):
Vout = Vin × Av - Solve for Input Voltage (Linear):
Vin = Vout / Av - Solve for Output Voltage (from dB):
Vout = Vin × 10(Av(dB) / 20) - Solve for Input Voltage (from dB):
Vin = Vout / 10(Av(dB) / 20)
Power Rearrangements
- Solve for Output Power (Linear):
Pout = Pin × Ap - Solve for Input Power (Linear):
Pin = Pout / Ap - Solve for Output Power (from dB):
Pout = Pin × 10(Ap(dB) / 10) - Solve for Input Power (from dB):
Pin = Pout / 10(Ap(dB) / 10)
Worked Problems with Strict Unit Tracking
The most common point of failure in amplifier tutorials is sloppy unit tracking. Below are two bench-realistic problems with explicit intermediate steps.
Problem 1: Voltage Gain in dB from Mixed Units
Scenario: You inject a 15 mVRMS sine wave into a microphone preamplifier. Your oscilloscope reads an output of 3.2 VRMS. Calculate the voltage gain in dB.
- Normalize Units: The formula requires identical units. Convert millivolts to Volts.
Vin = 15 mV = 0.015 V - Calculate the Linear Ratio:
Ratio = Vout / Vin = 3.2 V / 0.015 V = 213.33 - Apply the Base-10 Logarithm:
log10(213.33) = 2.329 - Multiply by the Voltage Constant (20):
Av(dB) = 20 × 2.329 = 46.58 dB
Answer: The amplifier provides 46.58 dB of voltage gain.
Problem 2: Output Power Calculation using dBm and dB
Scenario: An RF signal generator feeds an amplifier. The input power is -10 dBm. The amplifier datasheet specifies a power gain of 26 dB. Find the output power in milliwatts and Watts.
- Identify the Scale Types: dBm is an absolute power unit (referenced to 1 mW). dB is a relative ratio. You can add a relative ratio (dB) to an absolute power (dBm) to get a new absolute power (dBm).
- Add the Gain to the Input Power:
Pout(dBm) = Pin(dBm) + Ap(dB)
Pout(dBm) = -10 dBm + 26 dB = +16 dBm - Convert dBm to Milliwatts: The formula for dBm is
P(dBm) = 10 × log10(P / 1mW). Rearranging to solve for P:
Pout(mW) = 10(16 / 10) = 101.6 = 39.81 mW - Convert Milliwatts to Watts:
Pout(W) = 39.81 mW / 1000 = 0.03981 W
Answer: The output power is 39.81 mW (or roughly 40 mW).
Critical Unit Mistakes That Break Amplifier Calculations
Even seasoned engineers make unit errors when moving fast. Watch out for these three calculation killers:
Voltage gain uses
20 × log10, while power gain uses 10 × log10. If you accidentally use the power formula (10 log) for a voltage ratio, your calculated dB value will be exactly half of what it should be. As explained in All About Circuits' semiconductor textbook, the factor of 20 arises because power is proportional to voltage squared, and the exponent drops down during the logarithm operation.
Gain is a ratio, meaning the units cancel out. Therefore, you can use Peak-to-Peak (Vpp) for both input and output, or RMS for both. But if your function generator displays Vpp and your multimeter reads VRMS, your ratio will be skewed by a factor of 2.828 (which is 2√2). Always verify your measurement domain before dividing.
dB is strictly a ratio (Gain = 20 dB). dBm is an absolute power level relative to 1 milliwatt (Signal = -40 dBm). You cannot convert a voltage directly into dBm without knowing the impedance of the system, because
P = V2 / R. A 1V signal into 50Ω is a very different dBm value than a 1V signal into 600Ω.
Frequently Asked Questions
How do I calculate amplifier efficiency from power calculations?
Efficiency (η) is the ratio of useful AC output power delivered to the load versus the total DC power drawn from the supply rails. The formula is η = (Pout(AC) / Pin(DC)) × 100. To find Pin(DC), multiply your supply voltage (VCC) by the quiescent and load-dependent supply current (ICC). A Class-A amplifier typically maxes out around 25% to 50% efficiency, while a Class-D switching amplifier can exceed 90%.
Why do amplifier calculations use 20 log for voltage but 10 log for power?
This is rooted in Joule's first law and Ohm's law. Power is defined as P = V2 / R. If we take the power ratio of two signals across the same resistance, the R cancels out, leaving P2/P1 = (V2/V1)2. When you apply the base-10 logarithm to a squared term, the exponent moves to the front as a multiplier: log(x2) = 2 × log(x). Therefore, 10 × log(Pratio) becomes 10 × 2 × log(Vratio), which equals 20 × log(Vratio).
What happens to amplifier calculations if the input and output impedances are different?
If the input impedance (Zin) and output/load impedance (Zout) are not equal, the simple relationship between voltage gain and power gain breaks down. Voltage gain (Av) remains strictly Vout/Vin regardless of impedance. However, power gain must be calculated using the actual power values (P = V2/Z). For example, an op-amp might have a massive voltage gain, but if it is driving a high-impedance load, the actual power delivered (and thus the power gain) might be quite small. In RF design, where 50Ω matching is standard, this distinction is critical to prevent signal reflections and ensure maximum power transfer.
