What is Operational Amplifiers: Inverting vs Non-Inverting Gain and Virtual Ground?
Mathematical Foundation
Laws & Principles
- The 'ideal' op-amp assumptions used in these calculations: infinite open-loop gain, infinite input impedance, zero output impedance, and infinite bandwidth. Real op-amps (LM741, TL072, OPA2134) deviate from these, especially at high frequencies.
- Gain-Bandwidth Product (GBW): for a given op-amp, gain × bandwidth = constant. An LM741 (GBW = 1 MHz) at G=10 has only 100 kHz of usable bandwidth. At G=100, only 10 kHz. Choose op-amps with GBW >> gain × signal frequency.
- Supply rails limit Vout: the output cannot exceed ±(Vsupply − 1.5V) for standard op-amps, or almost ±Vsupply for rail-to-rail types. If the calculated Vout exceeds the supply, the output saturates at the rail.
- The inverting amplifier has a fixed, low input impedance of Rin (typically kΩ). This can load the source. The non-inverting amplifier has essentially infinite input impedance, making it ideal for high-impedance source buffering.
- Summing amplifier: the inverting topology extends to a summing amplifier by adding N input resistors all to V−. Vout = −Rf × (V1/R1 + V2/R2 + … Vn/Rn). This is the foundation of analog mixing consoles.
Step-by-Step Example Walkthrough
" Designing a microphone preamplifier: mic output = 5mV peak, need output = 500mV peak for ADC input. "
- Target gain: 500mV / 5mV = 100.
- Choose non-inverting (no phase inversion needed for audio).
- Formula: G = 1 + Rf/Rin = 100 → Rf/Rin = 99.
- Choose Rin = 1kΩ, Rf = 100kΩ (standard E24 values, ratio = 100, giving G = 101 — close enough).
- Verify GBW: using NE5532 (GBW = 10 MHz), max audio bandwidth = 10 MHz / 101 ≈ 99 kHz. Well above 20 kHz audio range. ✓
- Check supply: Vout_peak = 500mV. With ±5V supply, output has ±3.5V headroom. No saturation. ✓