Transimpedance Amplifier Calculator (Photodiode TIA Rf, Cf, Bandwidth)
By Santaji GadepatilinUpdated November 8, 2026
A transimpedance amplifier turns a photodiode's tiny current into a voltage, Vout = I × Rf, but the photodiode's capacitance and the op-amp's gain-bandwidth limit how fast it can be. This transimpedance amplifier calculator gives the output voltage, the achievable −3 dB bandwidth and the feedback capacitor for a flat (Butterworth) or 45° phase-margin response — see how a transimpedance amplifier works with a photodiode.
How to use the transimpedance amplifier calculator
Enter the photodiode current, the feedback resistor, the total input capacitance (photodiode junction capacitance at its bias plus the op-amp’s common-mode and differential input capacitance) and the op-amp’s gain-bandwidth product. The calculator returns the output voltage, the achievable bandwidth and two values of feedback capacitor: the larger gives a flat Butterworth response, the smaller is the minimum for a 45° phase margin (faster, with some peaking). To find Rf for a target output, press × on Rf.
Transimpedance amplifier formulas
Vout = Ip × Rf · f−3dB ≈ √(GBW ÷ (2π Rf Cin))
Cf (Butterworth) ≈ √(Cin ÷ (π Rf GBW)) · Cf (45°) ≈ √(Cin ÷ (2π Rf GBW))
These approximations hold when Cf is much smaller than Cin, which is the usual case.
Worked example
A photodiode with 20 pF total capacitance producing 10 µA, Rf = 100 kΩ and a 10 MHz op-amp:
- Vout = 10 µA × 100 kΩ = 1 V
- Bandwidth = √(10 MHz ÷ (2π × 100 kΩ × 20 pF)) = 892 kHz
- Feedback capacitor: 2.52 pF for a flat response (minimum 1.78 pF for 45° phase margin) — use the nearest standard value, e.g. 2.7 pF
Achievable bandwidth (20 pF input capacitance)
| Rf | GBW 1 MHz | GBW 10 MHz | GBW 100 MHz | Cf at 10 MHz |
|---|---|---|---|---|
| 10 kΩ | 892.1 kHz | 2.82 MHz | 8.92 MHz | 7.98 pF |
| 100 kΩ | 282.1 kHz | 892.1 kHz | 2.82 MHz | 2.52 pF |
| 1,000 kΩ | 89.2 kHz | 282.1 kHz | 892.1 kHz | 0.80 pF |
| 10,000 kΩ | 28.2 kHz | 89.2 kHz | 282.1 kHz | 0.25 pF |
Practical notes
- Values of Cf below about 0.5 pF are swamped by the resistor’s own parasitic capacitance (≈ 0.1–0.2 pF for an 0805 part); use a T-network or a lower Rf.
- Choose an op-amp with FET or CMOS inputs — bias current flows through Rf and appears directly as an output offset.
- Photovoltaic mode (zero bias) gives the lowest dark current and noise; photoconductive mode (reverse bias) gives lower capacitance and higher speed.
- Thermal noise of Rf is √(4kT/Rf) in current terms, so a larger Rf improves signal-to-noise as well as gain.
Frequently asked questions
What is the gain of a transimpedance amplifier?
The transimpedance gain is Rf in volts per amp — Vout = I × Rf. A 1 MΩ feedback resistor gives 1 V per µA.
What limits transimpedance amplifier bandwidth?
The input capacitance (photodiode junction plus op-amp input) and the op-amp GBW. With the right feedback capacitor the −3 dB bandwidth is about √(GBW ÷ (2π Rf Cin)), so raising Rf tenfold cuts bandwidth by about √10.
Why does a TIA need a feedback capacitor?
The input capacitance creates a rising noise gain that meets the op-amp's falling open-loop gain at a steep angle, so without Cf the circuit peaks or oscillates. Cf adds a zero that restores phase margin.
How do I get more bandwidth from a photodiode amplifier?
Reverse-bias the photodiode to reduce its capacitance, choose a smaller-area photodiode, use a higher-GBW op-amp with low input capacitance, or use a lower Rf followed by a second gain stage.