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Photodiode Transimpedance Amplifier

Calculate transimpedance amplifier (TIA) output voltage, bandwidth, and noise for photodiode signal conditioning.

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Formula

V_out = I_ph × R_f, BW = 1/(2π × R_f × C_f)

R_f— Feedback resistance (Ω)
C_f— Feedback capacitance (F)

How It Works

A transimpedance amplifier (TIA) converts the current output of a photodiode into a usable voltage. The op-amp holds the photodiode at virtual ground (zero bias, minimising dark current), and the feedback resistor R_f sets the gain: V_out = I_ph × R_f. The −3 dB bandwidth is limited by the feedback RC network: BW = 1/(2π × R_f × C_f), where C_f is the feedback capacitor added in parallel with R_f for stability. Without C_f, the total input capacitance (photodiode junction + op-amp input) can cause peaking or oscillation. A good design rule is to set the noise gain bandwidth equal to the op-amp gain-bandwidth product: C_f ≥ √(C_in / (2π × GBW × R_f)). The dominant noise source in high-gain TIAs is Johnson noise from R_f: e_n = √(4kTR_f), giving a noise spectral density in nV/√Hz. Choosing R_f involves trading off gain (higher R_f → more output voltage) against bandwidth and noise (higher R_f → lower BW, higher Johnson noise integrated over bandwidth).

Worked Example

Problem
Design a TIA for a photodiode with 5 μA full-scale current. Target output 1 V full scale and BW ≥ 10 kHz. What R_f and C_f are needed?
Solution
1. Required gain: R_f = V_out / I_ph = 1 V / (5×10⁻⁶ A) = 200 kΩ 2. Maximum C_f for 10 kHz BW: C_f = 1/(2π × 200×10³ × 10×10³) = 79.6 pF → use 68 pF (standard) 3. Johnson noise at R_f = 200 kΩ: e_n = √(4 × 1.38×10⁻²³ × 293 × 200×10³) × 10⁹ = 57.5 nV/√Hz 4. Check op-amp GBW: need GBW ≥ BW × R_f/R_in, choose an op-amp with ≥ 1 MHz GBW Result: Use R_f = 200 kΩ, C_f = 68 pF, op-amp with GBW ≥ 1 MHz (e.g., OPA657 or TLV2372).

Practical Tips

  • ✓Use a FET-input op-amp (e.g., OPA657, AD8065) for best noise performance — the low input bias current (< 10 pA) avoids adding to the photodiode's dark current.
  • ✓Place C_f physically across R_f on the PCB, not just in the schematic — stray capacitance from long PCB traces can cause parasitic oscillation at high gains.
  • ✓For wideband (> 1 MHz) TIAs, consider a transimpedance amplifier IC (e.g., MAX3864) which integrates the op-amp and feedback network for optimised high-frequency performance.

Common Mistakes

  • ✗Omitting the feedback capacitor C_f — the parasitic photodiode junction capacitance (even 10 pF) creates a resonant peak with R_f that can oscillate; always add C_f.
  • ✗Using a slow op-amp (< 1 MHz GBW) — the TIA bandwidth is set by min(1/(2πR_fC_f), GBW/noise_gain); a slow op-amp limits BW far below the RC cutoff.
  • ✗Choosing R_f too large for the desired bandwidth — 1 MΩ with 10 pF C_f gives only 15.9 kHz BW; verify the RC product before finalising R_f.

Frequently Asked Questions

In photoconductive mode (reverse bias), the photodiode junction capacitance is minimised and linearity is maximised. The TIA op-amp holds the cathode at virtual ground (0 V), providing reverse bias if the anode is at ground, while the photocurrent flows into the feedback resistor. Zero or reverse bias also reduces dark current compared to forward bias.
In photovoltaic mode (zero bias), the photodiode acts as a current source with very low dark current — ideal for low-light, precision applications. In photoconductive mode (reverse bias, up to −5 V typically), junction capacitance is reduced, giving faster response and better linearity, at the cost of slightly higher dark current. TIAs typically operate the photodiode at or near zero bias (virtual ground).
NEP = noise current / responsivity. The input-referred current noise is i_n = e_n_R / R_f = √(4kT/R_f) A/√Hz. Divide by the photodiode responsivity (A/W) to get NEP in W/√Hz. A lower NEP means the TIA can detect weaker light signals.

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