Photodiode Transimpedance Amplifier
Calculate TIA output voltage, bandwidth, and noise for photodiode circuits. Design transimpedance amplifier gain and feedback components.
Formula
How It Works
This calculator designs transimpedance amplifiers (TIAs) for photodiode signal conditioning, essential for optical communication engineers, laser power meter designers, and scientific instrumentation developers. A TIA converts photodiode photocurrent to voltage with gain Rf: Vout = Iph * Rf, where Iph is the photocurrent (typically 1 nA to 1 mA) and Rf is the feedback resistance (1 kOhm to 10 GOhm). The op-amp holds the photodiode at virtual ground, minimizing junction capacitance (Cj = 10-100 pF typically) and dark current. Bandwidth is limited by the feedback network: BW = 1/(2*pi*Rf*Cf), where Cf is the feedback capacitor required for stability. Without Cf, the total input capacitance causes peaking or oscillation. The stability criterion per Analog Devices MT-059 is Cf >= sqrt(Cin/(2*pi*GBW*Rf)). Dominant noise is Johnson noise from Rf: en = sqrt(4*k*T*Rf) = 4.07*sqrt(Rf) nV/rtHz at 25C using k = 1.380649e-23 J/K (exact SI, 2019 BIPM redefinition). Photodiode responsivity and NEP measurement methodology follows IEC 60747-5-5 (Semiconductor devices — Optoelectronic devices) and NIST Technical Note 2064 (Optical Power Measurement). TIA design best practices are documented in Analog Devices MT-059 (Transimpedance Considerations for High-Speed Amplifiers). A 1 MOhm Rf produces 129 nV/rtHz noise spectral density. Noise Equivalent Power (NEP) = in/Responsivity, typically 1-100 fW/rtHz for optimized TIAs.
Worked Example
Design a TIA for a Hamamatsu S5972 PIN photodiode (Cj = 10 pF, responsivity 0.65 A/W at 850 nm) to detect 0.1-10 uW optical power. Target 1V output at 10 uW, BW >= 100 kHz.
- Full-scale current: Iph = 10 uW * 0.65 A/W = 6.5 uA
- Required gain: Rf = Vout/Iph = 1V / 6.5 uA = 154 kOhm (use 150 kOhm standard)
- Maximum Cf for 100 kHz BW: Cf = 1/(2*pi*150k*100k) = 10.6 pF (use 10 pF)
- Check stability with OPA657 (GBW = 1.6 GHz, Cin = 4 pF):
- Johnson noise: en = sqrt(4*1.38e-23*298*150e3) = 49.8 nV/rtHz
- Current noise: in = en/Rf = 49.8 nV/rtHz / 150 kOhm = 0.33 fA/rtHz
- NEP = 0.33 fA/rtHz / 0.65 A/W = 0.51 fW/rtHz (excellent)
Practical Tips
- ✓Use FET-input op-amps (OPA657, AD8065, LTC6268) for best noise performance; low input bias current (<10 pA) avoids adding to photodiode dark current (1-100 nA typically) per Texas Instruments SBAA060
- ✓Place Cf physically across Rf on the PCB, not connected through traces; stray capacitance from 10 mm PCB traces (0.5 pF) can cause parasitic oscillation at gains above 1 MOhm
- ✓For wideband (>10 MHz) TIAs, consider integrated TIA ICs (MAX3864, AD8015) that combine optimized op-amp and feedback network for guaranteed stability and 100 MHz+ bandwidth
Common Mistakes
- ✗Omitting feedback capacitor Cf: parasitic junction capacitance (10-100 pF) creates a resonant peak with Rf; even 10 pF Cj with 1 MOhm Rf oscillates at 16 kHz without Cf per Analog Devices AN-1112
- ✗Using a slow op-amp (<1 MHz GBW): TIA bandwidth is min(1/(2*pi*Rf*Cf), GBW/noise_gain); a 1 MHz op-amp with 100x noise gain limits BW to 10 kHz regardless of Rf*Cf
- ✗Choosing Rf too large for bandwidth: 10 MOhm with 1 pF Cf gives only 16 kHz BW; verify the Rf*Cf product meets bandwidth requirements before finalizing component values
Frequently Asked Questions
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