Transimpedance amplifier calculator for a photodiode

What sets a transimpedance amplifier's gain and bandwidth

A photodiode is a current source: the current is proportional to the light, and the voltage across the diode should not move with it. A transimpedance amplifier gives it that. The photodiode drives the inverting input of an op amp, the non-inverting input sits at a fixed voltage, and a resistor R_F from the output back to the inverting input carries the whole photocurrent. The loop holds the inverting input at the non-inverting voltage, so the diode sees a constant bias and the output moves by I·R_F. TI's SBOA268 puts it as "The current to voltage gain is based on the feedback resistance. The circuit is able to maintain a constant voltage bias across the input source as the input current changes which benefits many sensors." So the gain, in volts per amp, is R_F itself, and choosing it is arithmetic: the output span over the full-scale current.

The bandwidth is where it gets interesting, because the circuit has a capacitor at its input it did not ask for. The photodiode's junction capacitance, the op amp's common-mode and differential input capacitance and the board all sit from the inverting input to AC ground. Call their sum C_i. The op amp sees its output fed back through R_F into C_i: a low-pass network in the feedback path, which adds phase lag to the loop much as a capacitive load on the output does. TI's SBOA122 writes this as a noise gain, the gain the op amp applies to its own input noise and the reciprocal of the feedback factor β: 1 + Z_F/Z_G, with Z_F = R_F ∥ C_F and Z_G the impedance of the input capacitance. It is 1 at DC, rises from a zero Z1 = 1/(2π·R_F(C_i + C_F)), and levels off at 1 + C_i/C_F above a pole P1 = 1/(2π·R_F·C_F).

Without a feedback capacitor, P1 does not exist: the noise gain keeps rising at 20 dB per decade until it meets the op amp's open-loop gain, which is falling at 20 dB per decade, and the two close at 40 dB per decade with little phase to spare. SBOA122: "only the feedback capacitor (C_F) and the source capacitance (C_S) are used for stability". On SBOA268's own parts, 100 kΩ, 6 pF of op amp input capacitance and a 1.2 MHz OPA170, removing the 150 pF feedback capacitor leaves a response with Q = 2.13, which peaks 6.8 dB at 532 kHz, from a loop with 26° of phase margin in a single-pole model of the op amp.

With C_F fitted, the circuit is a second-order low-pass. SBOA122 models the op amp with one pole and derives the closed loop exactly (its Equations 5 to 7), then simplifies: the natural frequency is F0 = √(Z1·GBP), and Q = F0/(Z1 + F_C), where F_C = GBP/(1 + C_S/C_F) is where the open-loop gain meets the noise gain's plateau. With the op amp treated as an integrator of gain bandwidth GBP, those two equations are exact, and the calculator uses them. SBOA122 then drops Z1 beside F_C to reach its "easier-to-use" Q = P1/F0; the calculator shows that too, and the difference between them is a measure of how far a design is from the note's assumptions.

Photodiode amplifier bandwidth: the widest flat response by R_F, C_i and GBW

The widest bandwidth a transimpedance stage can have without peaking, for a given resistor, input capacitance and op amp, and the feedback capacitor that gets it. Each cell is the −3 dB bandwidth at Q = 0.707, SBOA122's Butterworth response, with the exact C_F beneath it; fit the E12 value at or above that capacitance, which lowers Q and the bandwidth a little. C_i is the whole input capacitance, photodiode and op amp together.

The bandwidth goes roughly as the square root of GBW/(R_F·C_i), so each tenfold step in GBW, gain or capacitance moves it by about √10. With C_F kept in the sums the steps are not exact: at 1 MΩ and 10 pF, ten times the GBW buys 3.34 times the bandwidth, and ten times the gain costs a factor of 2.99. That is SBOA122's Equation 12, GBP = 2π·F_−3dB²·R_F·C_S, which the note offers both ways round: "Knowing the bandwidth required by the application, the photodiode capacitance, and the transimpedance gain specification, calculate the minimum GBP requirement for the amplifier", or "Knowing the amplifier, the transimpedance gain, and the photodiode capacitance, calculate the maximum achievable bandwidth". Equation 12 takes Z1 as 1/(2π·R_F·C_S), leaving the feedback capacitor out of it, which is SBOA122's own simplification "C_S >> C_F". Where that does not hold, as at low gain with little input capacitance, Equation 12 reads high: by up to 34 % in this table, at 10.0 kΩand 10.0 pF. The table and the calculator keep C_F in.

Worked example: TI's SBOA268, 0 to 50 µA into 0 to 5 V

SBOA268 is one of TI's Analog Engineer's Circuits: a photodiode-style current of 0 to 50 µA into 0 to 5 V, with a 10 kHz bandwidth, on ±15 V, around an OPA170. Its three design steps are "Select the gain resistor", "Select the feedback capacitor to meet the circuit bandwidth" and "Calculate the necessary op amp gain bandwidth (GBW) for the circuit to be stable". It takes the input capacitance as "Ci = Cs + Cd + Ccm", the source plus the op amp's differential and common-mode capacitance, with the source an ideal current source at 0 pF. The left column is the calculator's arithmetic; the right is what TI prints.

gain      R_F = (5 V − 0 V) / 50 µA                = 100 kΩ     TI: 100kΩ
C_F       C_F ≤ 1 / (2π × 100 kΩ × 10 kHz)         = 159 pF     TI: 159pF
          E12 value at or below                    = 150 pF     TI: 150pF
C_i       C_s + C_d + C_cm = 0 + 3 pF + 3 pF       = 6.00 pF    TI: 6pF
GBW       (6 pF + 150 pF) / (2π × 100 kΩ × (150 pF)²) = 11.0 kHz   TI: 11.03kHz
fitted    OPA170, GBW 1.2 MHz: margin over the floor = 109×      
corners   Z1 10.2 kHz, P1 10.6 kHz, F0 111 kHz
          F_C 1.15 MHz, Q 0.095
BW        transimpedance −3 dB, single-pole model  = 10.6 kHz   TI: BW = 10.57kHz (simulated)

Every printed figure is reproduced. The 150 pF is the E12 value at or below 159 pF, since the step is an inequality: a smaller capacitor puts the feedback pole higher, so the bandwidth is at least 10 kHz. The OPA170's 1.2 MHz, from the "Design Featured Op Amp" table, clears the 11.03 kHz floor about a hundredfold, and with so much GBW to spare the bandwidth is the feedback pole P1 itself, 10.6 kHz. TI's simulated response marks "BW = 10.57kHz"; the single-pole model gives 10.6 kHz, 0.4 % above it. TI's figure comes from a simulation of the circuit, not from a formula. The simulation's "G=100dB" is 20·log(100 kΩ): the gain in dB relative to 1 Ω. The OPA170's 8 pA bias current through 100 kΩ shifts the output by 800 nV.

The calculator's defaults are this example. Switch the capacitor choice to a flat response and the same parts give C_F = 3.76 pF, fitted as 3.90 pF, and a bandwidth of 431 kHz: the most this op amp can do with 100 kΩ and 6 pF, and 41 times what SBOA268 asked for.

The TIA feedback capacitor: two criteria, from two TI notes

The two notes size C_F for different reasons, and it is worth being clear which question each answers. SBOA268 sets C_F for the bandwidth the signal needs, C_F ≤ 1/(2π·R_F·f_p), and then asks how fast the op amp must be: GBW > (C_i + C_F)/(2π·R_F·C_F²). That floor is P1·(1 + C_i/C_F): the GBW at which the open-loop gain meets the noise gain exactly where the noise gain levels off. Above it, the loop closes at 20 dB per decade on the plateau. SBOA122 sets C_F for the response instead: "Maximum bandwidth will be achieved for a Butterworth response with Q = 0.707", which is Q = 1/√2 in its Equation 9, solved for C_F. Neither note prints that solution; with SBOA122's simplifications it is C_F = √(C_S/(π·R_F·GBP)), and exactly it has a second, small term the calculator keeps.

SBOA268's criterion is a floor, not a design point, and what it leaves depends on how much of the noise gain the input capacitance makes. At exactly the floor, in the same single-pole model:

SBOA268's own design is the first row: with 6 pF against 150 pF the noise gain barely leaves 1, the loop is close to a plain integrator and the margin close to 90°. What the floor does cost there is bandwidth: an OPA170 exactly at 11.03 kHz would give 7.02 kHz, not 10 kHz, because the op amp's roll-off and the feedback pole coincide. As the input capacitance grows, the same floor leaves less margin, levelling off near 52°, with Q near 1, where SBOA122's Equation 10 puts it, and 1.24 dB of peaking. So the floor is where a design starts to fail, and a real design wants the GBW some way above it, as the OPA170's hundredfold is. A C_F from SBOA268's step 2 that is smaller than SBOA122's flat value peaks, whatever the floor says; the calculator reports both capacitors and warns when the fitted one is below the flat one.

High-speed transimpedance design, from SBOA122

SBOA122 is written for wideband photodiode receivers, and its first practical point is about the amplifier: "a unity-gain stable amplifier is not necessary for transimpedance applications. In fact, it is recommended to use a decompensated amplifier instead, because these decompensated amplifiers offer better voltage noise specifications and larger gain bandwidth products than any compensated version." A decompensated part is stable only above a minimum gain (the OPA846 "is stable for gains greater than 7V/V"), and what it must be compared with is the noise gain at the crossover, 1 + C_i/C_F, which with a large photodiode and a small C_F is well above that.

Its worked circuit is 310 kΩ with 200 pF of source capacitance on ±5 V: "using Equation 12, the OPA846 achieves 2.1MHz while the OPA657 achieves 2MHz". With the GBPs the note gives, 1.75 GHz for the OPA846 and 1.65 GHz for the OPA657, Equation 12 gives 2.12 MHz and 2.06 MHz. The flat C_F for the OPA846 is 0.343 pF, and the simplified form gives the same to three figures: at 200 pF against a fraction of a picofarad, C_S >> C_F holds. That gives 2.12 MHz and a noise gain of 585 at high frequency. Its Figure 3 plots the same equation against gain for 10 pF, and at 1 kΩ its curves start where Equation 12 puts them: 249 MHz, 167 MHz, 162 MHz, 113 MHz for the OPA847, OPA846, OPA657 and OPA843. For §3's example of "R_F = 20kΩ, f_–3dB = 10MHz" the equation asks for 126 MHz of GBP, which all four clear, and the note says that there "other considerations such as noise or power dissipation must be considered before making the amplifier selection."

Noise is its Equation 13, the equivalent input noise current integrated to a post-filter limit F: i_EQ = √(i_B² + 4kT/R_F + (e_N/R_F)² + (e_N·2πF·C_S)²/3). The last term is the op amp's voltage noise times the rising noise gain, and with a large photodiode it dominates. SBOA122 at 1.4 MHz: "the OPA657 yields a 5pA/√Hz equivalent input noise current", "dominated by the third term", and "The OPA846 ... actually yielded a lower equivalent input noise current of 3pA/√Hz." With Table 1's data sheet values and 200 pF plus each amplifier's own capacitance, Equation 13 gives 5.0 pA/√Hz and 3.1 pA/√Hz; the third term is 5.0 pA/√Hz of the OPA657's total, while the OPA846's is led by its 2.8 pA/√Hz current noise, with 1.2 pA/√Hz from the capacitance term.

The note also solves for the gain at which a FET and a bipolar amplifier tie, and says that with a 10 MHz filter and a 10 pF diode "for a resistor lower than 2kΩ, the bipolar amplifier offers a noise advantage". Its Equation 15, as printed, drops the squares on 2π·F and C_S that Equation 14 has, and on these numbers has no real solution. Solving Equation 14 itself gives 4.25 kΩ with Table 1's source capacitances of 15.2 pF and 13.8 pF, and 2.08 kΩ with the 10 pF diode alone for both, which is TI's figure. The conclusion is unchanged: at gains well above a few kilohms the FET part is quieter.

And it explains why the quieter OPA846 is still not used at 310 kΩ: "The input bias current of the OPA846, 19μA, generates an output offset voltage with the feedback resistor of 310kΩ of 5.89V. Because the OPA846 is operating on a ±5V power supply, this offset voltage sends the output into saturation." 19 µA × 310 kΩ is 5.89 V. The calculator takes the bias current and the rails for the same check.

Where the model stops being valid

One pole in the op amp. The calculator, like SBOA122's Equation 1, models the op amp as a single pole, and in the limit of large open-loop gain as an integrator of gain bandwidth GBW. The phase margin it reports has nothing in it for any further pole the real amplifier has, so treat it as the best case, and check a design close to SBOA268's floor with the vendor's simulation model, as SBOA268 itself does.

The input capacitance is an estimate. C_i is the sum of three numbers, and SBOA122 notes that "C_CM and C_DIFF include both the board layout and the op amp parasitic capacitance". A photodiode datasheet states the reverse bias its capacitance is given at; read it at the bias the circuit uses. Everything else on the page scales with it.

DC errors. The model is small-signal. The input bias current flows through R_F and appears at the output, and SBOA268's first design note is "Use a JFET or CMOS input op amp with low bias current to reduce DC errors." SBOA122's §6 is the case where that decides the part.

Output swing. The output has to stay in the op amp's linear range: SBOA268's third note is "Operate within the linear output voltage swing (see Aol specification) to minimize non-linearity errors." On a single supply, zero current puts the output at the negative rail, and the second note is the fix: "A bias voltage can be added to the non-inverting input to set the output voltage for 0 A input currents."

The noise equation's assumptions. SBOA122 lists them: the evaluation "is not intended to be used as a spot noise equation for narrowband applications"; "The final signal bandwidth for both the transimpedance design and any post-filtering is greater than 10 times the 1/f noise corner frequency"; "The transimpedance bandwidth is set greater than the post-filtering"; and the non-inverting input's current noise is "either negligible or made negligible by providing adequate bypassing." Outside those, the noise rows of the calculator are an estimate at best.

Common transimpedance amplifier mistakes

• Leaving out C_F. The noise gain then rises until it meets the open-loop gain, and SBOA268's parts without their 150 pF peak by 6.8 dB.
• Counting only the photodiode. SBOA268's input capacitance is all op amp, 3 pF differential and 3 pF common-mode; SBOA122's Table 1 adds 5.2 pF and 3.8 pF to a 10 pF diode. Leave them out and the flat C_Fand SBOA268's GBW floor both come out too low.
• Taking SBOA268's GBW floor as enough. At the floor, its own example gives 7.02 kHz where it asked for 10 kHz. The floor is a stability limit; the bandwidth needs GBW well above it.
• Rounding the flat C_F down. Below SBOA122's value the response peaks; round up, which the calculator does. SBOA268's bandwidth capacitor is the other way about: rounding it down keeps the bandwidth above f_p.
• Assuming the op amp must be unity-gain stable. SBOA122: "a unity-gain stable amplifier is not necessary for transimpedance applications", as long as the noise gain at the crossover exceeds the gain it needs.
• A bipolar op amp behind a large R_F. 19 µA through 310 kΩ is 5.89 V, beyond a ±5 V output.
• Mixing up the two notes' symbols. SBOA268's C_s is the source alone and its C_i the total; SBOA122's C_S is the total. The calculator uses C_i for the total throughout.

Further reading

• TI SBOA268, Analog Engineer's Circuit: Transimpedance Amplifier Circuit — the three-step design for 0 to 50 µA into 0 to 5 V at 10 kHz: R_F, C_F for the bandwidth, and the GBW for stability.
• TI SBOA122, Transimpedance Considerations for High-Speed Amplifiers — the noise gain and its corners, the second-order closed loop, the Butterworth bandwidth equation, input-referred noise and the FET-versus-bipolar choice.
• Op amp gain calculator — the inverting and non-inverting stages, for gain after the transimpedance stage when R_F has been kept small for bandwidth.
• Op amp capacitive load calculator — the same loss of phase margin from a capacitor, on the output rather than the input.
• Op amp error calculator — offset voltage, bias current and gain error, the DC side of this page.
• ADC noise floor calculator — whether the amplifier's output noise is above or below the converter that digitises it.
• Active filter calculator — the Butterworth second-order response, Q = 0.707, as a filter in its own right: the post-filter that sets F in the noise equation.