Circuit Response
every model has an edge, and the figure carries it
Series

Transimpedance — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A photodiode's own capacitance sets the bandwidth, as its -0.50 power. computed by solving, not by drawing. The bandwidth of a 1.0 MΩ transimpedance stage against the capacitance of the diode driving it, with the feedback capacitor at each point bisected to give exactly forty-five degrees of phase margin on the solved loop. The classical expression √(GBW/2π·rf·cd) is drawn over it: the right shape, and conservative by about a fifth at every capacitance. The bandwidth falls as the -0.497 power of the capacitance — a square-root law, so a diode of four times the area costs half the bandwidth rather than three quarters of it. At 30 pF the compensation is 0.598 pF against the expression's 0.725, and the bandwidth 295 kHz against 220.

    Where the trouble is at the input

    Every other arrangement in this field has its difficulty at the output — a capacitive load, an isolation resistor, a load that draws current. A photodiode amplifier has it at the input, and the capacitance causing it is not a parasitic: it is the diode's junction, which is the price of its area, and area is what a photodiode is bought for. The feedback resistor's own noise is 127 nV/√Hz against the amplifier's 4, and the amplifier is still ninety-five per cent of the noise at a large diode.

    part 1 · feedback
  2. One capacitor moves three quantities, and the expression's own answer peaks by 1.18 dB. computed by solving, not by drawing. The bandwidth, the peaking and the total output noise of a 1.0 MΩ transimpedance stage against its feedback capacitor, swept from 0.30 to 4.20 times what the classical expression asks for. Over that factor of fourteen the bandwidth falls from 334 to 55.4 kHz, the total noise from 230 to 65 µV, and the peaking from 10.0 dB to nothing. The expression's own answer sits at 1.18 dB of peaking, 0.82× is where the loop reaches forty-five degrees, and √2× is where the response is flat — so the choice usually quoted as maximally flat is neither of the two conditions it is quoted for. The faint families are the same three quantities at 3 pF and 300 pF of diode: normalised this way they are one curve, so the diode sets the scale and the multiple sets the shape.

    The factor the expression leaves out

    The classical compensation for a photodiode amplifier is quoted both as the forty-five degree choice and as the maximally flat one, and it is neither: it leaves 1.18 dB of peaking and 52.4° of margin. Flat is at exactly √2 times it — fitted at 1.4186 against 1.4142, at every detector from 3 pF to 1 nF. And the third quantity the capacitor is supposed to trade, the noise, does not move at all inside the signal band: three compensations spanning a factor of fourteen give 8.37 against 8.36 µV in a 4.36 kHz measurement and 230 against 65 µV over the whole plane.

    part 2 · feedback
  3. The direct-current error a bigger feedback resistor does not fix. computed by solving, not by drawing, at direct current with the amplifier in the netlist. A 1.0 nA photocurrent into 100 MΩ gives -0.100 V. The bias current flows in the same resistor, so its contribution is the ratio of the two currents — 0.1000% at 25 °C, and the same percentage at 1 MΩ and at 1 GΩ, which is checked by changing the resistor rather than by reading an expression. The amplifier's 100 µV of offset behaves the other way: it is multiplied by one plus the resistor over the diode's own leakage, so it grows with the resistor and with temperature. Both terms double every ten kelvin, for two different reasons, and at 85 °C they are 6.40% and 0.740% against 0.100% and 0.110% at 25.

    The error a bigger resistor cannot help

    Every other quantity in a photodiode amplifier improves with a larger feedback resistor: the signal grows as R and the resistor's own noise as √R, so the ratio goes as √R. The bias current does not behave like either — it flows in the same resistor the signal does, so its contribution is the ratio of the two currents with the resistance cancelled, 0.1 per cent at 25 °C and the same 0.1 per cent at a megohm and at a gigohm. The offset behaves the other way, growing as one plus the resistor over the diode's own leakage, and at 85 °C the two are 6.40 and 0.74 per cent against 0.10 and 0.11 at room temperature.

    part 3 · feedback
  4. What a feedback tee buys, and the single thing it charges. computed by solving, not by drawing. Two 50 kΩ resistors with a 6.250 kΩ tap give 500.0 kΩ of transimpedance, which is R₁(1 + R₂/R₃) + R₂ solved on the netlist. That expression has no term for the noise gain, and the tap sets it to 1 + R₁/R₃ = ×9.00 — against exactly one for a single resistor of any value, because at direct current the source is a capacitor. One quantity then does all three things at once. The signal and R₁'s own noise are multiplied together, so the tee buys no signal-to-noise ratio at all: 1846 against 5582 in a hertz at a nanoamp for one resistor of the same transimpedance, a factor of 3.02 which is the square root of the noise gain. The amplifier's own voltage noise and offset are multiplied where they were not before. And the phase margin RISES, from 1.8° to 15.9°, because starting the noise gain high shortens its climb to the peak — so the tee is a compensation that charges for itself in noise, and a capacitor across the feedback is the same compensation for nothing. With no tap the three effects are absent rather than small.

    The tee that charges for its own compensation

    A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.

    part 4 · feedback
  5. At 50° of margin a capacitor across the resistor carries 85.2 µV and a tee 347 µV: the tee is dominated everywhere. computed by solving, not by drawing. A photodiode amplifier of 500 kΩ transimpedance on 33 pF of detector, behind a 10 MHz amplifier with 4 nV/√Hz of voltage noise, compensated either by a capacitor across a single feedback resistor, swept from none to 4 pF, or by a tee of the same transimpedance and factor 3 to 64. Each design's phase margin is drawn against its output noise integrated to its own bandwidth. At every margin a tee reaches, the capacitor reaches it with less noise: at 50.4° the capacitor's design (0.97 pF) carries 85.2 µV in 396 kHz and the tee (factor 32) 347 µV in 401 kHz, 4.1 times more. The ratio runs from 1.29 to 7.47 across the margins compared.

    The compensation that costs nothing

    A photodiode amplifier can be compensated with a capacitor across its feedback resistor, or with a feedback tee, whose raised noise gain also raises the phase margin. At one transimpedance, 500 kΩ on 33 pF of detector, each design's margin is drawn against its output noise over its own bandwidth. The capacitor dominates at every margin the tee can reach: at 50° it carries 85 µV where the tee carries 347, in the same 400 kHz. The two keep the same band to within 1.3 per cent at equal margin, so the tee's noise buys no bandwidth. The conjecture that the answer would turn on which noise dominates does not survive either. With the amplifier's voltage noise cut tenfold, the tee is still 5.6 times worse.

    part 5 · feedback
  6. A tee of factor 10 turns 1 µV/K of offset drift into 65.1 pA of error at −40 °C, 10 times the single resistor's. computed by solving, not by drawing. The direct-current error of a 10 MΩ photodiode amplifier calibrated at 25 °C, as the input current that would give the same output, against temperature: the part from the amplifier's offset drift of 1 µV/K (solid) and the part from its input bias current, 1 pA at 25 °C and doubling every 10 K (dashed), for a tee of factor 10 (heavy) and a single resistor (light). The offset reaches the output multiplied by the direct-current noise gain — one for the single resistor, since the source is a capacitor, and 10 for the tee — while the bias current sees the same transimpedance through both. The bias term overtakes the offset term at 85 °C with the tee and 40 °C with one resistor; below that the tee's error is the offset's, 10 times as large.

    The drift a tee multiplies

    A feedback tee reaches a large transimpedance with small resistors, and on a photodiode its price has been counted in noise: at direct current its noise gain is its own factor, where a single resistor's is one. The same noise gain multiplies the amplifier's offset. Calibrated at 25 °C and read at −40, a 10 MΩ amplifier with 1 µV/K of offset drift is 6.56 pA wrong through one resistor and 65.1 pA through a tee of factor ten — ten times, exactly, while the input bias current's error is identical through both. So the tee's drift penalty is hidden wherever the bias current dominates, which for a field-effect input doubling every 10 K is the hot end, and fully exposed at the cold end. The worst error over −40 to 85 °C comes within a factor of two of the single resistor's only above 10 MΩ for a tee of ten, and above 100 MΩ for a tee of a hundred.

    part 6 · feedback

All series