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I'm designing a transimpedence circuit with a photodiode.

  • Op-amp: LTC6268 (GBW = 500MHz)
  • Photodiode: SFH2700 A1 (C_j = 4.6pF, max current = 1.5uA)
  • Required bandwidth: 2MHz

Using the calculation from this question, I figured out that my R_f = 1.33M => C_f <= 59fF, which is super small. Should I install some 0.2pF capacitors in series there? Or any other recommendations? Here is my schematic: enter image description here

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  • \$\begingroup\$ Include the schematic. some 0.2pF capacitors in series there? In series WHERE? Again: include the schematic. \$\endgroup\$ Commented Mar 10, 2021 at 13:34
  • \$\begingroup\$ I added the schematic :D Sorry for not being clear :D \$\endgroup\$
    – Michael
    Commented Mar 10, 2021 at 13:42
  • \$\begingroup\$ Yes, 0.056 pf gives you a flat response in a simulator - however your real circuit will add additional capacitance in parallel with the photodiode. So you will require more than 56 femtofarads in parallel with Rf to compensate. \$\endgroup\$
    – glen_geek
    Commented Mar 10, 2021 at 14:07
  • \$\begingroup\$ What do you mean by "more than"? does it mean that I need feedback capacitor with value BELOW 56fF? If that's the case, how can I achieve it? \$\endgroup\$
    – Michael
    Commented Mar 10, 2021 at 14:23
  • \$\begingroup\$ The opamp and traces will have at least a pF of capacitance, so you can't achieve that. You probably need to use a different opamp or to change your design spec to something achievable with that opamp. \$\endgroup\$ Commented Mar 10, 2021 at 14:40

1 Answer 1

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You are going down the wrong path here.

When you have a transimpedance amplifier you must take into account the "noise gain" that the photodiode introduces by virtue of its 4.6 pF junction capacitance. At 2 MHz, that's an impedance of 17.3 kΩ and, it will make the noise gain of your op-amp this value at 2 MHz: -

$$ \text{Noise gain } = 1+\dfrac{1.33 \text{ MΩ}}{17.3\text{ kΩ}} = 77.87 = 37.8 \text{ dB}$$

This is the amplification value of internal noise within your op-amp (at 2 MHz) and quite often, you need to compensate this by putting a capacitor in parallel with your \$R_F\$ of 1.33 MΩ. However, let's continue the noise gain saga and ignore, for now, the parallel capacitor that you may need.

I'll begin with this simple diagram of where the op-amp noise exists and how this gets amplified by the photodiode capacitance: -

enter image description here

It's now easy to predict the noise gain at all other frequencies and this basically means that the amplification of noise will follow this trajectory (green line): -

enter image description here

At low frequencies the noise gain is unity (0 dB) but will begin to rise at about 23 kHz and, by the time you reach 2 MHz, the noise gain will be 77.87 (37.8 dB). It's a 20 dB per octave slope and will keep on rising until you hit the natural open-loop gain of the op-amp (the LTC6268 from which I stole and modified the above basic picture).

So, I added a green circle at 2 MHz and 37.8 dB and another at 200 kHz (a decade lower) at 20 dB less (17.8 dB). I then drew a straight line between those two points and noted the frequency at which it dropped to 0 dB (unity gain).

So, unless you can live with a big hump of noise in the region 23 kHz to about 300 MHz, you need to do something about it.

You can estimate the amount of op-amp output noise based on the voltage noise \$e_N\$ in the data sheet (4.3 nV per \$\sqrt{Hz}\$). If we say that the bandwidth is approximately 300 kHz to 30 MHz with a gain of about 30 dB we wouldn't be far wrong. So it's an approximate BW of 30 MHz and that means an equivalent noise voltage of 23.6 μV RMS at the input. Multiply this by 30 dB (31.6 in real numbers) and we'll see approximately 0.74 mV RMS of noise on the output.

That's about 5 to 10 mV p-p with a bit of handwaving.

If you can't live with that amount of noise then you have to lower the noise gain. If you can live with that amount of noise then you should still worry about parasitic capacitance on your PCB (maybe 0.5 pF) ruining your bandwidth.

Basically, 1.33 MΩ and 0.5 pF produce a BW of 239 kHz i.e. nowhere near the 2 MHz you require. I reckon you need to lower \$R_F\$ by at least a factor of ten.

You might decide that living with a noise gain of (say) 5 is just about OK. So, if 5 is the number (and you have to decide this based on how good a performance you expect from your circuit), then you want the numerator in the equation above to be no more than 4 times 17.3 kΩ. To get this, your feedback capacitor needs to be no smaller than one quarter of the 4.6 pF junction capacitance of the photodiode.

So, you'd go for something like 1.2 pF. But now, the problem is that you will only have a bandwidth of 99.7 kHz based on \$R_F\$ being 1.33 MΩ.

It's the game we play with photodiodes and transimpedance amplifiers and, you either go for a higher (worsening) noise gain and hence a smaller capacitor that can deliver a wider bandwidth OR you bite the bullet and reduce the 1.33 MΩ feedback resistor. So, if you reduced the feedback resistor by ten times, the noise gain without a feedback capacitor will be: -

$$ \text{Noise gain } = 1+\dfrac{133 \text{ kΩ}}{17.3\text{ kΩ}} = 8.69$$

So, this time when adding a capacitor of about 1.2 pF (an impedance of 66.3 kΩ), the overall noise gain would be close to around 5. And, 1.2 pF in parallel with 133 kΩ produces a bandwidth of 997 kHz.

Clearly, you are still not at the 2 MHz bandwidth you require so, inevitably, you must reduce your feedback resistor to about 50 kΩ

$$\color{red}{\boxed{\text{These are the games we have to play with transimpedance amplifiers}}}$$

High bandwidth and high gain comes with high noise. An inescapable truth.

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    \$\begingroup\$ Great answer. One thing I would add is that the 4.6 pF that determines the noise gain is not a constant, and can be reduced by a lot by applying a reverse bias to the photodiode. \$\endgroup\$ Commented Mar 10, 2021 at 16:24
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    \$\begingroup\$ Great answer. Thanks alot. I'm using a photoconductive mode (i.e reverse bias) I decided to reduce my feedback resistor to about 50k to achieve a maximum voltage output of about 100mV (this should be enough for detecting the illuminance for my application) \$\endgroup\$
    – Michael
    Commented Mar 10, 2021 at 18:28

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