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The labels on your expected frequency response graph and the schematic of what you simulated are too small to see, so I can only answer with general observations. The Analog Devices schematic show...

posted 3d ago by Olin Lathrop‭ · edited 2d ago by Olin Lathrop‭

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#3: Post edited by

Olin Lathrop‭ · 2025-05-27T11:14:55Z (2 days ago)

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The labels on your expected frequency response graph and the schematic of what you simulated are too small to see, so I can only answer with general observations.<ol>
<li>The Analog Devices schematic shows only 371 fF across the feedback resistor. Specifying such a small capacitance to 3 digits is silly when the inevitable parasitic capacitance could easily be several times that unless you very carefully use unusual construction techniques.
<li>Taking the 371 fF capacitance across 1 MΩ at face value (again, silly), the low pass rolloff frequency is 430 kHz. In reality, it will be less than that because of the real capacitance being higher, but not anywhere as low as 10 Hz.
<li>To get 10 Hz rolloff with 1 MΩ, you'd need 16 nF. That's much larger than stray capacitance, so would need to be deliberate. Again, because the schematic of what you're simulating is too small to read, I can't say what you may have done wrong.
<li>The capacitance of the photodiode doesn't matter because the voltage across it is constant. The cathode is held at 3.3 V, and the anode at 0 V. Since the voltage across the capacitor doesn't change, no current flows thru it, and it becomes irrelevant.
</ol>
<blockquote>I replaced the schematic with new one with improved readability.</blockquote>
My volunteer time here feels abused due to the initial unreadable schematic, so I'll be brief.
~~The Analog Devices schematic shows 371 fF, and yours 371 F. The first is silly, and the second absurd. You changed a value by 1016 and still expected it to work!!?~~

The labels on your expected frequency response graph and the schematic of what you simulated are too small to see, so I can only answer with general observations.<ol>
<li>The Analog Devices schematic shows only 371 fF across the feedback resistor. Specifying such a small capacitance to 3 digits is silly when the inevitable parasitic capacitance could easily be several times that unless you very carefully use unusual construction techniques.
<li>Taking the 371 fF capacitance across 1 MΩ at face value (again, silly), the low pass rolloff frequency is 430 kHz. In reality, it will be less than that because of the real capacitance being higher, but not anywhere as low as 10 Hz.
<li>To get 10 Hz rolloff with 1 MΩ, you'd need 16 nF. That's much larger than stray capacitance, so would need to be deliberate. Again, because the schematic of what you're simulating is too small to read, I can't say what you may have done wrong.
<li>The capacitance of the photodiode doesn't matter because the voltage across it is constant. The cathode is held at 3.3 V, and the anode at 0 V. Since the voltage across the capacitor doesn't change, no current flows thru it, and it becomes irrelevant.
</ol>
<blockquote>I replaced the schematic with new one with improved readability.</blockquote>
My volunteer time here feels abused due to the initial unreadable schematic, so I'll be brief.
The fact that your simulation is failing means something with the simulation configuration is bad. Try putting in a simple step of 1 µA and see what happens.

#2: Post edited by

Olin Lathrop‭ · 2025-05-27T11:09:04Z (2 days ago)

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The labels on your expected frequency response graph and the schematic of what you simulated are too small to see, so I can only answer with general observations.<ol>
<li>The Analog Devices schematic shows only 371 fF across the feedback resistor. Specifying such a small capacitance to 3 digits is silly when the inevitable parasitic capacitance could easily be several times that unless you very carefully use unusual construction techniques.
<li>Taking the 371 fF capacitance across 1 MΩ at face value (again, silly), the low pass rolloff frequency is 430 kHz. In reality, it will be less than that because of the real capacitance being higher, but not anywhere as low as 10 Hz.
<li>To get 10 Hz rolloff with 1 MΩ, you'd need 16 nF. That's much larger than stray capacitance, so would need to be deliberate. Again, because the schematic of what you're simulating is too small to read, I can't say what you may have done wrong.
<li>The capacitance of the photodiode doesn't matter because the voltage across it is constant. The cathode is held at 3.3 V, and the anode at 0 V. Since the voltage across the capacitor doesn't change, no current flows thru it, and it becomes irrelevant.
</ol>

The labels on your expected frequency response graph and the schematic of what you simulated are too small to see, so I can only answer with general observations.<ol>
<li>The Analog Devices schematic shows only 371 fF across the feedback resistor. Specifying such a small capacitance to 3 digits is silly when the inevitable parasitic capacitance could easily be several times that unless you very carefully use unusual construction techniques.
<li>Taking the 371 fF capacitance across 1 MΩ at face value (again, silly), the low pass rolloff frequency is 430 kHz. In reality, it will be less than that because of the real capacitance being higher, but not anywhere as low as 10 Hz.
<li>To get 10 Hz rolloff with 1 MΩ, you'd need 16 nF. That's much larger than stray capacitance, so would need to be deliberate. Again, because the schematic of what you're simulating is too small to read, I can't say what you may have done wrong.
<li>The capacitance of the photodiode doesn't matter because the voltage across it is constant. The cathode is held at 3.3 V, and the anode at 0 V. Since the voltage across the capacitor doesn't change, no current flows thru it, and it becomes irrelevant.
</ol>
<blockquote>I replaced the schematic with new one with improved readability.</blockquote>
My volunteer time here feels abused due to the initial unreadable schematic, so I'll be brief.
The Analog Devices schematic shows 371 fF, and yours 371 F. The first is silly, and the second absurd. You changed a value by 1016 and still expected it to work!!?

#1: Initial revision by

Olin Lathrop‭ · 2025-05-26T21:38:02Z (3 days ago)

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The labels on your expected frequency response graph and the schematic of what you simulated are too small to see, so I can only answer with general observations.<ol>

<p><li>The Analog Devices schematic shows only 371 fF across the feedback resistor.  Specifying such a small capacitance to 3 digits is silly when the inevitable parasitic capacitance could easily be several times that unless you very carefully use unusual construction techniques.

<p><li>Taking the 371 fF capacitance across 1 M&Omega; at face value (again, silly), the low pass rolloff frequency is 430 kHz.  In reality, it will be less than that because of the real capacitance being higher, but not anywhere as low as 10 Hz.

<p><li>To get 10 Hz rolloff with 1 M&Omega;, you'd need 16 nF.  That's much larger than stray capacitance, so would need to be deliberate.  Again, because the schematic of what you're simulating is too small to read, I can't say what you may have done wrong.

<p><li>The capacitance of the photodiode doesn't matter because the voltage across it is constant.  The cathode is held at 3.3 V, and the anode at 0 V.  Since the voltage across the capacitor doesn't change, no current flows thru it, and it becomes irrelevant.

</p></ol>

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