Silicon photocell

The selected silicon photocell is: 2CU84
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parameter description:
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Silicon photocell 2: VBWP34S, the parameters are as follows:
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Silicon Photovoltaic Cell: 5800B, under the test of daily light intensity, the current is about 1ua. Use a spotlight to hit it, the distance is 50ua when the distance is 0, and the distance is 10ua when the distance is about 1m. Measure its internal capacitance to be about 20pF.
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Three-stage amplifier circuit

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An amplifier is also required to provide the bias voltage:

Primary amplifier circuit

The following is a photodiode amplifier circuit from TI's official website.
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Its frequency response is as follows:

circuit analysis

We analyze this circuit:
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This circuit is a transimpedance amplifying circuit. The analysis of the transimpedance amplifying circuit is as follows:
Due to the virtual break characteristic of the circuit, the current generated by the photodiode under the illumination completely flows into the resistor R1. Due to the virtual short characteristic of the amplifier, the voltage on the left side of R1 is constant at 1.65V, then the voltage on the right side of R3 is equal to:
$V_{OUT}=V_{REF}+I_{PD}*R3$
For the above simulation circuit, our $I_{PD}$ It is 25uA, after being amplified by R3=52K resistor, the voltage change that can be generated is: 1.3V.
The simulation results are as follows:
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This is for a 200K signal.
We increase the signal frequency to 2M, and the simulation results are as follows:

It is found that the waveform is no longer a square wave, but has been distorted into a triangular wave.
In the feedback circuit, in addition to the feedback resistor, there is also a feedback capacitor C2:
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We calculate the capacitive reactance of C2 to 2MHZ signal:
$XC_{C2}={1 \over {2 \pi fc}}={1 \over {2\times3.14\times2\times10^{6} \times 3.3\times10^{-12}}}=24.1K \Omega$

Capacitive reactance of C2 to 200KHZ signal:
$XC_{C2}={1 \over {2\pi fc}}={1 \over {2\times3.14\times2\times10^{5} \times 3.3\times10^{-12}}}=241K \Omega$

Since C2 and R3 are placed together as a feedback loop, if the capacitive reactance of C2 is relatively small, the feedback current flows completely from C2 through the output terminal.
Therefore, for a 2MHZ signal, we can choose to increase the capacitive reactance of C2, or reduce the feedback resistor R3.

First, we increase the capacitive reactance of C2 and change the capacitance of C2 from 3.3pF to 0.33pF. The simulation results are as follows: Simulation

results:

It was found that the square wave returned to normal again. But at this time, the voltage at the reverse input terminal of the amplifier fluctuates greatly, which is already at the limit of the amplifier.

If we remove the feedback capacitor, all the current at the optical signal end will flow into resistor R3. In theory, this is true, but in practice there must be a feedback capacitor to prevent the self-excited vibration of the transimpedance amplifier. We test to remove the feedback capacitor:

Simulation results:
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We then use the method of reducing the feedback resistance so that the feedback current still flows into the resistance. Since the capacitive reactance is 24.1K, we change R3 from 52K to 5.2K, the gain becomes 10 times smaller, and the voltage fluctuation range changes from 1.3V to 0.13V. As follows:

Simulation results:

in conclusion

Therefore, we conclude that the gain of a transimpedance amplifier depends on the feedback resistance of the amplifier. In order to prevent self-excited vibration of the circuit, a feedback capacitor must be added to the feedback resistor. If the resistance value of the feedback capacitor is large, its capacitive reactance to the high-frequency signal is small. At this time, the feedback circuit does not flow through the feedback resistor, and the transimpedance amplification fails. It must be ensured that the impedance of the capacitive reactance of the capacitor to the signal is about 10 times that of the resistor.

PD sensor analysis:

The PD sensor itself has capacitance, and the larger the bottom area of the PD sensor, the larger the capacitance, and the more affected the high frequency. The analysis is as follows:
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Here, the internal capacitance of the simulated PD sensor is 3.3pF, and its capacitive reactance to the 2M signal is calculated as:

$XC_{C3}={1 \over {2\pi f c}}={1 \over {2\times3.14\times2\times10^{6} \times 3.3\times10^{-12}}}=24.1K \Omega$

The input impedance of the transimpedance amplifier is: 24.1K and 52K, which is equivalent to: 16.47K
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and the two are similar. If the internal capacitance of some PD sensors is large, such as changing 3.3pF to 33nF, its capacitive reactance is 0.00241K, which is much smaller than the input impedance of the transimpedance amplifier 16.47K, and all current signals are absorbed by its own capacitance: as follows: Simulation
results
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Reduce self-oscillation

We mentioned earlier that if the feedback capacitor is removed, it is prone to self-oscillation. However, with the addition of the feedback capacitor, its capacitive reactance is required to be large at high frequencies, so the capacitor requires its capacitance to be small. The capacitance value is usually 0.5PF minimum. So is there any way to remove the feedback capacitor without self-excited oscillation?
We use the following circuit:
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Self-excited oscillation will appear at this time:

We increase the internal capacitance of PD to form a low-pass filter relative to C3 and R3. Increase C3 to 33pF:
as follows:

Simulation results:

You can also add a low-pass filter to the inverting input of the amplifier:
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As shown above, C2 and R3 form a low-pass filter:
The -3dB cut-off frequency of the RC filter is calculated as follows:
$\over{2 \times \pi \times R \times C}}={1 \over{2 \times \pi \times 1 \times 10^3 \times 79.6 \times 10^ {-12}}}=2 MHZ$

The simulation results are as follows:
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5M circuit

With 10M RC low-pass filter
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Simulation results:
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remove the RC low-pass filter:

simulation results: also very good,

but at this time the PD capacitor is changed to 33PF, the result is very bad;