Photodiode

Hi Amit,

The purpose of this photodiode amplifier is to pass AC signals from the photodiode while rejecting constant background light levels. It is not possible to use conventional capacitor coupling at the input of this type of circuit because the photodiode would forward-bias, disturbing its proper operation.

T1 and R4 create a current source controlled by the output voltage of U2. This current is intended to balance (equal) the DC component of current flowing in the photodiode. The integrator  U2, C1, R2 control the current in T1. The voltage at the non-inverting input of U2 (a 2.5V reference source) biases the integrator. If, for example, the output of U1 is less than the 2.5V reference, the output of the integrator will ramp upward, increasing the current in T1. This will cause the output of U1 to ramp upward until it reaches a 2.5V equalibrium.

The combination of R2-C1 and gain through the feedback loop sets a low frequency cutoff. AC photodiode signals above this frequency are amplified by U1. DC or low frequency variation in background light level does not affect the average output voltage of U1.

Regards, Bruce.

Hello Amit,

U2 and T1 form a servo loop to keep the average output of U1 at the Vref level. This second loop cancels out the steady current caused by the ambient light while passing the AC signal.

U2 forms an integrator, which biases T1 to pull just enough current away from the diode to make the average value of U1's output equal to the Vref voltage on U2's + input.

The clue is in the text "Light signal is small AC component riding on large ambient light level". The integrators time constant is set so that it does not interfere with the small AC signal on the input. So the small AC signal will be present on the U1 output, but the large DC value that would also be present is nulled out by the servo loop. R2/C1 set the time constant and determine the lower AC frequency limit.

Of course, there are limits. If the sensor is saturated, the loop cannot work miracles. The "ambient" light can also contain AC signals (60/120/240/300Hz or 22/44KHz) that cannot be filtered out.

Regards,

Hello,

I have a question about "ambient" light that contains AC signals (60/120/240/300Hz or 22/44KHz).

The low frequency cuttoff of the feedback loop is about 16Hz in the above schematic [ f = 1/ (2*pi*1M*10n) = 16Hz) ].

In my application I use a laser that generates an AC signal of 100kHz.
Is it possible to increase the low frequency cuttoff of the feedback loop to around 60kHz, by using a capacitor (C1) of 2.7pF and filter out the frequencies of artificial light?

Frank,

The values shown in the figure at the top of this thread product a cutoff frequency of approximately 6.5kHz. The cutoff is affected by the transfer function (transconductance) of the voltage-to-current conversion in the T1 circuitry.

It is possible to change the cutoff to a higher frequency by altering the R2*C1 time constant. I would recommend a minimum value for C1 of 50pF, or so. Adjust R2 to a much lower value to get the desired cutoff.

The problem is that your cutoff of 60kHz very close to your interfering frequency of 44kHz. This approach only gives you a single pole response so you will get very little attenuation at 44kHz.

Regards, Bruce.

Thank you Bruce for your answer.

I can increase my AC signal to 200kHz at a max, so the cuttoff frequency can become higher.
How can the above schematic be changed to get a higher order (i.e. second order) for the feedback loop?

Regards,

Frank

Frank,

A higher-order response in this loop would be unstable. I would recommend a separate high-pass filter after this stage.

Regards, Bruce.

Bruce,

Thank you for your advice. Now, I can start prototyping.

Frank

Could you explan why light can also contain AC signals (60/120/240/300Hz or 22/44KHz),thank you

Please start your own forum thread.