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Richie;

It sounds as if your "DC current" might be ambient background illumination.

It is important to know the capacitance of your photodiode and if you are limited to certain power supplies.

Dear Neil

Yes! It can be considered as ambient light..

The capacitance of the photodiode is about 5pF...And the power is +-12V dual.

Thanks!

Richie

Richie;

Does your signal also extrend down to DC?

Excuse me Neil....

What do yo mean by extend down to DC?

The signal I need is actually a very small portion of the whole current generated by the photodiode.

The whole current is 8uA (DC+ac), the signal I want is only 0.3uA (ac).

Can I block the DC part out at the inverse port of the OpAmp? Or I have to use a second stage OpAmp

Thanks a lot

Richie

Richie;

What is the lowest frequency of your AC signal?

Neil

The frequency of the signal ranges from 1kHz to 10kHz.

Thanks.

Richie

Richie;

I think this circuit will work for your application (see attached TINA file).

It is a transimpedance amplifier (TZA) with 1E6 (120dB) gain. With a 300nA p-p sine wave input, you have 300mV p-p output and only about 450uV rms noise. You can improve your SNR by following the TZA with an AC coupled low pass filter. As you can see from the TINA noise curve, the noise drops to 150uV rms if you limit the noise BW to 10kHz. To prevent roll-off of your overall frequency response, you might want to set your LPF -3dB point to around 20kHz. The noise will be a bit higher but you will have a flatter response up to 10kHz.

There is no DC response.

Dear  Neil

Thanks very much for your design. Could I ask you three more questions?

1. Can I change R1 (feedback resistor) into 8M ohm? Then the Vp-p will be 3V p-p.

2. What if the input current of the photodiode is reduced to 40pA? Is it still suitable to block the DC current will a capacitor?

3. If the DC part of the input current is blocked by the capacitor, where does it go? Through the impedance of photodiode? Then the voltage across photodiode should be quite high.

Thanks a lot

Richie

1. Yes, Richie, you can change the feedback resistor to 8M but you will need to reduce the feedback capacitor to 0.2pF, about the parasitic capacitance of a 0603 surface-mount resistor. Two resistors in series will reduce the parasitic feedback capacitance.

2. The DC ambient light current can even go to zero-- no problem.

3. The DC portion of the photocurrent flows through the photodiode, bias resistor (R2), bias voltage source, and into ground. The photocurrent causes a voltage drop across R2 which subtracts from the bias supply voltage. Because of this, R2 cannot be made too large if you have a large DC photocurrent or the bias voltage across the photodiode may be too low (see your photodiode data sheet).

Dear Neil

Sorry for my late reply.

Just want to say that thanks a lot. I have learned some from your design.

Best regards

Richie

You are welcome, Richie.

I've learned a lot over the years and I'm glad to pass along some of that knowedge.

Hi Neil,

would  you  convert your design to pdf form?

(I've no TINA on my machine.)

Thanks 

Tamas

Tamas;

Here is that circuit in a .pdf file. I hope it helps.

I'm leaving the TI E2E Forum-- TI did not like some of my answers so I am not going to continue to provide their applications engineering support free of charge.

Neil,

thank you for your kindness.

Tamas

Dear Neil..

I have some more questions....

1.Will the AC part also go through R2? Since R2 is much smaller than R1.1M..

2.Will the capacitor C1 influence the input? Since the input could be very small, such as 30pA...

3.Is C2 big enough? If the power is not clean enough...

Thanks a lot. You are helping much to give what you know to the beginners. 

Best regards

Richie