Hello there,

My schematic is as follows

The idea is that AC and DC signals can be split into two outputs separately. The photodiode is Hamamatsu's S3071 (Dark current less than 10nA). The laser light produces  a DC signal of about 0.5 mA  superimposed with a AC signal (500KHz) of ~1 uA. The DC Output is OK. But the AC output noise reachs more than 1V. This noise is still there even without light. And the noise between AD8015 pin 6,7 is about one-twentieth of the output signal noise, so the noise should have nothing to do with the second amplifier OPA847.

So my question is: Where does this noise come from and how can I lower it?

Thanks a lot.

Parents
• Hi Jone,

Here are some inputs from our application expert:

You say the output has about 1V noise. We assume this is peak.

The gain on the OPA847 is 20 (RF=1k, RG=50 ohms).

This mean the RTO noise of the AD8015 is 1V/20=50mV Peak, equivalent to 50mV/6.6=7.6mV rms.

If instead of “about 1V”, the output had 890 mV noise, the 8015 output noise would be 6.8 mV rms.

From the AD8015 datasheet and calculations:

Note from Figure 2 of the AD8015 datasheet that the RTI current noise increases approximately 0.5 pA/SQRT(Hz) for every 0.5pF increase in external capacitance.

The terminal capacitance of the Hamamatsu S3071 photodiode is 15 pF.

At 100 MHz the RTI current noise would increase to 18pA/SQRT(Hz) because of the 15pF external capacitance coming from the Hamamatsu photodiode.

The TIA section of the AD8015 has a 10k feedback resistor. This will result in an RTO noise density of 10k x 18 pA/SQRT(Hz) = 180nV/SQRT(Hz). The AD8015 has a second gain stage with gain=3. This would then increase the noise density at the output to 3 x 180 = 540nV/SQRT(Hz).

Integrating over the 100 MHz BW of the AD8015, the rms noise at the output would be 540nV/SQRT(Hz) x SQRT(p/2 x 100 MHz)=6.8 mV rms.

Two conclusions; measurements correlate pretty well to the calculations and the results are as expected.

How to lower this?

The AD8015 is not the most appropriate choice for this application. The AD8015 was intended for high data rate communications where the terminal capacitance of the photodiode is low, most cases sub 1pf. Your application uses a high terminal capacitance photodiode and therefore the AD8015 is not an optimized choice.

The “total” transimpedance of your current setup is 10k x 3 (from the 8015) x 20 from the (OPA 847) for a total of 600k.

To minimize noise you need make the TIA feedback resistor as high as possible.

The table below shows approximate numbers to consider.

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 10 14 MHz 10 mV 60 91 mV peak AD8099 10 45 MHz 10 mV 60 144 mV peak

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 100 4.6 MHz 100 mV 6 16 mV peak AD8099 100 14 MHz 100 mV 6 48 mV peak

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 500 2.1 MHz 1 V 1.2 5 mV peak AD8099 500 6.6 MHz 1 V 1.2 29.5 mV peak

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 600 1.9 MHz 1 V 1 4.2 mV peak AD8099 600 6 MHz 1 V 1 29 mV peak

Thanks,

Kris

• Hi Jone,

Here are some inputs from our application expert:

You say the output has about 1V noise. We assume this is peak.

The gain on the OPA847 is 20 (RF=1k, RG=50 ohms).

This mean the RTO noise of the AD8015 is 1V/20=50mV Peak, equivalent to 50mV/6.6=7.6mV rms.

If instead of “about 1V”, the output had 890 mV noise, the 8015 output noise would be 6.8 mV rms.

From the AD8015 datasheet and calculations:

Note from Figure 2 of the AD8015 datasheet that the RTI current noise increases approximately 0.5 pA/SQRT(Hz) for every 0.5pF increase in external capacitance.

The terminal capacitance of the Hamamatsu S3071 photodiode is 15 pF.

At 100 MHz the RTI current noise would increase to 18pA/SQRT(Hz) because of the 15pF external capacitance coming from the Hamamatsu photodiode.

The TIA section of the AD8015 has a 10k feedback resistor. This will result in an RTO noise density of 10k x 18 pA/SQRT(Hz) = 180nV/SQRT(Hz). The AD8015 has a second gain stage with gain=3. This would then increase the noise density at the output to 3 x 180 = 540nV/SQRT(Hz).

Integrating over the 100 MHz BW of the AD8015, the rms noise at the output would be 540nV/SQRT(Hz) x SQRT(p/2 x 100 MHz)=6.8 mV rms.

Two conclusions; measurements correlate pretty well to the calculations and the results are as expected.

How to lower this?

The AD8015 is not the most appropriate choice for this application. The AD8015 was intended for high data rate communications where the terminal capacitance of the photodiode is low, most cases sub 1pf. Your application uses a high terminal capacitance photodiode and therefore the AD8015 is not an optimized choice.

The “total” transimpedance of your current setup is 10k x 3 (from the 8015) x 20 from the (OPA 847) for a total of 600k.

To minimize noise you need make the TIA feedback resistor as high as possible.

The table below shows approximate numbers to consider.

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 10 14 MHz 10 mV 60 91 mV peak AD8099 10 45 MHz 10 mV 60 144 mV peak

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 100 4.6 MHz 100 mV 6 16 mV peak AD8099 100 14 MHz 100 mV 6 48 mV peak

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 500 2.1 MHz 1 V 1.2 5 mV peak AD8099 500 6.6 MHz 1 V 1.2 29.5 mV peak

 RF (kΩ) BW Vout for 1uA Gain on the OPA 847 Output noise ADA4817 600 1.9 MHz 1 V 1 4.2 mV peak AD8099 600 6 MHz 1 V 1 29 mV peak

Thanks,

Kris

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