Simulation of Photodiode connection to ADA4350

Hi Mohi,

Several points related to your issues with the ADA4350 LTspice file you attached:

1. LTspice Oscillation: I can't tell what the screenshot instruction you attached for designing the feedback components (R1, and C1) comes from since it's not from the datasheet. However, if you want a Transimpedance gain of 200kohm, ADA4350 datasheet equation 8 shows you how to compute the feedback across it:

Plugging in:

RF= 200k, CS = 900pF, fGBW = 175MHz --> CF = 2pF

Once I changed CF (your C1) to 2pF (vs. 0.2pF you had used), the output oscillations you were seeing disappear.

Also, the gain is correct: Vout1 / I(R6) = 0.516V / 2.7uA = 191kohm (vs. 200k).

I have reduced the stimulus frequency to 100kHz (instead of your 2MHz) because I think at that frequency you're already bandlimited, which might make it look like the transimpedance gain (ohm) is off or lower than expected.

Here is the simulation file:

1016.ADA4350_Sinewave_Stimulus modified EZ 1_13_21.asc

2. Single Supply Operation: For single supply operation, I've modified your circuit and tied IN_P to some voltage (1.3V) to make sure I allow bipolar swing at the FET amplifier using single 3.3V supply. I've also reduced the input stimulus to 1uAp to avoid distortion. Gain expression computes correctly:

Vout1 / I(R6) = (1.51-1.1)V / 2uApp = 205k (vs. 200k).

Here is the simulation file:

1537.ADA4350_Sinewave_Stimulus modified Single Supply EZ 1_13_21.asc

Hopefully, all this helps.

Regards,

Hooman

Hi Hooman,

Thank you for your quick reply and corrections to my schematic. I really appreciate your time and efforts.

I will be using your first modified schematic for further discussion. Following are more of my questions:

1. The "design-gain x signal-bandwidth" should not exceed the GBW of the OpAmp. The rule does not seem to be applicable for transimpedance amplifier. The signal I need to process is 2MHz. What gain is achievable for this signal? For the transimpedance gain of 200k and the input signal of 100KHz, the design GBW = 200k x 100kHz = 20Ghz - I must be missing a point here.

2. What does the "The signal bandwidth" [1/(Rf x Cf)] mean mentioned in the paragraph following equation 4? For this design, it is ~2.5MHz but the input signal is limited to ~100kHz. Output starts to reduce above 100KHz of input signal.

3. If R4 is reduced, the signal start to become DC biased. Is it because of V3?

Hi Mohi,

You're welcome!

To answer your specific questions on the ADA4350 operation:

1. GBW vs I-to-V Gain: The GBW computation you've done is mixing parameters / units together. The 200kohm is the transimpedance gain in units of ohm. The GBW is in units of V/V*Hz or just Hz. The I to V -3dB BW you can expect will be the same as fN in equation 9:

Here are some examples:

Here is a simulation file that shows the AC small signal response for various values of RF (with the appropriate value of CF computed for each step), as reference:

1830.ADA4350_Sinewave_Stimulus modified AC EZ 1_14_21.asc

2. Signal Bandwidth: Figure 58 schematic shows that the photodiode current flows through the parallel combination of RF and CF to arrive at the output. Such an output voltage will roll-off beyond 1/2pi()RFCF (in Hz) thereby setting the bandwidth achievable. "1/RFCF" would be the bandwidth in rad/sec (not Hz). You can follow my description above to see what bandwidth you can expect for a given CS, and CF given that GBW is 175MHz.

3. Reducing R4 (shunt resistor across photodiode): If you reduce this large (2Gohm) resistor, you are stealing the signal current from the photodiode and diverting it to ground. Not sure why you're doing this. R4 should be what the Photodiode manufacturer specifies. It is generally very large so all photo current flows through the amplifier.

Regards,

Hooman

Hi Hooman,

I am back wit a couple of more questions regarding TIA's signal and SNR.

1. What is the lowest current level that can be amplified and digitized with this TIA.
Above 100k feedback resistor, I noticed that the value of feedback capacitor start to becomes unrealizable and the signal is unrecognizable in LTSpice.
I am given the task to detect 300nA current. Is it possible with any of the off-the-shelf TIA without using cryogenics?

2. I can sum the two noise sources at input of the TIA namely thermal noise from the feedback resistor and TIA's internal current noise. This is for the SNR calculation. How one add the internal voltage noise to the equation? Ultimately how do one calculate the internal resistance of the TIA? Is there a simpler way to calculate SNR?

Hi Mohi,

To answer your specific questions about the ADA4350 interface with your photodiode:

1. Minimum discernible input current from your photodiode: I'd use the information in Table 17:

This would be output noise. With a high impedance photodiode, the low frequency noise gain is 1V/V (as Figure 60 bode plot also shows).

The input noise current contribution will not be dominant, as stated in the paragraph just above Table 17:

"The effect due to the current noise is negligible in comparison".

You can then compare your output "signal" [ = i_Photodiode * RF] against the output "noise" [ = RMS sum of the RF noise and the VNOISE from Table 17]. You'll be able to detect the 300nA photodiode current as long this ratio is at least a few dB's.

2. Individual Noise Terms Effect / Contribution: As the datasheet states and as Table 17 shows, your dominant noise sources are RF thermal and the input voltage noise. Apparently, input current noise is of little significance here.

Regards,

Hooman

Hi Hooman,

Thanks for all your help and suggestions. You are a very good teacher. I now have much better understanding of TIA and especially ADA4350. Can't say yet I can compete with you 

I am including my latest LTSpice project (few changes from the previous project) along with a worksheet calculator and a HowTo doc. I am hoping it might help others who come across this post.

One comment: The noise calculation formulas reveals that one cannot go beyond a certain Rfb values. As the gain increase, so does the noise.

In my case, I found that 2nA(pp) input current and a 2M-Ohm Rfb gives about 4dB of SNR. I am hoping that ADC will be able to discern this signal - which is the next step of this project. Any suggestion for a good ADC - say 12 to 14 bits of ENOB?

6428.Photodiode.zip

Hi Mohi,

Thanks for the compliment :-)

I went through some of the spreadsheet. You've not used the noise_pp cell J6 anywhere but it usually has a 6.6x multiplying factor to RMS noise (not 2*SQRT(2)) as you've used). Also, I could not see why you've calculated Cfb in a separate worksheet and not just computed it interactively for any RF value. At least, that's how I'd have done it. And, you probably won't need to invoke LTspice to compute the output voltage separately because your TIA gain is known based on the feedback resistor you've chosen and you could just use the RF value in your spreadsheet directly. But, overall I think your approach works.

With regard to your question about selecting an ADC: I'm not really an expert on that, but I'll point you to the parametric selection guide here to do your own evaluation:

https://www.analog.com/en/parametricsearch/11007#/

Maybe somebody else on this forum who is more knowledgeable about ADC's can chime in?

Regards,

Hooman

Hi Hooman,

I back with another question.

In another thread related to ADC discussion, Sean suggested to use AD4008 or AD7685.

One of the advantages of using AD4008 is it has "The analog input incorporates circuitry that reduces the nonlinear charge kickback seen from a typical switched capacitor SAR input." as stated in the datasheet.

The ADA4350 has built-in driver circuit. Is using AD4008 gives any advantage in our case.

We will be sampling the signal only few times per second (~100SPS) but continuously. However, input to ADC can be as little as 10mV.

The reason to asking this question is the price difference between the two ADCs. 08 cost ~$30 while 85 is ~$15.

Thanks

Hi Mohi,

I'm not on expert on the ADC's and the driver you've noted. However, here is my 2-cents:

As far as driving the input of a SAR ADC like AD7685:

Usually an RC as shown in Figure 27 of AD7685 datasheet (33ohm series and 2.7nF shunt cap) alleviates most issues with charge kickback from the ADC frontend. You may want to make some measurements with the RC combination to see if you see any issues.

You may want to consult this excellent article (link below) from Alan Walsh which goes into the design of the front-end RC network for charge kickback mitigation:

"Front-End Amplifier and RC Filter Design for a Precision SAR Analog-to-Digital Converter."

I took Alan's idea and simplified it a bit here in this article:

The Perilous Path from the Transducer to the ADC: What’s an Engineer to Do?

There is also some help from ADI on the ADC driver Tool in configuring the input network here which does some of the tedious computations automatically for you:

Precision ADC Driver Tool

Hopefully this answers your question.

Regards,

Hooman

Thanks Hooman.

Once again, your answer are very through. I will go through both articles to help me understand the root of the issue.

Mohi

Hi Mohi,

Thanks for the compliment :-)

BTW, I've been told that the the AD4008 actually has a lower 1k list price of $4.95 vs the AD7685’s $6.78, and has better performance overall. So the $30 cost difference is questionable.

Also, some comments below per the Application eng. responsible for these ADC's for your reference:

"I will make the general comment though that since you're only sampling at 100SPS, that the RC filter recommended by the data sheet is probably much higher bandwidth than it would really need to be to ensure proper settling. The ADC Driver Tool should illustrate just how much easier it is to settle the inputs when sampling at 100 SPS vs. the 250 kSPS upper limit of the AD7685.

Outside of the improvement in specs (INL, SNR, etc.) one argument I would give to select AD4008 over AD7685 is that the AD4008 inputs being easier to drive means that for a given analog input bandwidth, you’ll be able to sample the AD4008 faster than the AD7685, which would allow for oversampling and averaging to reduce the overall noise of the measurement – but that only matters if you're really trying to implement a very low noise measurement."

Regards,

Hooman

Excellent suggestions. Thanks Hooman.

Excellent suggestions. Thanks Hooman.

Hi Hooman,

Attaching is a mostly completed circuit (yet to add FB components) - but this time it is from the schematic capture.

https://drive.google.com/file/d/1EV1c2_NZWZRGQg0iJ73Yo8QllCfITLQJ/view?usp=sharing

Would you please suggest any modification if necessary. 

One question about the DVDD: should it be connected to 3.3V to work with microcontroller which is a 3.3V?

Thanks,

Hi Mohi,

I don't see anything obvious, but to be honest I did not spend an inordinate amount of time reviewing your ADA4350 TIA design feeding into the ADC :-).

I also wanted to point out a TIA device Preliminary datasheet for LTC6563 with ADC driver built-in if you've not seen it before, for your reference, if you wish to go that route:

https://www.analog.com/media/en/technical-documentation/data-sheets/ltc6563.pdf

I'm hoping other people on the forum offer suggestions (if any) to your ADA4350 schematic.

Regards,

Hooman

Hi Hooman,

I got suggestion to use AD4000 in another thread. I am set for the ADC.

My primary question is to confirm if any errors in the attached schematic stand out?

Also confirming that for DVDD my use of 3.3V is correct since uC (DIO and SPI) is based off 3.3V.

drive.google.com/.../view

Thanks,

Mohi