I am building a system with AC coupling in first stage. I currently build the schematic attached, it doesn't work very well. The current gets divided between the opamp and supply and reduces my overall SNR. If I remove the R5, short C1 and everything works really well but this is not AC coupled and cannot be used in my app.
Would the schematic 2 work? I will build this at the bench and test but I wasn't sure so I like to ask the gurus here. This is a simple AC couple architecture, just tie the PD to opamp with a cap.
I'm not sure what you mean by "...it doesn't work very well." Your first circuit rolls off the low end frequency response at 1.6kHz but if you increase R5 to 100k it will extend the low end to 16Hz.
The second circuit had no return path for the photodiode bias.
Regards, Neil P. Albaugh ex-Burr-Brown
In the first circuit, I felt that the gain is weaker as not all AC current is pulled from the Opamp. When I look at the simulation, I also see that there is AC current on R5, so I guessed there is less signal for opamp to amplify since all current generated by the PD is split between opamp and the bias circuit and this would reduce SNR. Am I wrong?
In the second circuit, I tested this and it works. Since I am not interested in DC current fo the PD, does it matter? It still biases the PD and there is a return path for the AC current.
In the first circuit the photodiode load resistor, R5, is 100k but remember that it is shunted by the input impedance of the transimpedance amplifier. This impedance approaches zero ohms due to the negative feedback of the op amp.
The second circuit might work, sort of, but the photodiode is not being reverse biased the way it is drawn.
Sorry, I am not following Neil. When I simulate I see AC ripple on R5, which is essentially saying, the current generated by the PD is shared between R5 and the OPAMP. Wouldn't this reduce the opamp's input current and hence the SNR?
The second circuit is also not clear to me, I thought the biasing is correct however I am not fully sure, anyway I was just experimenting with it. Thx, Frank
I ran a Tina Transient Analysis on your circuit with the 1MHz sine wave increased to 15uA peak to get better resolution and looked at the current in series with R5 and in series with C1. The current in R5 is 0.061uA p-p while the current into C1 (the transimpedance amplifier input) is 13.74uA p-p. So the answer is, yes the photodiode current does divide between the bias resistor and the amplifier input but it does so as I suggested-- almost all the photodiode current flows into the amplifier since it is a very low impedance.
There is another effect that the DC background current causes, even with AC coupling-- the photocurrent from the ambient background light forward biases the photodiode, causing gain compression of the AC signal. With a 200uA DC current, you lose about half of your AC signal. This is due to the voltage drop across the 100k bias load resistor caused by the ambient light current. To prevent this, a smaller load resistor can be used; a 10k resistor maintains the photodiode in the reverse bias region but you do lose a bit more signal. A smaller load resistor also contributes more current noise so the SNR suffers. I don't think that you can get away without an optical filter to minimize the background illumination, Frank.
It is always a compromise. Photodiode transimpedance amplifiers always look so simple in the textbooks...................
One other thing that I did in simulating your circuit was to add a diode in the photodiode model, a 1N4148, and reduced the photodiode capacitance from 20pF to 16pF to compensate for the diode's 4pF capacitance. Normally the diode is not really necessary in a DC coupled transimpedance amplifier since the op amp feedback keeps the voltage across the diode close to zero volts but in your case that is not so. Adding the diode gives you a realistic picture of what happens when you have a large background light current in an AC coupled circiot.
Thanks for the explanation. Now, I get it. When I start building this, I thought this would be a piece of cake but hey. I am going through my third board and still not perfect.
I like to make 2 comments:
- If I change from AC coupled architecture to Opamp biased one, in other words, short the capacitor and remove the R5 so that the PD biased from opamp, I face a different challenge. My first stage gain has to be limited as now I take the DC in. Assuming I bias the PD with 1.8V (i.e. Opamp is biased at 1.8V), I get roughly 1.8V dynamic range in terms of gain in the first stage that is 9K for 200uA. I can later put a cap between first and second stage and eliminate DC and gain another 1000 times in the second stage. The problem is that the first stage needs to be so clean that the in the second stage, I amplify signal but not noise. Also 9K resistor will introduce more noise than say 100K. My signal is around 50nA range.
- I still think of the alternative architecture I mentioned earlier. Connect the PD to the negative terminal of the Opamp and keep the AC coupling capacitor. In this architecture, there can be no DC current flow due to cap, but the AC current should flow nicely. Potentially this also doesn't suffer from the resistor noise in the AC coupling circuit. I am not an expert on this matter but I cannot seem to find any serious issue with this circuit, other than no return path for the DC. Can you comment?
I attached a picture.
It is a tough tradeoff without using an optical filter. Your circuit will work but the DC biasing is not really doing anything without having a return. The photodiode is operating at zero bias. You still suffer the gain compression by high levels of DC background current.
I think you can see by now what an optical filter that attenuates the background wavelengths but passes the laser wavelength will do to improve your sensitivity.
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