Photodiode Transimpedance Amplifier

Thread Starter

Rishi1@

Joined Dec 28, 2018
19
Hi,
I am trying to observe interference from a 1550 nm LASER on Thorlabs (FGA21 - InGaAs Photodiode, 14 ns Rise Time). I have the photodiode connected in Photovoltaic mode across an OP27E Op-amp. When I observe the output signal from the Op-Amp there is a lot of noise and also there is a very large DC offset (the photodiode current is 2.38 mA). I can observe the amplitude variation when I distort the mirror alignment but I cannot observe a fringe pattern due to the noise. I have a few questions:

1) Due to large DC offset the maximum feedback resistance is limited to 4.7 Kohm (voltage supply +- 15V ). Should I just add an HPF to remove the DC offset followed by amplification stage or should I maximise the voltage output from the transimpedance amplifier and finally remove the dc offset.

2) Is it better to use AD549/AD8651?
 

ScottWang

Joined Aug 23, 2012
6,926
When I observe the output signal from the Op-Amp there is a lot of noise
Could you upload the waveform of noise as 800x600 x.jpg?
there is a very large DC offset (the photodiode current is 2.38 mA).
What is the DC offset (the current)?

According to the data of +15V power, Op27E (see the page 4, OUTPUT VOLTAGE SWING), 4.7K, photodiode(3V).

So I calculate the current of photodiode below.
Vout of OP27E = ±13.8V, so the Output of positive is +13.8V
I_Pdiode = (13.8V-3V)/4.7K =10.8V/4.7K = = 2.297 mA ≅ 2.3 mA
If what I calculated was right then 2.3 mA is very close to 2.38 mA.
 

Thread Starter

Rishi1@

Joined Dec 28, 2018
19
Thanks for you reply.

Could you upload the waveform of noise as 800x600 x.jpg?
I will upload the waveform soon.
What is the DC offset (the current)?
When I observe the photodiode output for the incident interfered beam the output is:
2.38mA*4.7K=11.186 V(dc offset).
(Laser Diode: LPS-1550-FC)
Using AC Coupling I can see the variation in the signal (when I malign the mirror or excite it with sinusoidal signal) on oscilloscope however due to the noise I cannot observe the fringes.

TIA_Rishi1.jpg
 

ebp

Joined Feb 8, 2018
2,332
To null the so-called "offset" (offset is not the correct term - it is simply a signal), add an equal and opposite current at the summing node (inverting input of op amp) of the amplifier. If you are getting +2.38 mA from the photodiode, then add -2.38 mA. This could be done by applying -2.38 V through a 1000 ohm resistor or -5 V through a 2100 ohm resistor - or any other convenient arrangement that will produce the required current. Care is required to minimize noise from the nulling source. I would use a precision voltage reference with suitable filtering. "Buried zener" references are generally less noisy than bandgap references. You could either use a voltage divider across the reference buffered with an op amp and fed to the summing node via a fixed resistor, or use a combination of a fixed and variable resistor from the reference (fixed resistor at the summing node end to isolate extra capacitance associated with variable resistor). Using the op amp circuit adds a bit of noise due to the amplifier but makes filtering easier since the load on the filter is very high impedance and the buffer can also be used as an active filter. If the circuit is surface mount, I would use thin-film resistors throughout for low noise and good stability. Using a potentiometer as a voltage divider instead of as a rheostat (that is use it across the reference with a buffer amplifier) helps minimize problems with stability with temperature. A wirewound potentiometer will give the lowest temperature coefficient of resistance. Cermet is not too bad. Carbon is terrible. Be very careful not to introduce more capacitance at the summing node.

If you attempt to do offsetting by adjusting the voltage on the non-inverting input of the amplifier, you totally change the behavior of the circuit because the photodiode is no longer operated with zero bias.
 
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Thread Starter

Rishi1@

Joined Dec 28, 2018
19
I am not sure why the current is so large (2.38mA) I mean a major contribution is from the constant part of interfered signal (but it should not be this large even if I add OpAmp noise) and the variable component, current is relatively very small (because the p-p of output using AC coupling is in mV so the dc offset is higher by a factor of 10000).

Now, this large current limits the amplification I can apply to using Transimpedane Amplifier. So what I don't understand is what will be the best approach to eliminate this 2.38mA.

Secondly, is it better to use a photodiode with much lower junction capacitance and AD549/AD8651 to eliminate the noise as much as possible.

I am hoping to use this https://www.analog.com/designtools/en/photodiode/ as a resource to develop the circuit.

I will upload the waveform from the current setup soon.
 

OBW0549

Joined Mar 2, 2015
3,506
I am not sure why the current is so large (2.38mA) I mean a major contribution is from the constant part of interfered signal (but it should not be this large even if I add OpAmp noise) and the variable component, current is relatively very small (because the p-p of output using AC coupling is in mV so the dc offset is higher by a factor of 10000).

Now, this large current limits the amplification I can apply to using Transimpedane Amplifier. So what I don't understand is what will be the best approach to eliminate this 2.38mA.
One approach to eliminating a large (and not necessarily constant or predictable) DC offset from a small AC signal is to null it out via a slow feedback loop. The diagram at the bottom of page one of this op amp data sheet shows this technique being used in a photodiode amplifier. I would imagine that R1, C1 and R5 would have to change to make the circuit suitable for your application, but other than that it should do what you need.
 

Thread Starter

Rishi1@

Joined Dec 28, 2018
19
(offset is not the correct term - it is simply a signal)
This signal is mostly likely above the saturation limit of photodiode and hence will always be constant? So what you are suggesting is to just remove this 2.38mA current by applying an negative current at the summing node,right?
 

Thread Starter

Rishi1@

Joined Dec 28, 2018
19
One approach to eliminating a large (and not necessarily constant or predictable) DC offset from a small AC signal is to null it out via a slow feedback loop. The diagram at the bottom of page one of this op amp data sheet shows this technique being used in a photodiode amplifier. I would imagine that R1, C1 and R5 would have to change to make the circuit suitable for your application, but other than that it should do what you need.
If I use this circuit the value of R1 (gain) will still be limited by the large 2.3mA current and maximum output of the OpAmp. So how is it any different from using a High Pass Filter.
 

OBW0549

Joined Mar 2, 2015
3,506
If I use this circuit the value of R1 (gain) will still be limited by the large 2.3mA current and maximum output of the OpAmp.
No.

In that circuit, the second op amp (the LT1097) acts, via the 2N3904 transistor, to draw away that 2.3 mA so it does not bias the signal, allowing a much higher gain to be used; the LT1793 never sees the 2.3 mA, so it can use a much larger gain-setting resistor. See the document @danadak linked to in post #7, on page 33.

So how is it any different from using a High Pass Filter.
It removes the need for any amplifying stage to have to deal with the 2.3 mA bias, by removing it right at the source.
 

ebp

Joined Feb 8, 2018
2,332
What is the frequency of the signal of interest?

OBW's circuit does exactly what I had suggested, but it does it automatically. It is not without its limitations. It nulls the long-term average of the input, so it is dubiously suitable if you are trying to resolve a small, very low frequency delta in the input. If your frequency of interest is at least a few hertz, it can work well, but is still limited because of the requirement for a long time constant which is demanding in terms of the capacitor used in the low pass filter of the nulling circuit (polypropylene is probably "best" in terms of electrical performance, but it might need to be physically very large; electrolytic types are compact but electrically horrible). For very low frequency signals riding on a DC level, digital techniques are probably superior but more complex. Whether digital or analog low pass filtering is used, the circuit will require some time to settle to steady state. I haven't really thought about it, but you might get faster initial settling with a combination of manual and automatic nulling (both of with could be incorporated with digital methods) - e.g. "manually" null 2.2 mA and automatically null the rest.

Properly done, the value of the DC that is being offset can be accurately recovered.

Note that the circuit on page 33 of danadak's link operates the photocell in photoconductive mode. It is the automatic version of what LesJones suggested.
 

Thread Starter

Rishi1@

Joined Dec 28, 2018
19
I am trying to use the intensity of interfered signal to study the ground vibration. So frequency range is from 2/3 Hz.
Sorry for the late reply the new year has been keeping me busy.
Now when I use the AC coupling on oscilloscope the remaining signal is few hundred micro volts while the dc signal is about 11V (2.5 mA*4700ohm).
 

danadak

Joined Mar 10, 2018
4,057
Scope AC response is typically 10's of Hertz at best, so no surprise your
signal is way down.

If you are doing this because of large DC offset on signal, use scope diff
mode and place a DC offset on one leg of diff input so you can bring screen
into visual range. Two channels, set to A - B on older analog scopes. Or DSO
use math capability to perform the subtraction of CM offset.

Both channels in DC coupled mode.

In older plugin scopes this was called a differential comparartor plugin.

Regards, Dana.
 

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Thread Starter

Rishi1@

Joined Dec 28, 2018
19
Thank you all I will try your suggestions and will inform you of the result.

Now I am thinking that maybe the large output current due to photodiode going over saturation limit. (which I did not check)

Is there a way to numerically calculate this value of saturation limit, I am using FGA 21 with a LPS-1550-FC - 1550 nm, 1.5 mW laser and CFC-8X-C collimator.
From the FGA 21 spec sheet the responsivity @1550 nm is 1A/W so I don't know why the output current is 2 mA.
 
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