Cascading two HF photo-amplification stages (opamp-based)

[nitin.jain]

Hi All,
I have made a photodetection circuit that is supposed to work in a broadband-freq regime (from a few kHz to around 100 MHz). I essentially subtract the currents of two PIN photodiodes (that are reverse biased), and amplify this difference (I_diff) in two stages - the first is essentially a transimpedance stage, while the second enhances the voltage gain.
This is also known as homodyning (pardon me, if you were already familiar with the term). Both stages are opamp-based inverting configurations. I've attached a pic of the schematic, conveying the main idea (opa 847 is the opamp I'm using).
Currently, I am trying to get the DC operation of the circuit to work. I've the individual stages running quite okay -- by adjusting the resistances at the non-inverting inputs (R4 and R10 in the pic), I can get the individual DC offsets below 40 mV (with no input signal i.e.).
However, if I cascade them, then the output of the second stage goes beyond 4.0 V ! It also seems to find some capacitance, because this voltage seems to discharge (very) slowly. And it probably is also loading the earlier stage.
Can anyone tell me if the resistive bridge between the two stages (please see image) could largely be the culprit for this behaviour? If yes, how should I choose the values so as to minimize? Currently, I've tried R19=R16=100 ohm, R2=50 ohm and R9=4.0K and R3=R5=1.8K.

I've designed (and am testing) this circuit on a PCB. I've tried to avoid all possible pitfalls I knew about -- all the components are SMD, I am using enough bypass capacitors for the opamp supplies, the supply tracks are on one side of the board and the other side is almost 80% GND plane. I can post an image of the board too, if required.

I would be very grateful to get your views and suggestions!

Thanks,
Nitin

 

[berkeman]

Hi All,
I have made a photodetection circuit that is supposed to work in a broadband-freq regime (from a few kHz to around 100 MHz). I essentially subtract the currents of two PIN photodiodes (that are reverse biased), and amplify this difference (I_diff) in two stages - the first is essentially a transimpedance stage, while the second enhances the voltage gain.
This is also known as homodyning (pardon me, if you were already familiar with the term). Both stages are opamp-based inverting configurations. I've attached a pic of the schematic, conveying the main idea (opa 847 is the opamp I'm using).
Currently, I am trying to get the DC operation of the circuit to work. I've the individual stages running quite okay -- by adjusting the resistances at the non-inverting inputs (R4 and R10 in the pic), I can get the individual DC offsets below 40 mV (with no input signal i.e.).
However, if I cascade them, then the output of the second stage goes beyond 4.0 V ! It also seems to find some capacitance, because this voltage seems to discharge (very) slowly. And it probably is also loading the earlier stage.
Can anyone tell me if the resistive bridge between the two stages (please see image) could largely be the culprit for this behaviour? If yes, how should I choose the values so as to minimize? Currently, I've tried R19=R16=100 ohm, R2=50 ohm and R9=4.0K and R3=R5=1.8K.

I've designed (and am testing) this circuit on a PCB. I've tried to avoid all possible pitfalls I knew about -- all the components are SMD, I am using enough bypass capacitors for the opamp supplies, the supply tracks are on one side of the board and the other side is almost 80% GND plane. I can post an image of the board too, if required.

I would be very grateful to get your views and suggestions!

Thanks,
Nitin

Welcome to the PF. A design goal of DC to 100MHz is quite ambitious. Why do you need such a wide range? I'd AC couple the two stages if you can remove the DC requirement.

 

[nitin.jain]

Hello!
Thanks very much your quick response.
I am using a pulsed laser with a repetition rate (fR) of ~76 MHz & need to resolve between consecutive pulses. So, I need the detector bandwidth to be > fR, thanks to Mr. Nyquist!

Practically speaking, a bit of aliasing wouldn't be frightfully bad, so it might suffice even if it's around 70 MHz... however, for now, I am grappling with the DC operation blues... high frequency stuff will be a pain for sure (and the forum will see more posts then, quite likely) but later.

-Nitin

 

[0xDEADBEEF]

Hello!
Thanks very much your quick response.
I am using a pulsed laser with a repetition rate (fR) of ~76 MHz & need to resolve between consecutive pulses. So, I need the detector bandwidth to be > fR, thanks to Mr. Nyquist!
[...]

So what's the problem with putting some 1 MHz high passes into your design and pulling the inputs to ground at dc frequencies?

 

[berkeman]

So what's the problem with putting some 1 MHz high passes into your design and pulling the inputs to ground at dc frequencies?

Exactly. There's no reason for this amp to have high gain at DC, and lots of issues with making that happen.

 

[nitin.jain]

I surely did intend to use a high pass filter at the final output of the (homodyne) detector, however, the current DC offset of ~4.0 V at stage II output is rather high too and should be addressed in my opinion, as it possibly stems from some loading. In other words, it isn't merely a simple DC gain...

Nevertheless, your idea still probably is worth a try.. so I'll try to replace the resistive-bridge by a HPF of ~1 MHz cut-off... will let you know if it helps!

thanks once more,
Nitin

 

[Bob S]

I did a SPICE design of a circuit for photodiode signal amplification that had an active dc zero feedback circuit. This circuit also had a narrow bandpass and high frequency cutoff. Please see my posts #15 and #16 in
https://www.physicsforums.com/showthread.php?t=323143
If I had a SPICE circuit or resistor values for your circuit I could run another SPICE.

 

[vk6kro]

That looks like a nice IC.
However it has
GAIN BANDWIDTH PRODUCT (MHz) 200 (min) 800 (typical) 1750 (max)
which means that for a bandwidth of 100 MHz you should be looking at gains of 2 to 17.5 with an average of 8 depending on your device.

So, with a conservative gain of about 5 per stage, you may find the devices a lot easier to tame than they are at present.

A start would be to put resistors from the supply lines to a potentiometer and connect the tap on the potentiometer to the non inverting input via a resistor. Bypass the tap point.
As a guess, I'd suggest two 15 K resistors and a 1 K potentiometer and the same resistor you are using now to the + input. This should let you balance the output with no input signal.

 

[nitin.jain]

Hi guys,
Thanks very much again for your replies. I was able to reduce the DC offset to the order of ~50 mV, by playing a bit with the resistors before as well as in the II stage - the non-inverting input resistor did really have a lot of effect.
I can post the latest values, if anyone is interested.

@vk6kro
Yes, OPA 847 looks like a good opamp to me as well. The gain values that you gave are however for 3 other OPA 84* ICs... OPA 847 as such has a 3.9 GHz BWidth.
Anyhow, my approach, which was more or less based on a single-photodiode design given in the datasheet http://focus.ti.com/lit/ds/symlink/opa847.pdf" is essentially: I see the first stage very critical as it has to tolerate the photodiode capacitance that forms a LPF plus screws the phase-margin (and this is worse in a Homodyne detector, as it sees two capacitors in parallel now)... so, my thought was to use a slightly small feedback resistance (than what is given in the example in the pg 10-11 of the datasheet), and adjust the DC offset. With the first stage having < 40 mV offset, I connected the second stage, where I can hopefully go a bit easy with the gain... but there again, I first got a DC balanced output... with this being ready now, I should be able to test the optical response.

Please let me know if you think there might be something wrong in this approach.

@Bob S
Thanks very much for your offer. I shall certainly look at your posts tomorrow, and let you know if I need help.

 

[vk6kro]

That chip has amazing bandwidth. My data didn't even hint at that 3.9 GHz figure.

I am puzzled at the terminology you are using.

You have two photodiodes in series so they form a voltage divider.
You have two voltage amplifiers with apparently a lot of gain (although R1 is not given a value). The second stage has a gain of about 26.


Homodyne detector is a valid term but it usually refers to a very different device to what you are showing in your diagram.

It seems very likely that the series diode arrangement would be able to introduce a DC component at the input which would get amplified through the opamps. A capacitor in series with R1 should help with this.

 

[nitin.jain]

The photodiodes might seem to be in series, but in a high-frequency (or even non-DC) picture, the +ve and -ve supplies will be essentially ground, so the photodiodes (along with their junction capacitance) will be in parallel.
About the use of R1, yes it creates a bit of confusion, but ideally speaking, it is supposed to be zero and the first stage is then a true "transimpedance" amp... and so, I am actually amplifying the difference of the two photocurrents. And as far as I understand, this is the principle behind a balanced optical homodyne detector. What I know is that the frequency mixing is not explicit here (as in the microwave implementation), so maybe that's what raises the confusion? The former as such is done by (optically) interfering the signal with the local oscillator.

On the circuit board, I am using a < 10 ohm SMD resistor for R1... the idea being to get another pseudo-feedback capacitance from it, so that it makes a high pass filter with the shunt resistance of the photodiodes... which will hopefully prevent the phase margin to go to 180 deg. This mightn't be a great idea, but I think I should be able to even use a 0 ohm resistor. But in any case, I'll be happy to get comments on any of the aspects.

warm regards,
Nitin

 

[vk6kro]

The first amplifier has a voltage gain of (R3 + R5) / R1. So, that is 3600 ohms / R1.
So, choose R1 and choose your consequences. That will be the amplifier's DC gain as well.

I would make R1 about 360 ohms and settle for a gain of 10.

The photodiodes are in series from a signal point of view.
Their source impedance is the result of putting them in parallel, but you presumably have different light sources striking the two diodes.
So at anyone moment, they have different resistances and will divide the supply voltage accordingly.
This will give you a varying voltage which will be amplified by the first opamp.


The single diode example used in the data sheet was not given a power supply and its resistance varied according to the light striking it. So, the gain of the amplifier varied with the resistance of the diode. This is quite different to your circuit, which does have power supplied to it.

 

[Bob S]

I attach a thumbnail jpg of what I would use to get rid of dc drift. I would have preferred something I could have pasted into SPICE. I don't have your resistor values, so my initial suggestions are R101 = 10 times R1, R102 = R103, R103 = R5 + R3, and integrator time constant = C times R103 = 1/w = 1/ 2 pi f_hipass.

 

[nitin.jain]

The first amplifier has a voltage gain of (R3 + R5) / R1. So, that is 3600 ohms / R1.
So, choose R1 and choose your consequences. That will be the amplifier's DC gain as well.

I would make R1 about 360 ohms and settle for a gain of 10.

The photodiodes are in series from a signal point of view.
Their source impedance is the result of putting them in parallel, but you presumably have different light sources striking the two diodes.

So at anyone moment, they have different resistances and will divide the supply voltage accordingly.
This will give you a varying voltage which will be amplified by the first opamp.


The single diode example used in the data sheet was not given a power supply and its resistance varied according to the light striking it. So, the gain of the amplifier varied with the resistance of the diode. This is quite different to your circuit, which does have power supplied to it.

The two diodes experience almost equal intensity (0.5*I_LO) at all instants, except when the signal (which is the light field of a quantum state and extremely weak, compared to the local oscillator) tips the balance very slightly in favour of one or the other. So, from symmetry arguments (assuming the two photodiodes are ~identical), the voltage at their intersection point should be almost zero, with tiny fluctuations maybe.
Nevertheless, I'll reiterate - the quantity of interest is the difference photocurrent, and hence the need of the transimpedance configuration.
Finally, the single diode example in the datasheet (FIGURE 3. Wideband, High Sensitivity, OC-3 Transimpedance Amplifier) is indeed reverse biased.

 

[vk6kro]

at all instants, except when the signal (which is the light field of a quantum state and extremely weak, compared to the local oscillator) tips the balance very slightly in favour of one or the other.

I can see that the first stage would probably oscillate in this configuration, but which oscillator are you talking about?

And what does "the light field of a quantum state" mean?

Are you still having trouble with DC levels of this amplifier or did you already fix that?

 

[nitin.jain]

I can see that the first stage would probably oscillate in this configuration, but which oscillator are you talking about?

And what does "the light field of a quantum state" mean?

Are you still having trouble with DC levels of this amplifier or did you already fix that?

Hi,
I guess I was a bit abrupt in the last post... forget the "the light field of a quantum state" statement... I've attached a pic now that shows optical homodyne detection (50:50 BS stands for a symmetric beam splitter, it reflects and transmits equal proportions of input light, i.e.). You've an electromagnetic "signal" field and a "local oscillator" field (LO)... they interfere on this BS, and the resulting outputs are detected by the photodiodes, and their respective photocurrents (i1 and i2) are to be subtracted. Now, if the signal field is both infrequent and extremely minute compared to the LO, it shouldn't be hard to understand that i1 and i2 will almost be equal and their difference shall tend to zero. Only when the feeble "signal field" appears, will this change because of interference.
So that's what I do, by putting those photodiodes in series, and then amplifying their current difference (Idiff in the figure).

Now, I thought getting a DC balancing would also indirectly imply a relatively predictable high-freq response... however, today I checked the overall electronic response using both an oscilloscope & spectrum analyzer... and it looks quite terrible! :(
The noise signal is huge (something like 2 V p-t-p), and the spectrum has numerous peaks... so as of now, I am out of ideas except for reducing the feedback resistance.

 

[nitin.jain]

I attach a thumbnail jpg of what I would use to get rid of dc drift. I would have preferred something I could have pasted into SPICE. I don't have your resistor values, so my initial suggestions are R101 = 10 times R1, R102 = R103, R103 = R5 + R3, and integrator time constant = C times R103 = 1/w = 1/ 2 pi f_hipass.

Thanks Bob. And apologies for not giving the resistance values earlier.
Anyhow, as of now, I think focusing on getting a good overall frequency response is a better idea, compared to doing just a DC balancing, as I've explained in the last paragraph of my previous post. So I'll probably be asking questions pertinent to that, soon.

 

[Bob S]

Your discussion implies that the noise signal is quite large, and I really don't understand where it is coming from. With the two photodiodes on (reverse biased), and with neither your signal nor LO on, is the noise voltage still high? What is the reverse diode current in your photodetectors, and what is the calculated shot noise? I looked up the noise current in your op-amps, and it is only a few pA per root Hz, so that is not the source, even at 100 MHz BW.

 

[nitin.jain]

Your discussion implies that the noise signal is quite large, and I really don't understand where it is coming from. With the two photodiodes on (reverse biased), and with neither your signal nor LO on, is the noise voltage still high? What is the reverse diode current in your photodetectors, and what is the calculated shot noise? I looked up the noise current in your op-amps, and it is only a few pA per root Hz, so that is not the source, even at 100 MHz BW.

Yes. It's something that has left me very puzzled as well. Getting that huge noise even without any optical fields impinging on the diodes, except for the stray light of the room doesn't make sense.
The spectral responsivity of the diodes is around 0.58 A/W at the desired wavelength of 790 nm - however, stray light would of course be quite wideband so I don't know how to estimate a reverse photocurrent (or the shot noise).
Apart from that, I think I've used enough bypass capacitors, bleed/ferrite resistors in supplying power to the opamps (via 7805 and 7905 voltage regulators, that have their own bypass capacitors as well). The diodes are also bypassed with two capacitors each. I also cut their leads to as short as possible, to minimize inductive coupling, before mounting on the PCB.

 

[Bob S]

You should try shielding the photodiodes from all light by using a light-tite enclosure and black cloth (e.g., black muslin). If you have worked with photomultipliers, you know how important light shielding is. It is very important to understand the noise source.

 

[nitin.jain]

You should try shielding the photodiodes from all light by using a light-tite enclosure and black cloth (e.g., black muslin). If you have worked with photomultipliers, you know how important light shielding is. It is very important to understand the noise source.

Hi,
Sorry for this late response, I was heavily occupied in other tasks. I'll return to working on the circuit today, and will try the above suggestion.

Thanks again,
Nitin