DC blocking the photodiode signal

[kplee]

Why did you choose that amplifier? Are you using it in powered or unpowered mode? It does not seem to be a good match for this application, unless you are planning on using the lock-in feature at some point.

I will use the lock-in at the end, actually. The thing is, If I use only lock-in without preamp I would not see the signal.

+ I have no choice but to use it because this is what our lab has. haha

 

[berkeman]

So presumably if you only use the 'scope, you see this?

• Input Coupling = DC, 1M Ohm, get 9Vdc offset and AC signal too small to see
• Input Coupling = AC, 1M Ohm, get 10mVpp square wave signal
• Input Coupling = AC, 50 Ohm, the square wave signal is too small to see

Is that right?

 

[kplee]

So presumably if you only use the 'scope, you see this?

• Input Coupling = DC, 1M Ohm, get 9Vdc offset and AC signal too small to see
• Input Coupling = AC, 1M Ohm, get 10mVpp square wave signal
• Input Coupling = AC, 50 Ohm, the square wave signal is too small to see

Is that right?

I am not sure whether I understand the coupling concept, but let me rephrase what you said in my version.

1. I coupled the PD signal directly to scope with BNC cable. I set the scope as DC coupling and saw ~ 3 Vdc offset (not 9, idk why)

2. I coupled the PD signal the same way as I did in 1). I set the scope as AC coupling and saw ~ 10 Vpp square waveform.

3. I coupled the PD signal to the scope through the T connection with the 50 Ohm termination at the third end. I set the scope as AC coupling and saw nothing.

Are we on the same page now?

 

[berkeman]

saw ~ 3 Vdc offset (not 9, idk why)

Does the PD battery need charging?

2. I coupled the PD signal the same way as I did in 1). I set the scope as AC coupling and saw ~ 10 Vpp square waveform.

10Vpp or 10mVpp?

 

[kplee]

Does the PD battery need charging?10Vpp or 10mVpp?

1) I am not sure but I think it is working at least.

2) Sorry, 10mVpp

 

[berkeman]

I mean, the test signal from photodiode made by test optical input.

Does that test signal come with any sort of data? What is the expected output from the PD when you use this test signal? The PD datasheet says 470mA/W of optical input power, and your 10mVpp into 1M Ohm implies 10nA...

 

[kplee]

Does that test signal come with any sort of data? What is the expected output from the PD when you use this test signal? The PD datasheet says 470mA/W of optical input power, and your 10mVpp into 1M Ohm implies 10nA...

How does that calculation happen? I thought it can be directly converted from the PD datasheet, which implies the PD current of ~ 4.7 uA ideally.

 

[berkeman]

How does that calculation happen? I thought it can be directly converted from the PD datasheet, which implies the PD current of ~ 4.7 uA ideally.

Sorry, where are you seeing that in the PD datasheet? I see this for the maximum linear photocurrent (your -21 model is the right-most column):

 

[berkeman]

How does that calculation happen?

$\frac{10mVpp}{1M \Omega} = 10nA$

 

[kplee]

$\frac{10mVpp}{1M \Omega} = 10nA$

Oh I see. I think I mixed up during the calculation. But somehow that doesn't make much sense since the optical input is around several uW, which corresponds to several uA. Anyway, I think that's not an important part for now. I can handle that later. The problem is how to match this signal with the DC blocker and the preamp.

 

[berkeman]

optical input is around several uW, which corresponds to several uA.

So $(10 \mu A) (50 \Omega) = 1mVpp$

You need to have a low resistance like $50 \Omega$ to ensure that the PD is reverse biased well. If you try to use a $1M \Omega$ measurement resistance, the PD will not be reverse biased and the frequency response will not be very good.

So if you use the DC blocking coax thing (which I think presents $50 \Omega$ on the input (female BNC side) and go into a good preamp, you should get a reasonable signal. You could also use a spectrum analyzer instead to take advantage of its better input sensitivity.

 

[berkeman]

BTW, I just checked my Tek 'scope which is a model similar to yours, and it does not seem to have the ability to switch the input coupling resistance from $1M \Omega$ to $50 \Omega$. My LeCroy 'scopes do have that ability.

 

[kplee]

So $(10 \mu A) (50 \Omega) = 1mVpp$

You need to have a low resistance like $50 \Omega$ to ensure that the PD is reverse biased well. If you try to use a $1M \Omega$ measurement resistance, the PD will not be reverse biased and the frequency response will not be very good.

So if you use the DC blocking coax thing (which I think presents $50 \Omega$ on the input (female BNC side) and go into a good preamp, you should get a reasonable signal. You could also use a spectrum analyzer instead to take advantage of its better input sensitivity.

Well, the thing is the signal already gets mixed up when passing through the DC blocker on the 50 Ohm side, as I showed you before (check the first pictures). I didn't even use the preamp yet.

 

[DaveE]

This is all really hard to follow, but it sounds to me like you don't have enough intensity on the detector in photoconductive mode to drive a 50Ω load, but you do have enough to see a signal into a high impedance (perhaps in photovoltaic mode).

As an aside, good communication skills about lab work is an important skill in EE world. The people you work for will need that, and it is likely to improve your own work. Slow down; document; ask good, complete questions.

 

[kplee]

This is all really hard to follow, but it sounds to me like you don't have enough intensity on the detector in photoconductive mode to drive a 50Ω load, but you do have enough to see a signal into a high impedance (perhaps in photovoltaic mode).

As an aside, good communication skills about lab work is an important skill in EE world. The people you work for will need that, and it is likely to improve your own work. Slow down; document; ask good, complete questions.

Okay. I was just not used to the language in this community since I am in physics and didn't have any chance to work on electronics before. I didn't mean to confuse you. I will keep it in mind.

About what you said, I am kind of confused because when we think about the Ohm's law the voltage at the high impedance like the one in the scope should not change no matter it is connected to the low load impedance like 50 Ohm. Based on that, I can assume that there should be no difference on the PD signal when I add the 50 Ohm termination on the third end of T connector, just like the last pictures I uploaded. Here are my questions:

1) What makes the difference on detecting the signal through the scope when the 50 Ohm termination is added?
2) If the signal is detected through the termination add-up, why does the signal only matter the 50 Ohm termination but not the high impedance in the scope?

 

[berkeman]

About what you said, I am kind of confused because when we think about the Ohm's law the voltage at the high impedance like the one in the scope should not change no matter it is connected to the low load impedance like 50 Ohm. Based on that, I can assume that there should be no difference on the PD signal when I add the 50 Ohm termination on the third end of T connector, just like the last pictures I uploaded. Here are my questions:

1) What makes the difference on detecting the signal through the scope when the 50 Ohm termination is added?
2) If the signal is detected through the termination add-up, why does the signal only matter the 50 Ohm termination but not the high impedance in the scope?

The PD is a high impedance current source, so the resistance you terminate it into determines the voltage level you will detect.

It might be possible to use a current mirror circuit to do a better job of detecting these small PD currents, but the best way is to use a good-quality preamp after the DC blocking circuit. The way that I usually implement photodiode detectors is with a negative voltage for the reverse bias...

https://www.electronics-tutorial.ne...current-to-voltage-converter/#google_vignette

 

[kplee]

The PD is a high impedance current source, so the resistance you terminate it into determines the voltage level you will detect.

It might be possible to use a current mirror circuit to do a better job of detecting these small PD currents, but the best way is to use a good-quality preamp after the DC blocking circuit. The way that I usually implement photodiode detectors is with a negative voltage for the reverse bias...

View attachment 335095
https://www.electronics-tutorial.ne...current-to-voltage-converter/#google_vignette

1) What I understand is that there are two terminations: one at the scope with high impedance, and the other at the 50 Ohm termination with low impedance. If this is correct, why does the low impedance affect much even in the presence of the high impedance which is used for the detection?

2) If I need a reverse bias operation, should I build my own circuit? I think that's too much thing for me... I am not confident enough to make a precise measurement circuit that is required on my experiment.

As you said, I am trying to do the best way, making a working DC blocking circuit before the preamp, and that's why I want to figure out how to make it. I actually tried to amplify the PD signal with/without DC blocker, and it didn't work for both cases. They didn't show any signal just like what's shown in the current DC blocking setup. I just want to know why the signal disappears.

 

[berkeman]

Have you looked at other photodetector modules? Your required bandwidth looks pretty low, and your power requirement is also low. You should be able to find an inexpensive PD module that works much better than this one, and gives you a $50 \Omega$ source impedance voltage source to drive the coax to the 'scope.

 

[Tom.G]

The SR554 preamplifier has a DC Input Impedance of 0.5 Ohm.
The AC Input Impedance is 0.5 Ohm and 0.5H (in series) in parallel with 1.6uF.

The bandwidth looks marginal for a 250Hz square wave, you will certainly lose the fast edges.

You did not say if you were supplying power to the preamplifier and using the builtin preamplifier for an overall gain of 500. If you are using the preamplifier, realize that the maximum input level 14mV RMS.

For your initial testing with the 'scope, instead of using the SR554 you can put a 0.001uF capacitor in series with the 'scope probe and the PhotoDiode circuit. This should greatly reduce any DC voltage drift you are seeing. However you can expect about a 50% slope on the flat parts of the nice squarewave. Using a larger capacitor value will flatten the squarewave tops but increase any interference fron DC voltage shift.

The SR554 User Manual is available at:
SR554m.pdf

Cheers,
Tom