Guitar - Photodiode Amplifier, Signal-to-Noise ratio

[dick_freebird]

Let me first recommend that you do not use
ambient or DC-biased illumination for the
photodiodes to receive. This just begs for
modulated-optical noise pickup from
lighting, windows (picking up vibrations and
turning them into reflection-modulated
light). Using a carrier frequency like 1MHz
(or higher) to drive D1-D6, a narrowband filter
and AM-demodulating would be better. This
is a common approach to IR beam-break
detectors and such, to tamp down ambient
light "spoofing" and any attempts to defeat
by a simple flashlight. Might be good to have
a DC "baseline" that keeps diodes slightly
lit, and a 1MHz square wave summed in
(laser diodes have a threshold current; LEDs
have their own "threshold" (Vf) and would
probably behave better if you didn't put them
full dark every half cycle).

Now your photodiodes (D7-D12) appear to
have no DC bias. That means they must
work in photovoltaic mode, but that will crush
optical-electrical voltage gain. You'd be better
off with a reasonable reverse bias and resistor
load (might make this adjustable, until you
find the range on your illuminator / filter /
demod) to use photoconductive mode for
much higher signal (hence better SNR).

 

[Yarrrrr]

Let me first recommend that you do not use
ambient or DC-biased illumination for the
photodiodes to receive. This just begs for
modulated-optical noise pickup from
lighting, windows (picking up vibrations and
turning them into reflection-modulated
light). Using a carrier frequency like 1MHz
(or higher) to drive D1-D6, a narrowband filter
and AM-demodulating would be better. This
is a common approach to IR beam-break
detectors and such, to tamp down ambient
light "spoofing" and any attempts to defeat
by a simple flashlight. Might be good to have
a DC "baseline" that keeps diodes slightly
lit, and a 1MHz square wave summed in
(laser diodes have a threshold current; LEDs
have their own "threshold" (Vf) and would
probably behave better if you didn't put them
full dark every half cycle).

Now your photodiodes (D7-D12) appear to
have no DC bias. That means they must
work in photovoltaic mode, but that will crush
optical-electrical voltage gain. You'd be better
off with a reasonable reverse bias and resistor
load (might make this adjustable, until you
find the range on your illuminator / filter /
demod) to use photoconductive mode for
much higher signal (hence better SNR).

Thank you for your reply!

I've been told enough times now to use modulation that I really need to figure that out for a future design.

Would you mind showing me with a schematic how I'd change my current arrangement for (D7-D12 ) to DC bias the diodes?

Google tells me I want to apply negative voltage to the diode, But I've only got +9 Volts. And biasing the op-amp at half that. Would using ground be the negative in my case? as it is relatively below my bias voltage.

 

[KlausST]

Hi,

Google tells me I want to apply negative voltage to the diode, But I've only got +9 Volts. And biasing the op-amp at half that. Would using ground be the negative in my case? as it is relatively below my bias voltage.

Voltage is a relative measure. Thus a voltmeter has two wires: the black one as reference, the red one as measurement input.

The photodiode won't find out whether you have true negative voltage (referenced to system_GND) or you use two random nodes.
What matters is simple the voltage difference between the both photodiode pins.

I'm sure there are many modulation / demodulation circuits around. Thus I see no need for an new one.
Please do some internet research on your own. Find one from reliable sources like: semiconductor manufacturers, universities...
Find one with good description ... this is one indicator for a reliable circuit.

Generally you may modulate the photodiodes by applying square wave to them.
For demodulation there are many circuits, like a envelope circuit or a controlled rectifier..

Klaus

 

[betwixt]

For Yarrrrr's sake, it think it would be useful to explain our thinking on this idea.

IR detectors are used all over the place, from TV remote controls to DVD disc reading and driving optical cables. All of these are prone to picking up light within their 'visual' range from other sources.

The usual way to avoid those other sources is not to use the light level picked up directly but to look for variations in a high frequency signal sent from the LEDs. Instead of driving them with DC, we drive them with some higher frequency so they (if you could see them) flash on and off rapidly. The frequency is usually above 35KHz but can run into many MHz or even GHz in some applications. As the frequency increases it tends to be harder to drive the LEDs so some compromise is needed.

The pick-up sensor sees the light pulses, detects their level and uses that as the output. The big advantage of doing it this way is the pulsed light beam is at a single known frequency so it is very easy to pass it through a filter where all other signals can be eliminated. Only the pulses, varying in amplitude according to the reflection from the strings will leave the filter and from the the amplitude can be recovered back to audio. Other sources do not pass through the filter.

You have an advantage in your system that the source of the pulsing and the detectors are in close proximity, that makes it easier to recover the audio because the detector can work synchronously with the emitter electronics.

I still have reservations that the reflected light may not accurately follow the vibration of the strings but that's a different problem.

Brian.

 

[Audioguru]

For low noise the photodiodes should have no bias voltage. See post #13.

Modulation is amplitude changes (tremolo) and frequency changes (vibrato). A string swinging around in an ellipse produces tremolo which does not change the tone of the guitar since different tones have different amounts of harmonics.
Adding many harmonics causes distortion like a buzzer (an electric geetar blasting acid rock noises).
Reducing many harmonics reduces distortion sounding like sinewave (flute).

 

[Yarrrrr]

At this point I can't tell if people are contradicting each other. I thought my current setup was photovoltaic like #13. But some people are telling me to reverse bias the diode for photoconductive operation to improve SNR.

 

[betwixt]

Shining light on any LED produces a tiny voltage but it can't sustain any current, in other words it needs a high resistance load across it. Equally, if reverse biased, the current that 'leaks' through it will increase with the amount of light falling on it. So it can be used in both modes. In general, the noise will be lower and the output higher in reverse biased mode.

Experiment: disconnect the wiring to one of the photo diodes and measure the voltage across it. You will probably see a few mV. Now measure the current flowing through it by changing to current mode, it will only be a few uA, you will see the overall power it can produce is tiny. This is a photovoltaic mode measurement.

Now measure the resistance of the photo diode. With the probes one way around you will probably see some resistance because the probe voltage is forward biasing the diode. Swap the probes over and you should measure a much higher resistance - but it will vary according to light level. This is photoconductive mode.

Note that in photovoltaic mode, the voltage can only be in one polarity so using V/2 at one end will not result in the voltage going above and below V/2, it can only go in one direction.

Brian.

 

[Audioguru]

At this point I can't tell if people are contradicting each other. I thought my current setup was photovoltaic like #13. But some people are telling me to reverse bias the diode for photoconductive operation to improve SNR.

I posted an article from National Semiconductor who says that reverse bias increases noise and photo-voltaic reduces noise. Sharp Electonics says the same.

 

[Yarrrrr]

I posted an article from National Semiconductor who says that reverse bias increases noise and photo-voltaic reduces noise. Sharp Electonics says the same.

I just tested grounding one side of the photodiodes, I had to flip their polarity in the circuit though to get any output at all. If what I did actually resulted in a photoconductive arrangement I can conclude that the SNR remained the same as before.

 

[Audioguru]

You can ground the anode of the photodiodes when you have a dual-polarity supply.
You say grounded "one side" of the photo-diodes but which side?
You "flipped" connected the cathodes to ground??
It does not make sense. Please sketch a schematic.

I think shorting together all the photodiodes causes shadows on some of them to affect the level of a string that you are playing. Use a separate opamp for each photodiode and mix them with the 2.2k resistors.

 

[Yarrrrr]

You can ground the anode of the photodiodes when you have a dual-polarity supply.
You say grounded "one side" of the photo-diodes but which side?
You "flipped" connected the cathodes to ground??
It does not make sense. Please sketch a schematic.

I think shorting together all the photodiodes causes shadows on some of them to affect the level of a string that you are playing. Use a separate opamp for each photodiode and mix them with the 2.2k resistors.

Having an op-amp per string would require me to design an entirely new PCB so that's not something I can test right now.

I think I had the photodiodes reversed the entire time before, but because the pins are connected to either side of the op amp inputs it seems to work just the same. I think originally the longer leg was towards the negating input of the op amps, this is getting confusing.

What I tried was grounding one pin as I don't have a negative voltage to bias with, but I had to flip the diode(from my initial wrong orientation?) to make that work.

I did another test as well, I completely removed two of the parallel op-amps and their respective resistors, Maybe I am missing something but that resulted in a significant increase of the noise floor, so they are effective?.

The red overlay is with only 2 op amps, and the grey with all 4 in the circuit.

I also tested to increase the gain of the second stage amplifier by switching the feedback resistor from 10K to 47K which resulted in about 2.5 dB SNR improvement, I'm not sure what that means, perhaps my audio interface is too noisy and the improvement came from me requiring less preamp gain.

 

[KlausST]

Hi,

Circuits have benefits and drawbacks.
When audioguru says that the non biased photodiodes are less noisy then this will be correct.
Non biased there will be much distortion, rectifying style. I can't imagine that it will work satisfactory.

Noise:
There will be voltage noise, current noise,
... noise from the photodiodes, resistors, Opamp
... noise from power supply or induced
... noise from the ambient light

When you say the noise does not change, then maybe the noise source is not the photodiode circuit, maybe the noise comes from elsewhere.

Thus I recommend to analyze the noise. Use your audio card, samplee the noise with still guitar strings in a dark room.
Then run an FFT on the sampled data and show the result. Free tools are available.

Klaus

 

[Yarrrrr]

Hi,

Circuits have benefits and drawbacks.
When audioguru says that the non biased photodiodes are less noisy then this will be correct.
Non biased there will be much distortion, rectifying style. I can't imagine that it will work satisfactory.

Noise:
There will be voltage noise, current noise,
... noise from the photodiodes, resistors, Opamp
... noise from power supply or induced
... noise from the ambient light

When you say the noise does not change, then maybe the noise source is not the photodiode circuit, maybe the noise comes from elsewhere.

Thus I recommend to analyze the noise. Use your audio card, samplee the noise with still guitar strings in a dark room.
Then run an FFT on the sampled data and show the result. Free tools are available.

Klaus

Based on the change in the noise floor when I removed two of the op amps, as seen on the last image in my previous post, wouldn't that mean that it is noise created by the circuit itself, if the summing of several op-amp stages reduce it?

The noise is broadband so I always assumed it was johnson-nyquist noise or op amp voltage/current noise, which led me to try to use as small resistors as possible, and parallel op-amps.

I doubt ambient light is a problem at all right now, completely covering the sensors does not change the noise in any way whatsoever.
Probably not power supply, same exact noise with battery power as with an external power supply.

Trying different photodiodes would be interesting, not sure what to look for though when it comes to photodiode noise. Lower noise op amps are for sure something I need to try, but the question is if I should keep a parallel setup or not.

 

[Audioguru]

Your photovoltaic (no bias voltage) photodiodes will probably sound the same for both polarities of the photodiodes. With the cathode feeding the inverting opamp then light causes the opamp output to swing from half the supply voltage to a higher voltage and if the photodiode feeds its anode to the opamp then the opamp output will go to a lower voltage.

The result is that the DC voltage changes in all your opamps since they are direct coupled. A pickup, microphone or vibration detector does not doo dat, they produce AC with no DC. The resulting DC will cause one-sided clipping producing even harmonics if clipping occurs from playing loudly.

A multimeter set to "diode-test" shows the polarity of a diode.

I think you have less noise with the photodiodes shorted together because they load down the output levels of the signals and the noise. A mixer does not load down signal levels. Your 2.2k resistors feeding the 2nd inverting opamp is a true mixer.

 

[c_mitra]

Could you explain why six amplifiers would be better? Wouldn't the signals and noise get summed all the same?

1. Each string vibration is recorded by one photodiode; the frequency and amplitude must be correlated with the photodiode output signal. Due to positioning (in)accuracies or signal amplifications etc (there can be many other reasons), we need to normalize the output from each string (as seen from the photodiode output). If you sum all the signals, this information may be lost.

2. The photodiode is not looking at the whole string; it is looking only at a very small part of it. For example, the photodiode may see at the middle of the string which is good for the fundamental but not for the odd harmonics. The positioning of the photodiodes is therefore very critical in my opinion.

3. The output from string instruments is polarised and this cannot be caught by a single photodiode; you need at least two per string in quadrature.

4. The phase of the outputs must be corrected before you add them and send it to the final amplifier. Each amplifier may have produced different phases because different strings may produce different frequencies.

5. The body of the instrument may act as a resonator and you need to apply frequency corrections for each photodiode because it catches only the motion of the string.

 

[betwixt]

Lower noise op amps are for sure something I need to try, but the question is if I should keep a parallel setup or not.

In my opinion, the benefits of summing random noise to attempt a zero average is not worth the effort. Yes, there will be some noise reduction but your underlying problem is trying to amplify such a tiny voltage from the outset. The more you have to amplify, the greater the resulting noise will be, that's electronics (or is it physics?) for you.

I think with a suitably designed optical sensor you should get tens if not hundreds of mV directly from the sensor, enough to drive a final power amplifier with little or no other amplifier stages. C_mitra's comments are very valid though, at least for acoustic guitars. I'm not sure how the solid body of an electric guitar influences the sound it makes.

Just an idea, if it is practical to implement - have you considered an interruptive sensor rather than a reflective one. I had the idea of using a slotted optocoupler with the string passing through the gap.

Brian.

 

[danadakk]

Some ref material -

http://www.analog.com/en/technical-articles/transimpedance-amplifier-noise-considerations.html

Battery noise - http://electronics.stackexchange.com/questions/221379/alkaline-battery-noise-varies-with-temp

http://tf.nist.gov/general/pdf/1133.pdf

http://www.edn.com/controlling-spot-and-integrated-noise-in-a-two-stage-transimpedance-design/

You have chosen a low noise OpAmp, but might help to do a complete noise analysis from
all sources, Battery, Bias Supply, Resistors, PSRR, C coupling (shielding).....a full and accurate analysis.

Be careful if doing spice noise sims, as the root model may not be adequately described to do so.
You can get a "measure" of that talking directly to manufacturer tech support.

For example the PSRR of this OpAmp at 1 Khz is ~ 90 to 100 db. So if your supply has 100 mV of noise
then output noise is ~ 1 uV, and thats RMS, not peak.....

Not all caps are ideal, and of course microphonic problems as well.

http://sh.kemet.com/Lists/Technical... Reduced Microphonics and Sound Emissions.pdf

There are techniques to get rid of low freq noise and offsets, like correlated double sampling, but
I think that would unduly compound a design like this.


Regards, Dana.

 

[Yarrrrr]

1. Each string vibration is recorded by one photodiode; the frequency and amplitude must be correlated with the photodiode output signal. Due to positioning (in)accuracies or signal amplifications etc (there can be many other reasons), we need to normalize the output from each string (as seen from the photodiode output). If you sum all the signals, this information may be lost.

2. The photodiode is not looking at the whole string; it is looking only at a very small part of it. For example, the photodiode may see at the middle of the string which is good for the fundamental but not for the odd harmonics. The positioning of the photodiodes is therefore very critical in my opinion.

3. The output from string instruments is polarised and this cannot be caught by a single photodiode; you need at least two per string in quadrature.

4. The phase of the outputs must be corrected before you add them and send it to the final amplifier. Each amplifier may have produced different phases because different strings may produce different frequencies.

5. The body of the instrument may act as a resonator and you need to apply frequency corrections for each photodiode because it catches only the motion of the string.

Are these issues that a magnetic pickup don't have?

--- Updated Feb 8, 2021 ---

In my opinion, the benefits of summing random noise to attempt a zero average is not worth the effort. Yes, there will be some noise reduction but your underlying problem is trying to amplify such a tiny voltage from the outset. The more you have to amplify, the greater the resulting noise will be, that's electronics (or is it physics?) for you.

I think with a suitably designed optical sensor you should get tens if not hundreds of mV directly from the sensor, enough to drive a final power amplifier with little or no other amplifier stages. C_mitra's comments are very valid though, at least for acoustic guitars. I'm not sure how the solid body of an electric guitar influences the sound it makes.

Just an idea, if it is practical to implement - have you considered an interruptive sensor rather than a reflective one. I had the idea of using a slotted optocoupler with the string passing through the gap.

Brian.

I have seen the interruptive design tried before, but it would be very finicky to design, and get in the way of playing.

The current reflective design can be seen in available products like the oPik by light4sound. They seem to have figured this all out.

 

[Yarrrrr]

A quick update on a few things I have tried that hasn't worked.

• Tried several different Photodiodes - negligible difference in sound.
  
  
• Different IR LEDs - negligible difference in sound.
  
  
• Tried a Phototransistor (WÜRTH 1540051NC2590)- no difference in signal strength, SNR or noise floor.
  
  
• Lower values for the feedback resistors - no SNR improvement ( as the phototransistor gave no stronger output ).
  
  
• Different resistor values for the voltage reference - no difference in noise floor.
  
  
• Using battery power instead of power supply - no difference.
  
  
• Tested a bunch of different Op amps(LM4562, OPA1612, OPA2192, OPA2209) - All resulted in negligible difference in noise floor or worse than NE5532.

If anyone has any ideas of why nothing I try changes anything, I'd love to hear them.

Current schematic:

 

[KlausST]

Hi,

If nothing brings an improvement ... maybe there are systematic mistakes or measurement mistakes.

Maybe...Unsuitable wiring, unsuitable PCB layout, unsuitable GND plane, unsuitable measurement equipment...

Klaus