A Wearable Sound-to-light Display

[ci139]

AC signal is from MIC? Is it fluctuating dc

is actually covered through and more concentrated at the end of the linked document for Grounded Source mode // Source-Follower mode // With separate connections to the S/D of the internal FET

a "balanced differential" input pre-amp

obviously and from this article comes out another potential improvement to the circuit - that is - an automatic Gain control - perhaps implemented on additional op amp(s) - to make the circuit invariant from the distance from the audio source
**broken link removed**
**broken link removed**

 

[ci139]

Ti's electret reference design https://www.ti.com/lit/ug/tidu765/tidu765.pdf

 

[audioguru]

I have used non-inverting opamps in many preamps for electret mics but I have never used TI's idea of a transimpedance amplifier. It is probably the best.

 

[MD MUBDIUL HASAN]

Can't you read what I said? "If this second opamp is changed to be inverting then its input safely stays at 0V and never goes negative."

Does it mean I need to connect first stage input to inverting input of second stage OPA? If so, what would be the pressure level of sound ? And how you will deal with negative feedback that is necessary.
If I did wrong to understand, pardon me!

 

[MD MUBDIUL HASAN]

is actually covered through and more concentrated at the end of the linked document for Grounded Source mode // Source-Follower mode // With separate connections to the S/D of the internal FET

obviously and from this article comes out another potential improvement to the circuit - that is - an automatic Gain control - perhaps implemented on additional op amp(s) - to make the circuit invariant from the distance from the audio source
**broken link removed**
**broken link removed**

Sounds meaningful, let me read.

 

[audioguru]

Does it mean I need to connect first stage input to inverting input of second stage OPA? If so, what would be the pressure level of sound ? And how you will deal with negative feedback that is necessary.
If I did wrong to understand, pardon me!

You have the second opamp as a non-inverting type so that the coupling capacitor feeds it signals from +4.5V to -4.5V. But signals more negative than -0.3V will damage the input.
So I recommend changing the second opamp to an inverting type with exactly the same AC gain then the (-) input pin of the opamp never goes negative due to the negative feedback. Then the opamp will not be destroyed and the sound pressure makes exactly the same output signal level as you want. The first opamp is not changed but I recommended using a non-inverting opamp type so that it has a high input resistance and does not load down the mic signal level but the Texas Instruments transimpedance circuit will probably be better.

 

[MD MUBDIUL HASAN]

You have the second opamp as a non-inverting type so that the coupling capacitor feeds it signals from +4.5V to -4.5V. But signals more negative than -0.3V will damage the input.
So I recommend changing the second opamp to an inverting type with exactly the same AC gain then the (-) input pin of the opamp never goes negative due to the negative feedback. Then the opamp will not be destroyed and the sound pressure makes exactly the same output signal level as you want. The first opamp is not changed but I recommended using a non-inverting opamp type so that it has a high input resistance and does not load down the mic signal level but the Texas Instruments transimpedance circuit will probably be better.

Yes, looking at your idea it might me effective, but it will became a "Current source for loads returned to ground".

 

[audioguru]

Yes, looking at your idea it might me effective, but it will became a "Current source for loads returned to ground".

No, the inverting opamp will have positive half-cycle outputs each time there is sound. The microphone produces AC output (fluctuating DC), the first opamp produces a fluctuating DC output and the input to the second opamp's input resistor is true AC. Since the second opamp is biased at 0V then its output is half-cycle positive DC fluctuations that turn on the transistor current sink.

Now I see that a peak detector will make the LEDs brighter because it will light them with DC, not with the DC fluctuations that turn off the LEDs half the time they should be turned on. Also a peak detector will turn on the LEDs long enough for you to see their brightness even when short duration sounds cause the LEDs to look dimmed.

 

[MD MUBDIUL HASAN]

No, the inverting opamp will have positive half-cycle outputs each time there is sound. The microphone produces AC output (fluctuating DC), the first opamp produces a fluctuating DC output and the input to the second opamp's input resistor is true AC. Since the second opamp is biased at 0V then its output is half-cycle positive DC fluctuations that turn on the transistor current sink.

Now I see that a peak detector will make the LEDs brighter because it will light them with DC, not with the DC fluctuations that turn off the LEDs half the time they should be turned on. Also a peak detector will turn on the LEDs long enough for you to see their brightness even when short duration sounds cause the LEDs to look dimmed.

The logic you have raised is reasonable, but the second OPAMP input (-0.3v) less negative does not mean the circuit will not work, perhaps after a long time the IC would be damage.
The fluctuating DC noise might be changed through a RC filter or they way Mr. ci139 has described.

If you treat the AC signal on the input of second opamp RC filter, then it would be meaningful, first reply from expert says, " it might be for DC biasing that is, 4.5 volt for second opamp.
I am not sure that, DC fluctuation could responsible for LED lighting, because at the beginning those are cancel out as RC filter in the second opamp picks a certain cutt off frequency!

 

[audioguru]

Why don't you follow the rules? The manufacturers of LM358 ICs say that the maximum allowed negative input voltage is -0.3V and an inverting opamp circuit for the second opamp is a simple way to do it.

What is "fluctuating noise"? You do not understand that the audio signals are a voltage that alternates up and down, so it is AC but you can say it is "fluctuating DC". The input and output of the first opamp already have RC highpass filters to block DC and pass AC.

The second opamp is biased at 0V because if it is normally biased at +4.5V for audio then the LEDs would be lit all the time, even when there are no sounds. With the input biased at 0V then the output is 0V with no signal so that the LEDs are turned off and goes positive and lights the LEDs on the positive half-cycles of the audio.

 

[audioguru]

The output signal from the first opamp does not need to be AC, it just needs to be the positive half-cycles. So bias the first opamp at 0V and remove the AC coupling capacitor from its output like this:

 

[MD MUBDIUL HASAN]

Why don't you follow the rules? The manufacturers of LM358 ICs say that the maximum allowed negative input voltage is -0.3V and an inverting opamp circuit for the second opamp is a simple way to do it.

What is "fluctuating noise"? You do not understand that the audio signals are a voltage that alternates up and down, so it is AC but you can say it is "fluctuating DC". The input and output of the first opamp already have RC highpass filters to block DC and pass AC.

The second opamp is biased at 0V because if it is normally biased at +4.5V for audio then the LEDs would be lit all the time, even when there are no sounds. With the input biased at 0V then the output is 0V with no signal so that the LEDs are turned off and goes positive and lights the LEDs on the positive half-cycles of the audio.

Thank you once again for your reply here.
Does it mean the previous circuit was wrong?
If so, the circuit should not work.

Anyway, I appreciate your answers.
Looking at your modified design, you have removed voltage divider means no +4.5v bias on non inverting input of first opamp. You have added a capacitor for DC power?
You have also removed the RC filter means, you are ignoring the AC signals that has been picked from MIC.
The reason behind adding RC filter is obvious, dont want to amplify the noisy signal !
You wont say the LED would glow all time, perhaps it will only lit up when transistor will turned on, that second opamp will amplify the audio signal.
Now the question arise is, if You make 0V in non inverting inputs, what will be gain equation for first opamp.

 

[audioguru]

Does it mean the previous circuit was wrong? If so, the circuit should not work.

Yes, it was wrong to drive negative voltages into the input pin of the second opamp. It might work for one minute or work for one hour before it is destroyed.

Looking at your modified design, you have removed voltage divider means no +4.5v bias on non inverting input of first opamp.

Yes, so that it can safely feed only positive voltages into the input in of the second opamp.

You have added a capacitor for DC power?

Yes, it creates a very low impedance for the DC power. All electronic circuits should have a supply bypass capacitor at the circuit to prevent oscillations and remove hum.

You have also removed the RC filter means, you are ignoring the AC signals that has been picked from MIC.

No, I removed the AC coupling capacitor and the 0V biasing resistor, they were not used as a filter. The capacitor blocked the 4.5VDC from the first opamp so that the second opamp could turn off the LEDs when there is no signal.
You do not understand that my new direct connection from the output of the first opamp to the input of the second opamp passes AC and DC.
The coupling capacitor that I removed blocked the +4.5V DC bias of the first opamp but passed AC. It was needed so that the input of the second opamp could be biased at 0V so that the LEDs could turn off when there was no sound and the AC was the positive parts of the audio to turn on the LEDs.

The reason behind adding RC filter is obvious, dont want to amplify the noisy signal!

No, the RC was not a filter.

You wont say the LED would glow all time, perhaps it will only lit up when transistor will turned on, that second opamp will amplify the audio signal.
Now the question arise is, if You make 0V in non inverting inputs, what will be gain equation for first opamp.

The simple circuit needs to amplify only the positive parts of the signal. The LM358 does this perfectly when its input is biased at 0V and when the input signal never goes negative more than 0.3V (the mic signal is about 0.02V). Its gain is the same as when it is biased at +4.5V. The gain of the first opamp is 470k/(2.2k + the 2.7k output impedance of the mic)= about 68 times. The second opamp has no gain, the output current level of the transistor is the same as the input voltage level to the second opamp.

 

[MD MUBDIUL HASAN]

Yes, it was wrong to drive negative voltages into the input pin of the second opamp. It might work for one minute or work for one hour before it is destroyed.


Yes, so that it can safely feed only positive voltages into the input in of the second opamp.


Yes, it creates a very low impedance for the DC power. All electronic circuits should have a supply bypass capacitor at the circuit to prevent oscillations and remove hum.


No, I removed the AC coupling capacitor and the 0V biasing resistor, they were not used as a filter. The capacitor blocked the 4.5VDC from the first opamp so that the second opamp could turn off the LEDs when there is no signal.
You do not understand that my new direct connection from the output of the first opamp to the input of the second opamp passes AC and DC.
The coupling capacitor that I removed blocked the +4.5V DC bias of the first opamp but passed AC. It was needed so that the input of the second opamp could be biased at 0V so that the LEDs could turn off when there was no sound and the AC was the positive parts of the audio to turn on the LEDs.


No, the RC was not a filter.


The simple circuit needs to amplify only the positive parts of the signal. The LM358 does this perfectly when its input is biased at 0V and when the input signal never goes negative more than 0.3V (the mic signal is about 0.02V). Its gain is the same as when it is biased at +4.5V. The gain of the first opamp is 470k/(2.2k + the 2.7k output impedance of the mic)= about 68 times. The second opamp has no gain, the output current level of the transistor is the same as the input voltage level to the second opamp.

Well said sir audioguru, I like your comment that is most important "The simple circuit needs to amplify only the positive parts of the signal".
Lets accept your design, and test it on project board.
If It work, I will make pcb.

Lets move on other circuit. I will come back soon.

 

[audioguru]

With very loud sounds and a new 9V battery, the 100 ohms "brightness" resistor will allow about 63mA in the LEDs which might be too much current for them. First use 220 ohms for the brightness resistor to limit the maximum halfwave current to about 29mA (the maximum allowed current in ordinary 5mm LEDs is 30mA).