Hi,

I want to determine the voltage across an electret condenser. In case of a loud sound in decibels what will be the voltage.

Please tell me the formula for it.

circuit is given below:

thanks

 

You have to check the datasheet. Usually the output from an electret is in the mV range. But there are a few variables including frequency response.

 

Why does your schematic show two 9V?
9V-1 should be marked +9V and 9V-2 should be marked 0V.

An electret mic draws a current of about 0.4mA so the DC voltage across the resistor is 0.4mA x 10k= 4V then the DC voltage across the mic is 9V - 4V= 5V.

At a distance of 10m from your mouth and if you are speaking in a normal conversation level the output AC from the mic will be about 10mV which is 0.01VC. +40dB louder will produce 1VAC.

 

Hi,

I want to determine the voltage across an electret condenser. In case of a loud sound in decibels what will be the voltage.

Please tell me the formula for it.

circuit is given below:

thanksView attachment 182077

@rahulb
Your question may have a couple of distinctly different answers depending on what exactly you are asking. Since you mention sound, I will assume you are talking about an electret microphone. The electret in such a microphone is usually a plastic film that has been specially treated to retain free electrons (i.e. an electric charge) for many years without recharging. The microphone comprises an electret film in close proximity to, but insulated from a usually flat metal surface, forming a capacitor with a permanent charge (the electret). When sound vibrates the film (membrane), the embedded charge moves closer and farther from the metal surface and in doing so creates a varying voltage between the film and the metal. This voltage is very small and requires amplification before it becomes usable as a microphone. That amplification is usually provided by a "junction field-effect transistor" (JFET) located within the microphone housing. The JFET requires a large (compared to the electret voltage) voltage to operate as an amplifier, and that voltage is most commonly provided by a small battery (1.3 to 3V). Thus a signal of (e.g.) 0.001V is amplified to a signal of (e.g.) 0.050V. That amplified signal is sufficient for common audio devices to amplify further until the signal can drive an earphone or loudspeaker. So, there are several voltages involved. A tiny voltage generated by the electret's movement due to sound vibrations. A larger voltage resulting from the amplification--this is the "output" of the microphone. Finally, a constant "DC" voltage that allows the amplifier to work. Manufacturers rarely specify the electret voltage, but they do almost always specify the level of the amplified voltage; usually in millivolts at a specific sound frequency and intensity. That voltage can also be converted to a logarithmic scale and given as a "dbm" value, again for a specific frequency and intensity of sound. Likewise, the battery voltage and type is generally specified in the datasheet for the microphone. Although I have given estimated voltage levels, each microphone design is a bit different and the actual voltages of a real device may be different than I have indicated. Look into the manufacturer's datasheet for the specific model microphone to get better data.

 

Why does your schematic show two 9V?
9V-1 should be marked +9V and 9V-2 should be marked 0V.

An electret mic draws a current of about 0.4mA so the DC voltage across the resistor is 0.4mA x 10k= 4V then the DC voltage across the mic is 9V - 4V= 5V.

At a distance of 10m from your mouth and if you are speaking in a normal conversation level the output AC from the mic will be about 10mV which is 0.01VC. +40dB louder will produce 1VAC.

hi,

please tell me, if a sound +40 dB will produce 1 volt AC, will the sound also have any effect on DC across the condenser which is 5v without any sound?


thanks

 

a sound +40 dB

The Bel, or decibel, is a ratio between two things.
So, Mary has twice as many apples as John has some meaning whereas Mary has twice doesn't.
The reference is commonly specified by a letter after db, e.g. dBV, dBm
See here: https://en.wikipedia.org/wiki/Decibel#Suffixes_and_reference_values

 

Why does your schematic show two 9V?
9V-1 should be marked +9V and 9V-2 should be marked 0V.

An electret mic draws a current of about 0.4mA so the DC voltage across the resistor is 0.4mA x 10k= 4V then the DC voltage across the mic is 9V - 4V= 5V.

At a distance of 10m from your mouth and if you are speaking in a normal conversation level the output AC from the mic will be about 10mV which is 0.01VC. +40dB louder will produce 1VAC.

I think you might have some extra zeros somewhere.

 

I need to add a reference voltage at non-inverting input of opamp such that the output of opamp will go high only in case of loud sounds. please check the circuit below

thanks

 

I need to add a reference voltage at non-inverting input of opamp such that the output of opamp will go high only in case of loud sounds. please check the circuit below

thanks

View attachment 182411

The DC level is not well controlled. So it would be better to AC couple the signal. Since the output is very low you might consider amplifying it first.

 

I think you might have some extra zeros somewhere.

Normal conversation level is about 72dB at 10cm and 40dB louder is 112dB. 40dB is 100 times the voltage.100 times 0.01V is 1.0V.

 

You show a comparator, not an opamp.
The DC from an electret microphone is different from each microphone even if they have the same part number. The DC voltage probably changes when the temperature changes.
The DC output of an electret microphone varies up and down with frequencies and levels of the audio. If your comparator has its DC reference voltage set for the loudness you want then the output of the comparator will vary at frequencies that occur. If the comparator output lights an LED then a steady signal will light the LED only for the halfwave inputs of the signal so the LED will look dim. But since the LED is fed with fluctuating levels then you probably will not see it blink.
You probably need to capacitor-couple the output of the mic to a "peak detector" that rectifies the signals and holds the peak level long enough to be seen brightly.

 

Normal conversation level is about 72dB at 10cm and 40dB louder is 112dB. 40dB is 100 times the voltage.100 times 0.01V is 1.0V.

I think most of them have a sensitivity of around 44 dB/Pa.
Or an output of around 6 mv at 94 dB.

 

It sounds a little like a peak light for the mic input. It would be better to have two opamps and have it trigger on both rails.

 

The DC level is not well controlled. So it would be better to AC couple the signal. Since the output is very low you might consider amplifying it first.
View attachment 182414

hi,

which software are you using for simulation?

 

hi,

which software are you using for simulation?

LtSpice.
https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html

 

You show a comparator, not an opamp.
The DC from an electret microphone is different from each microphone even if they have the same part number. The DC voltage probably changes when the temperature changes.
The DC output of an electret microphone varies up and down with frequencies and levels of the audio. If your comparator has its DC reference voltage set for the loudness you want then the output of the comparator will vary at frequencies that occur. If the comparator output lights an LED then a steady signal will light the LED only for the halfwave inputs of the signal so the LED will look dim. But since the LED is fed with fluctuating levels then you probably will not see it blink.
You probably need to capacitor-couple the output of the mic to a "peak detector" that rectifies the signals and holds the peak level long enough to be seen brightly.

I am using IC 311 opamp as a comparator. I will be using a monostable 555 circuit after the comparator to maintain my devices 'on' for a short time.
My idea is , in case of a loud sound, the comparator output will be high for a very short time, but the high output will be used to trigger the monostable circuit to maintain the output status for atleast 2-3 seconds.

I just need a way to make comparator's output high in case of a loud sound

 

Research peak lights. I have similar circuits in most of my audio gear. Peak lights are usually adjustable so you can set the level. When you have a VU meter the level that is at the ADC peak warns of too hot signal which is not good. Analog circuits are a little more forgiving and can be set lower and allow the signal to go into saturation.

Here’s a similar project. You just need a certain threshold. https://www.instructables.com/id/Mini-Decibels-a-Simple-Volume-Meter-With-an-Electr/

 

A 555 triggers when its trigger pin goes low, not high. It does not timeout while its trigger pin is low. Your comparator output will be frequently low and high during or between sounds.

 

A 555 triggers when its trigger pin goes low, not high. It does not timeout while its trigger pin is low. Your comparator output will be frequently low and high during or between sounds.

yes, thats why I want the comparator output goes high only in case of loud sounds. Loud sounds means like a cracker burst, a clap, a shriek etc a loud and sudden sound only.

 

yes, thats why I want the comparator output goes high only in case of loud sounds. Loud sounds means like a cracker burst, a clap, a shriek etc a loud and sudden sound only.