electret condenser microphone amplifier

Hello Vincenzo,

The circuit is comprised of two non-inverting gain stages with high-pass and low-pass filters built around the op-amps.

Gain for each stage is set by the non-inverting op-amp standard gain equation:  Gain = 1+Rf/Ri = 1+90/10 = 10V/V = 20dB
The two gains mutliply together for the total gain of:  10V/V * 10V/V = 100V/V = 40dB

Low-pass filtering is achieved by placing a capacitor in parallel with the feedback resistor in the op-amp circuits.  The equation to calculate the location of the cutoff frequency is:  Fcutoff = 1/(2*pi*Rf*Cf)
For the first stage Fcutoff = 1/(2*pi*90k*47p) = 37.625k
For the second stage Fcutoff = 1/(2*pi*90k*22p) = 80.381k

There is an error in the article regarding the high-pass filtering.  Both op-amp stages actually incorporate second-order high-pass filters, where the article only mentions a first-order high-pass filter.

The first high-pass filter is formed by DC-blocking filter on the non-inverting input with the parallel combination of the two DC biasing resistors. 
For the first stage this equals:  1/(2*pi*(100k||100k)*1u) = 1/(2*pi*50k*1u) = 3.18Hz
For the second stage this equals:  1/(2*pi*(100k||100k)*470n) = 1/(2*pi*50k*470n) = 6.77Hz

The second high-pass fitler is the one described in the article and it is a result of the input resistor on the inverting input and the DC blocking capactor to GND.
For the first stage this equals:  1/(2*pi*10k*10u) = 1.59Hz
For the second stage this also equals:  1/(2*pi*10k*10u) = 1.59Hz

The last filter mentioned in the article is the low-pass filter that is a result of the 2.2k biasing resistor for the Electret microphone and the 33pF capacitor between the non-inverting input and GND.
This filter equals:  1/(2*pi*2.2k*33p) = 2.192MHz

Please let me know if you have any other questions.

Regards,
Collin Wells
Precision Analog

Hello Vincenzo,

In general "audio frequencies" are considered to be 20Hz - 20kHz.  The filter cut-off frequencies were chosen so that the op-amp filters did not interfere with the bandwidth of the microphone that was used as an example in the article which had a frequency response from 20Hz to 20kHz.  The cutoff frequency of a filter is the frequency at which the gain of the system has decreased by 3dB from the flatband gain.  Placing the cutoff frequencies at 20Hz and 20kHz would have caused the gain to roll-off inside of the usable frequency response of the microphone which I did not want for this example. 

I mention a "rule-of-thumb" on page 8 of the article that states that when selecting the cutoff frequency for a filter, place the cutoff frequency 25% above/below the desired low-/high-pass filter frequency to achieve a ~0.25dB reduction at the desired frequency.  Therefore to achieve a high-pass filter where the gain begins to roll-off at 20Hz, design the filter so the cutoff frequency is at 20*0.25 = 5Hz.

Noise referred to output in Vrms is shown in Figure 34 of the article.  Follow the steps in the article to determine the total output noise in Vrms for your circuit.  To summarize, take the noise spectral density numbers for your system, multiply by the gain and then integrate over your usable bandwidth.  If you choose a low-noise op-amp then it's likely that the flatband noise will be dominated by the ECM itself.

For more information on noise, please look at the series below from Arthur Kay: 

Noise Article Series, Parts 1 - 9, EN-Genius Network http://www.engenius.net/site/zones/audiovideoZONE/technical_notes/avt_120108

Hope this helps,
Collin Wells
Precision Analog

Hello Vincenzo,

Yes, the op-amp circuit connected to the output of the electret microphone  (ECM) is an active band-pass filter.  The gain of the circuit is 10V/V or 20dB, and the circuit is designed to have a mostly flat response over the entire operating range of the chosen ECM which was from 20Hz - 20kHz. before rolling off.  Applying a bandpass filter limits the total noise of the circuit by minimizing the bandwidth that the noise power spectral density is integrated over. 

Regarding the 1.59Hz filters:  In order to achieve a flat frequency response down to 20Hz, the cutoff frequencies of the high-pass filters had to be chosen so that they were lower than 20Hz.  Remember that the cutoff frequency specifies the -3dB point of the filter.  Since the high-pass filter is a second-order (-40dB/decade), if the two cutoff frequencies were both set to 20Hz then there would be a 6dB reduction in gain at 20Hz.  In short, the 1.59Hz cutoff frequency was chosen so that at 20Hz there was only minimal reduction in the frequency response due to the op-amp filtering. 

The 2.192MHz filter is designed to reduce high-frequency EMI/RFI interference.

Cheers,
Collin Wells

Hi Vincenzo,

It was an effort to maintain a flat frequency response from 20Hz - 20kHz.  Choosing the same cutoff frequencies for both filters would have resulted in a greater roll-off around my desired flatband frequencies.

So to answer your question, it was a "method to project the roll-off of the Bode diagram"

In SPICE try simulating one circuit with the two cutoff frequencies set to the same value and a second circuit with the cutoff frequencies staggered.  This will help you better understand the difference between the two.  The same theory applies to the high-pass and low-pass filters so you really only need to experiment with one to see the effect.  

Use the low-pass filters and simulate one version with the cutoff frequencies both at 37.625kHz, and one with one cutoff frequency set to 37.625kHz and the other set to 80.381kHz.  Compare how the Bode plot of the two filters behave around 20kHz.  You should see that the version with both cutoff frequencies set to 37.625kHz rolls off more than the circuit with the cutoff frequencies staggerred.

Regards,
Collin Wells
Precision Analog

Hello Jaroslaw,

You've asked a number of questions so let me know if I miss any. 

First, if you're going to move to a +3.6V system later then you should probably design for it now because most of the standard audio op-amps will operate down to +5V with limited input range, but not much lower.  I would suggest you consider the OPA322 for this design.  If your audio requirements are not that high then consider a general purpose op-amp such as the LMV601.  If your audio requirements are very high then consider the OPA1662.

For your own reference, the link below contains a list of good audio op-amp upgrade options for many older op-amps:

http://e2e.ti.com/support/amplifiers/precision_amplifiers/w/design_notes/upgrading-op-amps-in-audio-equipment.aspx

1.)  The circuit shown should work fine but as mentioned above I would consider using an OPA322 instead of the OPA134 because the OPA134 won't work properly in a +3.6V system.

2.)  Yes, this op-amp is a dedicated audio op-amp and has excellent performance in the audio frequency band.  I'm not sure if your application requires such high fidelity but if it does then consider the OPA1662.

3.)  The circuit you've referenced is basically the same as the one I show in the article I wrote.  I used 100k for "R3 and R4" which provides the same function as the two 1M resistors.  If you're worried about loading down your microphone then consider the increasing the impedance to 1M as shown in Mr. Lathrop's post. 

The only other two differences between the circuits are the gain used and the fact that he chose not to AC couple between stages.  If you can tolerate the extra components then I always consider it "safer" to AC couple between all stages in an audio system and not have to worry about DC saturation issues.  You can change the gain in the circuit I showed by modifying the feedback resistor value. 

The note about taking time to stabilize is in reference to the RC time constant formed between the feedback resistor + input resistor and the AC coupling capacitor.  Upon circuit start-up, the output has to charge the inverting input to the same voltage as the non-inverting input and to do so it must charge a capacitor.  All non-inverting AC coupled systems will have this issue.  There isn't a really good way to minimize this issue other than to decrease the RC time constant which will in turn increase the high-pass filter frequency which may not be desired.

Hope this helps.