Clarification Regarding the MAX4466 application Circuit

Hi Avnendra, the voltage divider composed of the two 2kΩ resistors and 0.1µF capacitor is a DC bias for the electret microphone. The 0.01µF capacitor is an AC coupling capacitor (also referred to as DC blocking capacitor). It couples the AC signal from the microphone to the MAX4466 preamplifier. I hope that provides a detailed explanation of their function and purpose with respect to the circuit overall. See my responses to your enumerated questions below.

Are these components required solely for the Electret Condenser Microphone (ECM MIC) or are they also necessary for the MAX4466 amplifier?

The 2kΩ resistors and 0.1µF components are necessary for the microphone. The 0.01µF capacitor is necessary for the preamplifier.

Could you please confirm this [the role of the 1MΩ resistors] and explain the reasoning behind their inclusion in the design?

Correct. The two 1MΩ resistors are also a voltage divider. They divide the +5V bias to +2.5V. This provides a common-mode bias for the microphone's AC signal, allowing the AC level to vary ±2.5V without distortion before entering the preamplifier.

Regarding the use of an earphone microphone rather than an electret, I have to say that I am not familiar with earphone microphones. I imagine that there are various types, each with their own specifications. If you provide a part number for the microphone you intend to use, I can be more helpful.

Otherwise, you will need to assess the microphone's specifications (such as bias, impedance, and sensitivity) in the context of the MAX4466 datasheet EC Table, description sections, and application diagram. As long as the earphone mic is not wildly different than an electret, I think you should be able to use it with perhaps little or no adjustment to the applications diagram.

Hi KTarrats,

Thank you for your previous response, which helped me understand the role of the 2.2K resistor and 0.1uF capacitor in the Electret Condenser Microphone (ECM) circuit. I now recognize that these components are necessary for providing the bias voltage and AC coupling required by the ECM.

In my design, I am working with a 3.3V supply voltage, and I would like to simplify the circuit if possible. I have tested the ECM with a 2.2K ohm resistor and a 10uF capacitor for AC coupling, and it appears to be working fine with my earphone. I am wondering if it is feasible to use a single resistor for voltage biasing and a 10uF capacitor for AC coupling, as shown in the circuit below.

Additionally, I would like to confirm the common-mode biasing circuit. Can I use a single resistor with 3.3V, as shown below, or should I use the same voltage divider circuit as before? I assume that the AC level will vary between +3.3V and 0V with the proposed circuit. Is this sufficient for the preamplifier, or should I stick with the original voltage divider circuit for common-mode biasing?

Regarding the microphone in the wired earphone, it typically features an Electret Condenser Microphone (ECM) with the following approximate specifications:

• Sensitivity: -38db to -52db
• Impedance: 1.5K ohm  to 3.0K ohm
• Frequency response: 100Hz to 10KHz (some upto 15KHz)
• Operating voltage: 1V to 3.3V

Unfortunately, the manufacturer does not provide exact specifications, but I would like to know if it is possible to use this earphone with the proposed circuit. Please let me know if you require any additional information or if you have any concerns about the circuit.

Thank you for your time and expertise. I look forward to your response.

Regards,
Avnendra

A 3.3V supply is a good choice. MAX4466 accepts a range from 2.4V to 5.5V.

I think your choice of a single 2.2kΩ resistor to bias the microphone is fine. Your mic's specification is 1V to 3.3V. That 2.2kΩ resistor will be a pull-up to 3.3V.

A 10µF AC coupling capacitor is much larger than our application circuit's 0.01µF. However, that does not mean it is not advisable. Here are some things to consider:

Cons

• A larger capacitor is sometimes physically larger
• Often more expensive

Pros

• A larger capacitor blocks more DC. However, you are only blocking 3.3V so this benefit is perhaps not so valuable.
• Better noise attenuation
• Increased high frequency performance. This may also be a negligible consideration since you are working in the audio frequency domain

AVIKMR said:

Can I use a single resistor with 3.3V, as shown below, or should I use the same voltage divider circuit as before? I assume that the AC level will vary between +3.3V and 0V with the proposed circuit. Is this sufficient for the preamplifier, or should I stick with the original voltage divider circuit for common-mode biasing?

You should use a high impedance voltage divider, like the original two 1MΩ resistors in our applications diagram. You are partly correct in assuming that the AC level will have 0V to 3.3V to vary between. I say partly, because only the positive AC portion will be able to vary. You need a common-mode DC bias at the non-inverting input of the preamplifier from which the AC signal can swing up to 3.3V and down to 0V.

Thanks for the details on the earphone microphone you chose. I see no issues with it.

Hi KTarrats,

Thank you for the detailed explanation. I now have a clear understanding of the concept and will incorporate it into my design.

Best regards,
Avnendra

You are welcome Avnendra. Let me know if you have any other inquiries.

Hi KTarrats,

I am seeking clarification on a circuit analysis and would appreciate your feedback. Specifically, I have calculated the gain of the given circuit to be 11 V/V using the formula Av = 1 + Rf/Rin, where Rf = 100 kΩ and Rin = 10 kΩ. Based on this calculation, I estimate the -3 dB point to be approximately 54 kHz. Could you please confirm if my analysis is correct?

Additionally, I would like to understand the purpose of the 1 μF and 100 pF capacitors in this circuit. What role do they play in the overall design, and how do they contribute to the circuit's functionality? I would be grateful if you could provide insight into their significance.

Thank you for your time and expertise. I look forward to your response.

Best regards,
Avnendra

Hi KTarrats,

Any update on above Request.

Regards,
Avnendra

Hi Avnendra, your circuit gain calculation is correct. The pre-amp is a non-inverting op-amp circuit. Therefore, the gain is given by the inverting input resistor (Rinv) and feedback resistor (Rfb):

Gain = (Rinv + Rfb)/Rinv → 1 + Rfb/Rinv = 1 + 100/10 = 11

Keep in mind that the path gain, including the microphone, is determined using the mic's sensitivity (converted from dB to voltage) along with the circuit gain.

AVIKMR said:

Based on this calculation, I estimate the -3 dB point to be approximately 54 kHz. Could you please confirm if my analysis is correct?

If you do not mind, please show your work on this calculation. That corner frequency seems high to me. I think the pole of the transfer function is set by the feedback loop components, namely the 100pF capacitor and 100kΩ resistor:

fc = 1/2/π/R/C = 1/2/π/100e3/100e-12 ≈ 15.9kHz

This seems a reasonable corner frequency for a microphone input, if a bit low for audio-band signals which typically range from 20Hz to 20kHz.

AVIKMR said:

Additionally, I would like to understand the purpose of the 1 μF and 100 pF capacitors in this circuit. What role do they play in the overall design, and how do they contribute to the circuit's functionality? I would be grateful if you could provide insight into their significance.

The 1µF capacitor together with the 10kΩ resistor creates a high-pass filter. Note that this filter's corner frequency needs to be sufficiently low so as not to attenuate audio-band signals. We can easily check by calculating the zero:

fc = 1/2/π/R/C = 1/2/π/10e3/1e-6 = 6Hz

The 100pF capacitor is an amplifier stability component. It compensates for capacitance at the inverting amplifier input. As described above, it forms a pole with the 100kΩ feedback resistor.

Let me know if this answers your questions.

Hi Avnendra, your circuit gain calculation is correct. The pre-amp is a non-inverting op-amp circuit. Therefore, the gain is given by the inverting input resistor (Rinv) and feedback resistor (Rfb):

Gain = (Rinv + Rfb)/Rinv → 1 + Rfb/Rinv = 1 + 100/10 = 11

Keep in mind that the path gain, including the microphone, is determined using the mic's sensitivity (converted from dB to voltage) along with the circuit gain.

AVIKMR said:

Based on this calculation, I estimate the -3 dB point to be approximately 54 kHz. Could you please confirm if my analysis is correct?

If you do not mind, please show your work on this calculation. That corner frequency seems high to me. I think the pole of the transfer function is set by the feedback loop components, namely the 100pF capacitor and 100kΩ resistor:

fc = 1/2/π/R/C = 1/2/π/100e3/100e-12 ≈ 15.9kHz

This seems a reasonable corner frequency for a microphone input, if a bit low for audio-band signals which typically range from 20Hz to 20kHz.

AVIKMR said:

Additionally, I would like to understand the purpose of the 1 μF and 100 pF capacitors in this circuit. What role do they play in the overall design, and how do they contribute to the circuit's functionality? I would be grateful if you could provide insight into their significance.

The 1µF capacitor together with the 10kΩ resistor creates a high-pass filter. Note that this filter's corner frequency needs to be sufficiently low so as not to attenuate audio-band signals. We can easily check by calculating the zero:

fc = 1/2/π/R/C = 1/2/π/10e3/1e-6 = 6Hz

The 100pF capacitor is an amplifier stability component. It compensates for capacitance at the inverting amplifier input. As described above, it forms a pole with the 100kΩ feedback resistor.

Let me know if this answers your questions.