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Simple electret microphone circuit

What is an electret microphone

One of the most common microphone component that is used in the today's devices (eg. smartphone) is the electret microphone. The electret capsule contains, among other little details, an electret diaphragm and a JFET transistor module (eg. 2SK596). A cross-sectional drawing of an electret microphone is presented below:

Simple electret microphone circuit

Fig. 1 - The cross-section of am electret microphone

The electret microphone test circuit schematic is shown below:

Simple electret microphone circuit

Fig. 2 - The test circuit for an electret microphone

The electret microphone JFET amplifier has the Gate connected to a pick-up plate (which is pushed back and forth by the air), the Source is connected to the ground and the Drain is connected to the output pin. This configuration is known as common-source configuration, which requires an external power supply voltage Vcc.

Please note that the "capacitor" component shown in the electret capsule schematic above is represented by the pick-up plate + the electret material, which together create a capacitor.

The electret microphone operates in the range of 1.5-10V (typically 2V) and the current between the JFET's Source-Drain (ground-output) is usually max. 0.5mA. Our power supply Vcc is going to be at least +3V. In order to limit the output current we use a resistor R. The lower the resistor R the higher the output current.

We can calculate the value of R from the circuit supply voltage (Vcc), the microphone typical operating voltage (2V) and the microphone max. current (0.5mA):

R=\frac{V_{+}-2V}{0.5mA}

The sound signal has an AC form and this is the only signal we want to pass to the output. However, the microphone output signal amplitude would be influenced by the input DC power supply that we must use to power up the microphone. If we want to filter-out that DC signal we should use a capacitor. The capacitor must be chosen such that its impedance is much lower than the resistor R at audio frequencies. The microphone frequency f = 20-20000Hz. The resistor R and the capacitor C form a high-pass filter.

We could choose the capacitor C such that the filter's cutoff frequency is f=20Hz:

C<\frac{1}{2 \cdot \pi \cdot R \cdot f}=\frac{1}{2 \cdot \pi \cdot {\frac{V_{+}-2V}{0.5mA}} \cdot 20Hz}

Simple electret microphone circuit

Next we are going to create a small test circuit that will help us to test the electret microphone with the aid of a computer.

The required components are:

  • breadboard
  • electret microphone - 1 pcs
  • 1k Ohm 1/4W film resistor (ideally a 5K variable resistor) - 1 pcs
  • 1uF electrolytic capacitor - 1 pcs
  • PCB audio connector - 1 pcs
  • audio jack connector - 2 pcs
  • computer with an audio recording software (eg. Sound Recorder in Windows, Audacity on Windows/Linux/Mac)

 

Simple electret microphone circuit

Fig 3 - A breadboard setup for testing the microphone

First create the test circuit using the components listed above.

Insert one end of the stereo jack connector in the MICrophone port of the computer and the other in the PCB audio connector (TRS1) connector. Usually the computer MIC port provides the necessary power supply for the microphone (2-3 VDC).

Start your recording software and speak into microphone. When done save the recording and try to play the recorded audio file. If you are not happy with the audio quality try to adjust the R resistor. The lower its resistivity the greater the output signal. By using a variable resistor we can easily trim the resistor value until we get the expected audio quality.

How it works

Simple electret microphone circuit

Fig. 4 - Two overlapped layers of oscilloscope captures: one with silence (bright) and one capturing a sound (semi-transparent)

When we speak the sound travels through air (or another medium like water) then finally it reaches the microphone. When the vibrating air touches the microphone, its diaphragm (the pick-up plate) is going to be pushed towards the JFET's gate and so these two "plates" create an ad-hoc variable capacitor that will store the energy transported by the sound waves which are moving with a given frequency depending on sound (eg. the B note - Si in Latin - is 440Hz). The smaller the distance between the capacitor plates the higher the capacity. The higher the capacity the smaller the voltage (remember C=\frac{q}{V} where C is capacitance, q is the charge and V is the potential difference between the capacitor plates).

If we would monitor the voltage change between the diaphragm and JFET's gate we would notice a change in voltage (caused by the change of distance between them). Since the sound waves are sinusoidal the captured signal would have a sinusoidal (AC) pattern (see Fig 4. - the intense layer). However, if no sound is detected then the resulted voltage would be a steady line, ie. the DC voltage offset (see Fig. 4 - the semi-transparent layer).

However, the electret captured AC signal is very small (the order of millivolts) so by recording that signal without any amplification we would hardly hear anything useful by playing the record back.

Let's analyze the Fig. 4 above:

  • the cyan pattern captured in the 1V range shows the captured signal in two situations:
    • when total silence: this pattern look like a thin steady line and has the same amplitude as the power supply
    • when a 488Hz tone is played: it overlaps almost perfectly the signal captured while total silence condition; the reason for this is that the change in amplitude resulted while recording this 488Hz sound is so small (in the range of millivolts) that in the range of 1V it would be represented by one pixel at best
  • the yellow pattern captured in the 10mV range shows the captured signal in two situations:
    • when total silence: even if it's maximized by a factor of 100 it still looks like a steady (fat) line which has the same amplitude as the power supply
    • when a 488Hz tone is played: since it's zoomed by a factor of 100 we can easily see the change in amplitude (is the semi-transparent sinusoidal pattern); we can clearly see that when a sound is recorded the output signal amplitude is 30mV over the power supply voltage (a voltage change almost impossible to be noticed without an oscilloscope)

I have showed you these to help you understand the level at which the electrical signal is normally captured. Without amplification it can be hardly used as-is.

Take a look at the schematics of the test circuit above. The total voltage at the output pin is obtained by adding the input voltage Vcc and the microphone's output voltage mentioned earlier.

What happens where there is a total silence around the microphone (eg. the microphone is removed / turned off)?

Well, since the microphone does not produce any voltage the only voltage source is the default power supply voltage Vcc. Since this is not zero the output signal will have a constant amplitude given by the Vcc voltage. We can say that the output signal has the DC offset. However, we would expect/want that in this case the amplitude to be exact zero, ie. no offset. In order to remove the DC offset we have to extract the Vcc amplitude from the total output voltage. We do this by using the capacitor C which will filter out the DC offset. This is sometimes called "AC coupling".

The louder the sound, the higher the voltage at the JFET's gate. The higher the voltage at JFET's gate, the more current will flow between the JFET's source and drain terminals. However, an electret microphone cannot handle an unlimited quantity of current so a current limiting resistor is recommended, thus R.

A simple pre-amplifier circuit for the electret microphone

Simple electret microphone circuit

Fig 5. - The microphone pre-amplifier circuit

One of the most common amplifier is the class A amplifier which basically uses a single transistor in Common-Emitter configuration to produce a large output voltage from a relatively small input voltage. However, its efficiency is very poor (around 30%) in comparison with a class B amplifier which has an efficiency of +70% or with a class D amplifier which efficiency exceeds +90%.

For simplicity we will stick to the class A amplifier where a bipolar NPN junction transistor in a Common-Emitter configuration would do just fine, thus 2N2222A (a 2N3904 or any other NPN BJT would work as well).

How to couple the 2N2222A transistor to the circuit? Well, let's review quickly how a transistor works.

While in active mode, by applying a small current X with a voltage of (at least) VBE=0.7V at the transistor's Base terminal a larger current of the magnitude X*hFE will flow from the Emitter towards the Collector (where hFE is the transistor gain factor which can vary from 30 to +300). So the idea is to connect electret's output to the transistor's Base and to collect the amplified output at its Collector terminal.

Ok, now that we know what we have to do let's build and analyze our pre-amplifier circuit.

First the source of the electrons is the Emitter so we will connect the transistor's Emitter terminal to ground (GND). The Collector is the one which job is to "collect" the Emitter's emitted electrons; so the Collector terminal is going to be our new amplified output!

However, the current that the collector can handle is limited by transistor's design and according to 2N2222A datasheet its ICmax=0.8A. This means that we need to use a current limiting resistor (dummy load) between the Collector terminal and the power supply (Vcc), thus RL. Moreover, while the transistor is in forward-active mode the VC>VB>VE. Whenever the VB>VE and VB>VC the transistor enters in saturation mode (ie. acts like a short circuit) and the current freely flows between the Emitter and Collector. In order to limit the current through Emitter we need a current limiting resistor between the Emitter terminal and the ground, thus RE.

Since this MIC output has been previously AC-coupled it means that its voltage would equal with the electret generated voltage (tens of millivolts, hardly 0.7V). So we have to use an "external" power supply that can bias the transistor. Obviously the Vcc power supply is the natural choice. Keep in mind that the Base-Emitter voltage for the 2N2222A transistor is 0.7V and our power supply Vcc is going to be at least +3V. So we need a lower voltage for the job and as little current as possible because we really don't need any electron flow between Vcc and Base terminal. This can easily be achieved by creating a voltage divider circuit, thus R1 and R2.

Since the voltage between the Emitter and the Collector is going to have a DC offset we have to cut off that DC offset such that the amplifier would output only the amplified input AC signal. Although this might seem cumbersome there is in fact a simple passive component which can do exactly this, ie. blocks the DC while conducts AC. This component is known as capacitor. Keep in mind that this is true as long the voltage applied to the capacitor is under its rated breakdown voltage. If that voltage is exceeded the capacitor will be permanently damaged creating a short circuit.

So in order to filter out the DC on the output we need the capacitor C2. The C1 in this circuit is the same capacitor C described in the electret-microphone circuit above and its role is to cut off the DC offset of the input signal such that we work out only the AC component of the input signal.

Finally, we need one more component, a bypass capacitor between the Emitter terminal and the ground, thus CE. The role of this capacitor is to bypass entirely the RE resistor for AC signals, thus increasing the potential difference between the Emitter and Collector which results in a much higher voltage gain: since \beta=\frac{I_C}{I_B}=\frac{\frac{V_C}{R_L+R_E}}{I_B} then the smaller the RE the higher the beta. RE is smallest when it is zero, ie. when it is entirely bypassed as if it is zero. The value for this capacitor should be chosen such that its reactance represents 1/10th of RE at the lowest operating frequency.

The microphone-amplifier combined circuit should look like this:

Simple electret microphone circuit

Fig. 6 - The electret circuit combined with its pre-amplifier circuit

Now we know what parts we need, why we need them and how to connect them. The next logical step is to figure out what value should be used in order to obtain the the highest theoretical voltage gain possible.

Where to begin with? Well, let's enumerate the logical steps we have to follow in order to calculate all the components values:

  1. choose the right power supply (Vcc) for our circuit
  2. decide what's the voltage gain of your amplifier; note that the AC gain is Av=\frac{\Delta{V_{output}}}{\Delta{V_{input}}}
  3. identify the impedance of our input/output source/loads
  4. calculate the components values taking into account the limits imposed by the chosen transistor
  5. calculate the voltage gain at high/low frequencies

Few useful hints:

  • Bias the transistor to middle of the (preferably AC) load line; in case we don't know what the output load is then bias to DC load line instead:
  • to make sure that no current is flowing into the Base then R2 \leq h_{FE} \cdot \frac{R_E}{10}
  • the bypass capacitor should have a reactance of 1/10th of the bypassed resistor RE at the minimum useful frequency

1. Choosing the power supply

The electret microphone requires a voltage between 1.5-10V (normally 2V) and the 2N2222A transistor has a voltage drop of 0.7V. The rest of the circuit is going to have some resistance too so the power supply voltage should be between 3-9VDC. For our test circuit we choose \fbox{Vcc=9V}.

2. The open circuit voltage gain

The open circuit voltage gain from the input to the output when no load is connected must be at least 100, which can be rewritten as :

\fbox{A_{VOC}=-\frac{R_L}{r_e+R_E}=-100}, where 100 is the desired gain that we choose but which still must be within transistor's specs.

 

3. The input/output impedance

The input impedance is the impedance seen by the amplifier just before the capacitor C1 and it is given by the parallel connected resistor R1, R2 and the intrinsic transistor's impedance (which is (\beta+1)\cdot r_e):

Z_{in}=R_1 \parallel R_2 \parallel (\beta+1)\cdot r_e=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{(\beta+1)\cdot r_e}, assuming a \beta=100

The output impedance is the impedance seen by the output right after the C2 capacitor, that is the impedance of RL:

Z_{out}=R_L

We need to calculate the input/output impedance in order to be able to calculate the amplifier voltage gain for a given input/output source/load.

But first let's calculate the RL, R1, R2 and re values then we can calculate the input/output impedance.

4. Calculating the components given 2N2222A transistor

According to the 2N2222A datasheet we should limit the circuit's components such that:

  • IC cannot exceed 0.8A
  • hFE>=100 when 1 mA<=IC<=10 mA, which can be rewritten like \fbox{I_{C_{max}}=10 mA} => hFE>=100; \fbox{I_{C(SAT)}\approx I_{C_{max}}=10mA}
  • when VCE<1VDC the transistor enters the saturation mode; this means that the there is a voltage drop of 1V across the RE, ie. \fbox{V_{RE}=1V}
  • 0.6<VBE<1.2 VDC when IC=0.150ADC or VBE<2 VDC when IC=0.500ADC

When the transistor is OFF there is no current flow between Emitter and Collector which means that the voltage drop across the RL and RE is null. When that happens the voltage between the Emitter and Collector is equal to Vcc. The AC current changes direction at half-way of ICmax, which gives us the \fbox{I_C=\frac{10mA}{2}=5mA}

The transistor's intrinsic Emitter resistor re is directly proportional with its thermal voltage (for BJT this is VT=26mV) and inversely proportional with the current through Emitter IE (which is almost the same as the one through Collector, IC):

\fbox{r_e=\frac{V_T}{I_E} \approx \frac{V_T}{I_C}=\frac{26mV}{5mA}=5.2\Omega}

Since we know both the re and the AVOC we can determine the optimal value for RL. Keep in mind that for an AC signal the RE is bypassed and thus it can be considered zero:

\fbox{R_L=- (re+RE) \cdot A_{VOC}=- 5.2 \Omega \cdot (-100)=520 \Omega}

We want to bias this circuit in the middle of the DC load line such that we can output both edges of the audio signal. The saturation current would be:

I_{C(SAT)}=\frac{V_{cc}-(V_{R_L}+V_{CE}+V_{R_E})}{R_L+R_E}=\frac{9V-0.2V}{520\Omega+R_E} \Rightarrow 520\Omega \cdot 10mA+R_E \cdot 10mA=8.8V  \Rightarrow \fbox{R_E=\frac{8.8V-5.2V}{10mA}=360\Omega}

 

We said that R2 \leq h_{FE} \cdot \frac{R_E}{10} so we an determine the upper limit of R2: R2 \leq 100 \cdot \frac{360}{10}=3.6 k\Omega where hFE=100 when ICmax=10mA (see transistors's datasheet).

So let's take a smaller value than that, ie. \fbox{R2=3k\Omega}

The voltage across the Emitter V_E=I_E \cdot R_E \approx I_C \cdot R_E = 5mA \cdot 360 \Omega = 1.8V.

The voltage accros Base-Emitter is VBE=0.7V and VR2=VE+VBE=1.8V+0.7V=2.5V.

The R1 and R2 forms a voltage divider and thus VR2 can be calculated as:

V_{R2}=Vcc\cdot \frac{R2}{R1+R2} \Rightarrow R1=\frac{R2 \cdot (V_{cc}-V_{R2})}{V_{R2}}=\frac{3k\Omega \cdot 6.5V}{2.5V} \Rightarrow \fbox{R_1=7.8k \Omega}

Now that we know the values for the R1, R2, RL and re we can calculate the input/output impedance:

\frac{1}{Z_{in}}=\frac{1}{7.8k\Omega}+\frac{1}{3k\Omega}+\frac{1}{(100+1)\cdot 5.2\Omega} \Rightarrow \fbox{Z_{in} \approx 423\Omega}, assuming a \beta=100

 

\fbox{Z_{out}=520\Omega}

 

Ok, how about the coupling capacitors C1 and C2 and the bypass capacitor CE?

C_E=\frac{1}{2 \cdot \pi \cdot \frac{R_E}{10} \cdot f}=\frac{1}{2 \cdot \pi \cdot 36\Omega \cdot 20Hz} \Rightarrow \fbox{C_E=\approx 221 \mu{F}}

The capacitors C1 and C2 role is to decouple the AC signal from the input respectively from the output signals:

  • the C1 forms together with the audio source a low pass filter that will allow only the low frequencies from 0Hz to the cut-off frequency to pass while blocking the higher frequencies.
  • the C2 forms together with the output load a high pass filter that will allow only the high frequencies above the cut-off frequency to pass while blocking the lower frequencies.

Since the input electret microphone works in range of 20-20000Hz then the cut-off frequency for the capacitor C1 is:

f_c=\frac{1}{2 \cdot \pi \cdot Z_{in} \cdot C_1} where fc=20kHz. \fbox{C_1=\frac{1}{2 \cdot \pi \cdot 423\Omega \cdot 20000Hz}\approx 19nF}

and the cut-off frequency for the capacitor C2 is:

f_c=\frac{1}{2 \cdot \pi \cdot Z_{out} \cdot C_2} where fc=20Hz. \fbox{C_2=\frac{1}{2 \cdot \pi \cdot 520\Omega \cdot 20Hz}\approx 15.3\mu{F}}

Keep in mind that you can tune your circuit (the capacitors) such that the amplification gain in higher between certain frequencies than others. For instance, instead of choosing the corner frequencies 20Hz or 20kHz for the low pass filter and respectively high pass filter you could choose some other frequencies that fits your needs.

Ok, now that we know the parts and their values let's see how our test circuit would work:

Simple electret microphone circuit

Fig. 7 - Animation of the electret pre-amplifier test circuit

Note: the RM resistor in the animation above is just the microphone's current limiter resistor R as described earlier.

5. Calculate voltage gain

OK, now that the amplifier circuit is done we should calculate the voltage gain that we could expect.

The voltage gain is the ration between the output voltage and the input voltage:

V_{gain}=\frac{V_{out}}{V_{in}}=\frac{\Delta{V_L}}{\Delta{V_B}}=-\frac{R_L}{R_E+r_{e}}

So the voltage gain is directly dependent only on the chosen load and emitter resistors.

At higher frequencies the Emitter resistor RE=0 (is entirely bypassed by the Emitter capacitor EC), thus at higher frequencies the voltage gain is maximum:

\fbox{V_{gain}=-\frac{R_L}{R_E+r_e}=\frac{520\Omega}{360\Omega+5.2\Omega} \approx -1.4} at frequencies lower than the cut-off frequency .

At lower frequencies the Emitter resistor RE=0 (is entirely bypassed by the Emitter capacitor EC), thus at higher frequencies the voltage gain is maximum:

\fbox{V_{gain}=-\frac{R_L}{r_e}=\frac{520\Omega}{5.2\Omega} \approx -100} at frequencies higher than the cut-off frequency.

 

In the above animation when then input signal frequency is at 5kHz the voltage gain is V_{gain}=\frac{7.36V-5.35V}{10mV-(-10mV)}=100.5

Common application of an electret microphone

Simple electret microphone circuit

Fig. 8 - http://talkingelectronics.com/projects/Spy Circuits

Since the electret microphone is only 6-10 mm in diameter it is widely used as a component in a spy bug circuit: it's small, it's portable (works with a small 3V battery) and its transmission range reaches long distances (thousands of meters).

However, a classical (cordless) microphone used in a concert/show is nothing more than a (wireless) spy bug wrapped in a nice shiny metallic/plastic case.

In the next project I will create a small spy bug circuit that operates in the FM 87.5 - 108 MHz range because I intend to use a classical AM/FM receiver to pick-up the radio signal.

References:

Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.

 
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Simple electret microphone circuit

Eugen Mihailescu

Founder/programmer/one-man-show at Cubique Software
Always looking to learn more about *nix world, about the fundamental concepts of math, physics, electronics. I am also passionate about programming, database and systems administration. 16+ yrs experience in software development, designing enterprise systems, IT support and troubleshooting.
Simple electret microphone circuit

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