Simple electret microphone circuit – My *nix world What’s an electret microphone

One of the crucial widespread microphone element that’s used within the at the moment’s units (eg. smartphone) is the electret microphone. The electret capsule accommodates, amongst different little particulars, an electret diaphragm and a JFET transistor module (eg. 2SK596). A cross-sectional drawing of an electret microphone is introduced under:

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

The electret microphone check circuit schematic is proven under: Fig. 2 – The check circuit for an electret microphone

The electret microphone JFET amplifier has the Gate related to a pick-up plate (which is pushed forwards and backwards by the air), the Supply is related to the bottom and the Drain is related to the output pin. This configuration is called common-source configuration, which requires an exterior energy provide voltage Vcc.

Please observe that the “capacitor” element proven within the electret capsule schematic above is represented by the pick-up plate + the electret materials, which collectively create a capacitor.

The electret microphone operates within the vary of 1.5-10V (sometimes 2V) and the present between the JFET’s Supply-Drain (ground-output) is often max. zero.5mA. Our energy provide Vcc goes to be a minimum of +3V. With a view to restrict the output present we use a resistor R. The decrease the resistor R the upper the output present.

We will calculate the worth of R from the circuit provide voltage (Vcc), the microphone typical working voltage (2V) and the microphone max. present (zero.5mA): $R=fracV_+-2V0.5mA$

The sound sign has an AC type and that is the one sign we need to cross to the output. Nevertheless, the microphone output sign amplitude can be influenced by the enter DC energy provide that we should use to energy up the microphone. If we need to filter-out that DC sign we should always use a capacitor. The capacitor have to be chosen such that its impedance is far decrease than the resistor R at audio frequencies. The microphone frequency f = 20-20000Hz. The resistor R and the capacitor C type a high-pass filter.

We might select the capacitor C such that the filter’s cutoff frequency is f=20Hz:

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Easy electret microphone circuit

Subsequent we’re going to create a small check circuit that may assist us to check the electret microphone with assistance from a pc.

The required elements are:

• electret microphone – 1 pcs
• 1k Ohm 1/4W movie resistor (ideally a 5K variable resistor) – 1 pcs
• 1uF electrolytic capacitor – 1 pcs
• PCB audio connector – 1 pcs
• audio jack connector – 2 pcs
• pc with an audio recording software program (eg. Sound Recorder in Home windows, Audacity on Home windows/Linux/Mac) Fig Three – A breadboard setup for testing the microphone

First create the check circuit utilizing the elements listed above.

Insert one finish of the stereo jack connector within the MICrophone port of the pc and the opposite within the PCB audio connector (TRS1) connector. Often the pc MIC port supplies the required energy provide for the microphone (2-Three VDC).

Begin your recording software program and converse into microphone. When carried out save the recording and attempt to play the recorded audio file. In case you are not proud of the audio high quality attempt to regulate the R resistor. The decrease its resistivity the higher the output sign. Through the use of a variable resistor we will simply trim the resistor worth till we get the anticipated audio high quality.

The way it works Fig. four – Two overlapped layers of oscilloscope captures: one with silence (brilliant) and one capturing a sound (semi-transparent)

Once we converse the sound travels via air (or one other medium like water) then lastly it reaches the microphone. When the vibrating air touches the microphone, its diaphragm (the pick-up plate) goes to be pushed in the direction of the JFET’s gate and so these two “plates” create an ad-hoc variable capacitor that may retailer the power transported by the sound waves that are shifting with a given frequency relying on sound (eg. the B observe – Si in Latin – is 440Hz). The smaller the space between the capacitor plates the upper the capability. The upper the capability the smaller the voltage (keep in mind $C=fracqV$ the place C is capacitance, q is the cost and V is the potential distinction between the capacitor plates).

If we might monitor the voltage change between the diaphragm and JFET’s gate we might discover a change in voltage (brought on by the change of distance between them). Because the sound waves are sinusoidal the captured sign would have a sinusoidal (AC) sample (see Fig four. – the extreme layer). Nevertheless, if no sound is detected then the resulted voltage can be a gentle line, ie. the DC voltage offset (see Fig. four – the semi-transparent layer).

Nevertheless, the electret captured AC sign could be very small (the order of millivolts) so by recording that sign with none amplification we might hardly hear something helpful by enjoying the report again.

Let’s analyze the Fig. four above:

• the cyan sample captured within the 1V vary exhibits the captured sign in two conditions:
• when complete silence: this sample seem like a skinny regular line and has the identical amplitude as the facility provide
• when a 488Hz tone is performed: it overlaps virtually completely the sign captured whereas complete silence situation; the rationale for that is that the change in amplitude resulted whereas recording this 488Hz sound is so small (within the vary of millivolts) that within the vary of 1V it might be represented by one pixel at greatest
• the yellow sample captured within the 10mV vary exhibits the captured sign in two conditions:
• when complete silence: even when it is maximized by an element of 100 it nonetheless seems to be like a gentle (fats) line which has the identical amplitude as the facility provide
• when a 488Hz tone is performed: because it’s zoomed by an element of 100 we will simply see the change in amplitude (is the semi-transparent sinusoidal sample); we will clearly see that when a sound is recorded the output sign amplitude is 30mV over the facility provide voltage (a voltage change virtually unattainable to be observed with out an oscilloscope)

I’ve confirmed you these that will help you perceive the extent at which the electrical sign is generally captured. With out amplification it may be hardly used as-is.

Check out the schematics of the check circuit above. The whole voltage on the output pin is obtained by including the enter voltage Vcc and the microphone’s output voltage talked about earlier.

What occurs the place there’s a complete silence across the microphone (eg. the microphone is eliminated / turned off)?

Nicely, because the microphone doesn’t produce any voltage the one voltage supply is the default energy provide voltage Vcc. Since this isn’t zero the output sign could have a continuing amplitude given by the Vcc voltage. We will say that the output sign has the DC offset. Nevertheless, we might anticipate/need that on this case the amplitude to be actual zero, ie. no offset. As a way to take away the DC offset we’ve to extract the Vcc amplitude from the whole output voltage. We do that through the use of the capacitor C which can filter out the DC offset. That is typically referred to as “AC coupling”.

The louder the sound, the upper the voltage on the JFET’s gate. The upper the voltage at JFET’s gate, the extra present will stream between the JFET’s supply and drain terminals. Nevertheless, an electret microphone can’t deal with a vast amount of present so a present limiting resistor is really helpful, thus R.

A easy pre-amplifier circuit for the electret microphone Fig 5. – The microphone pre-amplifier circuit

One of the widespread amplifier is the category A amplifier which principally makes use of a single transistor in Widespread-Emitter configuration to supply a big output voltage from a comparatively small enter voltage. Nevertheless, its effectivity could be very poor (round 30%) as compared with a category B amplifier which has an effectivity of +70% or with a category D amplifier which effectivity exceeds +90%.

For simplicity we’ll persist with the category A amplifier the place a bipolar NPN junction transistor in a Widespread-Emitter configuration would do exactly nice, thus 2N2222A (a 2N3904 or some other NPN BJT would work as nicely).

How you can couple the 2N2222A transistor to the circuit? Properly, let’s evaluation shortly how a transistor works.

Whereas in lively mode, by making use of a small present X with a voltage of (at the least) VBE=zero.7V on the transistor’s Base terminal a bigger present of the magnitude X*hFE will move from the Emitter in the direction of the Collector (the place hFE is the transistor achieve issue which may differ from 30 to +300). So the thought is to attach electret’s output to the transistor’s Base and to gather the amplified output at its Collector terminal.

Okay, now that we all know what we now have to do let’s construct and analyze our pre-amplifier circuit.

First the supply of the electrons is the Emitter so we’ll join the transistor’s Emitter terminal to floor (GND). The Collector is the one which job is to “gather” the Emitter’s emitted electrons; so the Collector terminal goes to be our new amplified output!

Nevertheless, the present that the collector can deal with is restricted by transistor’s design and in line with 2N2222A datasheet its ICmax=zero.8A. Because of this we have to use a present limiting resistor (dummy load) between the Collector terminal and the facility provide (Vcc), thus RL. Furthermore, whereas the transistor is in forward-active mode the VC>VB>VE. Every time the VB>VE and VB>VC the transistor enters in saturation mode (ie. acts like a brief circuit) and the present freely flows between the Emitter and Collector. To be able to restrict the present by way of Emitter we’d like a present limiting resistor between the Emitter terminal and the bottom, thus RE.

Since this MIC output has been beforehand AC-coupled it signifies that its voltage would equal with the electret generated voltage (tens of millivolts, hardly zero.7V). So we have now to make use of an “exterior” energy provide that may bias the transistor. Clearly the Vcc energy provide is the pure selection. Needless to say the Base-Emitter voltage for the 2N2222A transistor is zero.7V and our energy provide Vcc goes to be a minimum of +3V. So we’d like a decrease voltage for the job and as little present as potential as a result of we actually do not want any electron stream between Vcc and Base terminal. This will simply be achieved by making a voltage divider circuit, thus R1 and R2.

Because the voltage between the Emitter and the Collector goes to have a DC offset we’ve got to chop off that DC offset such that the amplifier would output solely the amplified enter AC sign. Though this might sound cumbersome there’s in truth a easy passive element which may do precisely this, ie. blocks the DC whereas conducts AC. This element is called capacitor. Needless to say that is true as lengthy the voltage utilized to the capacitor is beneath its rated breakdown voltage. If that voltage is exceeded the capacitor will probably be completely broken creating a brief circuit.

So with a view to filter out the DC on the output we’d like the capacitor C2. The C1 on this circuit is identical capacitor C described within the electret-microphone circuit above and its position is to chop off the DC offset of the enter sign such that we work out solely the AC element of the enter sign.

Lastly, we’d like yet one more element, a bypass capacitor between the Emitter terminal and the bottom, thus CE. The position of this capacitor is to bypass completely the RE resistor for AC alerts, thus growing the potential distinction between the Emitter and Collector which leads to a a lot larger voltage achieve: since $beta=fracI_CI_B=fracfracV_CR_L+R_EI_B$ then the smaller the RE the upper the beta. RE is smallest when it’s zero, ie. when it’s totally bypassed as whether it is zero. The worth for this capacitor must be chosen such that its reactance represents 1/10th of RE on the lowest working frequency.

The microphone-amplifier mixed circuit ought to appear to be this: Fig. 6 – The electret circuit mixed with its pre-amplifier circuit

Now we all know what elements we’d like, why we’d like them and the way to join them. The subsequent logical step is to determine what worth ought to be used with a view to get hold of the the very best theoretical voltage achieve attainable.

The place to start with? Properly, let’s enumerate the logical steps we’ve got to comply with in an effort to calculate all of the elements values:

1. select the best energy provide (Vcc) for our circuit
2. determine what is the voltage achieve of your amplifier; observe that the AC achieve is $Av=fracDeltaV_outputDeltaV_input$
3. determine the impedance of our enter/output supply/masses
4. calculate the elements values bearing in mind the bounds imposed by the chosen transistor
5. calculate the voltage achieve at excessive/low frequencies

• Bias the transistor to center of the (ideally AC) load line; in case we do not know what the output load is then bias to DC load line as an alternative:
• to be sure that no present is flowing into the Base then $R2 leq h_FE cdot fracR_E10$
• the bypass capacitor ought to have a reactance of 1/10th of the bypassed resistor RE on the minimal helpful frequency

1. Selecting the facility provide

The electret microphone requires a voltage between 1.5-10V (usually 2V) and the 2N2222A transistor has a voltage drop of zero.7V. The remainder of the circuit goes to have some resistance too so the facility provide voltage must be between Three-9VDC. For our check circuit we select $fboxVcc=9V$.

2. The open circuit voltage achieve

The open circuit voltage achieve from the enter to the output when no load is related have to be a minimum of 100, which could be rewritten as : $fboxA_VOC=-fracR_Lr_e+R_E=-100$the place 100 is the specified achieve that we select however which nonetheless have to be inside transistor’s specs.

Three. The enter/output impedance

The enter impedance is the impedance seen by the amplifier simply earlier than the capacitor C1 and it’s given by the parallel related resistor R1, R2 and the intrinsic transistor’s impedance (which is $(beta+1)cdot r_e$): $Z_in=R_1 parallel R_2 parallel (beta+1)cdot r_e=frac1R_1+frac1R_2+frac1(beta+1)cdot r_e$assuming a $beta=100$

The output impedance is the impedance seen by the output proper after the C2 capacitor, that’s the impedance of RL: $Z_out=R_L$

We have to calculate the enter/output impedance so as to have the ability to calculate the amplifier voltage achieve for a given enter/output supply/load.

However first let’s calculate the RL, R1, R2 and re values then we will calculate the enter/output impedance.

four. Calculating the elements given 2N2222A transistor

Based on the 2N2222A datasheet we should always restrict the circuit’s elements such that:

• IC can’t exceed zero.8A
• hFE>=100 when 1 mA<=IC<=10 mA, which might be rewritten like $fboxI_C_max=10 mA$ => hFE>=100; $fboxI_C(SAT)approx I_C_max=10mA$
• when VCE<1VDC the transistor enters the saturation mode; which means the there's a voltage drop of 1V throughout the RE, ie. $fboxV_RE=1V$

When the transistor is OFF there isn’t any present movement between Emitter and Collector which signifies that the voltage drop throughout the RL and RE is null. When that occurs the voltage between the Emitter and Collector is the same as Vcc. The AC present modifications path at half-way of ICmax, which provides us the $fboxI_C=frac10mA2=5mA$

The transistor’s intrinsic Emitter resistor re is instantly proportional with its thermal voltage (for BJT that is VT=26mV) and inversely proportional with the present by way of Emitter IE (which is nearly the identical because the one by means of Collector, IC): $fboxr_e=fracV_TI_E approx fracV_TI_C=frac26mV5mA=5.2Omega$

Since we all know each the re and the AVOC we will decide the optimum worth for RL. Take into account that for an AC sign the RE is bypassed and thus it may be thought-about zero: $fboxR_L=- (re+RE) cdot A_VOC=- 5.2 Omega cdot (-100)=520 Omega$

We need to bias this circuit in the midst of the DC load line such that we will output each edges of the audio sign. The saturation present can be: $I_C(SAT)=fracV_cc-(V_R_L+V_CE+V_R_E)R_L+R_E=frac9V-0.2V520Omega+R_E Rightarrow 520Omega cdot 10mA+R_E cdot 10mA=8.8V$ $Rightarrow fboxR_E=frac8.8V-5.2V10mA=360Omega$

We stated that $R2 leq h_FE cdot fracR_E10$ so we an decide the higher restrict of R2: $R2 leq 100 cdot frac36010=3.6 kOmega$ the place hFE=100 when ICmax=10mA (see transistors’s datasheet).

So let’s take a smaller worth than that, ie. $fboxR2=3kOmega$

The voltage throughout 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=zero.7V and VR2=VE+VBE=1.8V+zero.7V=2.5V.

The R1 and R2 types a voltage divider and thus VR2 might be calculated as: $V_R2=Vcccdot fracR2R1+R2 Rightarrow R1=fracR2 cdot (V_cc-V_R2)V_R2=frac3kOmega cdot 6.5V2.5V Rightarrow$ $fboxR_1=7.8k Omega$

Now that we all know the values for the R1, R2, RL and re we will calculate the enter/output impedance: $frac1Z_in=frac17.8kOmega+frac13kOmega+frac1(100+1)cdot 5.2Omega Rightarrow fboxZ_in approx 423Omega$assuming a $beta=100$ $fboxZ_out=520Omega$

Okay, how concerning the coupling capacitors C1 and C2 and the bypass capacitor CE? $C_E=frac12 cdot pi cdot fracR_E10 cdot f=frac12 cdot pi cdot 36Omega cdot 20Hz Rightarrow fboxC_E=approx 221 muF$

The capacitors C1 and C2 position is to decouple the AC sign from the enter respectively from the output alerts:

• the C1 types along with the audio supply a low move filter that may permit solely the low frequencies from 0Hz to the cut-off frequency to move whereas blocking the upper frequencies.
• the C2 varieties along with the output load a excessive move filter that may permit solely the excessive frequencies above the cut-off frequency to move whereas blocking the decrease frequencies.

Because the enter electret microphone works in vary of 20-20000Hz then the cut-off frequency for the capacitor C1 is: $f_c=frac12 cdot pi cdot Z_in cdot C_1$ the place fc=20kHz. $fboxC_1=frac12 cdot pi cdot 423Omega cdot 20000Hzapprox 19nF$

and the cut-off frequency for the capacitor C2 is: $f_c=frac12 cdot pi cdot Z_out cdot C_2$ the place fc=20Hz. $fboxC_2=frac12 cdot pi cdot 520Omega cdot 20Hzapprox 15.3muF$

Needless to say you possibly can tune your circuit (the capacitors) such that the amplification achieve in greater between sure frequencies than others. As an example, as an alternative of selecting the nook frequencies 20Hz or 20kHz for the low cross filter and respectively excessive move filter you can select another frequencies that matches your wants.

Okay, now that we all know the elements and their values let’s examine how our check circuit would work: Fig. 7 – Animation of the electret pre-amplifier check circuit

Notice: the RM resistor within the animation above is simply the microphone’s present limiter resistor R as described earlier.

5. Calculate voltage achieve

OK, now that the amplifier circuit is completed we should always calculate the voltage achieve that we might anticipate.

The voltage achieve is the ration between the output voltage and the enter voltage: $V_gain=fracV_outV_in=fracDeltaV_LDeltaV_B=-fracR_LR_E+r_e$

So the voltage achieve is immediately dependent solely on the chosen load and emitter resistors.

At larger frequencies the Emitter resistor RE=zero (is completely bypassed by the Emitter capacitor EC), thus at larger frequencies the voltage achieve is most: $fboxV_gain=-fracR_LR_E+r_e=frac520Omega360Omega+5.2Omega approx -1.4$ at frequencies decrease than the cut-off frequency .

At decrease frequencies the Emitter resistor RE=zero (is completely bypassed by the Emitter capacitor EC), thus at larger frequencies the voltage achieve is most: $fboxV_gain=-fracR_Lr_e=frac520Omega5.2Omega approx -100$ at frequencies greater than the cut-off frequency.

Within the above animation when then enter sign frequency is at 5kHz the voltage achieve is $V_gain=frac7.36V-5.35V10mV-(-10mV)=100.5$

Widespread software of an electret microphone Fig. eight – http://talkingelectronics.com/projects/Spy Circuits

Because the electret microphone is simply 6-10 mm in diameter it’s extensively used as a element in a spy bug circuit: it is small, it is moveable (works with a small 3V battery) and its transmission vary reaches lengthy distances (hundreds of meters).

Nevertheless, a classical (cordless) microphone utilized in a live performance/present is nothing greater than a (wi-fi) spy bug wrapped in a pleasant shiny metallic/plastic case.

Within the subsequent undertaking I’ll create a small spy bug circuit that operates within the FM 87.5 – 108 MHz vary as a result of I intend to make use of a classical AM/FM receiver to pick-up the radio sign. 