The following picture shows NanoNote's audio circuit. It shows the mic, headphone and speaker's connections and how they're connected with the Intern Codec's analogic module (yellow blocks represent intern processor blocks).
JZ4720, JZ4740 and JZ4725 processors allow direct connection from headphones to amplifier's exits and the mic's direct connection. This chip has a BG amplifier at the mic's entrance, making unnecessary to amplify its coming signal. In addition, the MICBIAS pin gives the necessary polarized voltage for the correct mic operation. Modifying the SW1 and SW2 switches state we can:
SW1 closed: Lead the audio signals coming from the microphone or the LINEIN_L and LINEIN_R input lines (not available on NanoNote) directly to the output and depending on SW2 state mix them with the coming signals prom the digital/analogical conversers (DAC) (SW2 closed) or not (SW2 open).
SW1 open, SW2 closed: DAC's exit is connected to the output circuit, allowing the reproduction of DAC's data (i.e. playing an audio file).
SW1 and SW2 open: No signal is connected to output circuit.
Microphone, LINEIN_L and LINEIN_R lines enter directly to analogical/digital conversor (ADC) wich allows to record audio for its later process or reproduction. Signal gaining coming from mic must be set to 0 (CDCCR2.MICBG = 0) whenever working with LINEIN_L and LINEIN_R, otherwise their signals will mix.
As said before, the JZ4740 processor allows direct connection from headsets to exit pins. This reduces circuit complexity thus archieving a costs reduction. Next figure shows that every channel has a serial connection with a condenser (C59 / C52) blocking the DC component, a N-channel MOSFET transistor (Q9 / Q10) working as switch and a resistor (R37 / R38) limiting the circuit's exit current for possible short circuits coming from headphone's jack. R40 and R41 resistors are used to sign the headphone's presence or absence. The POP signal is used to control the Q9 and Q10 switch's state.
When POP has an high analogic value (3.3V in this case) the Q9 and Q10 transistors Gate-Source voltage is 3.3V. As the required voltage for the transistor to enter into saturation state is 1.3v they'll enter into saturation region (switch closed) so the circuit will look like this:
If POP ha a low logic value (0V) the Q9 and Q10 transistors Gate-Source voltage would be 0V, therefore the transistors would be in cut state (switch open) so the circuit would look like this:
When the headphone is plugged in (closing Q9 and Q10 "switches") the equivalent circuit is shown in the next picture. It can be observed that the signal going towards the headphones has only an AC component, given that DC signal is blocked by C59 / C63 condensers. BD8 inductance is used for ????????????????
When the headphone jack is removed (equivalent circuit is shown in the next picture) there is a value change in AMPIN node's voltage. Going from 3.3V (as seen in previous picture) to ~0V.
Speaker is handled by a two-channel potency amplifier, wich can be enabled using an enabler (pin 7 CE). By disabling the audio amplifier its spent potency is zero, thus reducing the potency expenditure and improving the battery charge time.
It is obvious that when the headphone is pugged in the amplifier must be disabled for the audio not to be played by both exits, the equivalent circuit is shown is showed in the next picture. We must check the CE signal state, which is determined by the Q6 transistor's state (P channel MOSFET). Its Gate-Source voltage is 3.3V because R36 and R42 resistors lead Gate and Source to 3.3V and 0V respectively; thus the transistor being in cut state, removing Q6 from the circuit and setting amplifier's signal CE to 0V disabling the circuit.
When the headphones are removed, is necessary for the amplifier to be able to be enabled when the user wants. The equivalent circuit resulting in removing the headphones is shown in the next picture. Again the amplifier's CE pin depends on the Q6 transistor state (which acts as a switch), in this state Q6 Gate's voltage is VAMPIN which can be calculated prom the resistive divisor from R36 and the parallel of (R40 + R37) with (R38 + R41) (next picture shows the equivalent circuit.
Source's voltages is given by the AMPEN signal, if AMPEN has an high logical value (3.3V) the Q6's Gate-Source voltage is VGate(0) - VSource(VAMPEN) = -3.3V, thus setting Q6 transistor into saturation state and CE pin to 3.3V; if AMPEN has a low logic value (0V) the Q6's Gate-Source voltage is 0V thus setting Q6 transistor into cut state (switch open) and CE pin's voltage into 0V.
This circuit is really simple because of the processor including the necessary support to feed and amplify the mic's signal. This circuit is shown next.