ADPCM Amplification
Before being mixed to the audio left and right signals, the ADPCM signal generated by the M5205 is biased and shrunk, reamplified and goes through a few low pass filters.ADPCM Filter Circuit
Biasing and Passive Low-pass FilterThe biasing is introducted by the R86 and R85 bridge divider conencted to 5V (see Figure 2). We will start with the calculation of the Thevenin equivalent ciruit at R85.
R_{TH_{R85}}=R84//R86//R85=\frac{1}{1/R84+1/R86+1/R85}=\frac{1}{1/270k+1/33k+1/180k}=25.277k\Omega
As you can see, the DAC output is divided by 10.682 so the signal is greatly shrunk by the biasing circuit and it is shifted up by 3.83V.
Now that we have determined the Thevenin equivalent, we can calculate the low pass filter including R57 and C23.
V_{C23}=\frac{Z_{C23}}{R_{TH_{R85}}+R57+Z_{C23}}\times V_{TH_{R85}}=\frac{1/jC23\omega}{R_{TH_{R85}}+R57+1/jC23\omega}\times V_{TH_{R85}}
V_{C23}=\frac{1}{1+j\left(R_{TH_{R85}}+R57\right)C23\omega}\times V_{TH_{R85}}At DC level, when j\omega=0, C23 acts an open circuit and therefore V_{TH_{R85}} is applied directly to V+. V_{TH_{R85}} will include 3.83V and DC bias added in software to VDAC which is typically 2.5V. Since VDAC is divided by 10.682, a 2.5V DC offset will give 0.234V. V_{TH_{R85}}=3.83+0.234\thickapprox4VLet's now work out the cut-off frequency of the low-pass filter.
\omega_{C}=\frac{1}{(R_{TH_{R85}}+R57)C23}\rightarrow f_{C}=\frac{1}{2\pi(R_{TH_{R85}}+R57)C23}=\frac{1}{2\pi\times(25.277k+1k)\times390\times10^{-12}}=15538HzFirst order filter with -3dB attenuation at f_{c}=15538Hz and -20dB per decade.
Positive Gain AmplifierIC114B has a negative feedback and the input voltage is connected to the positive input. Therefore it is configured as an amplifier with a positive gain. C52 is a fairly large capacitor connected to ground which ensure gain is only applied at AC level where the opamp becomes a simple follower at DC level. In other words, the ADPCM signal is amplified and bias voltage is preserved.At DC level, when j\omega=0, C52 is seen as an open circuit therefore:V_{IC114B+}=V_{R85}=V_{IC114B-}=V_{IC114BOUT}=4VAC level, IC114B acts as positive gain amplifer.
V_{IC114B-}=\frac{R88+1/jC52\omega}{R88+1/jC52\omega+R87}\times V_{IC114BOUT}When j\omega\rightarrow\infty, capacitors be come short circuits and we can determine the gain for high frequencies.
V_{IC114B-}=\frac{R88}{R88+R87}\times V_{IC114BOUT}=V_{IC114B+}=V_{R85AC}
Gain=\frac{V_{IC114BOUT}}{V_{R85AC}}=\frac{R88+R87}{R88}=\frac{10k+91k}{10k}=10.1
\frac{V_{IC114BOUT}}{V_{R85AC}}=\frac{R88+1/jC52\omega+R87}{R88+1/jC52\omega}=\frac{1+j(R87+R88)\times C52\omega}{1+jR88C52\omega}To determine the cut frequency of the high-pass filter we know that gain must be 3dB away from the final final. i.e. 10.1/\sqrt{2}=7.142
\left|\frac{V_{IC114BOUT}}{V_{R85AC}}\right|=\frac{\left|1+j(R87+R88)\times C52\omega\right|}{\left|1+jR88C52\omega\right|}=\frac{\left|1+j0.4747\omega\right|}{\left|1+j0.047\omega\right|}=\frac{\sqrt{1+0.4747^{2}\omega^{2}}}{\sqrt{1+0.047^{2}\omega^{2}}}
\frac{1+0.4747^{2}\omega_{C}^{2}}{1+0.047^{2}\omega_{C}^{2}}=7.142^{2}=511+0.4747^{2}\omega_{C}^{2}=51\times\left(1+0.047^{2}\omega_{C}^{2}\right)
0.4747^{2}\omega_{C}^{2}-51\times0.047^{2}\omega_{C}^{2}=51-1
\omega_{C}^{2}\left(0.4747^{2}-51\times0.047^{2}\right)=50
\omega_{C}=\sqrt{\frac{50}{\left(0.4747^{2}-51\times0.047^{2}\right)}}=21.065It is now possible to work out the cut frequency of the high-pass filter. f_{C}=\frac{\omega_{C}}{2\pi}=\frac{21.065}{2\times3.14}=3.354Hz
This opamp is configured as an MFB (multi-feedback) filter which is a second order low-pass filter but also no gain at DC. For the calculations, C53 is removed but its purpose is to remove any DC offset before it is fed to the opamp.When j\omega=0, C53, C21 and C22 act as open circuits. Therefore:
V_{IC114A-}=V_{ADPCMOUT}=V_{IC114B+}=V_{BIAS}=3.90VWhen j\omega\rightarrow\infty, C22 is a short circuit turning IC114A into a follower but at this stage C21 is shorting the input signal to ground so there will be no high-frequency output signal.To find the transfer function of this opamp circuit, we will break in down in 2 stages: calculation of voltage at C21 and then calculation of voltage at IC114A-. To simplify calulations, VBIAS biased will be switched off to make V_{IC114A-}=V_{IC114A+}=0V. With this simplification, the connection R56 and C22 will be tied to GND. V_{C21}=\frac{\frac{V_{IC114BOUT}}{R54}+\frac{V_{ADPCMOUT}}{R55}}{\frac{1}{R54}+\frac{1}{R55}+\frac{1}{R56}+jC21\omega}
V_{IC114A-1}=\frac{\frac{V_{ADPCMOUT}}{1/jC22\omega}+\frac{V_{C21}}{R56}}{jC22\omega+\frac{1}{R56}}=0
V_{C21}=-jR56C22\omega\times V_{ADPCMOUT}
\frac{\frac{V_{IC114BOUT}}{R54}+\frac{V_{ADPCMOUT}}{R55}}{\frac{1}{R54}+\frac{1}{R55}+\frac{1}{R56}+jC21\omega}=-jR56C22\omega\times V_{ADPCMOUT}
\frac{V_{IC114BOUT}}{R54}+\frac{V_{ADPCMOUT}}{R55}=-jR56C22\omega\times V_{ADPCMOUT}\times\left(\frac{1}{R54}+\frac{1}{R55}+\frac{1}{R56}+jC21\omega\right)
V_{ADPCMOUT}\times\left(\frac{1}{R55}+jR56C22\omega\times\left(\frac{1}{R54}+\frac{1}{R55}+\frac{1}{R56}+jC21\omega\right)\right)=-\frac{V_{IC114BOUT}}{R54}
\frac{V_{ADPCMOUT}}{V_{IC114BOUT}}=-\frac{1}{R54}\times\frac{1}{\frac{1}{R55}+jR56C22\omega\times\left(\frac{1}{R54}+\frac{1}{R55}+\frac{1}{R56}+jC21\omega\right)}
\frac{V_{ADPCMOUT}}{V_{IC114BOUT}}=-\frac{1}{R54R56}\times\frac{1}{\frac{1}{R55R56}+jC22\omega\times\frac{R55R56+R54R56+R54R55}{R54R55R56}-C21C22\omega^{2}}
\frac{V_{ADPCMOUT}}{V_{IC114BOUT}}=-\frac{1}{R54R56C21C22}\times\frac{1}{\frac{1}{R55R56C21C22}+j\frac{R55R56+R54R56+R54R55}{R54R55R56C21}\omega-\omega^{2}}Since R54=R55=R56, we will replace them with R instead to simplify the equation:
\frac{V_{ADPCMOUT}}{V_{IC114BOUT}}=-\frac{1}{R^{2}C21C22}\times\frac{1}{\frac{1}{R^{2}C21C22}+j\frac{R^{2}+R^{2}+R^{2}}{R^{3}C21}\omega-\omega^{2}}=-\frac{1}{R^{2}C21C22}\times\frac{1}{\frac{1}{R^{2}C21C22}+j\frac{3}{RC21}\omega-\omega^{2}}
H\left(j\omega\right)=-\frac{1}{1+j3RC22\omega-R^{2}C21C22\omega^{2}}Now that we have the transfer function, we can see that when j\omega=0, the DC gain is 1. \omega_{C}=\frac{1}{\sqrt{R^{2}C21C22}};\omega_{C}=2\pi f_{C};f_{C}=\frac{1}{2\times\pi\times\sqrt{8.2k\times8.2k\times11\times10^{-9}\times390\times10^{-12}}}=6463HzThe cut-off freuqency is 6463Hz and -40dB percade will be applied by the filter. The next stage is the Quality factor calculation but I can't remember how to do that. To get the value, I typed the component values in [http://sim.okawa-denshi.jp/en/OPttool.php||this link] and it gave me Q=1.2209.********