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Part Number: TPA6404-Q1
the ch1 of TAP6424 connect the load of 6.8uF&4ohm. customer use AC diagbostic,
set gain =1 I=10mA
read register 0x17 value is 1D, so the Impedance_CH is 29*0.059=1.711ohm.
use Channelx Impedance equlation of 22.214.171.124.3, the Channelx Impedance=1.711*2.371/(1*10)=0.4ohm. But, With 18.75kHz signal, the actual load impedance is 4.19ohm. there have 10times gap between caluation and actural value.
is there mistake? thanks
Please set to gain=4, 19mA.
a) Set AC_DIAGS_LOOPBACK (Reg0x16<7>) =0.
b) Apply 0dBFS 19K signal and enable AC load diagnostics (set Reg0x15<3:0> = 1 for CH1 to CH4; Set Reg0x15<3>=1 and Reg0x15<1>=1 for PBTL12 and PBTL34).
c) Read back AC_IMP (Reg0x17~Reg0x1A), AC_LDG_PHASE2 (Reg0x1B and Reg0x1C) and STI (Reg0x1D and Reg0x1E).
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