TAC5212: TAC5212 and OPA2189 review request

Hi,

Give me another day to review your schematic, sorry about the delay here.

-Mir

Thank you

I have a question ... What is the maximum BCLK frequency that TAC5212 can handle. Al

Also regarding AC-coupling capacitors. If you check the schematics i have 22u capacitors C210 (C213) on the input of the Single Ended to Differential OPAMP and C211 and C212 (C214 and C215) on input of TAC5212 pin 17 and 18 (15 and 16). I believe these work as AC-coupling, unless i misunderstood something. 

For the output ... TAC5212 pins 21 and 22 (20 and 19) are connected through 22u capacitors C216 and C217 (C219 and C220) and Differential to Single Ended opamp has on the output C218 (C221). I believe these work as AC-coupling, unless i misunderstood something. 

Can you please clarify a bit more your suggestions

Mirza

Hi Mirza,

You are correct, you have AC coupling caps in place, it's just not obvious from the schematic. The IN1P/M and OUT1P/M ports are not blue so they are harder to spot on the block diagram.

25MHz is the largest BCLK accepted by the device.

Best regards,
Jeff McPherson

Hello 

I have an other question if you can help me with. As mentioned before i will use TAC5212 in avionics. Sound quality is very important. I am wondering about the sound level on the input of TAC5212.

I am trying to determine the minimum input voltage level required for the TAC5212 to reliably digitize an analog signal. Since I am using a single-ended to differential operational amplifier for signal conversion, I assume that impedance matching between the microphone and the TAC5212 can be reasonably addressed by selecting appropriate resistor and capacitor values in the SE-to-DIFF conversion stage—although this approach may not be perfect, it should be sufficient for practical purposes. From my understanding, the minimum input voltage required for reliable ADC conversion is influenced by several key parameters, including the signal-to-noise ratio (SNR), dynamic range (DR), total harmonic distortion plus noise (THD+N), and common-mode rejection ratio (CMRR).

From the datasheet section 6.5 Electrical characteristics and my implementation (i have decided on 5.5kOhm) differential mode AC coupled, i read following:

Vfs = 2Vrms
SNR =  119dB "INxx differential AC-coupled input and AC signal shorted to ground, 0dB channel gain"
DR = 119dB "INxx differential AC-coupled input and –60dBFS AC signal input, 0dB channel gain"
THD+N = -98dB "INxx differential AC-coupled input and –1dBFS AC signal input, 0dB channel gain"
CMRR = 80dB "Differential DC-coupled input, 0dB channel gain, –6dBFS AC input, 1kHz signal on both pins and measured level at output"

Can you confirm if i am reading datasheet correctly for my implementation?

Here is my issue ... CMRR is given only for DC-coupled signals. Is it still the same for AC-coupled signals. Based on my choices and some math I get following, for minimum detectable voltage Vmin
​
Vmin​=Vfs/10^{SNR/20} = 2 / 10^{119/20} = 2.25uV

Including THD+N limits

Vmin_THD​=Vfs/10^{THD+N/20} = 2 / 10^{98/20} = 25uV

Considering CMRR

Vmin_CMRR = Vfs/10^{CMRR/20} = 2 / 10^{80/20} = 200uV

Is this correct or am i missing something? Based on this CMRR is the leading factor in determining minimum input voltage that can be meaningfully converted. Can you confirm this?
​

Thank you

Hi Mirza,

One quick point, it's a common misconception that audio circuits need to be impedance matched. In fact they should not be. Audio is transmitted not through power, but through voltage alone. So you'll find that input impedances to devices like ADCs are high while output impedances such as those from line drivers are low. This is so that the voltage divider formed by the two maximizes the signal delivered to the input. Just something to keep in mind while you're designing your input stage. Lots of helpful examples on this topic here: https://electronics.stackexchange.com/questions/6846/how-important-is-impedance-matching-in-audio-applications

Regarding the minimum detectable voltage, SNR should be the only relevant spec here. Since we're interested in very small signals, there will be virtually no harmonics meaning THD+N = SNR for Signal << Full Scale. For CMRR, the reduction is 80dB, meaning the -6dB signal applied by the test condition appears at the output at -86dBFS. Likewise, since we are considering very small signals, a common signal even as high as -40dBFs will get masked under the noise floor of the SNR. SNR dominates in this question.

Your math is correct though, 2.25uVrms differential would be equal to the noise floor. Anything higher will start to be distinguishable from noise.

Best regards,
Jeff McPherson