Part Number: PCM4202
Would you advise about below two questions?
1. There is a description of "Each analog input pair accepts a full-scale input voltage of
approximately 6.0VPP differential on page 13 of datasheet.
However when Vcc is 5V, input D range will be 5V.
In this case, we should input with D range from 0V to 5V.
Is this understanding right?
2. Although it is an analog input terminal, it is biased internally.
Is it OK to insert a coupling capacitor and input a signal that swings to +/- around zero volts?
1. Vcom on the inputs is roughly 2.5V. To achieve a 6VPP (or 3Vpeak) each single ended input will only need +-1.5V
lets take a sine wave example for simplicity, think of the positive peak.
Positive peak: IN1L+ will be vcom + 1.5 = 4V and IN1L- will be at VCOM -1.5 = 1V Total input voltage would be IN1L+ - (IN1L-) = 4 - 1= 3V peak.
negative peake: IN1L+ will be vcom - 1.5 = 1V and IN1L- will be at VCOM +1.5 = 4V Total input voltage would be IN1L+ - (IN1L-) = 1 - 4= -3V peak.
This makes 6VPP
2. See the below post, it seems that the inputs are not internally biased and do require VCOM sourcing from the VCOM pins.
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1.) Yes, the ADC accepts a full-scale 6Vpp differential input signal. This means each IN+ can accept a 3Vpp waveform and IN- accepts a 3Vpp waveform that's 180 degrees out of phase of the IN+ signal.
2.) The input impedance of the PCM4202 is not very high and we'd recommend that you use a circuit like the one featured in Figure 13 of the datasheet to drive the inputs of the PCM4202. This will allow you to AC couple your input signals into the OPA1632 and then use the VCOM output of the PCM4202 to drive the VOCM input of the OPA1632.
Regards,Collin WellsPrecision ADC Applications
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