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ADC124S101: what if #CS going high at falling edge of SCLK

Part Number: ADC124S101

Hello,

I'm driving the ADC using a system that can only change the output pin states synchronously to the falling edge of the SCLK clock. The SCLK is not stopped with the #CS going high, because other devices need to be scanned with same clock in parallel.

Now, the datasheet states:

"When CS is brought high, SCLK is internally gated off. If SCLK is stopped in the low state while CS is high, the
subsequent fall of CS will generate a falling edge of the internal version of SCLK, putting the ADC into the track
mode. This is seen by the ADC as the first falling edge of SCLK. If SCLK is stopped with SCLK high, the ADC
enters the track mode on the first falling edge of SCLK after the falling edge of CS."

What would be the behavior if both #CS and SCLK are changing at the same time? Would it be unpredictable, or is the previous value of SCLK (high) still in effect when #CS going high is recognized?

Any hints are highly appreciated.

Ralf

  • Hello Ralf,

    If CS and SCLK are acting the same, meaning the device sees a falling edge of CS and SCLK at the same time, the device will enter track mode at this moment. 

    Regards

    Cynthia

  • Hello Cynthia,

    thanks for your reply.

    Yes, for the falling edge of CS the datasheet somehow describes this case (Figure 6 with Tcsu and Tclh both equal to 0.5 Tsclk) and it is good to know that this is indeed a legal case.

    What is actually unclear to me is the rising edge of CS and the falling edge of SCLK at the same time, and how to interprete the quoted text from the datasheet in this case. Sorry I did not make that clear enough.

    Yours, Ralf

  • Hi Ralf,

    I understand now. if SCLK is changing states when CS is rising, but is rising at the same rate as CS then I would conclude the SLCK would be considered high when gated off. This would mean that when CS goes low, the device will enter track mode on the first falling edge of SLCK.

    This quoted text is to describe the internal workings of the device, but ultimately will not greatly affect you

    Regards

    Cynthia