Question
1)
In the manual there is an information that this IC has 10 bit control
register, but the timing diagram shows that I have to write 0x6040 ( 16 bit )
after power-up and 0x4040 every cycle. Is it a mistake in manual or how can I
write 16 bit in 10 bit register?
2)
As I realised, it is a sequence "1100000010"
with A0 = 1. But if I use a single 7812 is it correct to set A0 to 1?
Answer
1. The control register is a 10 bit register and the bits driven into the part
on the first 10 falling edges of SCLK following TFS going low are the bits
written to the control register. However, the control register itself in not
updated until the 13th rising edge of SCLK (Point B on Figure 18 in the
datasheet) so 13 SCLKs are needed for a full update of the control register. In
our "quick evaluation seup" on page 13 of the datasheet we use 16 bits as this
is a common standard for SPI communication. If 16 SCLKs are applied the unused
SCLKs are just ignored.
2. A0 is essentially the address of the part. It is both a pin on the part that
must be hardwired either high or low and one of the bits in the control
register. It allows two devices to share the same serial bus. If the external
pin is hardwired high - you must write "1" to the A0 bit in the control
register to communicate with the device. If the external pin is hardwired low -
you must write "0" to the A0 bit in the control register to communicate with
the device. Only the serial port of the device that received the last valid
control byte will attempt to drive the serial bus on the next serial read. In
your case when you are using a single AD7812, you may leave A0 as "0" as long
as the external pin is hardwired low.