0
\$\begingroup\$

Hello I am looking on some analog and digital multiplexers that have high supply voltage along with high amounts of channels and I've found some that are programmable either serially or parallely and I am wondering how I should interpret the timing characteristics in the datasheet.

So for example lets look at these two multiplexer circuits from Analog Devices and Intersil respectively, AD75019 and CD22M3494.

The data load time of the AD75019 is 52µs since the max serial clock speed is 5MHz and 256 bits need to be loaded in order to program it.

On the CD22M3494 however there isn't any load time parameter since it is programmed parallely, so the parameter that I should compare the AD75019 to is what?

I've looked at some other multiplexers that are programmed parallely as well and some of them have a parameter called "Switching frequency" which I would assume is the one parameter that I should look at in this particular case, but on the CD22M3494 however I can't seem to find any parameter for this.

Can anyone help me and point me in the right direction?

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

On page 8 of the datasheet for the CD22M3494 it says:

To make a connection (close switch) between any two points, specify an “X” address, a “Y” address, set “DATA” high, and switch “STROBE” rom low to high. To break a connection, follow this same procedure with “DATA” low.

According to the timing diagram on page 7, and the timing parameters in the tables on page 5 and 6, each connection will take a minimum of 120 ns:

t\$_{CS}\$ + t\$_{SPW}\$ - t\$_{DS}\$ + t\$_{PZH}\$ = 10 + 20 - 10 + 100 ns = 120 ns.

The actual time will depend on how fast you can generate the I/O signals, assuming you are going to be doing this in software with parallel GPIO port(s).

I would think the timing for most parallel switches would be in the same ballpark, certainly sub-microsecond.

\$\endgroup\$
4
  • \$\begingroup\$ A complete load of all X(n) to any Y(m) would take 7 times this, of course. \$\endgroup\$ Commented Jul 18, 2016 at 11:02
  • \$\begingroup\$ @PeterSmith Why is that? The X and Y addresses are both loaded in parallel. \$\endgroup\$
    – tcrosley
    Commented Jul 18, 2016 at 11:04
  • \$\begingroup\$ The X address and Y address select one each of the X rows and one of the Y columns on a per write basis. One write could (for instance) connect X0 to Y3. To also connect X1 to Y7 would need a new write. \$\endgroup\$ Commented Jul 18, 2016 at 11:07
  • \$\begingroup\$ @PeterSmith Agree, each connection will take the 120 ns, but I don't believe the OP stated how many connections he needs to make at one time. I have updated my answer to say "each connection". Thx. \$\endgroup\$
    – tcrosley
    Commented Jul 18, 2016 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.