5
\$\begingroup\$

Can someone help me to identify this VISHAY IC? Tips on how to do this are also welcome :). I could not find anything by searching with the numbers on top of the IC. The IC is used in a MKS 649B Pressure Controller.

unkown VISHAY IC

\$\endgroup\$
  • 1
    \$\begingroup\$ You need to consider it in context. What sort of connections does it have? What do you observe on it in operation. And why do you care? \$\endgroup\$ – Chris Stratton Aug 4 at 12:54
5
\$\begingroup\$

Been searching the internet top to toe for an answer too, I think the best I could come up with alongside the H bridge was this Vishay resistor network ic. The naming on the datasheet doesn’t line up at all so not convinced I’m correct, but just thought I’d share.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ The PCB layout for the MKS649B is shown on page 39 of the manual: idealvac.com/files/manuals/MKS-649B-MAN.pdf This denotes the 24 pin IC as 'RN1' - so I suspect you are correct, it is a resistor network IC. Maybe something made as a 'special' with non-standard value resistors - hence no standard part number on the device \$\endgroup\$ – Yellow Yeti Aug 4 at 12:57
4
\$\begingroup\$

I cant comment as I don't have enough points, so I will submit this as an answer: closest I could find. It is an IC made by Vishay with the correct package (SOIC-24(W)) as far as I can guess. And I only found it by using different search terms with some terms between hyphens.

I guess this could be the correct IC, as the one I found is an H-bridge driver, so this one could be used to power the pump of the pressurizer.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Good shout, but not a SOIC24 package. SOIC packages are 1.27mm pitch - pin pitch on the IC shown is far too small. \$\endgroup\$ – Tom Carpenter Aug 4 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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