I've been looking into different Analog Switch ICs and it seems they always allow a switching voltage between GND-VCC. Now the voltage I want to switch comes from a different source than VCC. Generally it should be aprox. the same as VCC but the exact voltage is always unkown and can be between 4.5-5.5v. Vcc is rated to be 5v. All switches I found are rated for a max. VCC of 5.5v or higher.

Now does it pose a problem if my switching voltage is slightly higher than VCC in some cases? I'm a bit confused that VCC is the limit when these ICs are just using MOSFETs internally.

The grounds of both voltage source are of course connected.

  • \$\begingroup\$ The datasheet will tell you \$\endgroup\$
    – PlasmaHH
    Commented Aug 10, 2015 at 14:28
  • \$\begingroup\$ Well the max. ratings usually state "VCC = 0.8v-5.5v" for the switching voltage. Which doesn't make it very clear to me wether 5.5v is the limit or actual VCC. \$\endgroup\$
    – timonsku
    Commented Aug 10, 2015 at 14:30
  • \$\begingroup\$ there seems to be a missing in the title \$\endgroup\$
    – user16324
    Commented Aug 10, 2015 at 14:32
  • 2
    \$\begingroup\$ Just be advised if the signal you are switching is applied before power to the analog switch. That may result in something called SCR latchup, which is bad. Alternately, can you use something like a BAT54 to clamp the input voltage to Vcc-0.6? \$\endgroup\$
    – cowboydan
    Commented Aug 10, 2015 at 14:52
  • \$\begingroup\$ Thanks I didn't know that. Yea I thought about a zener, though I would have liked to use as little as components as possible. \$\endgroup\$
    – timonsku
    Commented Aug 11, 2015 at 18:25

1 Answer 1


Of course it affects the signal if it is above Vcc.

Why do you think the fact that "these ICs are just using MOSFETs internally" is a reason that Vcc shouldn't be the upper limit for the signal?

Image borrowed from Maxim application note 5299:

Equivalent schematic of analogue switch

The surrounding n-substrate (or n-well) of each P-MOSFETs is connected to Vcc (see the "body" connection on the diagram above). So each p-MOSFET forms a diode from the p-channel to Vcc. If the signal is higher than Vcc + 0.6V the diode gets forward biased (turns on; i.e. signal current flows to Vcc).

Vcc + ca. 0.5V is already too high to have no current flowing through the diodes.

In addition there's probably a pair of Schottky-diodes for protection at each input and output: one going to Vcc and one coming from GND. Their forward voltage is even lower: 0.3V.

  • \$\begingroup\$ Thanks a lot for the detailed explanation! I was under the impression that they just pack up MOSFETs and include some protection. Only looked briefly into the schematic. \$\endgroup\$
    – timonsku
    Commented Aug 11, 2015 at 18:31

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.