Will I run into problems if I leave a few of the analog pins on a CMOS switch (such as the DG409) floating?

I know that CMOS digital inputs should never be left floating due to the fact that the transistors tend to both partially conduct (resulting in excessive power consumption and heat), but I'm not sure if this rule also applies to bidirectional analog pins, even if they use CMOS internally.

  • 1
    \$\begingroup\$ My prof. from university always says "If there's high impedance on a pin, don't let it floating". \$\endgroup\$ – Rohat Kılıç May 26 '17 at 8:57

The input of a digital CMOS gate is connected to the gates of (at least) a complementary pair of MOSFETs. If that input is floating the gates may have a potential which turns both transistors (partially) on, which must be avoided (see 1st subcircuit below).

This is not the case in an input/output of an analog CMOS switch (transmission gate). Here the I/Os are connected to the drains/sources of a complementary MOSFET pair and there is no danger of some harmful intermediary state (see 2ndt subcircuit below). Floating is no problem as long as min./max. voltage is not violated.

Of course the input controlling the CMOS switch has to be treated like any other digital CMOS input, i.e. don't let it float.


simulate this circuit – Schematic created using CircuitLab

(Note: in the 2nd subcircuits the substrate connections of the MOSFETs are not shown correctly as the schematic editor offers only MOSFETs with substrate connected to source; actually they are connected to V+ (P-MOSFET) and GND (N-MOSFET))


The DG408/9 data sheet says this: -

Signals on SX, DX or INX exceeding V+ or V- will be clamped by internal diodes

So, if S or D are left open-circuit I don't see much harm coming to them providing there is no chance of them receiving a current that might exceed the internal diode rating (30 mA).

Having said all of that, it's so easy to tie unused pins to a ground reference to "guarantee" the conditions on the pin that I can't see any reason not to.

  • \$\begingroup\$ The reason they can't be grounded is that I plan on using one of the switches to discharge a capacitor. When that mode it not selected, the switch needs to have no effect on the capacitor (i.e. it can't be connected to ground). \$\endgroup\$ – Caleb Reister May 27 '17 at 6:28
  • \$\begingroup\$ You need to check what the leakage current is of your analogue switch if you are using it in this type of circuit - it could easily recharge or discharge your capacitor. \$\endgroup\$ – Andy aka May 27 '17 at 8:07

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