I'm using this 4-channel ADC: MCP3422 -> Datasheet

I only need to use 3 of the available 4 input channels in total. Normally I would add footprints for pulldown resistors for example in case a floating condition causes problems. But this time I need the PCB space so I would like to free up some space.

So can I leave the two pins of one input channel floating? I need channel 1-3, but will leave both pins of the 4th channel unconnected. Can that cause issues? Any best practices?



  • \$\begingroup\$ Shouldn't cause any problems. Best practice, afaik, is to connect both of them to a known potential, either Vss or the midpoint of Vss and Vdd, if you have that reference voltage already available. Or, hey, maybe you could use the spare channel for something--monitor the circuit's input voltage perhaps? \$\endgroup\$ – Hearth Jan 30 '19 at 2:01

You should tie the pins to a known potential. From section 3.1 of the datasheet:

3.1 Analog Inputs

(CHn+, CHn-) CHn+ and CHn- are differential input pins for channel n. The user can also connect CHn- pin to VSS for a single-ended operation. See Figure 6-4 for differential and single-ended connection examples.

The maximum voltage range on each differential input pin is from VSS-0.3V to VDD+0.3V. Any voltage below or above this range will cause leakage currents through the Electrostatic Discharge (ESD) diodes at the input pins. This ESD current can cause unexpected performance of the device. The input voltage at the input pins should be within the specified operating range defined in Section 1.0 “Electrical Characteristics” and Section 4.0 “Description of Device Operation”.

Italic section is my emphasis. Leaving the pins floating implies that the potential could float to other ranges. Best to tie them to VSS or VDD to ensure they stay at a known level, and within the maximum range specified on the datasheet.

  • \$\begingroup\$ Thank you for your answer. So connecting these pins directly to VSS=GND, without a resistor can be done without a problem? \$\endgroup\$ – Henry Jan 30 '19 at 2:08
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    \$\begingroup\$ @Henry Yes, I don't see any reason that would cause any problems. \$\endgroup\$ – Shamtam Jan 30 '19 at 2:27
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    \$\begingroup\$ Never leave CMOS inputs floating. Unless you feel lucky (lol) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 30 '19 at 2:49
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    \$\begingroup\$ Floating pins will upset the internal amplifier, putting the opamp's output transistors into max UP or Max DOWN. FETs should quickly recover. Comparators, with massive differential-drives, can have tunneling thru the oxide into the underlying bulk and cause offsets that may linger. So tie them off. \$\endgroup\$ – analogsystemsrf Jan 30 '19 at 3:18
  • \$\begingroup\$ @analogsystemsrf: I read somewhere that in this case (If I connect the pins to GND or VDD I have to do it thru a 1k ohm resistor). But I can connect the ADCs unused channel pins (or address pins) just directly to VDD or GND (a known potential)? This is common practice? \$\endgroup\$ – Henry Feb 4 '19 at 0:44

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