Why does reverse supply polarity damage the ICs?

Question

Assuming you need for example an OR gate for your circuit project, you can use a 4071 IC or simply use two transistors as shown below:

But I wonder why the ICs are very sensitive to reverse supply polarity (as any other IC) but the transistor version circuit is not sensitive to supply reverse polarity connection (no damage.)

Note: I found another question which asked about what causes damage when the car battery is connected in reverse (but his question is about the car fuses not logic ICs): Why reverse polarity causes damage

score 3 · Accepted Answer · answered Sep 19 '19 at 14:48

The 4000 series ICs (and most modern ICs) use some form of CMOS topology. This use complementary MOSFETs to drive high and low, rather than the resistor-BJT arrangement shown in your diagram.

The problem with MOSFETs when it comes to reverse polarity is the body diodes within their structure. If we look at the structure of say a simple inverter, we end up with something like this:

The diodes aren't added intentionally, but rather come about due to the structure of a MOSFET (*).

These diodes aren't disastrous issue (**) in a CMOS gate if the source of the PMOS (to transistor) is at a higher voltage than the source of the NMOS (bottom transistor). The diodes are always reverse biased, so pose no threat.

If however we were to reverse the polarity and connect the PMOS source to a negative voltage with respect to the NMOS gate, then clearly there is now a direct current path through both body diodes. This causes a direct short-circuit across the power supply. Regardless of whether the transistors are switched on or off, a large current will flow, causing excess heating and damage to the circuit.

(*) You can search further for MOSFET body/parasitic diodes. (**) Ignoring possible latch-up issues.

OK. But the diodes across mosfet is for removing back emf when the load is inductor such as motor or relay, but for the mosfets of the logic ICs is it really necessary? — M.A.K, Sep 19 '19 at 16:01
@M.A.K That is a completely separate thing which involves adding large external diodes to perform a task. These diodes aren't *added* to do anything, they are an unwanted feature that exists due to the physical structure of the transistors (see the duplicate question). — Tom Carpenter, Sep 19 '19 at 16:08

score 2 · Answer 2 · answered Sep 19 '19 at 15:29

in the oft-used Nwell process (Pfets are in Nwells) with Nfets implanted directly on the P substrate, the Nwells must be tied to the most positive potential, or the various diode junctions become forward biased.

When the VDD and GND pins are reversed, given the large # of Nwell contacts and the large # of substrate contacts, there are hundreds if not thousands of diodes that become

forward biased

I recall a particular IC design-handbook I used that stated " a single contacts has 4,000 ohms of resistance".

Thus 4,000 contacts likely has ONE OHM of resistance, because of the plenteous intermixing of well-contacts and substrate-contacts.

score 1 · Answer 3 · answered Sep 19 '19 at 14:30

Generally, ICs contain protection diodes on pins which become forward biased in the event of a reversed supply. These are designed for only a few mA, but a reversed power supply is a low-impedance source that can deliver much more than that. The chip burns, whereas, a simple transistor circuit contains no such diodes and because of transistor symmetry, is not sensitive to reversed polarity.

Why does reverse supply polarity damage the ICs?

3 Answers3