Where does CMOS output short circuit limiting come from?

[Euler's Identity]

shorting cmos output

In a recent experiment I performed after fielding another's question in another thread (CMOS NAND), I wired a cd4001BE NOR gate as follows.

Vdd = 6V/5A source
Vss = 0v
A=B=X=Vss

To my surprise I found that the current drawn from the supply was only around 8mA.

6v/8mA=0.75x10^3 = 750Ω

Where did that 750Ω limitng resistor come from? In my research, the highest resistor I've found is 15-30Ω + Rds_on in series with a CMOS output (via cd4007 datasheet.)

What gives? Is an on p-channel MOSFET really 720Ω? (According to a cd4001 datasheet, the 1/4 cd4001 is actually a NAND with inverters on its inputs and output. So, it's really just a single on p-channel device when its inputs are low, not a series.)

I'm perplexed...

Added after 4 hours 7 minutes:

I'm still perplexed, been digging and digging.

Does anyone know?

 

[10kangstroms]

If you look at the CD4001 datasheet, you will see the pullup characteristics (see below). This is consistent with your measurement.

 

[Euler's Identity]

Thank you. I did see that graph, in miniature mind you, but I saw it. Likewise, I saw the internal schematic.

From these things and, as well, my experiment, I could only conclude that an on p-channel MOSFET is indeed over 700Ω from drain to source.

Isn't that rather high for a full on device??

Are we sure there isn't more to the internal circuit, something that's adding R?

 

[10kangstroms]

Thank you. I did see that graph, in miniature mind you, but I saw it. Likewise, I saw the internal schematic.

From these things and, as well, my experiment, I could only conclude that an on p-channel MOSFET is indeed over 700Ω from drain to source.

Isn't that rather high for a full on device??

Are we sure there isn't more to the internal circuit, something that's adding R?

Logic gates aren't designed to drive relays and motors - they're designed to drive other logic gates. When you consider that the input capacitance of a 4000 series gate is typically 5pF, a fanout of ten (50pF) would yield a risetime of around 50ns - plenty fast for the specs of the CD4000 series.

 

[Euler's Identity]

Vo ---- R ----- Vin ------ C ------ Vss

Vin = Vo(1 - e^-(t/τ))

Vin
----
Vo

=

1 - e^-(t/τ)

so

e^-(t/τ)

=

1 - Vin/Vo

inverting

-t/τ

=

ln(1 - Vin/Vo)

Therefore, w/R=750Ω and C=Cin=50pF,

-t = 750Ω*50pF*ln(1 - 2.5/5) = 37.5ns*-0.693 = -26ns

Ok, I'll buy it.

I never realized Ron was that large. Thanks!