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Consider the classic "pulse stretcher" circuit shown in Fairchild AN-140, Figure 12.

When the first gate drives low, the instantaneous capacitor discharge current will easily exceed the absolute maximum value for the gate sink current. Of course, the absolute maximum value is normally a DC rating, and the gate can actually withstand much higher currents for short periods. I guess this is the effect which this circuit relies on for successful long-term operation.

How can I determine if it's safe to instantaneously discharge 100nF 5V through an HC CMOS gate without a current-limiting resistor? I guess that is what the author is trying to establish with the "BE SURE THAT" equation below Figure 12. However, I don't understand why that equation is dependent on "t", and it seems to fail for any reasonable values of "C" anyway.

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  • \$\begingroup\$ Nothing instantaneously discharges unless someone can prove time is granular. \$\endgroup\$ – Andy aka Feb 7 '17 at 20:53
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    \$\begingroup\$ Yes, obviously it's limited by the gate resistance, capacitor ESR, etc... But I think you understand the premise - I'm trying to establish whether it's necessary to add a current-limiting resistor. \$\endgroup\$ – The Whole Hog Feb 7 '17 at 21:09
  • \$\begingroup\$ The equation depends on \$t\$ (the pulse length) because you need more capacitance for longer output pulses, the cap has to store the more energy the longer the output pulse should be. \$\endgroup\$ – auoa Feb 7 '17 at 21:14
  • \$\begingroup\$ Figure 31 of TI's SZZA008 shows that the typical current through a shorted HC output is about 45 mA or 75 mA. That's nothing. \$\endgroup\$ – CL. Feb 7 '17 at 21:36
  • \$\begingroup\$ In other words, 45 mA * 5 V = 225 mW; with an assumed package thermal impedance of 100 °C/W, this would result in a temperature increase of 22.5 °C (which would be safe even for a continuous current). \$\endgroup\$ – CL. Feb 7 '17 at 21:45
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This is a SAFE_OPERATING_AREA (pulse) question. The thermal timeconstant of silicon, at 1millimeter thickness, is 11.4 milliSeconds. If the heating event is at or shorter than 11.4milliSeconds, the heat remains within the die, not even having reached the metal flag/paddle under the die.

How fast does the 0.1UF discharge? The Requiv is 5v/50mA ==100 ohms. The RC timeconstant is 100 ohm * 0.1uF, or 10 microSeconds. In that time, the heat mostly remains within the top 30 microns of the silicon, and is at least 70 microns away from the flag/paddle. [10u cube tau 1.14uS, 100u cube tau 114 uS]

Lets use a volume of 30*30*30 microns, or roughtly 30,000 cubic microns. Using 1.6 picoJoule/micron^3 * Deg C, scaled by 30,000 we find the specific heat of our cube: 50,000 picoJoule, or 50 nanoJoule/30micron^3 * degC.

The energy stored in the cap is 1/2 * C *V^2 = 1.25 microJoules. Divide this by 50 nanoJoule, and we get 25 degree C rise, during that 10uS discharge.

For this answer, with all the heat trapped inside the silicon, the heat has not had time to even reach the bottom flag/paddle.

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How can I determine if it's safe to instantaneously discharge 100nF 5V through an HC CMOS gate without a current-limiting resistor?

The capacitor will get discharged through the resistance of the MOSFET so the discharge will not be instantaneous.

If I were to discharge something over a few uf, I would use a resistor. Otherwise it is fine to go naked.

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