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Theres ususally a current source coming out of it....just leave it open.
If you ground it, it will seem like an infinite capacitor...and the part will never get going.
If you leave it open...then its just a capacitor of vanishingly small value......still will be a capacitor though...just very small air dielectric capacitor!
May we be struck down in flames if we dare to think we can make a circuit without stray capacitors being all over the place.
I once saw a circuit in which there was a resistor with a value of 10K between pin 8 (softstart) and pin 16 (Vref). I spent weeks wondering why the use of this resistor, I could never understand. Does anyone have the idea why they put such a resistor?
I once saw a circuit in which there was a resistor with a value of 10K between pin 8 (softstart) and pin 16 (Vref). I spent weeks wondering why the use of this resistor, I could never understand. Does anyone have the idea why they put such a resistor?
If you are leaving pin 8 floating, the IC will start up “immediately“, as there is only the capacitance of the pin and the pad which has to be charged by the current source. If you ground pin 8, the IC is turned off, as the resulting voltage level at pin 8 is LOW, see page 7 of the datasheet. As already mentioned by @dick_freebird, the 10k resisor will lead to a even faster turn on of the IC (no discrete capacitor), as additinal current is provided to charge the pin & pad capacitances. Practically, a HIGH voltage level is applied at pin 8 immediately.
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