I am learning to design buck converter. And using LM70880-Q1 buck. There is power good pin is available. I want to make use of this pin. It is a open collector output that goes low if the Vout is outside of the range.similaly vice versa. How calculate the pull up resistor for this power good pin? Available data: VCC IS 3.3V and V_OL-PG :0.4 (TEST CONDITION:open collector,I_PG=2mA)
2 Answers
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The ratings are given at 2mA current. You have current, voltage, and need resistance, so use Ohm's law.
You can either use that value, or some other value. The point is you have to be within range in respect to currents and volages, for this chip and the next chip that utilizes the PG signal.
You have to be between maximum current where you put a smaller resistance, or if you want to use less current with higher resistance, you may be limited by leakage current of your chips or rise time from the resistance and capacitances combined.
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Available data: VCC IS 3.3V and V_OL-PG :0.4 (TEST CONDITION:open collector,I_PG=2mA)
Where did you get this information? The datasheet on the webpage is suspiciously empty.
The numbers tell you that at 3.3V and if you sink 2 mA into the pin, then the voltage at the pin is guaranteed to be at most 0.4V. I.e. the lowest possible resistor is $$R = U / I = (3.3\rm V-0.4\rm V)/2\rm{mA} = 1.45 \rm k\Omega$$ You might use an even lower resistor until you reach the maximum current of the pin (should be given somewhere else in the datasheet) but at this point the voltage can be higher than 0.4V.
Another piece of information can be found on the evaluation board: They use a 100kΩ resistor.
So, if you just want to connect the pin to read the PG status, e.g. with a multimeter, use any kΩ resistor at hand. 10 kΩ is usually a sane value unless you have special requirements (rise time, current, power consumption, ...). If you want to connect an LED, make sure that you adjust its current to be below 2 mA to stay within specs.
