Converting Open-Drain to Active-Low

Question

My question is very similar to Converting Open-Drain to Active-High, however instead of an active high, I want an active low.

I am trying to interface TI BQ51050B (QI receiver + LiPo Charger) and TI TPS63051 (3.3V Buck-Boost) with a 3.7V LiPo. The BQ CHG pin is an open drain output pin, and the TPS EN pin is an input (1: enabled, 0: disabled).

My thinking is that when the battery is charging(CHG-> GND), I don't want it to be attached to any load(EN->LOW), and when the battery isn't charging(CHG->Hi-Z), It should be attached to the load(EN->HIGH).

Input(CHG)	Output(EN)
GND	LOW
Hi-Z	HIGH

I was thinking of just using the circuit in the previously mentioned post with a NOT gate attached to the output which would give:

^{simulate this circuit – Schematic created using CircuitLab}

Using the CircuitLab simulation, I get the desired results on EN (3.7V when CHG is Hi-Z, 11mV when CHG is shorted to GND).

However, I was wondering if there is a better way to do this. If this is the best way, would the NXP BCW29 work for Q2 and the NXP PMBT3904 work for Q3?

Answer 1

An open drain just means that a MOSFET internally connects the pin to ground inside the IC:

^{simulate this circuit – Schematic created using CircuitLab}

The output being Hi-Z just means that the MOSFET is switched off. When the MOSFET is switched on, the output is pulled to ground.

If you want to turn the Hi-Z state to a logical HIGH, all you need is a pullup resistor:

^{simulate this circuit}

When the output is in a Hi-Z state, the MOSFET is off and the resistor pulls the output voltage up to 3.3V. When the output is in an active-low state, the MOSFET is on and pulls the output voltage down to 0V.