AnsweredAssumed Answered

Ideal diode load switch turn-on speed?

Question asked by petec on Jun 13, 2018
Latest reply on Jun 25, 2018 by Sal.Afzal



I am planning to use an ideal diode to reduce power dissipation, while having the ability to turn on/off the ideal diode, as a kind of load-switch.


The input voltage is somewhat lower(lets say 5% lower) than the output voltage, so the ideal diode wouldn't be on at all times. Only when a transient loads the output, it would drop enough to turn on the ideal diode and supply current doing the transient.


In 1-FET ideal diodes, the FET body diode handles the initial current, until the FET turns on and "shorts" the body diode. However, I would like to use a 2-FET solution, as I also need to be able to completely disconnect the input from the output.


I was looking to use the LTC4359 as shown in figure 6 in the datasheet(without C1 and R4), but I am concerned about the turn-on speed of the system. As far as I can see, the max gate current supplied by the LTC4359 is 10uA(typ). The total gate charge of the FQA140N10 and FDMS86101 is about 340nC(250nC+55nC), so with 10uA, it would take about 34ms to fully turn on the FETs. Since the body diodes are purposely opposite facing, they can't take care of the initial transient, before the FETs turn on.


In figure 16, a 100k resistor is placed across Q2(The switch-FET) in order to bias the source pin. How does this decrease gate ramp time? It lifts the Q2 source and gate to the level of its drain, but how does this reduce the turn-on time, when the gate-source voltage is still zero?  How can I estimate which turn-on time a certain value resistor will result in?


Thank you.