# Ideal diode load switch turn-on speed?

Hi,

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.

• Hello Petec,

Your observation is very astute. While it may not take 34ms for the FETs to turn on, based on my simulations, it will take somewhere between 10-15ms for the LTC4359 to be able to drive MOSFETs with that high of a gate charge. One think you could try to do is to use a skyhook circuit with a flyback converter to provide pullup on the gate pin.

Regarding the 100k ohm resistor - the gate pulls down to SOURCE. If the 100k ohm resistor was not present, then during every turn on cycle the gate would potentially travel from 0V to whatever threshold is required to enhance the FET. With the 100kohm resistor present, then the gate only has to travel the threshold distance of the FET to enhance it.