FT232R, 0.5V voltage drop over MOSFET?

I have made a circuit based on the FT232R data sheet example "Figure 6.3 Bus Powered with Power Switching Configuration". Basicly it uses a IRLML6402 (P-Channel MOSFET) tp power up the rest of the circuit once the FT232R has request > 100mA current.

The voltage drop over the IRLML6402 is 0.5V, I really require a voltage drop of 0.25V or less. There must be a better alternative to this? 0.5V seems higher then what I expected from the IRLML6402 recommended by FTDI.

The circuit draws around 250mA and can also be self powered. I noticed that while in self powered mode, the the IRLML6402 doesn't seem to stop current going through the reverse direction back to the USB VCC. For some reason, I expected the MOSFET to act as a diode and stop current flowing back though to USB VCC when in self powered mode.

Any help or advice would be greatly appreciated.

I rebuilt the circuit and it worked fine (so, I must have just had something the wrong way around) with only around 30mV voltage drop (does that sound about right?).

I modified the cct diagram out of the FT232R data sheet and added in back to back MOSFETS.

I am trying to have two MOSFETS act as diodes to protect each supply from driving the other.

The 5V VCC also powers the rest of the circuit.

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I don't understand how to manage a 0.5V voltage drop across IRLML6402 in regular circuit operation. The schematic and datasheet can't answer it, only a measurement in your circuit.

I am trying to have two MOSFETS act as diodes to protect each supply from driving the other.

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But it doesn't work this way. By supplying 5V to one side, both FETs will turn on completely.

Thanks, just trying to understand why this wouldn't work, the above diagram and the one in this post should be equivalent except one uses MOSFETS to do the job of the two diodes right?
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the above diagram and the one in this post should be equivalent except one uses MOSFETS to do the job of the two diodes right?

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No. The MOSFET shorts the diode for Vcc > Vgs,th. So if one side is supplied with a voltage above 1 - 1.5 V, both MOSFETs turn on.

Thanks,
So the second diagram with the diodes is ok, but the first diagram with MOSFETS in place of diodes is not ok?

I found this on the internet showing MOSFETS being used in place of diodes, does this look correct?
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All diagrams are O.K., e.g. for a low drop battery reversal protection. But they can't be used for a two-way diode network, because the MOSFET will also turn on, if the voltage is supplied at the load side.

Thanks, I finally get it .

So do you think I should scrap them 2 MOSFETS and should just use 2 diodes like that in the second diagram?

I have tried Schottky diodes (SS12 E3 **broken link removed** to be exact) in my circuit, the voltage drop I get is around 0.3V - 0.4V, I also tried 1N5817 Schottky diodes giving around 0.3V voltage drop. The circuit is used to program 5V Flash ROMS, to erase and program the Flash ROMS, voltage can not full below 4.5V.

As the USB voltage can go as low a 4.75V, the maximum voltage drop I can have is 0.25V (which would be right on the borderline).

Can anyone suggest an alternative diode that may be suitable? or do I have to use the MOSFETS (but in some other configuration)?

Maybe my best option is to just use the 1 MOSFET for switching the USB VCC and 2 Diodes for power source selection, as long as I can find a diode with a voltage drop of less than 0.25V @ 250mA.

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I don't think, that you'll find a suitable low drop solution with diodes. Generally, it's not impossible to use "zero drop" transistor switches, but the generation of the gate signal would be more complicated than just connecting gate to ground. You should get a reliable information about the active supply source and controld both switches accordingly.

This diode appears to have voltage drop of less than 0.25V @ 250mA.

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I imagine this is the best diode I will find.