Hello,

I am working on a schematic for a USB mux with multiple voltage input sources.

Voltage can come from USB-C (V_USB1) or a separate input source to a LDO (+5V).


When USB-C (V_USB1) is powered, the U5 usb mux select pin should stay high. When +5V is supplied, voltage shouldn't go through D4 to V_USB1, so that U5 select pin would stay low. From my limited knowledge, the schottky diodes D1 and D4 that was chosen should protect voltage up to 10V, but I am getting 5 volts from (V_USB1) when only (+5V) is supplied. Am I missing anything here?



Thanks!

 

Am I missing anything here?

D4 is shorted or installed backwards

 

schottky diodes, have a good leakage current,
IF you put a load of say 10 mA on vusb_1, say a 500 ohm resistor to ground, does the circuit work ?

 

schottky diodes, have a good leakage current,

I suspected the same.
At 5 volt reverse voltage the leakage could be as high as 2ma or 2.5K ohms.
There is 4.4K load across the output of the V-USB1 supply. That would leave appx 3 volts leakage at that supply line.

 

I will test it with a 500ohm resistor as soon as I can. Meanwhile, should I use another diode for this application?

 

I will test it with a 500ohm resistor as soon as I can. Meanwhile, should I use another diode for this application?

Some have lower leakage, but not that much lower,
FYI, There are chips that provide this USB power switch and isolation, all in smd packages, have a look on distributors sites like digikey, rs components etc,

 

I have decided to go with the TPS2116, prioritizing +5V_M2 supply. Hope this works!

 

I'd suggest a pull up on the St pin, allows you to debug with a scope.
I assume you copied the reference design ti provide ?

 

Good idea. Yes I did.

 

When measuring the voltage of an otherwise open circuit you DO need to consider the current, or lack of current draw of the measuring device. If that schottky diode has a one megohm effective leakage resistance at 5 volts reversed bias, and your digital meter presents a ten megohm load, you will read close to five volts, But if you put a five kilohm resistor load, which would draw one milliamp from a 5 volt source, you will not read much voltage at all.
Only "perfect" diodes have ZERO reverse leakage.

 

When measuring the voltage of an otherwise open circuit you DO need to consider the current, or lack of current draw of the measuring device. If that schottky diode has a one megohm effective leakage resistance at 5 volts reversed bias, and your digital meter presents a ten megohm load, you will read close to five volts, But if you put a five kilohm resistor load, which would draw one milliamp from a 5 volt source, you will not read much voltage at all.
Only "perfect" diodes have ZERO reverse leakage.

Devices like the tps selected, are back to back fets,
They appear as an almost perfect diode,

 

I will test it with a 500ohm resistor as soon as I can. Meanwhile, should I use another diode for this application?

You might try a transistor-as-diode... it'll have a slightly higher threshold on forward voltage than a Schottky diode, but its reverse current will be extremely low (in the pA, for the right transistor).

I'm assuming D1 comes from a battery, and D4 comes from a USB wallwart? If so, you could leave D1 as a Schottky, and it'd trickle-charge the batteries back up to 5 V.

 

You might try a transistor-as-diode... it'll have a slightly higher threshold on forward voltage than a Schottky diode, but its reverse current will be extremely low (in the pA, for the right transistor).

I'm assuming D1 comes from a battery, and D4 comes from a USB wallwart? If so, you could leave D1 as a Schottky, and it'd trickle-charge the batteries back up to 5 V.

FYI, chips such as the tps2116 used , are 4 back to back fets, that perform the " ideal" diode function.