For this case:But the battery voltage also shifts to 9V (which should not be the case)
Pin 5 of LTC4412 has 3.47V
It must be connected as it is, otherwise the substrate diode would short the switch.Meanwhile shouldnt MOSFET source be connected to positive. Just a a bit curious why its done that way
As FvM said: the FET works in both directions. Don´t worry anymore.What worries me is the connection of battery at drain
Hi,
i hope this is not related to LTC4412....
There are a lot of issues.
* it will not fully switch ON battery --> a lot of dropout.
* when auxilary is ON, then battery current flows through bjt...
As FvM said: the FET works in both directions. Don´t worry anymore.
Klaus
I tried the simulation with a 1 ohm V1 battery source (high for a battery) and the circuit still did the switch-over properly.A requirement for successful switchover is a sufficient low battery impedance. It's of course achieved with a voltage source representing the battery.
The MOSFET ON resistance or battery impedance value is actually of little consequence as far as the circuit switch-over is concerned.Yes, with reasonable MOSFET parameters. As said, the controller needs to sense up to 32 mV difference voltage across the FET. If you select a MOSFET far below 10 mohm Rdson, 1 ohm battery impedance can be too much. The same with a much higher battery impedance.
I think the problem (as noted in my post #11) was the fact that the STAT line can typically only sink 10µA (6µA worst-case) which means the STAT line wasn't going to zero with the 470kΩ shown in the data sheet schematic with a 9V supply (apparently that value was designed for a less then 5V supply switch-over or the data sheet writer missed that IC limitation).Yes, I misunderstood the circuit operation. Thanks for insisting.
Means switchover would occur even with a very high impedance battery. There must be some hidden fault in the originally described circuit.
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