I'm working on a project that will be battery or USB powered. The system load is 3.3V, so I'll be running the input power sources through an LDO regulator. The battery will be a 110 mAh lithium polymer battery at ~3.7V. The design does have diodes on both inbound power sources leading up to the regulator, however, do I need to be concerned about leak current through the diode that will flow from the USB 5V source to the ~3.7V battery? I have attached a diagram to help illustrate the setup.
No, not at all.
What diode are you using ? you can check the datasheet for the leak current.
The 1N4007 has 5uA.
To charge a battery of 110mAh with 10% it would take 3 months. But the leak current of the battery is larger than that.
I think the voltage drop over the diode is your problem.
A li-ion battery will vary from 4.2 down to 3.0 or lower.
With a normal diode you need 3.3 (Vout) + 0.6 (0.5 or 0.6 diode voltage drop) + 160mV (LDO voltage drop) = 4.0 V
So if your battery is still almost full at 4.0 V, the circuit stops.
It appears to be somewhere in the nano-Amp range for voltages below 20V if I'm reading that right. My application will normally run at about 10mA-35mA with occasional bursts to 135mA when writing to an SD memory card.
Thanks for the tip on the Schottky diodes. That looks like a nice improvement to the circuit to get a little more out of that battery. I'm looking at this one:
At 135mA the voltage drop over the BAT30 is between 500 and 580 mV. That's very large for your circuit.
You could use mosfet switches to simulate diodes.
There are also ic's for switching voltages without voltage drop. Search for "ideal diode".
For example the LTC4411 is made for a situation as you have !
As a bonus you get a status line telling which power source is used at the moment.
Thanks for turning me on to those mosfet switches. Those would certainly be "ideal", but the cost is quite a bit more than the alternative and I don't think my budget will allow for it. They are certainly a great product though.
