I have a 5.14V @ 2.5A wall wart connected to the bq24075EVM and a 2600mAh Li-Ion battery. I need to be able to drive a 2A load. I see that the AC path (Iin) can only provide 1.5A and that the battery can supplements the rest (up to a theoretical 4.5A).
At 0A load, Vout = 5.10
At 2.47A load (1 Ohm resistor), Vout = 3.67V
AT 1.5A load (2 Ohm resistor), Vout = 4.06V
Is Vout being averaged between Vac and Vbat?
What should my theoretical Vout be under these conditions?
When the battery is out and running a 5.14V at the Power path, I get Vout = 3.5V @ 1.44A load (2 Ohm resistor).
Shouldn't I be getting Vout ~= 5V at Iinmax?
The input should provide all the current until the OUT pin drops below the battery voltage.
When the system load increases and pulls OUT to 4.3V the charging current is reduced if pull lower, the charge current will be zero, the system current will be exceeding the input current and the OUT pin will drop below the battery and the BAT FET will turn on (only once the OUT voltage is below the battery votlage).
The input FET has a certain RDSon and that drops the voltage from the input.
You probably have an combination of things happening: 1) what is the input voltage at the IC; (2) what is the OUT voltage; (3) Subtract Vout from Vin and divide by the input current to get the RDSon of the input FET; (4) your load resistors are also changing values with the higher current (3.5V/1.44A = 2.43 ohms).
The RDSon is the main reason for the voltage drop between the input and output and is a function of the temperature of the IC....is it heat sunk to 2 oz copper (>2in^2), with multiple vias?
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