can you tell me how this circuit work when input voltage from 2.7V~4.5V but constant output 3.3V, as we know ,Charge Pump can increase output .but when input is 3.3V~4.5V,how to work?
This DC-DC converter IC is regarded a LDO+Charge Pump,when input is greater than output by
about 100mV it operates as a low dropout regulator(LDO),when output is greater than input it
automatically switches into charge pump mode to boost input to the regulated output voltage.
This DC-DC converter IC is regarded a LDO+Charge Pump,when input is greater than output by
about 100mV it operates as a low dropout regulator(LDO),when output is greater than input it
automatically switches into charge pump mode to boost input to the regulated output voltage.
hello Malcolm:
You're right, you can see the curve when the input voltage is lower than output, the device worked on charge pump mode and the efficiency is decrease as the input voltage increase. When the input voltage reach to V_out+100mV, the device changed to work on LDO mode, ang the efficiency is jump to higher and decrease as input voltage increase.