I have a device that operates with 3.2 V MAX, and I wish to power it from a standard 3.7 Li-Ion battery. A fully charged battery=4.2V.
I used a simple LM317 circuit to drop voltage to 3.2V and it worked perfectly.
Now the problem is as the battery starts to drain, the output voltage starts to become lower, so subsequently the LM317 output voltage will be lower as well, for example
Batt=3.7V,
LM317=2.5V
In which this case the voltage is not enough to keep the device up and running.
So what I need is some sort of dynamic regulator to keep the output voltage fixed at 3.2V until the battery voltage drop to 3.2V as well.
Advice on good ways to achieve this would be appreciated.
Best Answer
Maybe something like this is what you really need: -
It's an LT3127 buck-boost regulator that can deliver the required output voltage (3.3 volts in the example above) from input voltages ranging from 2.9 volts to 5.5 volts. It can supply 1 amp too. The 320k and 182k resistors set the output voltage and, for example, with a 330k and a 220k, Vout will regulate at 3.0 volts.
Power efficiency is pretty good too: -
Because it's a switching converter, it doesn't burn off the excess input-output voltage as heat but rather takes less current than the load when the input voltage is higher than the output voltage. For instance, when supplying a load of 300 mA, the power out is (say) 3 volts x 0.3 amp = 0.9 watts. The efficiency implies that the input power will be 0.95 watts when the battery voltage is 4 volts. This means the current taken from the battery is only 240 mA (and not 300 mA+ as you would get with a linear regulator). This means an extended battery life.
Also, because the regulator will work at lower input supply voltages than the output voltage, battery life will be further extended but, take care not to over discharge your battery.
Here's a link to the search engine results from Linear Technology. Similar in principle to searching on TI's site too.