0
\$\begingroup\$

I'm designing a circuit that will run from 3V battery. The battery can get as low as 2V and I want to use it as much as I can.

I'm using a PIC24 that operates with the integrated regulator for VDDCORE disabled with a VDD from 2.0V to 3.6V but VDDCORE goes from 2.0V to 2.75V max, so I can't power the microcontroller directly from the battery as I will be exceeding the VDDCORE allowed voltage.

I'm planning to use an LDO as the MCP1703 with an output voltage of 2.5V to power the microcontroller VDD and VDDCORE. My current consumption won't be higher than 50mA. My question is, when the battery voltage goes below 2.5V (plus the dropout voltage) will the regulator act as a switch bypassing the input voltage to the output plus a dropout? How high would be the dropout?

\$\endgroup\$
2
  • \$\begingroup\$ If the data sheet doesn't tell you then contact the supplier. Some regulator data sheets will show the drop in output voltage when the input is below the normal level - usually in a graph but i don't think this device does. \$\endgroup\$
    – Andy aka
    May 1, 2016 at 16:56
  • \$\begingroup\$ I couldn't find it in the datasheet that's why I'm asking :/ \$\endgroup\$
    – Andres
    May 1, 2016 at 16:57

2 Answers 2

Reset to default
2
\$\begingroup\$

MCP1703 datasheet figure 2-31 shows the output tracking the input down to well below 2.0V.

enter image description here

However this is at a load current of only 1mA. Figure 2-17 shows short circuit current vs input voltage. Here we see that it can only deliver a few milliamps at 2V.

enter image description here

This poor low voltage performance is probably caused by the pass FET having a relatively high Gate threshold voltage, which may be an inherent result of its 16V rating.

I would use the MCP1700, which is specified for 200mA output at 2.3V. It can only handle 6V maximum, but this is well above your battery voltage.

\$\endgroup\$
2
  • \$\begingroup\$ I have trouble finding similar figures in the MCP1700 datasheet. Where do I see the 200 mA value when using a MCP1700-250? In the Features Section of its datasheet I see '200 mA Output Current for Output Voltages < 2.5V' - is that what you are referring to? And does it apply in general, i.e. also when the regulator is in its dropout zone? \$\endgroup\$
    – maxschlepzig
    Feb 27, 2022 at 10:28
  • \$\begingroup\$ The MCP1700 is rated for operation down to 2.2V when regulating at 1.2V, with load current up to 200mA. Unfortunately it doesn't show dropout voltage vs load current at 2.2V, but it can't be >1V at 200mA. Assuming linear resistance that means you should expect <0.2V drop at 50mA down to 2.2V input, ie. 2.2V in, 2.0V out. \$\endgroup\$
    – Bruce Abbott
    Feb 28, 2022 at 3:42
0
\$\begingroup\$

I've designed stuff that's battery powered and I think a boost converter would allow you to use a lot more of the battery up before your circuit stops working. I was successfully able to use the NCP1402. It's available with a 2.7 volt output. With it, you could reconfigure your project to operate on 1.5 volt power - perhaps two 1.5 volt cells in parallel rather than series if that's how you're powering it currently.

To directly answer your question, though, the MCP1703 datasheet shows a minimum input voltage requirement of 2.7 volts - which implies that it won't function properly (what that actually means is unknown) below that.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ It very much depends on current. Boost converters usually have relatively bad efficiency at low currents (e.g. a few dozen μA). A LDO with its 16.6% voltage drop (-> 83% efficiency) will probably be better and cheaper for this application. \$\endgroup\$
    – Michael
    May 1, 2016 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.