Electrical – LDO regulator below minimum input voltage

battery-operatedldolow-power

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?

Best Answer

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

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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.

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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.

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