From: Jonathan Kirwan
Subject: Re: Question about 6V (4 X 1.5V battery) to 5V for MCU operation
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NNTP-Posting-Date: Mon, 16 Dec 2002 02:53:11 GMT
Organization: AT&T Broadband
Date: Mon, 16 Dec 2002 02:53:51 GMT
On Sun, 15 Dec 2002 18:31:28 -0800, Jonathan Kirwan
>On Mon, 16 Dec 2002 09:28:27 +0800, "Reala" <-> wrote:
>>I would like to design a device with 4 batteries for power source.
>>It is about 1.5V X 4 = 6V. Normally, MCU runs at 5V.
>>What is the normal method for this issue?
>>Should I add a diode IN4001 for voltage drop (6V - 1V =5V)?
>>Is this suitable enough for simple application? Any idea?
>>or we do not need to concern this issue because of the internal resister of
>>(The actually voltage of the battery is about 1.3V....)
>If you select an LDO (and I think a few can operate at under
>100mV of headroom, perhaps even 50mV), you can operate down to
>5.1V or even a little less on the series battery. That'll get
>you more than 1/2 of their energy. If you are looking for
>operating down to say 3.2V to get all of the energy drained out,
>you'll need a buck/boost switcher. I think maxim has one. And
>they also have an apnote with a schematic for a low quiescent
>current design for one, as well, using the MAX539 and an HC132
>(I think it is TA-14, Figure 6.) It delivers 5V.
Found the apnote at:
Look at Figure 6, there.