Saturday, July 6, 2013

Browse » home» » » » » » Simple 9V battery replacement circuit

Simple 9V battery replacement circuit


This power supply circuit contains two charge-pump DC-DC converters which can delivery up to -10V for a 2,0V to 3,6V input voltage battery. For a current consumption of 7mA the output voltage it will drop to 9V, but this is enough to power a portable handheld instrument. The overall efficiency it will be up to 74% for a 2V input voltage and 60% for a 3.6V lithium polymer battery cell.

First charge-pump DC-DC converter based on MAX 619 from Maxim , C70851 smd code , delivers a regulated 5V +/- 4% output at max. 50mA. To maintain the greatest efficiency over the entire input voltage range, the MAX619s internal charge pump operates as a voltage doubler when input voltage ranges from 3.0V to 3.6V, and as a voltage tripler when it ranges from 2.0V to 2.5V.

The second charge pump converter use a TC682 circuit from Microchip and it provides an inverted doubled (-10V) output from a single positive supply (+5V regulated). An on-board 12kHz (typical) oscillator frequency provides the clock and only 3 external capacitors are required for full circuit implementation. Low output source impedance (typically 140Ω), provides output current up to 10mA.

10V DC-DC converter placement of circuit componentsComponents placement on printed circuit board
More voltage convertors:
    5V dc-dc converter this circuit can deliver over 1.6A at 5V and still work at 2.0V
    Max761 boost converter module from 5V to 13.5V or 12V, ideal for Flash Memory Programming

No comments:

Post a Comment

Note: Only a member of this blog may post a comment.