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Subject: 
Re: Using the DC to DC converter
Newsgroups: 
lugnet.robotics.handyboard
Date: 
Thu, 18 Mar 1999 13:14:24 GMT
Original-From: 
FThompson9@^avoidspam^aol.com
Viewed: 
969 times
  
In a message dated 3/18/99 4:56:07 AM Central Standard Time,
philchia@singnet.com.sg writes:

The sensor's current consumption maximum is  35mA.
If the output Watt of the DC-DC convertor at the 12v source is 1W, can I
said that the the current is   1/12 = 83mA   (P=VI)  ?


Yes, P=VI.  When the Specification sheet says the converter can source 1W,
they mean at most 1 Watt.  When your sensor says it consumes 35mA, this is
again the max and only if the 12V is supplied.  If less than 12V is supplied
it will probably draw less current, and it probably will not work.  When you
connect the 12V DC-DC converter to the sensors, they will only draw 35 mA from
the converter, this only means that the converter will run cooler and will not
exceed its maximum.  You don't have to do anything else.  If your converter is
100% efficient, it will draw (35mA * (12V/5V)) = 84mA) from the battery.  It
will probably draw a little more from the battery since it probably is not
100% efficient.

My battery source is 5V,1.3Ah before going thru the convertor.

I have heard that if the batter is NiCd, you should charge at C/10
(capacity/10).  Your capacity is 1.3Ah so you charge at 130mA.  I have also
heard that it is best to discharge at this rate, although NiCds will supply a
great deal of current if you let them.  With some 100mA going to your
converter, this leaves 30mA for the rest of the circuit.  If your 5V stuff
draws more, it will still work, but it will reduce the number of
charge/discharge cycles on the NiCd.  If you wish to consider a separate 12V
battery for your sensor, you only need to have a common ground for everything
to work.

Pherd



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