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Old 23rd August 2004, 04:09 AM   (permalink)
Default What does "low dropout" mean?

Hi,
I've looked for volatge regulator, and encountered term "low dropout volatge regulator" (for e.g. LM2940). What doe it mean actually?

Thanks,
banh is offline  
Old 23rd August 2004, 05:16 AM   (permalink)
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Its the voltage differential between the input and output voltage.
If, for example, you want to use a 5 volt regulator on a 6 volt battery. If the dropout voltage is 1.5 volts this is not goin to work as the regulator only gets 4.5 volts input.
So, a low dropout voltage regulator of, say, 0.5 volt dropout, would work in that example.

Hope tht explains it.

Klaus
Klaus is offline  
Old 23rd August 2004, 05:48 AM   (permalink)
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thanks,
so " dropout" is just sth we want to minimize correct, not a parameter?
banh is offline  
Old 23rd August 2004, 02:29 PM   (permalink)
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Quote:
Originally Posted by banh
thanks,
so " dropout" is just sth we want to minimize correct, not a parameter?
The dropout is a characteristic of the regulator. Your voltage source would have to satisfy this requirement of the regulator in order for the regulator to function properly.
checkmate is offline  
Old 23rd August 2004, 10:54 PM   (permalink)
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Another way to look at "dropout" is as the "headroom" that a given regulator requires. As has already been explained, it is the additional voltage over the desired output that must be available at the regulator input for the regulator to function properly. As to minimizing this value in circuit design (and as a general rule), the lower the input voltage is for linear regulators the better, so long as the input voltage is at least at the output voltage plus the headroom. Any additional increase in input voltage will need to be dropped by the regulator, resulting in additional heating there, and thus increasing the need for regulator cooling.
ChrisP is offline  
Old 24th August 2004, 05:49 AM   (permalink)
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thanks a lot
banh is offline  
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