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about losses in regulator

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why does i/p voltage of any regulator always more than its output? may be it is because of losses - but what kind of losses ????????
 

The input voltage of any linear regulator is higher than the output voltage for one simple reason. The regulator is acting like a large automatically adjusting variable resistor. This is all a linear regulator really does. Lots of people think it some kind of magic inside, but it really just acts like a resistor. This is why linear regulators always dissipate heat. The more current you pull from the regulator and the larger the differential between the input and the output the greater the heat dissipated.

On the spec sheets, you see that their must be a minimum differential between input and output. This is often called the drop-out voltage. If the input is less than (the output voltage + the drop-out voltage), then the regulator stops working properly and the output voltage will droop. This voltage is the internal bias voltages needed to get the transistors to act like the auto adjusting variable resistors. Newer designs reduce this drop-out voltage to less than 0.5 volts. Older designs often required 3 or more volts.
 

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