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I have studied that regulators like 7805 are used to remove the ripples present in the DC power supplies. But while searching for information about 7805, I came across the circuit below. The circuit shows an LM7805 converting 9V supply to 5V. This is something like stepping down the voltage. can a DC regulator really step down a DC voltage (because I've studied only about their ripple removal function)?

LM7805

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3 Answers 3

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Yes they can. They sort-of put themselves between the IN and OUT pins as a resistor. They watch the out pin. If the voltage is > 5V they make the resistor value higher, if the output voltage is < 5V they lower the resistor value. Thus they maintain the 5V at the output.

Like a real resistor, the difference between input and output voltage (multiplied by the current supplied by the OUT pin) is dissipated as heat. Hence such a linear regulator is not very power-efficient (a switched mode regulator is more complex, but much more power efficient).

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I would say that the primary purpose of linear regulators like the 7805 is to reduce an unregulated power supply output to a fixed voltage - removing the ripple comes as a byproduct of the voltage regulation.

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  • \$\begingroup\$ How do you define an unregulated power supply? \$\endgroup\$ Dec 24, 2013 at 17:18
  • \$\begingroup\$ An unregulated power supply is one whose output voltage changes in proportion to its input voltage. \$\endgroup\$
    – Dave Tweed
    Dec 24, 2013 at 17:24
  • \$\begingroup\$ ...or because the load current changes. \$\endgroup\$
    – Joe Hass
    Dec 24, 2013 at 18:23
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A simple voltage divider can "step-down" a voltage. For example:

schematic

simulate this circuit – Schematic created using CircuitLab

However, the 5V output of the circuit above is unregulated. This means that

(1) the output voltage will vary in proportion to the input voltage

(2) the output voltage will vary with output current

Now, imagine that R1 in the schematic above were variable and controlled in such a way that, within a certain range of input voltage and output current, the output would be 5V regardless.

A simple implementation of such an idea is:

enter image description here

The transistor and base bias circuit form a dynamic voltage divider with the load resistance, "dropping" the input voltage down to the desired output voltage and maintaining that voltage relatively constant with respect to variations in input voltage and output current.

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