I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.

  • \$\begingroup\$ With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching. \$\endgroup\$ Feb 20, 2019 at 20:32

3 Answers 3


No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.

\$P_{in} = P_{out}\$

but for a linear regulator it looks like this:

\$V_{in} \times I_{in} = (V_{out} \times I_{out}) + [(V_{in} - V_{out}) \times I_{out}]\$

The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:

\$V_{in} \times I_{in} = V_{in} \times I_{out}\$


\$I_{in} = I_{out}\$


No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.

However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.

A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)

A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)

They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.

  • 3
    \$\begingroup\$ "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook \$\endgroup\$ Feb 21, 2019 at 10:16

since power = voltage * current.

It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.

In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.


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