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.
Electronic – Will linear voltage regulator step up current
linear-regulatorpower supply
Related Solutions
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).
I believe I may need a buck-boost however, because the components will need at least 1.8 V and 3.3 V, and my 3.6 V battery could drain below that
First hit on Linear Technology might be useful: -
Try going to LT to get an idea and use this parameter search engine.
for the 1.8 V rail it's 1.154 mA
Use a linear regulator - it's just not worth the effort I believe to use a switcher from 3V3 to 1V8. However, if current was over 50mA I might start considering a switcher.
Generally, you can assume a buck regulator will give you about 90% power efficiency near full load and a buck-boost will be about 85%. Ball-park figures.
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Best Answer
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}\$
Therefore:
\$I_{in} = I_{out}\$