A friend of mine told me to put a diode between the LDO's GND pin and the circuit's GND. He said the LDO will regulate it's fixed voltage plus the diode drop. Is it true? How does it work? I am using the MCP1703 regulating 5V, but I wanted it to regulate 5.2V~5.3V (a Schottky drop), so doing so with a Shottky will work?

How does this affects the thermal dissipation calculations? Ot it's just a matter of changing the Vout in the formula?

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    \$\begingroup\$ You can do this, but it will not be very accurate and the output voltage will drift with temperature. Much better to just use an adjustable LDO, of which there's hundreds to choose from. \$\endgroup\$ – The Photon Jan 7 '13 at 5:01

Yes, if you raise the LDOs GND pin, the output will rise accordingly. This is because the LDO now see's a higher voltage between it's reference point and ground.

All a fixed regulator is, is a standard regulator with the programming resistive divider inside the IC, rather than outside it.

enter image description here

If you look at the above image (imagine the resistive divider is inside the IC), you can see if you raise the bottom of the Vref pin, the output voltage will have to rise in order to match the divider input of the comparator with the divider input. As Photon comments though, it's not so accurate though to use a diode (unless you use a more complex circuit which can compensate for temp variations, etc, but then you may as well buy an adjustable regulator)

In the LM7805 datasheet, there is an example of how to create and adjustable regulator by driving the reference input with an opamp:

Reg Fixed Adj

And a slightly simpler version which uses a fixed divider and no opamp:

Reg Fixed Adj 2

For this version, you can use the formula provided - if you use a 5V regulator and want 5.3V, then:

5V * (1 + (3kΩ / 50kΩ)) = 5.3V

So R2, = 3kΩ, and R1 = 50kΩ (if you want exact you need to take into account the error from the Iq*R2 bias current, which is also given in the formula - the above is just a basic "near enough" example)

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Yes, it will work, but keep in mind that the quiescent current of the LDO (the current that flows out of the Gnd pin) is only about 2.0 µA, so be sure to evaluate the forward voltage drop of your diode at this current to see whether it gives you the drop you want.

A better approach is to set up a voltage divider with two resistors between the LDO output and ground, with 10× to 100× the IQ of the regulator flowing through them. Connect the Gnd pin of the LDO to the tap on the voltage divider. The LDO will regulate the voltage across the upper resistor to its specified value, and since the same current flows through the lower resistor, too, you can precisely control the voltage offset that it creates.

For example, if the upper resistor is 5V / 200µA = 24.9 kΩ, you can set the output voltage to 5.25V by making the lower resistor 0.25V / 200µA = 1.24 kΩ.

You can still calculate the dissipation of the regulator as (VIN – VOUT) × IOUT.

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    \$\begingroup\$ A better approach still is to use the diode, and control its operating current by adding a resistor to Vout. \$\endgroup\$ – Brian Drummond Jan 7 '13 at 11:40

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