6V to 5.2V linear voltage regulator

Hello!

I would like to know how it is possible to make a linear regulator from 6-V to 5.2V. I have already made a little board like this (mine is with 6-V to 5-V regulator) on Youtube:


But the problem I'm having is that my device needs 5.2V and 5.0V is not enough. I want to make a linear voltage regulator, not any switching one.

Thank you!
, Pete.

Use a low dropout adjustable regulator, such as an LM1086?
Set the adjustment resistors to give 5.2V

Most electronics should stand a 10% variation from nominal voltage, or at least 5% anyway...

rjenkinsgb said:

Use a low dropout adjustable regulator, such as an LM1086?
Set the adjustment resistors to give 5.2V

https://www.ti.com/lit/ds/symlink/lm1086.pdf?ts=1649826595675&ref_url=https%253A%252F%252Fwww.google.com%252F

Most electronics should stand a 10% variation from nominal voltage, or at least 5% anyway...

Click to expand...

That regulator has a dropout of 1.5V @ 1.5A load current, so may not work for this application.

petelms, what is your maximum load current?

Thanks for the answers! My load current will be at least 1.5 amps. 3 amps would be the best.

One method is to have the ground connection at a slightly higher voltage, and use a fixed voltage regulator. To get 5.2 V from a 5 V regulator, you need to arrange 0.2 V on the ground pin. A common way to do that is with a voltage divider from the output. The impedance of the voltage divider needs to be kept low enough that the current in the ground pin of the regulator does not affect the voltage too badly.

For instance, this regulator https://uk.farnell.com/microchip/mic29500-5-0wt/ldo-fixed-5v-5a-to-220-3/dp/2851565
has a low drop-out voltage of 370 mV and a typical ground current of 37 mA with a 3 A load.

If you had a voltage divider made from a 75 Ω resistor and a 3 Ω resistor, that would give around 0.2 V on the ground pin. The 37 mA ground current would raise that voltage by about 0.11 V at 3 A load. The ground current is approximately proportional to the output current, so you could add around 40 mΩ in series with the output if the change in voltage is a problem.

The 75 Ω resistor will generate about 330 mW of heat. You should add some capacitance across the 3 Ω resistor.

If you use higher value resistors in the potential divider, the heat will be reduced but there will be a larger change in voltage due to the regulator's ground current.

This is just one way of obtaining a non-standard voltage. It may not be the best.

petelms said:

Thanks for the answers! My load current will be at least 1.5 amps. 3 amps would be the best.

Click to expand...

OK, you will also definitely need a heatsink on the regulator at that, as well as being a high current LDO type, such as the the one Diver suggests..

I was basing things on the board in the photo having no heatsink, so only being suitable for fairly low current.

One method is to have the ground connection at a slightly higher voltage, and use a fixed voltage regulator.

Click to expand...

The MIC2930x family are 3A LDO regulators and come in both fixed and adjustable versions.
They're spec'd at 600mV max dropout voltage

A 6V to 5.2V regulator could be done with a decent PNP pass element in a Darlington configuration. Or a low gate threshold mosfet. The reference could be a Zener, a TL431, etc. An LM358 would do for a voltage and current regulator. It won't be very efficient at light loads, but modern power PNPs are plenty good for 0.5V or less C-E saturation at currents of several amps. An LM358 is not a stellar performer at 6V, but it would do the job adequately, and is dirt cheap. Most off-the-shelf LDOs will have that circuit beat by a long margin in every spec imaginable, but if you want to have some fun designing one - go for it! If you don't need a very accurate current limit, then TL431 would work as a voltage regulator all by itself - no additional op-amps needed.