linear regulator with high current capacity [closed]

Question

Recently I have seen that adding 7805 in parallel is not really a good idea as the load on each regulator may NOT share equal load, is there a regulator whose output voltage is 5V and can supply at least 2A of current under full load ?

Answer 1

tl; dr: unless you need an ultra quiet analog supply, forget linear regulators. There’s almost no reason to use a 7805 except out of habit.

The 7805 has an overhead voltage of at least 2V, more typically 3V. That means dissipating at least 4W, more typically 6W, if it were handling 2A. That’s kind of a lot.

By the time you figure out a heatsink, mounting, cooling, packaging, etc... the costs and system footprint really begin to add up. That cheap-and-cheerful 7805, LM317, or external pass transistor doesn’t seem so cheap anymore, does it?

If your design is already using the 7805, and need to increase the current without changing your layout, consider the DCDC drop-in replacements available for it. These will be more efficient and shed less heat. Here’s a whole range of them from MPS. https://www.monolithicpower.com/en/mezd71202a-a.html Other vendors make them as well. These will require minimal to no heatsink.

Doing a new design? 1 to 2A buck DCDC switching ICs are very inexpensive, less than 20 cents in volume. Diodes, Inc, Richtek and others have a good selection of low cost ones - this is a very competitive space for power ICs.

Space constrained? MPS makes even more compact buck solutions that integrate that inductor right into the IC package, ranging up to 6A in a 4x5mm package. I use these for M.2 cards, that’s how small they are.

Answer 2

The usual solution to increase linear regulator current is to use an external pass transistor. These kind of circuits are described in regulator datasheet.

Another solution is to not use a linear regulator at all and perhaps look at switch mode regulators.

Answer 3

My recommendation is to use the OKI. 3-Terminal TO-220 DC-DC regulator.

It is a drop in replacement for the linear LDO and more efficient and not expensive.

X—x-x-x-x

By entering parameters like All TO-220 packages or 5V 2 to 3 A you can easily find solutions from a variety of disti’s.

e.g.

https://www.digikey.ca/en/products/filter/pmic-voltage-regulators-linear/699?s=N4IgjCBcpgbFoDGUBmBDANgZwKYBoQB7KAbRABYwBmMcgDhAMptgFZGLq4B2D5ueEy6xyfLtwZCa3QZ2m8pYAJwAmMSyXrlEALoEADgBcoIAMqGATgEsAdgHMQAXwJgVm6CGSR02fEVIgKgAMYKxKoi40rDQcrqxB3EGxsNxhonogRibm1vZOLtzc7kiomLgExJBkKuRKYNy6BsaQZpa2Ds4gALRqHl6WAK5%2BlWTsGVktTp29ZAAyACIA8iA6jkA

Then use a heat sink rated for 2x the Vdrop max x 2A= Pd to avoid a dreaded max temp. input unregulated voltage drops with rising current so the Vdrop min must be satisfied by the LDO . BJT types are 1.5V to 2V while FET types are <1V typically but may cost more.

Matching LDO voltages ensures parallel current sharing is equal but thermal connections between all of them improves power sharing and cooling.

• e.g. of load regulation error is 1% and voltage tolerance is 1% you may have to add a small series R to each that equals the mismatched voltage and shared current to normalize them at the tradeoff of increased Rs/load = max load regulation error.

Say 1% V = 50mV and 1% Load error = 50mv with a 2.5 Ohm=5V/2A load implies source R =25 mOhms so adding 25 mOhms in series reduces the max variation in half meaning the sharing improves 50%. Now determine how many are sharing you need and add extra to account for imbalance to avoid thermal runaway from hogging. If load drops the highest regulator within the initial tolerance, then the others will share current. So adding this Rs improves the sharing but with slight added drop in voltage but still within 5%.

ok I lost you... but that’s how it’s done.

you see the same in power amps after complementary Darlingtons. With 0.1 Ohm or less added to each driver. This normalizes or linearizes the output impedance. Although Amps use negative feedback to then lower the impedance again, which you cannot do with a 3 terminal regulator.