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I have been looking at a lot of different regulators. Generally speaking switching regulators will give me a wide range of input voltages at the out put amps I am looking for with added efficiency. However, the simplicity of a linear regulator is really appealing for the project that I'm looking at. The problem that leaves me with is that most linear regulators will do between 0.5 and 1.5 amps. Is there a way to hook up two regulators so that they provide my 5 volts with their combined amp output like with batteries?

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    \$\begingroup\$ What input voltage? \$\endgroup\$ Feb 22, 2011 at 3:56
  • \$\begingroup\$ The input will be 9v \$\endgroup\$ Mar 1, 2011 at 17:32

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You can use an external pass transistor with a fixed 7805 regulator, see for example page 14 of National Semiconductor's datasheet.

It's also possible to use the LM317 to build tracking regulators that can be connected together so that they share the power.

But the reason the integrated packages don't go much above 1A is the dissipation due to the product of the difference between input and output voltages times the current. Switching regulators are much better for higher currents than linears because of this thermal problem.

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  • \$\begingroup\$ For a linear regulator, something like this is probably the best way to do it. An alternative, with no short-circuit protetcion, however, is using a TL431 with an external pass transistor. Cf. fig. 24 or fig. 19 in the TL431's data sheet (focus.ti.com/lit/ds/symlink/tl431.pdf). What is ist input voltage (range?) for your application? Remember that the linar regulator has to be big enough to get rid of the dissipated power P_V (surface area or heat sink!). P_V,max = (V_in,max - V_out) * I_out,max. \$\endgroup\$
    – zebonaut
    Feb 22, 2011 at 10:23
  • \$\begingroup\$ I think the diagram is exactly what I am looking for. Could you provide some information on how to set up the equations under it? My input will be between 9-5v and my max draw will be 2.5a \$\endgroup\$ Mar 4, 2011 at 3:03
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Ti's UC385-ADJ may do the trick. You didn't specify the input voltage.

You don't wish to parallel regulators; it isn't that simple.

You might consider Point of Load (PoL) switching regulators. They are just about as simple since they have the parts on-board.

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If you want to stay with linear there's the LM323. I can deliver up to 3A, and can dissipate up to 30W, thanks to a TO-3 package.

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Depending on the input voltage, a linear regulator is more complex in my opinion due to the cooling. Sometimes it's hard to fit a giant heatsink in your project than it is to add a simple switching supply.

You could try paralleling LM317's to the desired current. However, they are rated at 1.5A max, but that's with a very low input-output difference. If that becomes above 15V, it's degrading quickly (to about 0.4A at 40V difference).

Maybe you should look at a LM2596. You only need a 5pin regulator, an inductor and a diode. If you keep in mind to place the diode and inductor close to the regulator, it's relatively easy. And ofcourse some input/output caps, but you would use them on a LM317 too if you want to drive 2A.

Watch out that the On/off pin on the device can't handle >25V inputs, whilst the regulating part can go up to 45V.

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You could take a look at L200 regulators. They should be capable of providing 2A if the difference between input and output voltage is lower than 20 V, if I remember correctly. Control circuit is also relatively easy to build.

You will need a big heatsink if you want to run it at full power

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  • \$\begingroup\$ It also has adjustable current limiting. \$\endgroup\$ Sep 3, 2013 at 6:07
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This might do the trick:

http://www.fairchildsemi.com/ds/LM/LM7812.pdf

See fig.14. It uses a simple resistor and a PNP transistor to increase the output current capability of a standard regulator. Even the required formulas are included.

regards

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  • \$\begingroup\$ If you look further down the page you'll see a variant which has short-circuit protection (fig.15) \$\endgroup\$
    – stevenvh
    Jun 3, 2011 at 16:11

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