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I'm trying to design a variable power supply, and need some clarification regarding using voltage regulators in parallel as independent voltage sources.

As this is a power supply, one of the outputs may not be used.

My question is, if say V2 is set to 1.2V (the minimum voltage), will it sink too much current so that V1 will not be able to draw 10W (ideally)?

The voltage regulators will probably be LM317's and not intended to be tied together.

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V2 will sink very little current unless there's a load attached (similar for V1).

Your two voltage regulators are in parallel, so their branch currents add (see: Kirchhoff's Current Law).

So if V2 is attached to a load which will draw 0.4A at 1.2V, then that means V1 can draw at most 0.4A regardless of what voltage V1 is set to.

Note that linear regulators (like the LM317) have a drop-out voltage, so your output channels will only be able to regulate up to about 2V below the input voltage (the 12V output from the boost converter).

V1 will not be able to draw 10W because the single boost converter is supplying the two regulators; it is only rated to supply 10W total. There is 4.8W drawn by V2 (12V*0.4A, note that this is 12V because V2 is a linear regulator), thus V1 is limited to 5.2W. edit: However, this limit is above the 4.8W limit of what V1 nominally could draw because of the specifications on the output limits of 12V and 0.4A. So even if V2 was not present, V1 could never deliver 10W.

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  • \$\begingroup\$ If there is no load to V2, and assuming the dropout voltage is 0, then can V1 ideally draw 10W? \$\endgroup\$ – tgun926 Dec 12 '14 at 9:52
  • \$\begingroup\$ No, because 12V * 0.4A = 4.8W. That is all V1 can ever draw, even though the boost converter can supply 10W. \$\endgroup\$ – helloworld922 Dec 12 '14 at 10:13
  • \$\begingroup\$ Where does the 0.4A figure come in? In your first line you're saying V2 will sink very little current unless there's a load, so where is the 4.8W going? \$\endgroup\$ – tgun926 Dec 12 '14 at 10:19
  • \$\begingroup\$ The 0.4A is on your drawing as the load current - so max load power would by 12*0.4 \$\endgroup\$ – user1844 Dec 12 '14 at 10:34

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