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There is a variable load that draws current between 1mA up to 1.5A. I want to supply the load with almost constant 4.2V(precision not improtant). If I only have 5V supply I thought I can perform the regulation as follows:

enter image description here

But how can a proper Rs be calculated in this case? Or will this never work?

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  • \$\begingroup\$ Why not a 5V regulator and a series diode? \$\endgroup\$ – JonRB Nov 1 '19 at 13:08
  • \$\begingroup\$ "precision not improtant" - 5V is only 19% higher than 4.2V, so... \$\endgroup\$ – Bruce Abbott Nov 1 '19 at 18:21
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To get 4.2V from 5V, that doesn't have to be too accurate, a series silicon diode would put you right in the same ball park. It would have to dissipate the same power as the LDO in Stiddly's answer.

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For 1.5A to go into your load you'd need an Rs of (5-4.2)/1.5 = .53 Ohm. Then calculate the Pd in the Zener at low load. You'd still have that 1.5A flowing through the zener now so Pd is 1.5A*4.2V = 6.3W... That's a lot.

Meanwhile, an LDO would have a maximum power loss of 0.8V*1.5A = 1.2W. 4.2V isn't a standard value, but an adjustable 5V would do just fine.

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    \$\begingroup\$ Thanks for the warning, would this then work for my purpose?: ti.com/lit/ds/symlink/lp3966.pdf It says Fast LDO Voltage Regulator, 3A Adjustable, 1.2 → 5 V, \$\endgroup\$ – pnatk Nov 1 '19 at 12:02

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