10
\$\begingroup\$

In the following circuit two DC sources with different voltage and current are connected in parallel using diodes. Can I assume that the load works fine if either of the source is available (S1 or S2) and when both sources are available, then power is supplied from S1(15V) source and S2(12V) remains idle?

When S1 alone is available, considering D1 drops 0.51V and Regulator giving 75% efficiency, I would still get 3.3V, 400mA output from regulator (i.e 15 - 0.51 = 14.49V, P = 14.49 * 0.133 * 75% = 1.44W, Regulator Output = 1.44/3.3 = 0.436A). Is this calculation correct?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
  • \$\begingroup\$ As long as you don't mind the drop in the Schottky's all is well. You could also probably arrange to have a FET in place of D2, and have it turn on only when S1 is absent. \$\endgroup\$
    – user57037
    Commented Nov 12, 2020 at 8:33

1 Answer 1

Reset to default
16
\$\begingroup\$

When both sources are available, only S1 will power the load, D2 will be reverse biassed and S2 will remain idle.

When S1 alone is available, considering D1 drops 0.51V and Regulator giving 75% efficiency, I would still get 3.3V, 400mA output from regulator (i.e 15 - 0.51 = 14.49V, P = 14.49 * 0.133 * 75% = 1.44W, Regulator Output = 1.44/3.3 = 0.436A).

This calculation is correct, within normal engineering rounding. Congratulations for avoiding inappropriate precision.

If your load is indeed a filament bulb, then be aware that when cold it will draw perhaps 10x the current it does when hot. Whether it gets up to running temperature will depend on what S1 and your regulator do when overloaded briefly. If your regulator current limits at close to 400 mA, and has a large enough reservoir capacitor on its input to avoid collapsing S1, then eventually the bulb will warm up, and all will be well. If it's 'clever' enough to shut down on overload, then it may never light the bulb. If it has a high current limit, then S1 will be asked to provide a higher output current for a moment, which it may fail to do.

\$\endgroup\$
4
  • \$\begingroup\$ Sorry the bulb is only used as an indication. Anyway the explanation is really helpful. Is my assumption of source selection correct (first question)? \$\endgroup\$
    – Zac
    Commented Nov 12, 2020 at 7:46
  • \$\begingroup\$ Thanks Neil. Do you think this is a good solution that can be used in production? I am trying to avoid relays and changeover. \$\endgroup\$
    – Zac
    Commented Nov 12, 2020 at 8:07
  • \$\begingroup\$ @Zac subject to adequate fusing of S2 if required, and a capacitor on the input of the regulator if it's not already there, and any build requirements unusual to your target market (medical, military, space, explosive atmosphere etc) then there's nothing wrong with that simple circuit. If you want to waste slightly less power, you could use active FETs instead of diodes, but that's a complexity/performance tradeoff. \$\endgroup\$
    – Neil_UK
    Commented Nov 12, 2020 at 11:12
  • \$\begingroup\$ Since you mentioned about using FETs, can you add a diagram showing the same? I may still prefer diodes considering the simplicity, however it may help others. \$\endgroup\$
    – Zac
    Commented Nov 17, 2020 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.