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I am trying to create a UPS circuit purely to provide battery power when main input power is lost. I think I've figured out part of the circuit that stops the battery input when the main power is still on:

schematic

simulate this circuit – Schematic created using CircuitLab

But, as you can see, when main input is lost, the battery will end up providing voltage to Q2, ultimately grounding Q1 ... Creating some sort of paradox there Lol ... I'm thinking that there is another high side switch, or PNP that I need to add to this circuit for it to work. This is all theoretical, and Main Input = Battery Input.

UPDATE

From the comments and answers, I think I've found that diodes could solve this circuit ... Here's my new schematic ... It seems to work in my head Lol:

schematic

simulate this circuit

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  • \$\begingroup\$ You have linked the main input to the load .Most schemes use diodes to make it easy to design and less likely to blow stuff up . \$\endgroup\$
    – Autistic
    Nov 1, 2015 at 2:20
  • \$\begingroup\$ I am a newbie to electronics ... I thought transistors had all the answers Lol ... I guess Diodes is the thing I need to investigate ... \$\endgroup\$ Nov 1, 2015 at 3:19

1 Answer 1

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Do you need to use switching transistors? You can accomplish the same switching by using just two diodes. Select the diodes to safely pass the highest amount of current expected, and the diode's reverse voltage to withstand the highest voltage expected. (Including a fuse on the battery side is also a good investment in case of any problems.) .See below.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Its getting more like it . \$\endgroup\$
    – Autistic
    Nov 1, 2015 at 3:25
  • \$\begingroup\$ Hmm ... This seems to just put the 2 inputs in parallel. Would this not double the current to the load? \$\endgroup\$ Nov 1, 2015 at 3:42
  • \$\begingroup\$ Looking good, a Schotky type diode on the battery side may be helpful if you want to minimise the voltage drop. \$\endgroup\$
    – KalleMP
    Nov 2, 2015 at 7:15
  • \$\begingroup\$ @Stretch, Effectively the two power sources are in parallel but as long as the Main Input voltage is above the battery voltage then current is only drawn from the Main Input. \$\endgroup\$
    – Nedd
    Nov 2, 2015 at 23:03
  • \$\begingroup\$ Using diodes with low Vf (eg. Schotky) could be helpful if your input voltages are already close to the minimum voltage needed at the load. Any reason your updated schematic still shows the transistors? These aren't needed unless you have additional isolation requirements. Add to your question if needed, if satisfied accept the response to show the question as answered. \$\endgroup\$
    – Nedd
    Nov 2, 2015 at 23:04

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