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I need a 2 kind of voltage for my project - 24V/2A and 5V/2.5A, and I found that a good way to ensuring high current with 78xx is to add a bipolar transistor - I wonder if I connect two such current boosters in order to use my current single voltage output power supply, so:

  • Is there some flaws in the circuit I have drawn, and will it be able to fit the requirements ?

enter image description here

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    \$\begingroup\$ Any reason you're not using a switching converter? Q10 is going to be getting very hot... \$\endgroup\$
    – Hearth
    Commented Jun 26, 2021 at 13:53
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    \$\begingroup\$ Sorry, I don't understand. What does emh mean? \$\endgroup\$
    – Hearth
    Commented Jun 26, 2021 at 14:00
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    \$\begingroup\$ The 78xx are linear regulators. They waste a lot of power, especially if the input is that much higher than the output--you're looking at 5/24≅20% efficiency. If you use a buck converter instead, you can avoid this power loss and get more than 90% efficiency. \$\endgroup\$
    – Hearth
    Commented Jun 26, 2021 at 14:04
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    \$\begingroup\$ Here's a link on DC-DC converters ; that will be cheaper than the heat sink. \$\endgroup\$
    – bobflux
    Commented Jun 26, 2021 at 14:13
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    \$\begingroup\$ I didn't notice it before, but the left half of the circuit isn't going to give you a good 24 volts either, not unless your 27 volts is very stable. The 78xx requires a lot of dropout voltage compared to LDO regulators, so if that 27 is the average or peak output of a transformer+rectifier, your output voltage will likely dip at the lower points. You can still use a linear regulator for this, just make sure you pick one so that the lowest the input voltage ever goes is still greater than 24 volts plus the dropout voltage of the regulator. \$\endgroup\$
    – Hearth
    Commented Jun 26, 2021 at 16:39

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enter image description hereYour Circuit needs a little modification but it will work. A Switching type power supply would be more efficient but keeping with your original design. I have used this arrangement many times before. Use an NPN Power Transistor instead of a PNP add a diode to the Ground terminal of both regulators, the diodes will add 0.7 Volts to the regulator output making up for the 0.7 Volts drop across the Emitter Follower, NPN transistor. The NPN will carry the full load and will require a heat sink. While you could run the 5 Volt regulator off of the 24 Volts output that would increase its' current load quite a bit and would need a larger Heat Sink.

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  • \$\begingroup\$ I'm pretty sure this is a worse design than the original... Why would you use an emitter follower here and have the output be outside the control loop? The design in the original question will work better than this one. \$\endgroup\$
    – Hearth
    Commented Jun 27, 2021 at 3:58
  • \$\begingroup\$ If you want to talk to the original engineer of this circuit they would not like your response since it was used for over 20 years in their design. \$\endgroup\$
    – user66377
    Commented Jun 27, 2021 at 4:01
  • \$\begingroup\$ I'd like to see their explanation of the merits of this circuit over the PNP variant that actually has the output as part of the control loop. \$\endgroup\$
    – Hearth
    Commented Jun 27, 2021 at 4:07
  • \$\begingroup\$ Why are you arguing about something I know works. Then explain why the other circuit is better when it is taking its input from the unregulated side of the circuit. \$\endgroup\$
    – user66377
    Commented Jun 27, 2021 at 4:12
  • \$\begingroup\$ Why are you arguing about something I know works. and Then explain why the other circuit is better when it is taking its input from the unregulated side of the circuit. The base of the PNP transistor goes to the 27 volt source instead of the regulate output. \$\endgroup\$
    – user66377
    Commented Jun 27, 2021 at 4:18

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