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I present the following circuit found in the datasheet of MC7824 page 24/29 but without any explanation (just the schematic). It seems very strange for me on how to get a switching regulator by using a linear regulator.

Here is the circuit: enter image description here

By taking a general look to the circuit, we can say that it is a non-synchronuous buck converter if we remove the MC78XX. And the LDO plays the role of an error amplifier and PWM module (Strange in terms of the internal design of LDO).

Please how does the LDO controls the output voltage of the buck converter as well as the switching frequency of the transistor?

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    \$\begingroup\$ With a dropout spec somewhere in the 2-3V range, a 78XX is not an LDO. Not all linear regulators are LDOs. \$\endgroup\$
    – brhans
    Commented Dec 30, 2019 at 10:14
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    \$\begingroup\$ It seems they incorrectly shorted pin 1 and 2 of the LM78xx: there are 2 dots between the 470Ω and the 0.5Ω. Probably, the wires connection right of the zener shouldn't be there. \$\endgroup\$
    – Huisman
    Commented Dec 30, 2019 at 10:30
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    \$\begingroup\$ Interesting circuit! I assume the switching must be triggered by the overcurrent protection of the MC78XX. \$\endgroup\$
    – Stefan Wyss
    Commented Dec 30, 2019 at 11:05
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    \$\begingroup\$ HAH! Most people try to STOP the LM78XX oscillating ... here it seems to be a feature! \$\endgroup\$
    – user16324
    Commented Dec 30, 2019 at 13:01
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    \$\begingroup\$ This appears to be a duplicate of previous question:" 78xx switching regulator question" \$\endgroup\$
    – SamGibson ♦
    Commented Dec 30, 2019 at 16:30

1 Answer 1

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How does the LDO controls the output voltage of the buck converter as well as the switching frequency of the transistor?

Can't tell you – they're tying together input voltage and ground, so this circuit is operating the 78xx outside its operational boundaries; so, this definitely depends on the properties of one or two specific 78xx implementations, and quite possibly won't work with newer 78xx models – remember, the 78xx is fourty years old now, and people didn't have good components, so they hacked together whatever worked with the parasitic properties of what they had.

Even if the Vcc == GND regime was specified in the datasheet, the temporal behaviour specifications of the 78xx are very vague, anyway, so this is really not a case of "designed, tested, worked".

You say in your profile you want to become an SMPS expert – so don't try to recreate really obsolete stuff like this.

Update: Sam pointed out (comment below) that the VCC-to-GND connection is an artifact of TI's scan, and that they are not actually tied together; well, that leaves us still with the underdefined / vendor-specific temporal/stability behaviour that makes this design highly undesirable and unreliable.

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  • \$\begingroup\$ So the circuit is not practical anymore and it does not work as nowdays linear regulators. Thanks for your contribution. \$\endgroup\$
    – learn design
    Commented Dec 30, 2019 at 11:28
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    \$\begingroup\$ that's not what I said. I said it might possibly not work. \$\endgroup\$
    – Marcus Müller
    Commented Dec 30, 2019 at 11:31
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    \$\begingroup\$ To add a bit to what Marcus said - anytime you use a device outside of it's intended operating range or conditions, you are taking on risk that in most cases is not warranted. The design may, or may not work. It may work at room temp, then fail when it gets hot (or cold). It may work with one part, then fail with another. It may work today, then stop working tomorrow. No need to put yourself through this kind of grief. \$\endgroup\$
    – SteveSh
    Commented Dec 30, 2019 at 13:15
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    \$\begingroup\$ Hi Marcus, "they're tying together input voltage and ground" FYI the diagram in the question which shows that connection, is misleading. If we check equivalent figure 18 in this Fairchild LM78xx datasheet or figure 19 in this ST L78 series datasheet, we see that the regulator input (pin 1) and ground (pin 2) are not meant to be connected in that application circuit. \$\endgroup\$
    – SamGibson ♦
    Commented Dec 30, 2019 at 16:26
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    \$\begingroup\$ @SamGibson oh! that "kind of" obsoletes that part of my answer, but the fact that this only becomes a switching regulator due to instability doesn't change. \$\endgroup\$
    – Marcus Müller
    Commented Dec 30, 2019 at 16:29

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