1
\$\begingroup\$

I have the following circuit in my breadboard. My intention is to split a +24VDC into +12VDC and -12VDC. I'm using MC79M12 with this datasheet for the negative output and UA7821 with this datasheet for the positive voltage output. My goal is to power a circuit with op-amps from a +24VDC.

I was hoping to get +12VDC at node a and -12DVC at node b to power the op-amps relative to GND. But the voltage source goes from CV to CC automatically and the voltage regulator gets heated pretty quickly (to the point of feeling it by the smell). I don't measure -12VDC at node b; in fact, I measure a positive voltage, which is pretty small. I can imagine that the regulator is maybe reversed-biased and I'd need a -VCC for it to work... but theoretically it seems plausible to work since my GND is the 24/2 = +12V.

I've connected the input pin of both regulators to the +24VDC input voltage; I've connected the ground pin of both regulators to the common ground of the circuit; I've connected the output pin of the LM7812 regulator to the positive terminal of +12VDC output I've connected the output pin of the LM7912 regulator to the negative terminal of -12VDC output. Finally, connected loads and capacitors as suggested in the datasheet (I took into account the polarity of C4 and C5, which have + side to the GND).

But it is not working. What am I doing/thinking wrong here? If it is not possible doing it this way, how could I split the +24VDC into +12VDC and -12VDC?

Thank you very much!

enter image description here

\$\endgroup\$
11
  • \$\begingroup\$ What is D2 for? \$\endgroup\$
    – Theodore
    Apr 3, 2023 at 21:17
  • 1
    \$\begingroup\$ @ludicrous They don't specifically need a +ve and a -ve to work, it depends what kind of circuit you want to build out of an op-amp. \$\endgroup\$
    – Justme
    Apr 3, 2023 at 21:38
  • 1
    \$\begingroup\$ @ludicrous I have no clue why you selected those voltages, but you could. But simpler to use just 24V and 0V, but you can select any voltages you want and they don't have to make sense. And you can still have 0V as ground. \$\endgroup\$
    – Justme
    Apr 3, 2023 at 21:51
  • 1
    \$\begingroup\$ @ludicrous people have been making single supply op-amp circuits since 1960s, I bet whatever problem you may have understanding op-amps, it's solved already. \$\endgroup\$
    – Justme
    Apr 3, 2023 at 22:05
  • 2
    \$\begingroup\$ As others have said I think this is just a misunderstanding of what analog supplies mean or do. In a dual supply circuit with +/- 12V, Ground is said to be the center point between these supplies. Many opamp circuits need this "center" node in some form. You can convert any such circuit to single supply though; if your opamps are supplied with just +24v, then you need to create +12v as your center. This arrangement is usually called a virtual ground and often you will want to buffer it, eg with an extra opamp. All voltages in the circuit will be shifted identically. \$\endgroup\$ Apr 3, 2023 at 22:56

1 Answer 1

4
\$\begingroup\$

That won't work, I'm afraid. A look at the 79xx datasheet should show you that it regulates an already negative power supply - and you don't have one. It can't magically convert positive supply to negative.

You need a DC-DC converter or a negative voltage buck converter.


Your MC79M12BT is very unlikely to have survived.

\$\endgroup\$
3
  • \$\begingroup\$ what about if I do a voltage divider with two equal resistors and connect the middle node into ground? Would that work? \$\endgroup\$
    – ludicrous
    Apr 3, 2023 at 21:19
  • 1
    \$\begingroup\$ For reference? Yes... For using as power supply? No..just think, load will change resistance \$\endgroup\$
    – Selvin
    Apr 3, 2023 at 21:22
  • \$\begingroup\$ Why can't I use a voltage divider as a power supply? \$\endgroup\$
    – Matt S
    Apr 3, 2023 at 23:26

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.