I am attempting to implement a linear dual power regulator circuit using the 7805 and the 7905, as shown below.

It almost works, except that the voltage difference from the ground rail to the positive rail sits at around 4.73 volts instead of the expected 5.

The difference between the ground and the negative rail on the other hand is just as expected, 5 volts and change.

Swapping out the 7805 for a new one changed nothing, nor did raising or lowering the voltage being fed into the circuit, nor changing the values of the capacitors. Could somebody shed some light on what might be happening here?

enter image description here

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    \$\begingroup\$ What is V1? Your schematic shows 1V, but I am guessing it is not the case? \$\endgroup\$
    – Eugene Sh.
    Nov 29, 2021 at 20:40
  • \$\begingroup\$ whoops. yeah that's irrelevant, circuit lab threw it in there for me and I neglected to remove it. \$\endgroup\$ Nov 29, 2021 at 20:41
  • \$\begingroup\$ Do you know the 78xx regulator needs 2V higher input then output? \$\endgroup\$
    – user208862
    Nov 29, 2021 at 20:46
  • \$\begingroup\$ We need more information. What is the input voltage? Do you have a load (and if so what is it)? \$\endgroup\$
    – Null
    Nov 29, 2021 at 20:49
  • \$\begingroup\$ I've tried input voltages from 14v-20v, and the difference between positive and ground remains the same. There is no load. \$\endgroup\$ Nov 29, 2021 at 20:51

2 Answers 2


Since 0V is floating, it depends on the load ratio or quiescent current ratio of +/- LDO's. You need to buffer this with a biased Op Amp that can handle much (100x) more current than your single-ended load. Yet OA's tend to be differential loads, so that is not a big issue.


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Works like a charm! Thank you! \$\endgroup\$ Nov 30, 2021 at 2:50
  • \$\begingroup\$ If you have high load step pulse currents, and Vin was low, then split C3 into 2 separate caps to ground to decouple the inputs better. But I showed the minimum filtering while the linear LDO's will reduce ripple well into the 100 kHz range somewhat reducing due to GBW of internal error amplifier. So really big uF outputs are only for max currents. Then you may want to add reverse signal diodes to protect against Vout> Vin during turn off. \$\endgroup\$ Nov 30, 2021 at 4:15

Q1 & Q2 cater for larger but unequal load currents.

For large currents, heat sinks required for Q1, Q2 and RG1.



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