Trying to figure out a +/-15V rail in a power supply - Page 1
Forum section: electronics / projects

author: Yamin
author_rank: Frequent Contributor
timestamp: January 25, 2025, 03:06:21 pm
content: Hi guys, I am attaching part of a schematic from a trainer board. I was hoping you could help me figure out the following.1.Is there an advantage of having two 7815 for the dual voltage instead of using a 7915 for the negative side? would it need to be a center tap transformer in that case?2.Why is the output of the 'positive supply' shifted by 0.7V (by having the diode on the gnd pin)?3.What maybe the reason that C9 is grounded like that and C7 is not? Why couldn't C9 be the same way as C7?I feel like these are simple questions but I couldn't figure them out.Thanks for the help.

author: ledtester
author_rank: Super Contributor
timestamp: January 25, 2025, 04:56:14 pm
content: Quote from: Yamin on January 25, 2025, 03:06:21 pmA previous discussion:https://www.eevblog.com/forum/beginners/diode-in-gnd-pin-of-voltage-regulator/I imagine you measured the 0.7 increase without a load on the +15 output. Note that the BY251 will somewhat compensate for this increase - but only when there is current through it.QuoteOne possible reason is that it simplifies the BOM (bill-of-materials) and could reduce the cost since you are buying more 7812's.The transformer has two independent windings which could be linked together to form a center-tapped secondary.From what I understand, this use of a positive regulator to regulate a negative voltage only works because you have two independent windings. It wouldn't work if you only had one center-tapped winding.QuoteAs I see it, C9 and C7 both are smoothing capacitors for the 7812's -- they are connected between the input and the ground of the regulator. The "ground" of V2, however, is really the cathode of V36.

author: Xena E
author_rank: Frequent Contributor
timestamp: January 25, 2025, 05:33:40 pm
content: The positive regulator is biased up by 0.7 volts to compensate for the series diode in the output. That diode is there to protect the regulator from situations where the output may be maintained after the supply is switched off, for example when a large electrolytic capacitor is placed across the output.The capacitor question is answered if you look carefully at the way the circuit is arranged, the positive regulators bridge rectifier negative output is 'ground' however the negative regulators bridge rectifiers negative outputisthe circuit negative output, and obviously cannot be also grounded, as the regulator being used is a 78xx and itsoutputis at ground potential.

author: Yamin
author_rank: Frequent Contributor
timestamp: January 26, 2025, 03:37:36 pm
content: Thanks guys for the responses. Couple of follow ups1. If the diodes at the output are for protection, why couldn't they have just placed it across the 'OUT' and 'IN' terminals like shown in the application notes of the regulator?2. Why is there no such protection on the negative side? should be there no?3. I don't understand why the output of the positive supply is shifted by a diode drop and just shifted back using a diode again - my initial thought was could the whole setup have some sort of isolation related scenario?Thanks again for the help

author: Xena E
author_rank: Frequent Contributor
timestamp: January 26, 2025, 04:26:30 pm
content: Points 1 and 3 you make are possibly connected as your schematic shows another output branched off the positive supply. As its only a part schematic I can't commentPoint 2, if you think about the way the positive regulator is arranged to provide the negative output, all that a more negative output will do is take the input of the regulator negative, which within reason will not danage the device from reverse flow when the supply is turned off. It is basically regulating its ground return.X