# PSU First stage with IRCL Voltage rails swapped

The image below shows the first stage of a power supply I am trying to design and build. The transformer is a toroidal type with two 18V coils on the secondary and a 230V winding on the primary.

The resistors and the relays are active inrush current limiting. The relay should short out the resistors 2 seconds after power on to give the capacitor bank enough time to charge up without setting my diodes on fire.

In the drawing below, I am expecting VM2 to read just under 18V and VM1 to read just under 36V. I am however getting a higher voltage on VM2 (about 27V) and a lower voltage on VM1 (only 11V)

simulate this circuit – Schematic created using CircuitLab

Note:

• I have not activated the relay so all the current is still flowing through the resistors (which, I think, explains most of the voltage drop) but why the voltage swap.
• The resistors are actually a set of 10 470ohm 3W resistors in parallel for each of R1 and R2 in the diagram below.
• C1 is 14 470microF 50V Caps in parallel (6.58mF - not micro)
• C2 is 6 more 470microF 50V caps in parallel (2.82mF - not micro)
• Transformer: (I have connected Red and Orange together to form the centre tap)
• PRI: 230V/50-60 (Blue and Brown wires)
• SEC: 18V/80VA (Black and Red wires)
• SEC: 18V/80VA (Orange and Yellow wires)

What am I doing wrong?

Update based on Neil_UK post:
What I need out of this stage is one +18V rail (the bottom one in the diagram above) and one +36V rail (the top one in the diagram above) or as close as possible.

simulate this circuit

Neil, you've been very helpful and I will redo the circuit, but I'm keen to fully understand this before I continue, if you don't mind.

I've redrawn the diagram so that the individual components can be seen. I understand that D4 and D7 are in parallel on the output of the Rectifier, but I don't see the parallel part of the input signal. Are you referring to D3 and D5? They have separate outputs. I get that the caps AC couple them, but there should be no AC component on that side of the Bridge Rectifiers.

You said: "Consider what happens when the common AC input is negative, and BR1's AC input is positive. Draw it out." Would that be solvable by switching the wires around on one of the secondary winding to bring it back in phase (I get that I would have to be sure that this is the case first).

The first time I turned this thing on, my diodes actually blew quite dramatically. Investigation lead me to install the Inrush Current Limiting circuitry because with the amount of capacitance I have I calculated that I would be sucking roughly 160A-370A through my diodes when I first turn it on. These diodes have a max surge current of 70A each. But now I'm wondering if this was not caused by the fact that they are cross coupling (Still need to understand that as explained above, please ;-) The parallel nature of D4 and D7 should not cause that as they are parallel so should provide slightly higher current (assuming they are thermally locked and all, which they aren't, but for the purposes of this discussion...)

Another update for Neil Okay, so since I can not spot the difference between the two images I drew, I'll include an actual screen grab of my schematic with the transformer coils drawn in the way I connected them. (I simply turned the image around in the second image showing the common in the middle but I thought the circuit was the same as the first.)

Here is what I have done: (I've also updated the output voltages to compensate for the gain in peak voltage. I.E. (1.4141*36)-1.4=49.5V and (1.4141*18)-1.4=24V)

• Swop? Swoped? do you mean swamp or swoop or maybe swap? Why the resistors before the full bridge rectifiers? Your only burning up power and attenuating your voltage. If your trying to generate a positive and negative voltage, ditch the resistors, and put the full wave rectifiers on each coil. If you must have resistors put them after the rectifiers and before the caps. Why such huge caps? Commented Feb 17, 2016 at 7:34
• Swapped... :/ The power supply will eventually need to supply 3A per rail, so the large caps where required to keep the ripple voltages below the drop-out voltages for the rectifiers in later stages. The resistors are required to limit the inrush current due to the large caps. Main reason for the placement of the resistors was because it was an after thought after I set my diodes on fire during initial testing. They will be moved in the final design to go between the bridge rectifier and the caps. Commented Feb 17, 2016 at 9:33
• if you want 18 and 36v rails, you need roughly 13v rms secondaries, not 18v, because you gain sqrt(2) getting to the peak voltage. And you need to lose all those extraneous resistors, and the spare bridge rectifier, and the (presumably) accidental cross connection between your rectifiers. See my answer below, two rails, one of +24v, one of +48v, adjust the transformer voltage down to suit. Commented Feb 17, 2016 at 16:32

A very complicated circuit that does not do what you'd like it to do (I assume, as you don't tell us what you want it to do, just what you've done).

I assume you want two DC outputs, one with full wave rectified 18v on it, one with full wave rectified 36v. So roughly 24v and 48v supplies. Am I correct?

One problem is that the BRx components contains 4 diodes. Unfortunately, you have paralleled one AC input on the transformer side, and one DC output on the output side, and they are cross coupling. Consider what happens when the common AC input is negative, and BR1's AC input is positive. Draw it out.

There is a second complication, you don't need all these resistors. Bridge rectifiers are built for the purpose, they withstand capacitor filter inrush. Get a data sheet for your rectifiers, and look up the 'single cycle surge' specification. You may be surprised at the large number. You should find that for 'few amp' sized rectifiers, the switch on surge that they are rated to withstand is 30 to 100 times their continuous current. They really are as tough as old boots.

simulate this circuit – Schematic created using CircuitLab

This is all you need. I've drawn out what's inside the bridge rectifier as discrete diodes, I have labelled the equivalent bridge connections. Fewer components, less diode drop.

If you take a load current from just C1, it will still give you balanced transformer load, which is important, trace it out. C1 and C2 can have different values, if you want to draw different currents. That's OK.

You may need to think a bit to change the capacitor values from those you had, as they are connected in series, not parallel. Each output capacitor gets charged up to sqrt(2)*18v - a diode drop, about 24v or so. They are connected in series, so you can take an output of 24v or 48v. The input transformer provides isolation, so you can ground any of the capacitor output terminals.

• See "Update based on Neil_UK post" in original question above. Commented Feb 17, 2016 at 15:17
• Neil, you've been very helpful and I will redo the circuit, but I'm keen to fully understand this before I continue, if you don't mind. (See update above) Commented Feb 18, 2016 at 6:12
• It would help if you redrew your circuit accurately. Your seperate diodes circuits has a nice balanced arrangement, and so even though it has twice as many diodes as necessary, there are no hazards. The centre tap of the transformer goes to the common AC terminal of the rectifiers. Each rectifier+capacitor sees its own half of the transformer. Your original bridge diagram is different. BR1 sees the whole of the secondary voltage, and your common terminal at at one end of the secondary, not in the middle. Redraw one so they match please. Commented Feb 18, 2016 at 9:11
• Could it be that I just have one of my two secondary's connected the wrong way around and thus resulting in them being out of phase and therefore causing the cross coupling you referred to? Commented Feb 18, 2016 at 19:11
• It's disappointing you can't spot the difference in topology. Where the two bridges are commoned at the AC side, your discrete diagram shows that node connected to both secondaries, but your original and your latest diagrams shows that node connected to a single secondary. Your discrete diagram gives you 2 indepedant +18v outputs only. I'll print out the latest diagram for myself, annotate it, and get back to you with chapter and verse, though why you can't take my hint about which transformer polarity causes problems and do the work yourself I don't know! Commented Feb 18, 2016 at 19:30