Transformer For A Rowe Wre Wallbox

I just got a Rowe Wre Wallbox and it has it's transformer missing.Does anyone know a suitable one for replacement?I believe the input voltage is 28 volts ac with a center tap on it's primary of 6 volts ac for it's lamps.The secondary is fed to a rectifier and comes out 13 volts dc.Any help would be appreciated.Thanks

Hello Bob,

I do not have a drawing for the WRE only one for the WRA & WRC and can tell from what you describe the transformers are not the same. The alternative if you can obtain a transformer is to have yours rewound.

Regards
Alan

Alan Hood
ami-man
UK

Hello rpucci

I also have no wiring diagram for the WRE. I think the 13 VDC are needed for the electronic components. The best solution would be, of course - as Alan suggests - if you could get an original transfomator. If you can not get one, it should be possible to produce 13 VDC from the 28 VAC with a simple power supply. For this, only a half-wave rectifier could be used, because I suspect that one wire from the 28VAC is connected to ground. But that should be no problem, because the current requirement is not very large and the voltage is probably stabilized by a voltage regulator.

The 6 VAC are actually only needed for the lamps. These you could replace with 28 V types, as the known Seeburg lamps # 1819 (28V / 0.04A). If you do it this way, you do not need a transformer.

Best regards - charly49

Thanks Alan and Charly49.

I'm going to try to attach the diagram of the WRE.

Bob

Hello Bob,

Thank you for posting the drawings, I will have a look at them and see if I can find a solution, but after saying that charly49 has come out with a good solution. I would keep the replacement 28 volt lamps at a very low wattage and protect with a suitable sized fuse connected in line. To do this however you may need a good 28 volt supply and low loss cable (hefty current wise) or you may run into problems. When we had machines on site and took over an installation we always had to rewire the wallbox cable because the 28 volt cable size would be too small and with these older wallboxes you would see the cabinet lights dim on selection.

Regards
Alan

Alan Hood
ami-man
UK

Hello Bob, hello Alan

I don't believe that problems are to be expected with the voltage drop in the cable. The standby current does not change significantly when a selection is made, because there is no motor and no coil in the credit unit to operate. The lighting dimming during the selection process therefore exists only in older WRA or WRC wallboxes, not on WRD or WRE.

The only problem I see is the rectifier. The bridge rectifier can not be used because power ground and logic ground are connected. A bridge rectifier would therefore require a galvanically isolated input voltage. Therefore, only a half-wave rectifier can be used. To prevent the voltage drop during the missing half-wave, presumably the charging capacitor has to be increased to 6000-10000 MFD. Probably it is necessary that the voltage is limited to 13 volts (for example by means of a zener diode). If necessary I can make a drawing.

Best regards - charly49

Thanks Charly49

Can you make a drawing?Thanks

Regards

Bob

Hello Bob

I made two drawings. For the first it would be necessary to know the power consumption of the circuit in order to calculate the components properly. Therefore, I have been thinking a bit and came to a better solution.

The second circuit has several advantages. It operates largely independent of the load and is protected against short circuit and thermal overload. For a simple fixed-voltage regulator, the input voltage is too high. Therefore, I have chosen an adjustable voltage regulator with floating operation. If the voltage regulator heats up (depending on the power consumption of the electronics) in operation, then he should be screwed to a heatsink.

Best regards - charly49

Hi Charly49

Thanks for the diagram.I'll go with the second one.I'll have to get the parts I need and I'll keep you posted.

Thanks

Best Regards

Bob

Hi Charly49

One more question.

I see by your diagram,I will be by-passing the power supply board in the WRE.Is that correct? Thanks

Bob

Hello Bob

The Power Supply Board is no longer needed and can therefore be removed. Instead, the new circuit is used. The arrangement of the connectors (J202 / P202) is the same as drawn on the original schematic.

Best regards - charly49

Hi Charly49

I finally got around to building the circut.I,m trying it on a breadboard before I permently put it on a PCB.I tried it and it blew the fuse,so I figuared I had a short.Checked it out then tried again.Fuse blew again.Am I doing something wrong Charly49?thanks

Regards

Bob

Hello Bob

I'm sorry, I made a mistake while drawing. The terminal pins of the voltage regulator are mislabeled. The input is pin 3, the output is pin 2 (the center pin). In the drawing it is just the opposite. I hope that the voltage regulator has survived this abuse. It might be that the voltage regulator or the protection diodes were destroyed.

I hope you can forgive me.

Best regards - charly49

Hi Charly49

Are all those capacitors electrolytic's?I tried it again,and the fuse blew again.I didn't replace any parts from the first time that the fuse blew.Do you think i should replace some components?I do understand some electronics.thanks

Regards

Bob

Hello Bob

I have spent some time to build the circuit for testing. It works well. I have made a photo of the experimental setup, on which the correct connections of the voltage regulator can be seen.

It is very likely that something has been destroyed with the wrong connection of the voltage regulator. The capacitors are of course electrolytic capacitors. They have a positive and a negative terminal. If connections are reversed, they will be destroyed within a short time. Also at the diodes is to ensure the correct polarity. The cathode terminal is marked with a white ring. Wrong polarity may destroy the diode itself and other components. You should check the correct polarity of the electrolytic capacitors and diodes. Furthermore, you should measure with an ohmmeter if the electrolytic capacitors and diodes are shorted and replace them if necessary.

When testing the circuit I've found yet following: The capacity of the first capacitors must not be 4700μF, 2200uF are enough. But it would be better if he would have 64 volts operating voltage, because the voltage is very close to 40 volts and should have a safety reserve. As can be seen in the photo, I have mounted a small heatsink on the voltage regulator. Whether it is necessary depends on the power requirements of electronics. In testing, I connected a 12V/5W lamp (current about 0.4 A) and for that it was necessary. If the voltage regulator does not run hot when in use, it can be omitted. The voltage regulator has an overheat protection, which shuts down before it gets too hot.

If you must use a heatsink, it should be noted that the mounting tab is connected to the pin 2 It is therefore essential to use a mounting kit, so that the voltage regulator is isolated from the heat sink. There is also a version available with insulated mounting tab (ISOWATT220 case).

The 28 volt lamps need something more current than specified. Depending on the current needs of the electronics and of the lead resistances, it may be that the rated current of 4/10A - fuse is exceeded. In this case, you can increase the value to 6/10A.

I hope, this will help you.

Best ragards - charly49

I forgot the photo

Hi Charly49

Thanks for all your help.I really appreciate it very much.I will do what you said and get higher voltage capacitors,and check the diodes.I will keep you informed.

Did you say that you attached a photo?If you did,I didn't get it.

All the best

Bob

I received the photo's

Thanks

Bob

Hi Charly49

Just want to bring you up to date.The power supply works.I get 13.5 VDC out.Right now it is mounted on a breadboard,but I want to transfer it to a PCB and mount it permently in the wallbox.Do you have any suggestions on what kind Printed Circut Board to get?

The reason why it didn't work I had it hooked up wrong and it kept blowing fuses.I didn't have the logic common hooked up to power common.

Regards

Bob

Hello Bob

I am pleased that the circuit works. For the final version, there are various ways, depending on your possibilities. The easiest way is to build the circuit on a hole matrix board or a strip board. But you can also make an etched circuit board. Various methods for this are to be found in the intenet. It is also possible to make a drawing with a layout program and with this, a PCB can be produced professionally.

Best regards - charly49

Hi Charly49

Just to bring you up to date.I finally got the power supply mounted in the wallbox.I have been busy doing other things.I have 13 vdc at pin 4 on J1/J2 connector but the display doesn't light.The 5 volts is on and the select led on the computer board flashes.The lamps work ok.The regulator's heat sink gets very warm.

Do you think the cpu on the board could be bad?Any advise.I'm going to try and trouble shoot it any advise you can give me would be helpful.Thanks

Bob

Hello Bob

Unfortunately, I have no service manual of the WRE - wallbox. Therefore, it is difficult to say what might be the cause. In most cases, the CPU is not the cause, but peripheral devices (such as display drivers, or similar). I think you should first examine these components.

Best regards - charly49

Hello Bob,

Remind us again which Rowe Ami jukebox you are attaching the WRE to?

Regards
Alan

Alan Hood
ami-man
UK

Hi Alan

It's good to hear from you.My jukebox is a R-86.

Im using the power supply that Charly49 designed.The 28 volt lamps work good.The 5 volt led on the wallbox computer stays lite and the select led on the computer flashes on and off.I also have 13 volts dc going to pin 4 on the J1/P1 connector on the computer board.

I haven't hook it up to the jukebox.I just tried it with the power supply.The data line is dis-connected and maybe that's why.

Regards

Bob