NeoVibe questions

[piperbob]

I have some questions on completing the build for my NeoVibe that one of you experts could help me on.
1. On my lamp (Radio Shack mini lamp 12volts/24 ma) there is a white and red wire. Looking at the schematic I see no polarity noted. Should the lamp be installed with polarity in mind or not?
2. Any advice on powering the unit with AC or DC? I'm kind of leaning towards AC with a transformer inside an aluminum box that will house everything. Any thoughts on a wallwart to DC than cord to unit? Any advantages/disadvantages of one or the other?
3. As a temporary power source can I put 2-9 volt batteries in series as a power supply?
Any advice would be appreciated.

[R.G.]

1. On my lamp (Radio Shack mini lamp 12volts/24 ma) there is a white and red wire. Looking at the schematic I see no polarity noted. Should the lamp be installed with polarity in mind or not?

No polarity is needed or implied with filament style bulbs. It needs NOT to be some kind of LED bulb replacement.

2. Any advice on powering the unit with AC or DC? I'm kind of leaning towards AC with a transformer inside an aluminum box that will house everything. Any thoughts on a wallwart to DC than cord to unit? Any advantages/disadvantages of one or the other?

The original 'vibe had a transformer inside the box and an AC powewr cord coming out. I recommend that you NOT do that because of the danger inherent in AC power wiring. An AC output wall wart removes the danger completely. As to AC versus DC from the wart: The PCB is designed for an AC input, like the original. If you use a DC input, it will always be "rectified" to the proper direction by the rectifier bridge, so as long as the DC or peak AC is on the order of 18-24V, it will work fine. Note that this is 18-24Vdc and 13-15Vac because the peak value of the AC waveform is bigger than the RMS value that's quoted in the rating.

If you use a DC power supply, you must choose whether to use an isolated DC power jack or not. Because of the rectifier bridge, you cannot use a nonisolated power jack and the on-board rectifier bridge at the same time. You must either use an isolated (plastic busing style) power jack or else remove the rectifier diodes from the board and connect + to +DC and - to -DC from the power jack. Using the rectifiers and a non-isolated power jack simultaneously will result in overloads on the wall wart and some release of smoke, as well as another string of questions here.

3. As a temporary power source can I put 2-9 volt batteries in series as a power supply?

Yes, you can. However, they will likely not last long, possibly only a few hours. The filament light bulb eats a lot of current, as do the many transistor stages.

[piperbob]

Thanks RG! One more question :oops:  

1.)Lets say I use a Dunlop UniVibe 18vdc transformer (5.5mm barrel, center negative).  Should I have an insulated from chassis adapter plug? If so should the positive and negative power go to the points on the bridge diode (DB1) marked "K" + & "J" - ? or should the diode bridge be removed and the + & - go directly to the soldering pads that go to C24/25?

2.) If I temporarily hook up 2-9 volt batteries in series should the + & -  go to K & J or the C24/25 side?

3.) Is the problem with the outside of the power jack being positve then conducting thru the chassis and mixing with neg. voltage from the board or pots?

Being a piping/mechanical designer for 30 years I look at everything as a mechanical problem. Sometimes the electrical practice and theory give me a little trouble :roll:

[R.G.]

1.)Lets say I use a Dunlop UniVibe 18vdc transformer (5.5mm barrel, center negative). Should I have an insulated from chassis adapter plug?

All DC supplies follow the rule: if it's DC, and you leave the rectifier bridge on the board, use an insulated adapter socket.

If so should the positive and negative power go to the points on the bridge diode (DB1) marked "K" + & "J" - ? or should the diode bridge be removed and the + & - go directly to the soldering pads that go to C24/25?

Either of these are OK. The problem is that there is a diode bridge there that expects AC. One end of the diode bridge is connected to signal ground, which is then connected to the enclosure for shielding. If you supply DC to the plug, especially if the center is negative, then you are supplying the + power to the chassis through the plug, and the - power to the chassis through the diode bridge. So the power supply is trying to bring the voltage across the bottom diodes up to the power supply voltage. The diodes are trying to hold it down to 0.7V. Whichever one wins, the unit still doesn't work.

In fact, if you have a DC adapter plug with center negative, you still can't use a non-isolated DC jack socket because now you're connecting the + on the outside of the adapter to the chassis, and the power supply is dead shorted.

You have to use an isolated adapter for AC; for DC, you can use an isolated adapter or an adapter that gets the - voltage on the part that contacts the enclosure. This is not optimal for ground reasons, but you usually get away with it. If you have DC available and isolated from enclosure, you can leave the diode bridge in. If you have DC available and it contacts the enclosure on the power jack coming in, you really should remove the diode bridge.

2.) If I temporarily hook up 2-9 volt batteries in series should the + & - go to K & J or the C24/25 side?

It doesn't matter as long as (a) you don't mind the 1.4Vdc that the diode bridge will eat as it "rectifies" the DC if you hook up to K and J and (b) there isn't that connection to the enclosure at the DC power jack.

3.) Is the problem with the outside of the power jack being positve then conducting thru the chassis and mixing with neg. voltage from the board or pots?

Yes.

[piperbob]

Thank you again!