0
\$\begingroup\$

This question already has an answer here:

I've been in the process of restoring a jukebox. Recently I figured out that the junction box, which holds quite a bit of resistors fuses and capacitors, has three large capacitors from the 60-ies... Needless to say they have to be replaced with more recent ones.

They are:

  1. 50 WVDC 100 MFD wurlitzer part no. 73862

enter image description here

  1. 50 VDC 500 MFD wurlitzer part no. 71816

enter image description here

  1. 400 VDC 0.5 MFD wurlitzer part no. 73099-140

enter image description here

This replacement thing however has prooven to be quite the challenge.

I have bought the replacement capacitors which are on their way, however I overlooked an important bit... According to the schematic drawing (see attachment below) only one capacitor has to be polarized, the other two simply have the icon for capacitor with their respective values.

Yet two (possibly three) capacitors that are currently inside the jukebox are polarized, and I bought the replacements based upon their values.

  1. The 50 WVDC 100 MFD (paper), also has plus signs next to the word AMCON,
  2. The 50 VDC 500 MFD (yellow), carries plus signs, and a ribbon,
  3. The 400 V 0.5mfd (wax) has a ribbon, I assume this is negative.

Any thoughts on polarity, replacements and my original question if polarized caps can be used in these locations, are very welcome :)

I've attached a complete schematic and pictures of the current caps in place below:

Currently the 50 WVDC 100 MFD cap that is in place looks like this:

enter image description here

Currently the 50 VDC 500 MFD cap that is in place looks like this:

enter image description here

Currently the 400 VDC 0.5 MFD cap that is in place looks like this:

enter image description here

The electrical schematic: enter image description here

\$\endgroup\$

marked as duplicate by Tom Carpenter, Leon Heller, Turbo J, MCG, Chupacabras Jun 18 '18 at 11:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
\$\begingroup\$

This question has been answered previously here:

"Can" and "should" are two things. Should you do this? No: this use is outside the specified operating parameters of ordinary electrolytic capacitors. You seem to understand this already. Can you do it? Yes, as the video demonstrates. To understand why requires some understanding of what's inside the capacitor.

A capacitor is two conductors (usually plates) separated by an insulator. The more surface area, and the closer together they are, the higher the capacitance. Electrolytic capacitors have a thin film rolled up in the can. This film is covered in a thin oxide layer, and the thinness of this layer is what gives electrolytic capacitors their high capacitance relative to their size.

This oxide layer is created by the chemistry of the materials in the capacitor, and the polarity of the voltage applied to each side of the film. A voltage applied in the correct direction builds and maintains the oxide layer. If the polarity is reversed, the oxide layer dissolves.

If the oxide layer dissolves, you no longer have an insulator between the two plates of the capacitor. Instead of two plates separated by an insulator, you have two plates separated by a conductor. Instead of a device that blocks DC, you have a device that conducts it. Basically, you have a wire in a can.

Usually, when you encounter this failure mode, a large current flows, rapidly heating the internals of the capacitor. The expanding fluid and gas ruptures the safety vent or the can explodes.

Why then, does the capacitor in this example not explode?

The reverse polarity voltage is never applied for very long, and never without a correct polarity voltage applied soon after to repair whatever damage was done. The oxide layer doesn't dissolve instantly when a reverse voltage is applied; it takes time. The time depends on the voltage applied, the size of the capacitor, the chemistry, etc, but half a cycle of 50 Hz AC is probably not long enough to cause serious damage. When the other half of the cycle comes around, the oxide layer is restored.

Any fault current is significantly limited by the series resistors. With those resistors in series, the power available to heat the capacitor is small. There simply isn't enough power available to catastrophically destroy the capacitor because most of the available energy goes into the resistors. Perhaps you just warm the capacitor slightly. When the voltage reverses direction, the oxide layer can reform.

Probably you still damage the capacitor eventually, to some extent, but it is operational enough for the demonstration.

\$\endgroup\$
  • \$\begingroup\$ If it is already answered, don't copy the contents of the already answered question. Simply flag the question as a duplicate. \$\endgroup\$ – Tom Carpenter Jun 18 '18 at 10:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.