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does a choke have polarity?

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Exceptions...

Actually, I have a couple of chokes that are marked to delineate the rectifier end. I assume that this is because the insulation to chassis is better at this end than the other (this would be significant in a choke input supply).

Otherwise, there isn't any "polarity" in chokes.
 
A choke might have a polarity so that the magnetic flux lines are in a certain direction relative to the chassis or some wires nearby.

Also, the core might possibly be designed to react a certain way to the movement of flux in one direction or the other.

Generally, I'd imagine that if the choke itself has no polarization marks, it isn't intended to have polarization and thus should be used how the builder likes.
 
I'd imagine 1) capacitance from end of winding closest to core to the core itself, which is usually chassis-grounded; 2) insulation as EC8010 says; 3) flux orientation is a possibility, though generally a stray field from anything is considered detrimental, but whatever.
Polarity only really applies in transformers or tapped chokes which act as such (autotransformers).

Tim
 
Chokes that are intended for audio signal usage do have a right way and wrong way to connect them. When used as a plate choke you want the "start" (closest to the core) end of the winding to be at the lowest AC potential, usually an AC ground. The "stop" end of the winding should be connected to the plate of the tube.

The same thing applies to SE transformers.

Gary
 
Gary P said:
Chokes that are intended for audio signal usage do have a right way and wrong way to connect them. When used as a plate choke you want the "start" (closest to the core) end of the winding to be at the lowest AC potential, usually an AC ground. The "stop" end of the winding should be connected to the plate of the tube.
Along the same lines, with a dc choke-input filter you would run the outer end of the winding to the rectifier because of it's lower capacitance to ground, and the inner end to the filter caps. Maybe not a a huge difference to be had, but it costs nothing and might be worthwhile if you are chasing the last little bit of performance...
 
I see the reasoning in the last two posts, but I have to say I'm unconvinced. Consider. A layer of insulation is needed before the choke winding starts, but a single layer would be insufficent to be certain that the higher contact pressure as the wire bends over the corners didn't break through. I would expect at least two layers of reasonably stiff paper. We're now looking at a capacitor with very low area (just the four sides of the core) and quite thick dielectric. A quick calculation using two layers of 7 thou paper on a 2" square core 2" high suggests 750pF. Agreed, it's more than the capacitance from the outer winding to core, but I really don't believe it's going to have much effect at 50Hz (or even 60Hz!).
 
On a different note, how do you best measure the value of an known DC choke? I have several of them (salvaged from UPS etc)
This is one is from an Emmerson Power 1.5kW UPS.
 

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The best way would be to apply 1kHz from an oscillator via a resistor set to roughly halve the output voltage from the oscillator. Then, keeping oscillator voltage constant, adjust frequency and plot a graph of choke voltage against frequency. The gradient can then be used to find the inductance.

Alternatively, you just use a component bridge. The first method alerts you to errors due to parallel capacitance.
 
EC8010 said:
The best way would be to apply 1kHz from an oscillator via a resistor set to roughly halve the output voltage from the oscillator. Then, keeping oscillator voltage constant, adjust frequency and plot a graph of choke voltage against frequency. The gradient can then be used to find the inductance.

Alternatively, you just use a component bridge. The first method alerts you to errors due to parallel capacitance.

How about using a resistor in series with the choke connected to the sine generator, and adjusting freq until the volt drop across the R is equal to the drop across the choke (note: this is NOT where the midpoint = half generator voltage!). Then R=2.pie.f.L so L=R/2.pie.f?

Jan Didden
 
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