Noob Choke Question

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schmidlin

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Joined
Jul 7, 2008
Messages
165
Location
Ohio
Can you measure the DC resistance on a choke and use that measurement to accurately calculate current draw over it?

Thanks in advance.
 
> calculate current draw over it?

No. How could you? "Current draw" is whatever you CONNECT to the choke.

You can get a rough clue to what current is safe. Estimate surface area. From this you know roughy the degrees C per Watt of internal heat. Pick a likely temperature rating. Work out the permissible internal Watts loss. Figure what current in the measured DCR will cause that much heat.

This assumes The Designer got the heat factors to balance with the inductance factors. And that it is a "normal" choke. Obviously a choke for low duty-cycle pulse-work might have a good inductance at some high current, but would over-heat if that current were constant.
 
schmidlin said:
Can you measure the DC resistance on a choke and use that measurement to accurately calculate current draw over it?

Thanks in advance.


Are you asking - can it(DCR) be treated as a resistor - measure a drop and then calculate current?  If it's a mystery choke I would do a test jig circuit with some very low safe draw, say 3 to 5ma.  If choke DCR measures 520r then A/B it with a 520r resistor in same position and check drop difference.  That would seem to give you a real picture of the in circuit IR drop, current and dissipation numbers.  

EDIT - If you're confident it's at least up to doing say 20ma - then test at low and then a higher current draw and see if observed drop changes.
 
lassoharp said:
schmidlin said:
Can you measure the DC resistance on a choke and use that measurement to accurately calculate current draw over it?

Thanks in advance.


Are you asking - can it(DCR) be treated as a resistor - measure a drop and then calculate current?  If it's a mystery choke I would do a test jig circuit with some very low safe draw, say 3 to 5ma.  If choke DCR measures 520r then A/B it with a 520r resistor in same position and check drop difference.  That would seem to give you a real picture of the in circuit IR drop, current and dissipation numbers.  

EDIT - If you're confident it's at least up to doing say 20ma - then test at low and then a higher current draw and see if observed drop changes.

Yes!  It's not a mystery choke and I'm not trying back-engineer it.  It's a working preamp unit and all the current (sans heaters) runs thru it, so I thought maybe this would help easily tell me the unit's current draw.  But I guess it's *no* and I can live with that.

Many thanks for all your responses!!
 
> It's a working preamp unit

Is it a "well designed" unit? Or some home-brew?

If a paid designer put that choke there, it is just-about big-enough for the job it is doing.

Give-or-take. If you need one box of 13mA, and your distributor has 20mA, that's cheaper/quicker than a custom-order 13mA special-design. In a one-off you might drop a 50mA on there. If production was in the millions, you would custom-order a 13mA job. But mostly a "well designed" unit won't have a choke much/any better than needed.

> measure what voltage drops over it

There's _two_ possible limits.

First, it can overheat and smoke from simple DC current. You can estimate this from DC resistance and size. Or pull some current for many minutes and feel it. A choke out in the open may get warm but not HOT. (In-chassis, other stuff heats it, so it should not cook itself too.)

Second: the whole reason to use a choke is high impedance to 120Hz buzz. But iron can only take so much ampere/turns of DC flux before it turns limp and saturated. When well saturated, it works no better than a resistor, but more expensive. So you want to stay well below saturation current. That's not a "DC measurement", nor easily tested with a DMM.

There is another way. There is not a ton of design freedom in a classic choke. Look in Hammond's listings. Find the core with similar size and weight. Find the model with similar DC resistance. Your mystery choke may have similar inductance and rated current.
 
I think he just wanted to know the current draw of the unit without adding in a resistor to measure across... at least that's what I'm reading.
 
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