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Author Topic: Power supply ?  (Read 3214 times)
KB3HG
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Posts: 404




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« Reply #15 on: January 23, 2012, 08:08:26 AM »

A link for resistance http://hyperphysics.phy-astr.gsu.edu/hbase/electric/resis.html

Anyway since  I=E/R and from the link above(well actually from a class of years past) Resistance of a wire was determined  by the formula R= rho L/A. Length and area determine the resistance. This being said, the current running through a wire will develop a voltage drop. I squared R is the power consumed. E=IR , P=IE, P=I IR.
Now when considering a power supply and a radio. Typically two wires power the radio, for simplicity one red and one black Ideally the same length.  Two wires two voltage drops under load. the bigger the load the greater the drop. So what is read at the supply terminals is not necessary the same voltage supplied to the radio. For the purest, a supply will have sense lines to compensate for the voltage drop at the load.
Your losses increase if you increase you length or decrease your cross sectional area.

example:
 I used to have a Clipperton  2M amp. A 4cx250B  was the tube,  my buddy incorrectly loaded the amp. Well the shunt precision resistor across the meter gave out. I think it was some odd value like 0.00909 or something similar, its been decades, anyway being a ham I did the math and ended up using about 1 inch of wire to make a similar shunt. the meter was 250 ma max. and my shunt was close enough to the specified value that meter worked and I didn't burn the tube out. So this esoteric theory does come some use every now and again.

73,
Tom Kb3hg

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AC5UP
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Posts: 3927




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« Reply #16 on: January 23, 2012, 08:22:56 AM »

Actually, the V in your equation is voltage drop across the piece of wire in question and no, it is not negligible for heat dissipation purposes.  Much simpler to understand using I squared R.  Assuming that the regulator senses output voltage at the output terminals and 50A continuous output current.

Oh, I'm remembering a tad.  Resistance of AWG 12 at about 90F is about 1.7 Ohms per 1000 feet.  So, 4 inches worth at 50A is about 1.5W assuming a continuous 50A if my calculator is correct.  And, you'll at least smell the PVC.

One click up you'll see a good comment from KB3HG........ I do this as well so don't take it as criticism, but in the case of a voltage drop in a DC power supply or wiring harness the drop is over the total length of the circuit. Positive and negative legs combined. The drop may not be symmetrical but there will be some drop on both sides unless one is a perfect conductor....  Wink
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N4CR
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Posts: 1691




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« Reply #17 on: January 23, 2012, 05:28:01 PM »

I've edited the post N4CR took issue with... Should have used the word 'power' instead of current.

AA4PB: Thanks for adding to the discussion with your response. It helped me understand the nit that was being picked.
N4CR: Avoid any work that involves politics, sales or public relations.

Thanks for making the correction.

Is that an attempt at character assassination because I pointed out your factual error?

If so, I'm happy to report that I have no desire to become a politician who's stock in trade is trading lies, distorting the facts and capitulating.

And I've excelled at sales throughout my life. I had one electronics store owner say to me after I was on his sales floor for a month that he thought I could probably sell corn to Indians.

Public relations firms attempt to sell their version of the truth. Another thing I'd not willingly do since it's generally lies and distortions.

0 for 3. You could have been 0 for 4 but you forgot marketing.
« Last Edit: January 23, 2012, 08:48:12 PM by N4CR » Logged

73 de N4CR, Phil

Never believe an atom. They make up everything.
W6RMK
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Posts: 659




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« Reply #18 on: January 25, 2012, 02:32:24 PM »

Remember that the current capacity of a wire is determined by both the gauge and the length

No it's not. Current capacity of a wire is determined by it's cross sectional area and the material the wire is made of. Heat dissipation may be limited by insulation and conduit so that derates the current capacity.

Length has nothing to do with it. Length has effect on Power delivery and voltage drop, not current capacity.

Actually, both have an effect, although not necessarily as you expect.  [To establish my bona fides here, I used to blow wires up for a living, and the physics of melting and exploding wires is actually fairly complex]

Fusing current (which is sort of the upper bound for what you're talking about) is approximated by using a formula that includes diameter, resistivity, melting point and time.  There's an assumption in that formula about heat transfer to the surroundings, and that's where the physical shape/length starts to have an effect.  The time factor has to do with getting the heat to the surroundings, by conduction to still air is the usual assumption.  That is, the formula assumes an infinitely long wire in still air.

(By the way, the NEC thermally based ampacity limits are based on the same sorts of assumptions: 100% duty cycle, still air, very long wire)

This also assumes that electromagnetic forces aren't relevant (no kiloamp lightning strokes, for instance).

In any case, fusing current for AWG16 copper wire at "room temperature" for a 5 second current pulse is 123 Amps (Onderdonk formula).. that's for a wire with less than half the cross section of your AWG12 example, and to a first order you can scale with area: about 250A fusing current.  That's a lot more than the 50 Amps.  But remember that's the "infinitely long conductor in free air".

Length starts to get involved too, because for a lot of circuits (e.g. the leads on components and most circuit interconnects whether PC board traces or hookup wire), the primary heat path is conduction *along the length of the wire* (because air is actually a good insulator).  So a short wire will fuse at a higher current than a long wire, because the "ends" of the wire are usually near something that serves as a heat sink.

Copper is about 16000 times better at conducting heat than air, so this length effect is pretty strong.  It has to be a pretty long skinny wire before the fusing current is as low as the one from the formula.

And this applies for just looking at heating in an insulated wire too, by the way.  A lot of times, it doesn't matter if the wire is insulated or not, because the thermal path to the end is SO much lower resistance than out through the side.

So, you're both sort of right, and both sort of wrong.

And what people do is use empiricism.  They put in a wire that is probably big enough, and test it, and if it works, call it done.  The wire is often a LOT smaller than you'd expect. (because, after all, material is money)

Another trap is when someone blindly replaces the wire, not realizing that it has some secondary function.  (deliberately small fusible links in auto wiring, for instance)
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W6EM
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Posts: 878




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« Reply #19 on: January 25, 2012, 09:29:51 PM »

Interesting exchanges.

I made my comment as a caveat emptor, more than much else.  If there was a lack of understanding that even an infinitesimally short wire has the propencity to dissipate heat from passing current, it was a teaching moment as well.

It had nothing to do with the return path within the case of the power supply (the black wire) or voltage drop in both of the power leads to the radio.  The model in question used multiple paths for the ground return inside the case.

Ampacities of insulated conductors are calculated and tabularized by the NEC under a variety of geometries to avoid the consequences of overloads.  Amateurs seldom think about the consequences.  Or, worse, assume that what they buy is properly designed such that no overloads will occur.

One of the serious consequences of heating PVC insulations to temperatures of only slightly above 100C is the release of anhydrous hydrochloric acid.  It will make a corroded piece of junk out of pc boards and terminals in the vicinity if the heating continues for very long.

I never alluded to getting anywhere near the fusing temperature of copper, which is about 1083C.  Just wanted to make the point that the power supply should be one from a high quality manufacturer.  The one I cited in my example was not.  It had some other serious shortcomings too, including the lack of emitter ballast resistors to properly balance the eight paralleled output transistors and enough transient over voltage protection.  I fixed all that by essentially rebuilding it into a reinforced Astron.

Hooking up a $1K plus radio to a $50 power supply just may be a tad risky.  And if you think the typical crowbar SCR will save you from a regulator failure, I've fixed enough blown supplies to see quite a few fused-open SCRs and open pc board circuit traces to them. (translation: likely cooked radios from overvoltage).

73.

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KE7FD
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Posts: 169


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« Reply #20 on: January 26, 2012, 08:27:10 AM »

More is better but be reasonable with your choice.  The Astrons are excellent (I own two) but hare hernia makers.  I now use a 36 amp switching power supply recommended here on eHam Review section (from eBay).  It's more than I need but that's OK. Don't squeeze any power supply too hard unless you like generating heat and causing problems.  Buy more than you need so if you decide to toss something else in the mix, you're already set with enough juice.

IMHO,
Glen - KE7FD
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N6AJR
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Posts: 9921




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« Reply #21 on: January 28, 2012, 01:02:16 PM »

Like I said I have both, switchers and linear.  All of those cute little supplies sent with you Icom 746/756 etc, I think they are IC 125's, are switching amps, you have probably used them for ever and not even realised that.  the question asked was about getting a power supply for his shack, and I still say you need a  35 amp or so for a standar shack and will have the ability to have several radios and accessories hooked up, with out needing to buy a new supply. also the supply willlast for ever as it is not being pushed to the limit when in use.
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