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 Post subject: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 10:18 pm 
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:arrow: Restored a Grundig SO122us. Just finished AM/FM alignment. Everything is working very well, but I noticed the R71 for the 240/220v source is very, very hot. I'd expect it to be hot, but here's the nooby, "is it too hot?" question:

Image

Image
https://imgbb.com/

Schematic shows a 1k ohm 4W resistor. I calculated the power to be close to 2W, so since I had the space and one on hand, I replaced it with a 1k 10W sand filled wire-wound resistor. Now, I've checked all voltages and they're spot on, about 40V across the resistor on "FM", and to even be sure I checked current: 0.048A max on "FM"... That's just under 2W, but the resistor is scalding hot. I don't have an IR thermometer, but with a thermocouple I measured about 65C.

Anyway, I can't find the temp curve for this resistor (NTE10W210)... Best I could find online is that its max temp is 155C. But at about 20% of its power rating, should it really be so hot? Almost halfway to its max temp? If so, I'm going to move it away from anything I think it could melt.

Thanks
-Spencer


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 10:25 pm 
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Hi Spence,

Resistor dissipation, temperature, etc are all based on free air circulation.
That's suspended from a thread (or wire) in the middle of your livingroom.

Under a radio chassis with other proximate components and zero air circulation, all bets are off.

For any given resistor in a circuit, the dissipation will be identical regardless of power rating.
The difference is the surface temperature resulting from that dissipation.

If your part is too hot, put in a larger part.

- Leigh

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 10:54 pm 
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Many of those sand resistors also acted as fuses, 2 watts on a 4 watt resistor would not be a problem. Sand resistors do run hot. You are dissipating the same amount of heat using a 4 watt or a 50 watt resistor, the only thing you have accomplished is is possibly removing a fuse.

Dave


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 11:16 pm 
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Dissipation was definitely the wrong term... Of course the dissipation would be the same regardless of rating, but I'd expect the surface temperature to be much lower for a 10w resistor at 20% vs. a 4W resistor at 50%. Anyway, hadn't considered the fuse aspect... It was originally a wire-wound 4W, I don't think send filled. didn't think it was serving as a fuse as the circuit is fuse protected... But now you made me want to put a 4W back in there.


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 11:23 pm 
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Pez-diSpencer wrote:
But now you made me want to put a 4W back in there.
I didn't see anyone suggest doing so (other than as a fuse surrogate, which it's not).
Resistors used as fuses are usually 1/2-watt or smaller.

I would suggest a 20-watt to keep the surface temperature down.

And definitely move it away from those electrolytics. You'll cook 'em.

- Leigh

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 11:29 pm 
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Pez-diSpencer wrote:
Dissipation was definitely the wrong term... Of course the dissipation would be the same regardless of rating, but I'd expect the surface temperature to be much lower for a 10w resistor at 20% vs. a 4W resistor at 50%. Anyway, hadn't considered the fuse aspect... It was originally a wire-wound 4W, I don't think send filled. didn't think it was serving as a fuse as the circuit is fuse protected... But now you made me want to put a 4W back in there.

Nope, the amount of waste heat generated is exactly the same for the two parts. The 10W part may be a bit more efficient in radiating this heat since it is physically larger than the 4W part, but the wattage rating is merely an indicator of how much power the resistor can deal with before it gets into self-destruct territory, not how effective it is at radiating the heat to the environment.

If your measurements confirm the appropriateness of a 4W part (as the design documentation indicates), don't use a higher-wattage-rated part. 4W should be fine (its around double the actual dissipation, which is OK). The power-distribution schemes for these radios are usually set up to minimize damage in the event of a short circuit. Increasing the power rating of a resistor may ensure that "collateral damage" should a short occur may be much greater than if the original-rated resistor had simply been allowed to burn out.


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 11:33 pm 
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lorenz200w wrote:
If your measurements confirm the appropriateness of a 4W part (as the design documentation indicates), don't use a higher-wattage-rated part. 4W should be fine (its around double the actual dissipation, which is OK). The power-distribution schemes for these radios are usually set up to minimize damage in the event of a short circuit. Increasing the power rating of a resistor may ensure that "collateral damage" should a short occur may be much greater than if the original-rated resistor had simply been allowed to burn out.

The fuse-surrogate resistors were commonly used in AA5 and similar bargain-basement sets.

I've never seen such a technique used in a transformer-powered fuse-protected set.

A sand resistor rated 4 watts cannot dissipate anywhere close to that power safely under a cramped chassis, such as that shown in the op's photo.

- Leigh

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Fri 09, 2018 11:53 pm 
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I quite often use the chassis mount resistors to replace a field coil and mount them on a panel with some area (Thermal grease under them like a transistor). You will find in some sets with "Back bias" & some with resistors in the plate circuit of the rectifier (commonly here 6X5). The resistor is of the lowest wattage they can get away with. Never heavy them up: Better that they burn with a fault than a transformer or the tube.

Marc


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 12:05 am 
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Leigh wrote:
lorenz200w wrote:
If your measurements confirm the appropriateness of a 4W part (as the design documentation indicates), don't use a higher-wattage-rated part. 4W should be fine (its around double the actual dissipation, which is OK). The power-distribution schemes for these radios are usually set up to minimize damage in the event of a short circuit. Increasing the power rating of a resistor may ensure that "collateral damage" should a short occur may be much greater than if the original-rated resistor had simply been allowed to burn out.

The fuse-surrogate resistors were commonly used in AA5 and similar bargain-basement sets.

I've never seen such a technique used in a transformer-powered fuse-protected set.

A sand resistor rated 4 watts cannot dissipate anywhere close to that power safely under a cramped chassis, such as that shown in the op's photo.

- Leigh

It's not a "fuse surrogate" as in a AA5 set. It's a "managed failure" design feature intended to prevent a cascade failure. Can't say that I've ever seen a fusible resistor per se used in a Grundig transformer-operated chassis. But it's very common in German tube radio circuit design to partition the B+ bus using relatively-low-value, relatively-high-wattage series resistors. Non-fusible resistors will uneventfully burn open just fine if subjected to current far above their rating.


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 12:15 am 
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lorenz200w wrote:
But it's very common in German tube radio circuit design to partition the B+ bus using relatively-low-value, relatively-high-wattage series resistors.
So partitioning the high-voltage supply system to lower-voltage branches is unique to German sets? I'll admit that I've never seen that, since I don't work on German sets.

But I certainly have seen dropping resistors in B+ distribution systems in US sets. In these sets the resistor values and power ratings are as required by the location in the circuit.

- Leigh

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 12:33 am 
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Leigh wrote:
lorenz200w wrote:
But it's very common in German tube radio circuit design to partition the B+ bus using relatively-low-value, relatively-high-wattage series resistors.
So partitioning the high-voltage supply system to lower-voltage branches is unique to German sets? I'll admit that I've never seen that, since I don't work on German sets.

But I certainly have seen dropping resistors in B+ distribution systems in US sets. In these sets the resistor values and power ratings are as required by the location in the circuit.

- Leigh

I don't think that this practice was unique to German sets but their engineers seemed to have the blessing of their management to "do the right thing" versus "do what gives the minimum recurring cost to the company consistent with few warranty returns".

Grundig is a very good example of a German manufacturer that exercised "zone control" over their B+ bus, but most other contemporary German radio makers did so as well. USA makers, not so much.

Not exactly on topic, but somewhat related: Grundig is the only radio manufacturer I have ever seen that used a fuse block protecting not only the AC mains line (yawn- all Euro sets did this) but also main B+, the 6V heater bus, and the B+ branch feeding the output transformer. Apparently this was a bit of cost overkill even for the tolerant Grundig company beancounters, since it only lasted for about one model season.


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 2:46 am 
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No these resistors are not the same as those used in AA5 radios, these have an actual fuse function. These were very common in television sets. I have replaced lots of these
Attachment:
Wire-wound-Fusible-Resistor-1W-5-2W.jpg_220x220.jpg
Wire-wound-Fusible-Resistor-1W-5-2W.jpg_220x220.jpg [ 11.46 KiB | Viewed 232 times ]

Attachment:
fuse resistor.jpg
fuse resistor.jpg [ 5.74 KiB | Viewed 232 times ]


Here is a discussion on ARF concerning these resistors
http://antiqueradios.com/forums/viewtop ... 9&t=327441

Dave


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 3:21 am 
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Leigh wrote:
...
And definitely move it away from those electrolytics. You'll cook 'em.

- Leigh


Yes, I was expecting heat but not as much as I got...To be clear, the resistor is actually further away from those electrolytics than the picture makes it look (about an inch)... But based on this discussion, I think I will move it and go back to the 4W rating... I think there is room along the side of the chassis (above the rectifier in the photo) to mount the resistor that will allow for better heat dissipation without damaging anything else.

Originally, that resistor was the only thing occupying the space of that corner under the chassis... and now I know why.The electrolytics replace the cap can, of course, and the original selenium rectifier was mounted on the exterior side of the chassis.

This radio was a garage find, dead from a short that had started a small fire inside the thing-- a victim of the phantom repair person. Was a lot of work to resurrect, so I like the idea of the added short protection! (I'm trying not to be another phantom).


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 5:11 am 
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As to resistors seemingly running TOO hot, one must remember that resistors are not like diodes or capacitors. Many are rated for a max temp of 250 C.

https://www.vishay.com/docs/30204/rsns.pdf

Personally, I don't like to generate those high temperatures, so:

--use a larger, higher power rated resistor with a greater surface area to dissipate the heat and thus keep the resistor cooler
--use a resistor with a heatsink that will transfer the heat to a larger metal surface. Be careful in using a chassis as a heatsink as you might "cook" a nearby component.
--use resistors in series or parallel to spread the heat. If you need to dissipate 5 watts, you might use a combination of say, three 5-watt resistors to spread the heat around.

Rich


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 5:35 am 
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Glue or clamp the resistor to the chassis and if you want to feel safer, split the resistor into two equal of half the original value.

...or buy one of those that are specifically designed to be bolted to the chassis.

Attachment:
wirewound_aluminium.jpg
wirewound_aluminium.jpg [ 3.51 KiB | Viewed 207 times ]

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 6:41 am 
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Rich, W3HWJ wrote:
As to resistors seemingly running TOO hot, one must remember that resistors are not like diodes or capacitors. Many are rated for a max temp of 250 C.

https://www.vishay.com/docs/30204/rsns.pdf

Personally, I don't like to generate those high temperatures, so:

--use a larger, higher power rated resistor with a greater surface area to dissipate the heat and thus keep the resistor cooler
--use a resistor with a heatsink that will transfer the heat to a larger metal surface. Be careful in using a chassis as a heatsink as you might "cook" a nearby component.
--use resistors in series or parallel to spread the heat. If you need to dissipate 5 watts, you might use a combination of say, three 5-watt resistors to spread the heat around.

Rich


I think the crux of my question may have not been clear... I wasn't concerned about the capacitor failing or overheating, I was concerned that it was hotter than it should be... As in, doing more work than it should be... I wanted to make sure I wasn't missing something... But with 30-40V across the resistor, the correct voltage readings on either side, and .048A current... Everything seems to be right. I just didn't think the resistor should get so hot at 20% of its max rating; however, I think I just underestimated how much 2W of heat was.


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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 7:19 am 
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The resistor in the photo now appears to positioned
close to the cabinet bottom. Where was it originally ?
I
Is there a fire retardant ?

Could you use a tubular resistor, held on top of the chassis,
in a vertical position with a bolt ?
Attachment:
Tubular resistor.JPG
Tubular resistor.JPG [ 9.76 KiB | Viewed 190 times ]

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 8:36 am 
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A little different perspective: 65 C is not particularly hot. In fact, for a power resistor, it's basically lukewarm.

Keep in mind that power transfer to the environment goes roughly with the temperature difference between the surface of the resistor and the air around it (for a constant surface area). If there were free air flow and the ambient temp were 25C, then 65 C would have a 40C difference. At 155C, that difference would be 130C, a little over 3X higher. What's probably happening here is that the ambient is more like 40C, due to constricted air flow, so you've only got about 25C of difference. Without going into too much detail, you can see it could make sense that it reaches this temp at only 20% of its rated power.

In any case, 65C is plenty cool (neighboring parts will be significantly cooler than the surface of the resistor), and it would be moderately crazy to suggest a 20 watt resistor is needed here. In fact, a 5 watt part would do just fine. Unless the neighboring parts are virtually in contact with the resistor, heating will be close to the same regardless of the rating of the resistor. Once you get away from very close proximity, the main factor is actual power dissipation, and not the rating of the part. Surface temp of the resistor itself is different, but overall heating of the neighboring environment is about the same.

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 Post subject: Re: Calculating resistor heat dissipation
PostPosted: Feb Sat 10, 2018 9:33 am 
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I have serviced radios where field coils & chokes have had resistors used instead of chokes: Very common. Some of these were mounted topside, not an idea I was in raptures over, but you had to make an effort to get bitten. Heat rises, so provided there is reasonable provision for air flow in the pan and the resistor is placed (even near the bottom: good spot) so it gets a good airflow around it: All should be well.

Current squared multiplied by resistance = Watts. That should be a guide as to what is expected of the resistor.

Marc


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