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#1 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Can we estimate the running temperature of the feedback resistor when a continuous known dissipation is passing and we know maximum dissipation and maximum DeltaT?
If we now pass a transient, can we estimate the metal film peak temperature from these periodic transients? Let's give some data to start with. max Pd 600mW max DeltaT 150Cdegrees continuous operating dissipation 10mW. Peak dissipation 100mW. temp coef 50ppm. |
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#3 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Hi,
can we do something similar to heatsinks and active devices? Rth r-a ~ 250C/W. Double Rth for very low dissipations? Halve Rth for short duration transients? This would be very much guesswork. There must be something better. |
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#4 | |
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Banned
Join Date: Apr 2005
Location: Stockholm
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Andrew,
Quote:
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#5 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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I'm not worrying. I like to be informed.
I also like using 600mW resistors and not afraid to use two in parallel for 1.2W rating. Where I'm going with this is to estimate the resistor value change with operating conditions and from that decide whether 1.2W is adequate in critical locations, like the feedback loop. Looking at Lumba's post, I see nothing more than what has already been posted many times on this Forum. I come here to learn, not for cheap jibes. |
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#6 |
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diyAudio Member
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Sure one can do this even so far as doing finite element analysis so that you even have a spatial pattern of thermal distribution
Only question is, can you get somebody interested to do it? As crude rule for worst case scenario you can multiply the temp coeff by 150°C to get maximum change under allowed conditions. And since that's still tiny and real world currents will lead to much smaller temperature elevations anyway I would not consider that a relevant factor. Have fun, Hannes
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#7 |
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Banned
Join Date: Apr 2005
Location: Stockholm
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Andrew,
that`s not a jibe. To reduce distortion in resistors increase their physical size. Go for say one-tenth to one-fifth of max rating. |
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#8 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Ha & Lumba,
your estimations are too simplistic. A full temperature change due to a transient will lead to a 0.75% dynamic change in resistance for a 50ppm resistor. I cannot believe that leads to good reproduction. Let's assume for a moment that the Rth r-a does double when continuous dissipation is 2% of rating. That would give DeltaT ~ 250C/W * 12mW ~=3Cdegrees. With 50ppm that is a DeltaR~=+0.015%. Not too bad from cold to operating temp. Iif the ambient inside the amp has gone up from room temp by 10Cdegrees then DeltaT becomes 13Cdegrees and DeltaR is now 0.065%. Now look at transient signals ~10dB up from continuous/average. Convert to peak value adds another 3dB. We have dissipation ~54mW. If the transient Rth r-a were halved compared to the steady state the DeltaT becomes ~ 125C/W * 54mW ~=8Cdegrees. Leading to DeltaR ~0.034%. Add that to the previous operating temp DeltaR gives ~0.1% change in resistance. Now what if the transient were 20dB above average. The peak dissipation is up to around 380mW and DeltaT becomes +48Cdegrees. That leads to a total change in resistance of ~ 0.3%. What does that do to output distortion? It also confirms the advice given by many to over-rate critical resistors. I suspect it also explains the views of some that amps can sound different when fully warmed up compared to say 10minutes after switch on. We could double the rating of the resistor, but is that enough? I can guess just as well as anyone else. I would like to know, rather than guesstimate. Even an estimate would be better. Informed intuition from experienced builders/designers are probably much better than my quesstimates. |
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#9 | |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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Quote:
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#10 |
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diyAudio Member
Join Date: Jul 2003
Location: Bath, UK
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Interesting thought Andrew, and similar to ones I've been thinking about recently.
Most series feedback loops however are ratiometric, a proportion tapped by two resistors configered as a voltage divider. Its the difference in the instantaneous values of the two resistors that matters. Perhaps more complex but more accurate than simply using 'bigger parts ' would scale the resistors' tempco in inverse proportion to R^2; that way, the ratiometric change should be is zero over the range of allowable dissipation. |
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