| davidallancole |
| Has anybody tried the new ThermalTrak transistors from On Semiconductors? They have a diode for biasing built right into the package. |
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| Stocker |
I was going to tell you to search, but then I did it and nobody had used them yet and reported back about it. I did find one interesting application note where the temperature monitoring was done with the inbuilt diodes in series on paralleled transistors, but that's about it.
somebody actually HAD some of these, but didn't report back what they did with them...? |
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| DoomPixie |
why not get some and have a play around uourself and see what you can get out of them? or are they stupidly expensive? they sound pretty cool though.. :)
Owen |
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| davidallancole |
| They kind of are expensive. About $10 a piece from Digikey. Plus I don't have any equipment yet for building amps. |
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| DoomPixie |
fair enough, they are pretty expencive. They do sound pretty cool though, hopefully someone out there has played around with them, i will be keeping an eye on this thread, could turn out to be quite interesting.
Owen |
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| Leolabs |
| Sanken does manufacture those kind of transistors,should be nothing special.:xeye: |
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| lineup |
Interesting devices!
Diode that senses the temperature, like a transistor in a VBE multiplier,
has its own 2 pins.
So the ThermalTrak is a 5 pins device.
See attached image.
----------------------
Complementary ThermalTrak™ Transistors
NJL3281D(NPN) NJL1302D(PNP)
BIPOLAR POWER TRANSISTORS
TO-264, 0.625 °C/W
15 A, 260 V, 200 W
fT 30 MHz
http://www.onsemi.com/PowerSolution....do?id=NJL1302D |
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| clem_o |
Unfortunately they don't seem to be optioned for free sampling.
I wonder how well they match off with standard MJL3281/1302? Since most amplifier designers will go with a VBE multiplier (and perhaps use several output transistors in parallel), maybe it will be viable to get just one NJL device, use its thermal diode for bias tracking, and use standard MJL stuff for the rest - cost savings, assuming these transistors are reasonably matched...
Cheers!
Clem |
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| anatech |
Hi Clem,
Good thought for a manufacturer who wants to mystify the average service person. It adds a skew to the inventory and may become unavailable. Standard methods are reliable and so most manufacturers will be reluctant to use them.
If it's needed to track temperature that closely in an amplifier, I feel it's a marginal design.
Now if a replacement from NTE shows up ... Sad state of affairs for many repair shops.
-Chris |
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| clem_o |
Hi Chris,
Actually it seems a good idea to put the diode into the package, (check Doug Self's site and book - he's done some studies on the issue of thermal tracking) - seems there's quite a bit of thermal lag using the more usual mounting methods. Self recommends mounting the sensor (assuming plastic pack output transistors) on the front (plastic) of the outputs, as this actually heats up faster and is a better representation of the die's temp.
I guess if one is really trying to keep his outputs optimally biased (exactly Class B, avoiding gm doubling) this would be worth a try...
Yeah, you're right about having to deal with more parts!!
Cheers
Clem |
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| anatech |
Hi Clem,
Absolutely. That is the only way to get a true reading of actual die temperature.
I see problems with multiple devices, even an NPN, PNP pair becomes problematic. I guess this is one of those theory vs reality things. Simply stated another way, reality bites.
-Chris ;) |
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| Arius |
I'm the one with a bunch of the ThermalTraks. Things are going slow as I have a bunch of things to do (besides work). However, I am happy to say that my first prototype is almost ready. Preliminary measurements and listening tests show it to be pretty good.
I'll share more soon. Stay tuned. I did ask in a previous post if there were any interest in the ThermalTraks.
Price-wise, they're expensive at Digikey. OnSemi's volume quote is cheap at $2.20 but Digikey sells them for like $7. |
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| lineup |
| quote: | Originally posted by Arius
I'm the one with a bunch of the ThermalTraks. Things are going slow as I have a bunch of things to do (besides work). However, I am happy to say that my first prototype is almost ready. Preliminary measurements and listening tests show it to be pretty good.
I'll share more soon. Stay tuned |
One thing I have thought about, when using a BD139 in diode configuration
or some other diode,
for using mV base-emitter drop to measure temperature.
How much current is good to run through them?
Should not be too much, it will make them warm themselves.
But not too little.
Is this a trial and error situation or are there any good target values? |
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| Arius |
Lineup,
I'd treat the BD139 as a silicon PN junction diode. In that case, excite it with 10uA. You should get about 2.3mV/degC sensivity with that.
Since we're talking about power amps, many amps do not use the typical BD139 mounted on the heatsink to sense temp per se. Rather, as you probably already know, it is used as an amplified VBE generator which then biases the output stage while compensating for temperature.
The ThermalTraks are supposed to replace this temp dependent bias generator circuit (their internal diodes, that is). |
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| lineup |
thanks Arius
Yes, I was thinking in pure heatsink temp measuring 'sensor',
maybe with some LEDs to show how warm a Class A is.
With the last LED lighting when amp has reached full working temp.
Now I know the sensor only takes a very little current.
I will probably use another 'sensor' to set room temperature as a zero level reference.
One interesting fact with heatsinking Class A:
The more power output ( across load )
the less power and heat in output transistors and heatsink.
Heatsink is as coldest, when playing at maximum Watt power!
:) |
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| Arius |
Hi Lineup,
Hehe, for what you are trying to do, it seems like ...
LM35CZ + (optional gain) + LM3914 will do the job nicely.
The LM35 temp sensor is even available in a TO-220 package for a couple of bucks. Perfect for mounting to a heatsink. |
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| clem_o |
Or, just use one of those cpu temp sensors from a dead semi-modern motherboard - these are temperature-dependent resistors, and usually have about 10K ohms at 25 deg C. May not need amplification anymore since the R change versus temp would be large... They are pretty linear I think... you can find these on power-supply fans that change speed with temp of air (though some use a diode)...
Cheers!
Clem |
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| ostripper |
These are "scary" proprietary devices , would be hard to find
in some parts of the world ,and could be emulated by proper
design. (Vbe tranny or diodes bolted right on ,or close ,to output)
Maybe it is good for Sony, JVC to use these devices for
thermally challenged consumer garbage , but DIY'ers should
stick to common devices.:(
Absolutely , thumbs down..OS |
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| jacco vermeulen |
| quote: | Originally posted by ostripper
use these devices for thermally challenged consumer garbage |
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| guitar_joe |
i read the silicon chip mag issue with that module.
can't really say im too interested.
my main diy power amp uses the 2 of the older 1996 silicon chip 185wrms modules and two individual 300va tranny power supplies with 16000uF of caps PER RAIL. :)
these new amps would be perfectly suitable for that chassis however these amps run cool and clean so no point upgrading them.
these thermal track power transistors would be great for under engineered stuff and amps that really get a work out i think. :) |
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| homemodder |
Hi ostripper
I disagree with you, I think these devices have a lot of advantages and no disavantages, there are better power transistors available, it would be great if they incorporated this tehnology. I believe sanken also have some devices like this, will still look into it.
Jacco nice pic, no tell us what that system consists of. Are those DIY amps???
Guitar Joe, simple is better if it has same performance, and even more so when it has performance advantages. |
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| jacco vermeulen |
| quote: | Originally posted by homemodder
system consists of. |
Classe Omega preamp and Mr. Charles Hansen's GMAFB MX-R power amps.
Ya Kneu :the ones with ThermalTrak devices, 16 per channel.
(the screw type PS-connectors on the pcb are cool, i use them as well, manufactured by Riacon)
For DIY sniffers : a piccy of the Omega PS board with the first stage +/- series regulator for 1 channel, second stage is a separate shunt regulator (lateral mosfet) on a heatsink. |
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| jackinnj |
| quote: | Originally posted by ostripper
These are "scary" proprietary devices , would be hard to find
in some parts of the world ,and could be emulated by proper
design. (Vbe tranny or diodes bolted right on ,or close ,to output)
Maybe it is good for Sony, JVC to use these devices for
thermally challenged consumer garbage , but DIY'ers should
stick to common devices.:(
Absolutely , thumbs down..OS |
I disagree. On Semi keeps devices in their catalog for a long time. The ThermalTrak transistors are available from a wide number of sources, and they are seemingly easier to implement than the diode-incorporated Sanken's.
And the benefit -- as the thermal compensation is immediately adjacent to the junction the time for which compensation is effected is much faster than the temperature change crossing over the heat sink.
You could explore the benefit empirically and report on your results -- just set up an amplifier with Thermal Track devices -- for one set of experiments use a conventional VBE multiplier with the ThermalTraks, for the second hook up the incorporated diodes -- and report back to us with your results. |
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| jacco vermeulen |
| Keep that car wrecking woman away from my (chevy) suburban. |
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| pliedtka |
Keep that car wrecking woman away from my (chevy) suburban.
2002 Chevrolet Suburban mileage 54,000: Third and forth gear went out at 39,000 miles. Had transmission rebuilt. Old transmission had intermittently noticeable 1-2 shifts. Rebuilt has good 1-2 shifts and violent 1-2 shifts. Truck usually starts out with good 1-2 shifts and migrates into violent 1-2 shifts, but there doesn't seem to be any solid consistency to this pattern. One transmission mechanic using his Sun diagnostic unit saw a 1-2 error with a shift rate of .28. MAP, TBS, EGR, and everything else ok. Truck starts fine and engine has been checked. All other electrical seems to be fine.:smash: |
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| ostripper |
| quote: | | I disagree. On Semi keeps devices in their catalog for a long time. The ThermalTrak transistors are available from a wide number of sources, and they are seemingly easier to implement than the diode-incorporated Sanken's. |
I totally
agree with the thermal compensation superiority of these devices ,
but not everyone lives in NJ,USA , some of our foreign comrades
have a hard time finding MJE 340's, so I like to stick to common
parts. Most of my amps use "street parts" recycled from
the dumpster(except for capacitors ).
Also,with the diode on the die wouldn't there be "cycling"
or a thermal feedback effect by having an OVER responsive
compensation scheme.?
I can't find them at newark.com,farnell has them on backorder
,google search :thermaltrak
no dealers, google search 2sc5200 ..dealers galore
I,m in the USA but I still don't get instant gratification!!!
I,d rather superglue the diode right to the face of a $1 2SC5200
before I.ll go in search of exotic parts.(still I'd like to try them..
,any dealers??):) |
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| jackinnj |
| Avnet has them in stock. |
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| janneman |
| quote: | Originally posted by jackinnj
I disagree. [snip]The ThermalTrak transistors are available from a wide number of sources, and they are seemingly easier to implement than the diode-incorporated Sanken's.
[snip] |
I disagree too ;) . The ThermalTraks are more difficult to use than the Sankens. The Sankens' diode tempco is carefully matched to the transistors' (darlingtons) tempco at the recommended bias current.
In contrast, when you use the Onsemi ThermalTraks, YOU have to take care that the tempco tracking is set to the right value to avoid over- or undercompensation.
Jan Didden |
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| hitsware |
Why do the Sankens have 5 diodes
on the PNP and 1 on the NPN ? |
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| hitsware |
| quote: | Originally posted by hitsware
Why do the Sankens have 5 diodes
on the PNP and 1 on the NPN ? |
bump |
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| janneman |
| quote: | Originally posted by hitsware
Why do the Sankens have 5 diodes
on the PNP and 1 on the NPN ? |
The 5 diodes are schottky's. The reason for that particular organisation is that the total string of diodes comes close to the correct bias for the Darlingtons. But, more importantly, at 40mA bias and 2.5mA through the diodes, the tempco of the diode string matches the tempco of the Darlingtons. No messy bias transistors on the heatsink, but still almost perfect thermal stability.
Jan Didden |
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| jackinnj |
| quote: | Originally posted by janneman
I disagree too ;) . The ThermalTraks are more difficult to use than the Sankens. The Sankens' diode tempco is carefully matched to the transistors' (darlingtons) tempco at the recommended bias current.
In contrast, when you use the Onsemi ThermalTraks, YOU have to take care that the tempco tracking is set to the right value to avoid over- or undercompensation.
Jan Didden |
The diodes of the Sanken's are directly connected to the base, the ThermalTrak's aren't -- I think that this limits the possibilities of their deployment. I believe that this is dicsussed on one of the memorialized Bob Cordell threads.
ThermalTraks are easier to obtain in the U.S. There is only one U.S. distributor of the Sankens, and that distributor is difficult for little guys to deal with so I have purchased them from the U.K. Getting a couple of samples from Allegro took a half-year. |
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| tryonziess |
| Jack, Does not Ampslab carry the Sankens. His service and e-mail response is also pretty good. |
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| jackinnj |
| Ampslab -- I haven't purchased anything from him. He is probably much more prompt than me :) |
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| syn08 |
| quote: | Originally posted by jackinnj
There is only one U.S. distributor of the Sankens, and that distributor is difficult for little guys to deal with so I have purchased them from the U.K. |
Not sure who you are dealing with in the US, but I found http://www.tenfourltd.com/ as a very reliable source for Sanken devices. They currently don't have STD03N/P in stock (quite unusual) but it's certainly worth asking for a quote. It's not going to be cheap, that's the reason I gave up using them quite some time ago. While they have certain advantages, I do not think they are worth the money you can purchase them in North America.
The only thing I'm missing in the Onsemi ThermalTracks is the Sanken devices pinout, allowing a so much better PCB layout. |
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| janneman |
Yes that pinout makes it so much easier. I in particular liked the previous Sanken versions, the SAP15 and 16 devices, as they even had a 0.22 ohms Re integrated. They stopped making those because apparently the SOA of the NiChr Re's was less than the transistors' SOA so the resistors blew before the active devices.
Anyway. One other purported disadvantage of the Sankens is that you can't directly access the output devices' base terminal, to put in a charge-suck-out resistor. I didn't find any practical disadvanteges from it; these are quite fast devices to begin with.
I had the privelege to discuss this particular point with D Self at the Amsterdam AES earlier this year, and he told me not to worry about that. Which I don't ;)
Jan Didden |
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| hitsware |
>The 5 diodes are schottky's.
>The reason for that particular
>organisation is that the total
>string of diodes comes close
>to the correct bias for the Darlingtons.
OK. Thank You.
Do you mean that all 6 diodes
(both xsistorsworth)
in series are ~right for for the
total biasing?
Doesn't that imply both xsistors
are thermally as one?
So you couldn't have them on
separate heatsinks ....... ???? |
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| janneman |
| quote: | Originally posted by hitsware
>The 5 diodes are schottky's.
>The reason for that particular
>organisation is that the total
>string of diodes comes close
>to the correct bias for the Darlingtons.
OK. Thank You.
Do you mean that all 6 diodes
(both xsistorsworth)
in series are ~right for for the
total biasing?
Doesn't that imply both xsistors
are thermally as one?
So you couldn't have them on
separate heatsinks ....... ???? |
Yes all diodes need to be in series. See app info attached.
I haven't thought about separate heatsinks, but it seem obvious that the two xsistors should be on a common heatsink for the same temp of all junctions and diodes.
Jan Didden |
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| hitsware |
Thanks again. I'll stick with lateral mosfets. Much more
elegant solution to thermal tracking. :) |
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| anatech |
Hi Jan,| quote: | | In contrast, when you use the Onsemi ThermalTraks, YOU have to take care that the tempco tracking is set to the right value to avoid over- or undercompensation. |
I can't understand the problem. On one hand, your design is fixed by the part. Even the operating points are fixed by the part. On the other hand, you have complete flexibility exactly the way you've done things for years. The big difference is that you now have access to something very close to the exact die temperature and you can use that sensor any way you want.
If you can design with standard transistors, the On Semi parts pose zero problem. The Sanken parts lock you into very limited design options.
Just my thoughts on that matter.
-Chris ;) |
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| janneman |
| quote: | Originally posted by anatech
Hi Jan,
I can't understand the problem. On one hand, your design is fixed by the part. Even the operating points are fixed by the part. On the other hand, you have complete flexibility exactly the way you've done things for years. The big difference is that you now have access to something very close to the exact die temperature and you can use that sensor any way you want.
If you can design with standard transistors, the On Semi parts pose zero problem. The Sanken parts lock you into very limited design options.
Just my thoughts on that matter.
-Chris ;) |
Well, there isn't a problem really, I think. The discussion was about the tempco matching between the diodes and the xsistors. You are right that the Sankens are less flexible in how you want to configure your output stage, but you can also turn that into an advantage: integrated Darlingtons & bias diodes & pin-out to simplify pcb layout. I wasn't going to try to press anyone into a specific direction. I think ;) .
It's just that if you decide to use integrated bias diodes, you might as well take the ones that have an almost zero tempco to begin with. But sure, using the OnSemi's you can also get there with a little bit more effort. Up to the designer.
Jan Didden |
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| DouglasSelf |
I've just been looking at the ThermalTrak devices, and I was very disappointed by the ON-semi application note. It keeps banging on about thermal runaway, which hasn't been a problem since we took up silicon. In ** years I have only ever seen one case of it, and that was in a remarkable design that used a thin metal panel as a "heatsink". The power transistors were mounted on black crackle paint! I should never have left it on over lunch...
Other worries; the majority of the THD tests were done with no load, which is so wrong it really makes you wonder what the author was thinking.
It is claimed (with no evidence) that no bias adjustment is required, which seems to imply that the integral diodes have some magical way of knowing about the Vbe's of the driver transistors.
Now I am prepared to believe that these are magnificent power devices, but the quality of this app note bothers me. What do other people think? |
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| janneman |
Indeed, neither OnSemi nor Sanken is very generous with application information, but the situation you paint is really dismal. It does reflect my earlier feeling that those ThermalTrak's were designed in great haste to jump on this particular bandwagon by plunking what appears to be run-of-the-mill 1N400x diodes on the transistor die.
I have *somewhat* better info from Sanken, which I only once found on the 'net and now seems to have disppeared. See attached.
Jan Didden |
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| DouglasSelf |
Hi Jan, how are things.
Thanks for the SAP app note. I can tell you that Sanken's doubts about the durability of the built-in emitter resistors were well justified. Cambridge Audio found that out the hard way. (not my designs, I hasten to add) :)
I really must try the ThermalTrak devices out some time. I would be very interested in seeing how fast the integral diode responds compared with a sensor on top of the transistor.
BTW, the ThermalTrak specs I have seen say nothing at all about the diode- no ratings, nothing. A bit worrying. |
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| gaetan8888 |
Hi
I know the roender amp using Onsemi thermal track.
I did found few of those transistors, but I still hesitate to use 3 in parallel, like in the roender amp, to get the right number of diodes for the correct bias. And anyway I don't have enough Onsemi thermal track transistors to do like in the roender amp.
Anyone try it using only one transistor per side with a VBE and only one thermal track diodes per side ?
My diy amp work super and it have a Self type II output and I would like to try the thermal track transistors on my amp.
Thank
Bye
Gaetan |
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| stinius |
If you are using a predriver and a driver, I think you have to use three ThermalTrak per side.
I'm sure everybody has seen this app. note, but I post it anyway.
Stinius |
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| jneutron |
| quote: | Originally posted by DouglasSelf
I really must try the ThermalTrak devices out some time. I would be very interested in seeing how fast the integral diode responds compared with a sensor on top of the transistor. |
Certainly faster. On top, you have the diffusion velocity through the plastic and the mass of the sensor device.
Junction to backside for a typical die is 100 uSec to 1 millisecond. I don't think they are bonding a diode to the top of the chip, I think they patterned the diode into the basic device. Epoxy bonding to the top surface of the transistor die would be a temp and power cycle nightmare..ripping the aluminum off the die..
Laterally, I would expect the temperature settling time to be under a second or so..as a first guess. But boy, that datasheet has bupkus for data, eh?
They should at least provide an isolation voltage figure from diode to transistor. Is it made isolated tub out of nitride, reverse biased pn isolation..what?? It may go into crowbar conduction if the CE voltage reverses on transients.
Why not test the response? Put 10 milliamps in the diode forward bias, then apply a square wave dissipation to the transistor..watch the die Vf settle.
| quote: | Originally posted by DouglasSelf
BTW, the ThermalTrak specs I have seen say nothing at all about the diode- no ratings, nothing. A bit worrying. |
Actually, it has Vr of 200 volts, and average rectified current as 1 ampere. It would have been nice had they categorized those numbers under the heading "integral diode parameters" We can't even tell if it runs 2.2 mV per degree C..:confused:
And I read somewhere that it's supposed to be a fast diode..
Horrible datasheet..
Cheers, John |
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| JPV |
do we need stable bias current with temp in the output stage? NO
This seams provocative but here is why (IMHO):
Self is writing in his book that an optimal Vq ( voltage drop on Re the ouput emitter resistor) would be 26mV for minimum distortion ( in Complementary EF stage).
This is empirically found but there is more to that if you understand why.
In a previous thread I wrote a memo explaining better the original paper ( difficult to read because typo errors) of Oliver ( from HP) on this subject.
He studied analytically the non linear output resistance of a class B amplifier. It is the non linear variation of this ouput resistance with output signal which is the cause of crossover distortion.
He showed that the distortion is minimum if: gmRe =1
where gm is the transconductance of the transistor at bias Io.
What does it mean?
gmRe=1 is equivalent to (Io/Vt)Re=1 where Vt is the thermal voltage kT/q
This is the same as Io Re=Vt or the voltage drop on Re = 26mV !
But this means that if the temperature increases, so does Vt and the current should follow proportionally if we want to stay in the minimal distortion condition !!!
So, the Vbias tracking circuit should have a different Tempco than the Vbe of the outpu transistor.
Lets evaluate this Tempco for Temp from T1=300°K to T2= 375°K. for Thermaltrak transistor
first the condition IoRe=Vt means that Io2/Io1 = T2/T1
Io1 = bias at T1 and Io2 is bias at T2. This is very important In our case Io2 = 1.25 Io1 and Vt ( at T2) = 26mv * 1,25 = 32.5
If the circuit is symetrical we may calculate the Tempco of Vbias for one half. The final Vbias Tempco is twice the result.
Vbias (at T1)= Vbe ( at T1) + Io1 Re
Vbias (at T2)= ( Vbe ( at T2) + Io2 Re + Delta Vbe )
(Delta Vbe) is the increase of Vbe at constant temperature T2 to take into account that Vbe operates at a higher current Io2.
Using the diode equation Io2/Io1 = exp ( Delta Vbe/Vt)
so (Delta Vbe) = Vt ln ( Io2/Io1 )
in our case Delta Vbe = 32.5 * ln ( 1.25)= 7.25 mV
Then
Vbias ( T1) = 0.6v + 26 mV = 0.626
Vbias ( T2) = 0.6V - (2mV * 75) +32.5 mV + 7.25 mV
= 0.450 V + 39.75 mV
And Tempco Vbias =( Vbias (T2) - Vbias(T1) )/ (T2-T1)
If we replace the values Tempco = -2mV/°C + (39.75 - 26)/75
Temco = 2mV/°c - 0.183 mV/°C = - 1.817 mV/°C
NOW the good point
If we look at the specs of the diode in the thermaltrak data sheet
At low current 1ma in the diode, Tempco is about 1.81mV/°C
So by using a Vbe multiplier like the Leach one and force 1mA in the diode leg, we can adjust for the right operating point at room temperature and have the two diodes tracking optimally !!!
So Thermaltraks OK ??
JPV |
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| gaetan8888 |
| quote: | Originally posted by JPV
...
So by using a Vbe multiplier like the Leach one and force 1mA in the diode leg, we can adjust for the right operating point at room temperature and have the two diodes tracking optimally !!!
So Thermaltraks OK ??
JPV |
Hello
Like that ?
Thank
Gaetan |
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| gaetan8888 |
| quote: | Originally posted by stinius
If you are using a predriver and a driver, I think you have to use three ThermalTrak per side.
I'm sure everybody has seen this app. note, but I post it anyway.
Stinius |
Hello
I do not use a pre-driver, here is the kind of output I use in my diy amp.
My previous diy amp was a cfp output but there was oscillations, so I did come back to a standard type II output.
Thank
Gaetan |
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| JPV |
| quote: | Originally posted by gaetan8888
Hello
Like that ?
Thank
Gaetan |
Yes but with two diodes in serie one for the NPN and one for the PNP.
The resistors must of course be calculated for the 1mA current and this depends on the current source feeding the Vbe multiplier.
If you have predriver(s) you must include diodes to take the Vbe of thedrivers into account. Here de diodes of the correspondent thermatraks ( if the driver is thermaltrak) will not track aswell and induce an overall error but these transistors should run cooler
JPV |
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| gaetan8888 |
| quote: | Originally posted by JPV
Yes but with two diodes in serie one for the NPN and one for the PNP.
The resistors must of course be calculated for the 1mA current and this depends on the current source feeding the Vbe multiplier.
If you have predriver(s) you must include diodes to take the Vbe of thedrivers into account. Here de diodes of the correspondent thermatraks ( if the driver is thermaltrak) will not track aswell and induce an overall error but these transistors should run cooler
JPV |
Hello
Yes, I know, I've forgot to draw the second diodes, I did replace my image by a corrected version.
I do not use a pre-driver, you can see the type of output I use in my second images in my other post.
Thank
Gaetan |
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| JPV |
| quote: | Originally posted by gaetan8888
Hello
Yes, I know, I've forgot to draw the second diodes, I did replace my image by a corrected version.
I do not use a pre-driver, you can see the type of output I use in my second images in my other post.
Thank
Gaetan |
my comment were about the drivers and you have them. They will drift too and there we need an exact compensation.
One possibility is to use themaltraks as drivers too and use their diodes in serie with the two others in the Vbe multiplir. Because you need more voltage drop, you will need a diode ( 1Nxxx) in the base of each driver to compensate and be able to bias the Vbe multiplier transistor in the active region
The 2 extra diodes in the Vbe multiplier must track the driver
drift. One possibility is to // each diode with a resistor and have this diode working at let say 0.1 mA. The drift should then be 0.2mV/°C higher. This with a well cooled driver should make it.
You can try even 0.01ma which should give you another
-0.200mV/°C extra drift. Of course the room temperature Vbe is lower but with a well calculated potentiometer and resistors it should be possible to adjust the pot. This is without having experienced it. It is perhaps totally wrong for a reason I didn't see. Measurements and experimenting is required.
JPV |
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| DouglasSelf |
Just a thought- does any one know when the original Sanken SAP transistors were introduced?
I seem to recall it was about 15 years ago, but I could be quite wrong. Any info would be much appreciated.
By the way, one advantage of all temperature-sensing output transistors that no-one seems to have mentioned is that it saves the time required to mount a bias sensing component, which can be an awkward business. In production that might help offset their greater cost.:) |
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| janneman |
| quote: | Originally posted by JPV
do we need stable bias current with temp in the output stage? NO
This seams provocative but here is why (IMHO):
Self is writing in his book that an optimal Vq ( voltage drop on Re the ouput emitter resistor) would be 26mV for minimum distortion ( in Complementary EF stage).
This is empirically found but there is more to that if you understand why.
In a previous thread I wrote a memo explaining better the original paper ( difficult to read because typo errors) of Oliver ( from HP) on this subject.
He studied analytically the non linear output resistance of a class B amplifier. It is the non linear variation of this ouput resistance with output signal which is the cause of crossover distortion.
He showed that the distortion is minimum if: gmRe =1
where gm is the transconductance of the transistor at bias Io.
What does it mean?
gmRe=1 is equivalent to (Io/Vt)Re=1 where Vt is the thermal voltage kT/q
This is the same as Io Re=Vt or the voltage drop on Re = 26mV !
But this means that if the temperature increases, so does Vt and the current should follow proportionally if we want to stay in the minimal distortion condition !!!
So, the Vbias tracking circuit should have a different Tempco than the Vbe of the outpu transistor.
Lets evaluate this Tempco for Temp from T1=300°K to T2= 375°K. for Thermaltrak transistor
first the condition IoRe=Vt means that Io2/Io1 = T2/T1
Io1 = bias at T1 and Io2 is bias at T2. This is very important In our case Io2 = 1.25 Io1 and Vt ( at T2) = 26mv * 1,25 = 32.5
If the circuit is symetrical we may calculate the Tempco of Vbias for one half. The final Vbias Tempco is twice the result.
Vbias (at T1)= Vbe ( at T1) + Io1 Re
Vbias (at T2)= ( Vbe ( at T2) + Io2 Re + Delta Vbe )
(Delta Vbe) is the increase of Vbe at constant temperature T2 to take into account that Vbe operates at a higher current Io2.
Using the diode equation Io2/Io1 = exp ( Delta Vbe/Vt)
so (Delta Vbe) = Vt ln ( Io2/Io1 )
in our case Delta Vbe = 32.5 * ln ( 1.25)= 7.25 mV
Then
Vbias ( T1) = 0.6v + 26 mV = 0.626
Vbias ( T2) = 0.6V - (2mV * 75) +32.5 mV + 7.25 mV
= 0.450 V + 39.75 mV
And Tempco Vbias =( Vbias (T2) - Vbias(T1) )/ (T2-T1)
If we replace the values Tempco = -2mV/°C + (39.75 - 26)/75
Temco = 2mV/°c - 0.183 mV/°C = - 1.817 mV/°C
NOW the good point
If we look at the specs of the diode in the thermaltrak data sheet
At low current 1ma in the diode, Tempco is about 1.81mV/°C
So by using a Vbe multiplier like the Leach one and force 1mA in the diode leg, we can adjust for the right operating point at room temperature and have the two diodes tracking optimally !!!
So Thermaltraks OK ??
JPV |
Jean-Paul/Pierre,
VERY interesting reasoning. I must look into that better, sorry I reacted so late. Hmm. What does anybody else think about this?
Jan Didden |
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| janneman |
| quote: | Originally posted by DouglasSelf
Just a thought- does any one know when the original Sanken SAP transistors were introduced?
I seem to recall it was about 15 years ago, but I could be quite wrong. Any info would be much appreciated.
By the way, one advantage of all temperature-sensing output transistors that no-one seems to have mentioned is that it saves the time required to mount a bias sensing component, which can be an awkward business. In production that might help offset their greater cost.:) |
Hello Doug,
I bought my first ones in 1997 I think, at that time they had several types like SAP12, SAP15, SAP16.
Jan Didden |
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| JPV |
| quote: | Originally posted by janneman
Jean-Paul/Pierre,
VERY interesting reasoning. I must look into that better, sorry I reacted so late. Hmm. What does anybody else think about this?
Jan Didden |
I copied this post in the thermaltrak thread from B Cordell.
There are some other little developments there and a reaction from B Cordell agreeing with this
Do you believe that the spice models are accurate enough to simulate this? I asked the question to Andy_c in the spice thread but I have not yet an answer.
Jean-Pierre |
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| DouglasSelf |
| Thanks Jacco, I hadn't realised it was so long ago. |
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| DouglasSelf |
| quote: | Originally posted by gaetan8888
Hi
I know the roender amp using Onsemi thermal track.
I did found few of those transistors, but I still hesitate to use 3 in parallel, like in the roender amp, to get the right number of diodes for the correct bias. And anyway I don't have enough Onsemi thermal track transistors to do like in the roender amp.
Anyone try it using only one transistor per side with a VBE and only one thermal track diodes per side ?
My diy amp work super and it have a Self type II output and I would like to try the thermal track transistors on my amp.
Thank
Bye
Gaetan |
Hi Gaetan
It is expensive to use more NJL output devices just to get the right number of diodes, and in my opinion the application note misses the point utterly by trying to compensate driver temperatures with output temperatures. That app note really is a bit of a disgrace.
My first thought was the attached circuit, which is I think what you might be looking for.
I would be glad to have people's comments on it.
Douglas |
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| JPV |
| quote: | Originally posted by DouglasSelf
Hi Gaetan
It is expensive to use more NJL output devices just to get the right number of diodes, and in my opinion the application note misses the point utterly by trying to compensate driver temperatures with output temperatures. That app note really is a bit of a disgrace.
My first thought was the attached circuit, which is I think what you might be looking for.
I would be glad to have people's comments on it.
Douglas |
Do you expect the thermal delay to be less critical for the drivers?
What do you think of the possibility to control the Vbe tempco of the diodes by bleeding of current via // resistor and adjust it to have minimum distortion at high temperature as I proposed in the other thread on thermaltracks
Thanks for your opinion
JPV |
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| Bob Cordell |
| quote: | Originally posted by DouglasSelf
Hi Gaetan
It is expensive to use more NJL output devices just to get the right number of diodes, and in my opinion the application note misses the point utterly by trying to compensate driver temperatures with output temperatures. That app note really is a bit of a disgrace.
My first thought was the attached circuit, which is I think what you might be looking for.
I would be glad to have people's comments on it.
Douglas |
Hi Doug,
You are so very right about the app note being wrong to lump in driver Vbe with power transistor Vbe, as if all were alike. The circuit you show looks like it should work.
One of the key things to bear in mind is the tempco matching of the ThermalTrak diode as compared with the tempco of the ThermalTrak power transistor; one has to be very intentional about the current one runs through the diode to get the desired tempco, as junction drop tempco is always a function of current density.
Another approach is to build the tracking diodes into a separate Vbe multiplier so that their net effective tempco (mv/C) can be multiplied by the amount desired while running the nominal VAS bias current through the diodes. Then a second Vbe multiplier can be used to mop up the remaining bias spread needed by the drivers et al. Lots of different possibilities here.
Cheers,
Bob |
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| DouglasSelf |
| quote: | Originally posted by Bob Cordell
Hi Doug,
You are so very right about the app note being wrong to lump in driver Vbe with power transistor Vbe, as if all were alike. The circuit you show looks like it should work.
One of the key things to bear in mind is the tempco matching of the ThermalTrak diode as compared with the tempco of the ThermalTrak power transistor; one has to be very intentional about the current one runs through the diode to get the desired tempco, as junction drop tempco is always a function of current density.
Another approach is to build the tracking diodes into a separate Vbe multiplier so that their net effective tempco (mv/C) can be multiplied by the amount desired while running the nominal VAS bias current through the diodes. Then a second Vbe multiplier can be used to mop up the remaining bias spread needed by the drivers et al. Lots of different possibilities here.
Cheers,
Bob |
Hi Bob
From reading the other thread I gather that you have determined that the transistor Vbe temperature coefficient is -2.14 mV/ ºC while the diode has -1.7 mV/ ºC, and the best voltage matching occurs when the diode current is a quarter of the transistor Ic. That suggests that my circuit will leave the output devices somewhat under-compensated, even if the respective currents are scaled correctly. Guessing at a quiescent Ic of 100 mA, that means 25 mA through the diodes, which suggests a VAS buffer would be required.
I think you're right- some multiplication of diode tempco is required.
Douglas |
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| gaetan8888 |
Hello
ONsemi ap notes are like their transistors spice models... not alway reliables.
Do their distortions measurements of the VBE bias and Thermal Track bias are reliables ?
And does using a Thermal Track bias in a amp really made a sonical difference compared to VBE bias ?
Thank for your ideas and advices Doug, Bob and JPV
Bye
Gaetan
 |
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| DouglasSelf |
| quote: | Originally posted by gaetan8888
Hello
ONsemi ap notes are like their transistors spice models... not alway reliables.
Do their distortions measurements of the VBE bias and Thermal Track bias are reliables ?
And does using a Thermal Track bias in a amp really made a sonical difference compared to VBE bias ?
Thank for your ideas and advices Doug, Bob and JPV
Bye
Gaetan
|
The THD measurements in the app note are worthless because most them appear to have been made with no load. What ONsemi are thinking of I do not know.
Douglas |
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| gaetan8888 |
| quote: | Originally posted by DouglasSelf
The THD measurements in the app note are worthless because most them appear to have been made with no load. What ONsemi are thinking of I do not know.
Douglas |
Hello
Oops, I've see that graph today, I should have seen that.
Did you done distortions measurements on amps using the VBE bias and Thermal Track bias to made comparaisons between them ?
Thank
Gaetan |
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| megajocke |
| Something similar to this could be used to multiply the temperature coefficient of the thermal traks. For a 2-stage darlington output the multiplication will be about 1.5 making the temperature coefficients about the same. For a triple darlington the multiplication will probably be too high - but this can be lowered by inserting diodes or diode connected transistors (probably 2) in series with R1. These can be used to sense driver transistor temperature. |
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| megajocke |
| Or you could do this - making temperature coefficient of thermaltrak diodes adjustable. Pot turned upwards gives no multiplication, downwards multiplication increases. Bias voltage needs readjustment though when that is adjusted. |
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| JPV |
| quote: | Originally posted by megajocke
Or you could do this - making temperature coefficient of thermaltrak diodes adjustable. Pot turned upwards gives no multiplication, downwards multiplication increases. Bias voltage needs readjustment though when that is adjusted. |
Is the simple Vbe multiplier like the Leach one not the easiest and perhaps the best because very low interraction
The potentiometer in the lower leg of the base tunes the Vbias and the diodes in the upper leg tune the drift of the drivers and of the output transistors for minimal distortion at high temperature. This is easy done by adjusting // bleeder resistors.
JPV |
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| megajocke |
| It is easy but if you have only 1 pair of outputs in a smaller amp than the Leach it might be possible the temperature coefficient of the diodes is not enough to compensate. Best would be to try it. 4 diodes like the Leach might be overcompensated. |
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| JPV |
| quote: | Originally posted by megajocke
It is easy but if you have only 1 pair of outputs in a smaller amp than the Leach it might be possible the temperature coefficient of the diodes is not enough to compensate. Best would be to try it. 4 diodes like the Leach might be overcompensated. |
But that is the main point. By letting the diodes working at lower current you can adjust the tempco to the tempco of the transistors.
The amount of output transistors do not matter. You take the 2 diodes of one pair is serie with the diodes of the driver pair. They will exist in a small amp or in a big amp with many ouput pairs in //
The optimal Vbias is set by the potentiometer in the base leg. Of course you will need two normal diodes in each base of each driver to bias correctely the Vbe multiplier and not overcompensate. In this configuration the Vbias is > 4 voltages drops of diodes + Vbe, depending on the multiplication factor.
JPV |
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| DouglasSelf |
| quote: | Originally posted by gaetan8888
Hello
Like that ?
Thank
Gaetan |
Hi Gaetan
I had a quick look at this circuit today, and I'm not sure it will do the job. The current through the diodes is only 1.2 mA, and their voltage drop (which with this circuit is simply added to the output voltage, and not multiplied in any way) will therefore be much too low to cancel the output device Vbe's. Bob Cordell's figures show that 25 mA gives a good match for 100mA in the outputs.
Minimum output voltage is 4.13V, which seems too much; that's seven Vbe's.
Presumably the transistor in the Vbe multiplier is thermally coupled to the drivers? You mentioned something about two more diodes, so I may have that wrong.
Douglas |
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| DouglasSelf |
| quote: | Originally posted by JPV
Do you expect the thermal delay to be less critical for the drivers?
What do you think of the possibility to control the Vbe tempco of the diodes by bleeding of current via // resistor and adjust it to have minimum distortion at high temperature as I proposed in the other thread on thermaltracks
Thanks for your opinion
JPV |
As to the first question, I think that in an EF output stage the driver dissipation varies much less with output level than it does in the output devices, (see Fig 13.1 in the 4th edition of Audio Power Amplier Design) so it should be much less of a problem.
Not quite sure I understand the second question. Are you saying that the diode tempco will vary with the current through it? Apart from anything else, Bob Cordell's graph (which I take the liberty of attaching) says otherwise.
Douglas |
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| DouglasSelf |
| Another attempt to attach Bob's tempco graph. |
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| anatech |
Hi Douglas,
One manufacturer I know of used glass signal diodes between the C and E leads on the driver transistors. In this way they could track the die temperature better in the driver stage. A little heat sink grease blocked air currents from altering the temperature of the signal diodes. The Vbe multiplier transistor was mounted on the heat sink between the output devices.
From what I have seen in service, a slight overcompensation for bias control seems to work the best. Some of this is related to the heat sink size and air flow also. So an over compensated amp on my bench may well be under compensated in the customer's set up. Even the top cover position will change the bias tempco as everything else runs warmer with the top on.
I raise these points because we tend to talk about output circuitry in isolation when in fact we need to consider the entire system. In this case, all changes in the circuitry operation due to temperature rise. This can easily take a critically compensated output stage and turn it into a thermal runaway victim. I have seen this too many times over the years.
-Chris |
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| JPV |
| quote: | Originally posted by DouglasSelf
As to the first question, I think that in an EF output stage the driver dissipation varies much less with output level than it does in the output devices, (see Fig 13.1 in the 4th edition of Audio Power Amplier Design) so it should be much less of a problem.
Not quite sure I understand the second question. Are you saying that the diode tempco will vary with the current through it? Apart from anything else, Bob Cordell's graph (which I take the liberty of attaching) says otherwise.
Douglas |
1. Yes the tempco of the diode is changing with the current see post 127 and following in B cordell thread on thermaltrack.
I derived this simple equation:
tempco variation = k/q * ln ( I2/I1) where k/q = 0.086 mV/°K
this gives the usefull rule of thumb :
tempco variation = - 0.2mV/°C per decade decrease of current
The graph from B. Cordell do not show it because you need at least a decade of current drop.
If you look at the thermaltrack spec ( last figure) the diode tempco variation can be estimated and is 0.2 mV/°C for a decade drop of current
2. Then by bleeding current from the diode you adjust the tempco.
The Vbe of the diode with bias current is irrelevant because you can get the right bias in a Vbe multiplier by adjusting the multiplier factor.
Having the diodes in the upper leg of the base and the Vbe multiplier potentiometer in the lower leg allows to design and tune for the optimal bias at room temperature.
Then by adjusting the // resistors on each couple of diodes you can adjust the tempco of the diodes to have a better tracking
3. Large transient T° mistraking of output transistor Vbe will create evidently bursts of crossover distortion due to transient under or over bias. This is a non linear phenomenon difficult to model and to measure but we know it exists.
With the help of Thermaltracks, we can have a by fare better temperature tracking. Then the question arise: what tempco to utilise and how?
In the other thread, the thermaltracks where criticized because their tempco is not the same as the Vbe tempco of the ouput transistors. By adjusting the current in the diodes, you can tune the tempco to the exact value required.
Would this be optimal? No
In your book you gave the optimal Vq of about 26mV. In Oliver's paper ( I got the reference from your book thanks ! ) it is shown that the optimal bias is for :
gm* R = 1 this means Io * R = Vt = 26mV at room temp !!!!
where R is the resistance looking in the emitter of one transistor at bias, Io is the bias current and Vt is tehe thermal voltage.
Io * R is also your Vq if R is dominated by the emitter resistor.
This shows that at high temperature, the Vq so Io should rise to maintain optimal bias. This is of course a secondary effect but if we can get a good thermal tracking, then it may become a first order effect certainly at the low currents where crossover happens
3. Conclusion
By tuning the Vbias at room temperature for minimum distortion ( using a spectrum analyzer ). Then tuning the diodes tempco to minimize the distortion at high temperature, we can have a better amplifier.
The only reactions I had were from B. Cordell and Janneman. Thanks to them !!
I can be wrong but it seams that people are more interested in sophisticated circuit topologies or defending strange third order phenomena like cables, jfets and so on.
I am interested by your opinion on the reality of this. I hope to be able to measure and apply thsi in a near future.
JPV |
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| JPV |
Reading the Thermaltrack transistors NJL3281D spec, fig 15 diode forward voltage/
at 1 Amp Vf = 0.93 V at -25°C
Vf = 0.75 V at 100°V
then tempco at 1 Amp = (0.75-0.93)/125 = -1,44 mV/°C
at 1mA Vf = 0.59 V at -25°C
Vf = 0.34 v at 100°C
then tempco at 1mA = ( 0.34 - 0.59) /125 =- 2mV
Delta tempco = - 0,56mv/°C for 3 decades of current
This is about -0.2mV/°C per decade of current
JPV |
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| PMA |
| A colleague of mine made a nice video on crossover distortion change with load (temperature). He prepared optimally biased class B, with quite right thermal coupling of Vbe multiplier. He measured on oscilloscope screen, with fundamental notched. Changed load in sudden steps, i.e. changed tepmerature dynamically. One could see slowly changed crossover residual. To me, class B would never hit optimum, for many reasons. |
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| roender |
| quote: | Originally posted by PMA
... optimally biased class B, with quite right thermal coupling of Vbe multiplier. He measured on oscilloscope screen, with fundamental notched. Changed load in sudden steps, i.e. changed tepmerature dynamically. One could see slowly changed crossover residual. To me, class B would never hit optimum, for many reasons. |
Hi Pavel
It is not the same case as if you have a thermal tracking jonction insight the power BJTs.
But you might be right ... at step response the thermal tracking Vbe monitoring device will respond with some lag. On the other hand, there is no step like change in the normal audio signal, so a device with a monitoring diode very close to his BE junction will provide a minimal lag between audio signal and Vbe thermal correction signal.
Cheers,
Mihai |
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| JPV |
| quote: | Originally posted by PMA
A colleague of mine made a nice video on crossover distortion change with load (temperature). He prepared optimally biased class B, with quite right thermal coupling of Vbe multiplier. He measured on oscilloscope screen, with fundamental notched. Changed load in sudden steps, i.e. changed tepmerature dynamically. One could see slowly changed crossover residual. To me, class B would never hit optimum, for many reasons. |
We all know that crossover is changing with temperature.
The point is " quite right thermal coupling ". How was it done without on the chip sensor.
You do not give any technical comment on the content of my post.
By paralleling many ouput transistors optimally biased even in a transient way we could reach very good performance where the crossover distortion vanishes or at least be very low.
JPV |
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| albin |
what about Erik Margans use of diodes to control Iq?dare I ask Douglas.I built his test circuit but replaced the bias transistor with two diodes in series.they worked, I thought, but I could be wrong,
Please ignore or not
regards
max albin |
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| janneman |
| quote: | Originally posted by PMA
A colleague of mine made a nice video on crossover distortion change with load (temperature). He prepared optimally biased class B, with quite right thermal coupling of Vbe multiplier. He measured on oscilloscope screen, with fundamental notched. Changed load in sudden steps, i.e. changed tepmerature dynamically. One could see slowly changed crossover residual. To me, class B would never hit optimum, for many reasons. |
I think you missed the point of JPV's post. He argues that it is possible to all but eliminate that thermally-induced operating point transient due to sudden load changes.
What do you think of his reasoning?
Jan Didden |
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| PMA |
Mihai,
there is nothing like infinitely fast step in audio signal, I agree. Let's say it is 17us. The thermal settling would be in tens of ms - 1000 to 10000x longer.
I have never seen perfect thermal sensing and FB yet, resulting in absolutely perfect bias regulation. |
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| janneman |
PMA,
What would you guess that the thermal time constant is from applying a load to the reaction of the Vbe and the bias?
Jan Didden |
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| PMA |
| Similar (i.e. that long one), but not absolutely same in response as sense/FB circuit. |
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| janneman |
| quote: | Originally posted by PMA
Similar (i.e. that long one), but not absolutely same in response as sense/FB circuit. |
But the sensor is on the same die. If you don't know what the numbers are, how can you know they are different and how long they actually are?
What happened to your sharp reasoning mind?
And of course that 17uS is nonsense.
Jan Didden |
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| PMA |
We have some practical results with ThermalTrak.
17us is a nonsense as thermal time constant, it is the rise time of el. signal step (tr10%-90%) through 20kHz BW limited system, quite exactly. |
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| jneutron |
| quote: | Originally posted by janneman
But the sensor is on the same die. If you don't know what the numbers are, how can you know they are different and how long they actually are?
What happened to your sharp reasoning mind?
And of course that 17uS is nonsense.
Jan Didden |
Jan, from the other thread, somebody mentioned a run of devices that had the incorrect diode put in. That would mean the die is an independent descrete device..
From the junction to the bottom of the die, a typical number I've seen used for transient thermal is 100 uSec to 1 milli. Junction to case was in the 10 to 100 mSec. For a die that is epoxy bonded to the top of a bipolar junction, I'd expect the time constant to be in the 10 to 100 millisecond range.
It's easy enough to check, why hasn't anybody done so?
Cheers, John |
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| DouglasSelf |
| quote: | Originally posted by anatech
Hi Douglas,
One manufacturer I know of used glass signal diodes between the C and E leads on the driver transistors. In this way they could track the die temperature better in the driver stage. A little heat sink grease blocked air currents from altering the temperature of the signal diodes. The Vbe multiplier transistor was mounted on the heat sink between the output devices.
From what I have seen in service, a slight overcompensation for bias control seems to work the best. Some of this is related to the heat sink size and air flow also. So an over compensated amp on my bench may well be under compensated in the customer's set up. Even the top cover position will change the bias tempco as everything else runs warmer with the top on.
I raise these points because we tend to talk about output circuitry in isolation when in fact we need to consider the entire system. In this case, all changes in the circuitry operation due to temperature rise. This can easily take a critically compensated output stage and turn it into a thermal runaway victim. I have seen this too many times over the years.
-Chris |
Hi Chris
That's a most interesting idea. Am I right in thinking that the idea is to sense the temp of the collector lead, because that will be directly connected to the metal area on the package?
I have only seen one case of thermal runaway in decades of amplifier work. I think you would need a seriously under-sized heatsink for it to happen. My last commercial design had the sensor on the drivers rather than the main heatsink (for reasons I won't bore you with) which may not be ideal, but in severe testing there was never a hint of runaway- and that was with the additional heating from Class-XD operation.
Douglas |
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| Wavebourn |
| If thermal runaway is the rare case, oscillations of thermal compensation causing bias instability with the real musical signal are very common. |
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| ostripper |
| quote: | | posted by douglasself..I have only seen one case of thermal runaway in decades of amplifier work. I think you would need a seriously under-sized heatsink for it to happen |
I think you have described most HT receivers on the market
today! :bigeyes: (thank god they have thermal cutoff in their protection circuits) .
The idea of putting the Vbe on the collector leg is cool:cool:
I'll try it.(poor mans way of getting thermaltrak performance)
BTW.. Thanks,Mr. Self, for the BEST book on amplifier design
and for the "blameless design" which with proper layout
and device selection is one killer amp. I look forward to the latest
edition (I only have Ed.3).
Thanks,OS |
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| DouglasSelf |
| quote: | Originally posted by JPV
1. Yes the tempco of the diode is changing with the current see post 127 and following in B cordell thread on thermaltrack.
I derived this simple equation:
tempco variation = k/q * ln ( I2/I1) where k/q = 0.086 mV/°K
this gives the usefull rule of thumb :
tempco variation = - 0.2mV/°C per decade decrease of current
The graph from B. Cordell do not show it because you need at least a decade of current drop.
If you look at the thermaltrack spec ( last figure) the diode tempco variation can be estimated and is 0.2 mV/°C for a decade drop of current
2. Then by bleeding current from the diode you adjust the tempco.
The Vbe of the diode with bias current is irrelevant because you can get the right bias in a Vbe multiplier by adjusting the multiplier factor.
Having the diodes in the upper leg of the base and the Vbe multiplier potentiometer in the lower leg allows to design and tune for the optimal bias at room temperature.
Then by adjusting the // resistors on each couple of diodes you can adjust the tempco of the diodes to have a better tracking
3. Large transient T° mistraking of output transistor Vbe will create evidently bursts of crossover distortion due to transient under or over bias. This is a non linear phenomenon difficult to model and to measure but we know it exists.
With the help of Thermaltracks, we can have a by fare better temperature tracking. Then the question arise: what tempco to utilise and how?
In the other thread, the thermaltracks where criticized because their tempco is not the same as the Vbe tempco of the ouput transistors. By adjusting the current in the diodes, you can tune the tempco to the exact value required.
Would this be optimal? No
In your book you gave the optimal Vq of about 26mV. In Oliver's paper ( I got the reference from your book thanks ! ) it is shown that the optimal bias is for :
gm* R = 1 this means Io * R = Vt = 26mV at room temp !!!!
where R is the resistance looking in the emitter of one transistor at bias, Io is the bias current and Vt is tehe thermal voltage.
Io * R is also your Vq if R is dominated by the emitter resistor.
This shows that at high temperature, the Vq so Io should rise to maintain optimal bias. This is of course a secondary effect but if we can get a good thermal tracking, then it may become a first order effect certainly at the low currents where crossover happens
3. Conclusion
By tuning the Vbias at room temperature for minimum distortion ( using a spectrum analyzer ). Then tuning the diodes tempco to minimize the distortion at high temperature, we can have a better amplifier.
The only reactions I had were from B. Cordell and Janneman. Thanks to them !!
I can be wrong but it seams that people are more interested in sophisticated circuit topologies or defending strange third order phenomena like cables, jfets and so on.
I am interested by your opinion on the reality of this. I hope to be able to measure and apply thsi in a near future.
JPV |
Hi JPV, sorry I couldn't get back to you yesterday.
My first attempt at applying the ThermalTraks was to assume that the diodes would be used to attempt to directly cancel the transistor Vbe voltages,which means running the diodes at 25 mA (for transistor Ic=100mA) and accepting that the tempco would not match exactly. The compensation should, I think, still be much better than conventional methods but it is certainly not theoretically perfect. This is the approach I used in the attached circuit, which I think of as Plan A. I have never tried to run a VAS at 25 mA,and I fear the linearity might be worse. Certainly TO-92 transistors will get too hot for amplifiers above 100W/8R, which is a problem as the bigger devices tend to have less beta. A VAS buffer may be needed.
I accept what you say about increasing diode tempco's by reducing the current. I have just reread RA Pease's thoughts on it; http://www.national.com/rap/Story/vbe.html and and he certainly quotes - 0.2mV/°C per decade decrease of current. Our diodes have a tempco of 1.7 mV/degC and the transistor Vbe has a tempco of 2.14 mV/degC at 25mA (Bob Cordell's figures) so diode tempco needs to be increased by 0.44 mV/degC. The diode current therefore needs to reduced by 2.2 decades, or 158 times. That gives us a diode current of 158 uA. (not a typo) This is much too low for a VAS operating current so we need a circuit that will replicate the diode voltage at much greater current.
The lower current is clearly not going to give us enough diode voltage to cancel the transistor Vbe's directly. However we cannot use a conventional multiplier circuit as in multiplying the voltage we also multiply the tempco. What is needed is to add a fixed voltage to the diode voltage. If we assume we are using an EF type 2 output, then we also need compensation for the driver Vbe's which can be done with a conventional Vbe multiplier circuit, as used in Plan A.
Another point is that a diode current of 158 uA is quite small, and whatever circuitry we use will have to have its base currents looked at carefully if it is not be inaccurate.
It therefore looks as if we need a system like that shown conceptually in the second diagram. Vdriver compensates the driver Vbe's. The fixed voltage is derived from black box called Vfix, and the diode voltages are replicated by a black box called Vdiodes. I am calling this approach Plan B. It should give much better compensation than Plan A, but it's going to take a bit of designing...
What do people think?
I'm not sure if I can put two pictures in one post, but I'm about to find out.
Douglas |
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| DouglasSelf |
| Apparently not. Here is Plan B. |
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| megajocke |
But if we put the diodes in the upper leg of the Vbe multiplier base circuit (like leach) their tempco won't be multiplied. And 160µA sounds like a reasonable current to have flowing there, just use a buffer for the Vbe multiplier transistor and it will have enough beta to be stable with vas current.
Or is your worry that when doing this the tempco of the vbe multiplier transistor part will be too high, decreasing idle current with increased ambient/driver temperature? But this should be pretty stable.
edit:
"But this should be pretty stable." was referring to the ambient temperature which won't vary as much as output transistor case temperature. |
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| gaetan8888 |
| quote: | Originally posted by DouglasSelf
Hi Gaetan
I had a quick look at this circuit today, and I'm not sure it will do the job. The current through the diodes is only 1.2 mA, and their voltage drop (which with this circuit is simply added to the output voltage, and not multiplied in any way) will therefore be much too low to cancel the output device Vbe's. Bob Cordell's figures show that 25 mA gives a good match for 100mA in the outputs.
Minimum output voltage is 4.13V, which seems too much; that's seven Vbe's.
Presumably the transistor in the Vbe multiplier is thermally coupled to the drivers? You mentioned something about two more diodes, so I may have that wrong.
Douglas |
Hello Douglas
I alway use EF output stage in my diy amps and I use oversize heasink for the drivers, so the dissipation on the drivers varies much less, in that way I only thermally couple the transistor in the Vbe multiplier to the output transistors.
Thank
Bye
Gaetan |
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