Old school at the limit, inspired by the CD thread, a standard amp of the last millennium flowed from my fingers ... a kind of Santa Claus surprise today!
36 Peak Watts at 8 Ohm
Current Dumper without dump, only NFB.
TIP3055, BD139, BD140, BC5xx ... thats all we need.
Happy DIY,
HBt.
36 Peak Watts at 8 Ohm
Current Dumper without dump, only NFB.
TIP3055, BD139, BD140, BC5xx ... thats all we need.
Happy DIY,
HBt.
The fascinating thing about the old, very simple designs is, among other things, the fact that they can also be realized with the other conductor types, i.e. as boys and as girls. They are therefore gender-appropriate and work optimally biased with an Iq of around 60mAdc. A bootstrap is completely missing, which limits us, but secures us in a different way - the straightforward idea remains.
"Bub & Babe" with only 5 watts of losses; a B mode, which actually belongs to class AB.
HBt.
"Bub & Babe" with only 5 watts of losses; a B mode, which actually belongs to class AB.
HBt.
What's the use of R1? It cannot improve thermal stability with the emitter of Q2 connected to its upper side.
There was no particular reason for this, it was solely due to my dimensioning method (from right to left) and the five minutes.
But this will certainly benefit skeptics such as Mr. X.
There are said to be audio enthusiasts who are very skeptical of the traditional bootstrap (i.e. sound-damaging). I am not one of them.
😉
regards,
HBt.
The lack of current-sensing resistors around the loop formed by Q1, Q2, Q4, and Q6 means that this is another amplifier whose bias current will vary enormously with the temperatures of the transistors.
Ed
Ed
I have been waiting for this.
Q6 and Q7 sit on a common heat sink, Q4 is located entirely in a matching hole (with thermal paste) between the candidates.
Now there are four BJTs left, which need to be cleverly thermally coupled ...
I'm not afraid of any problems.
HBt.
Psst wahab predicted an extreme drift in the rapid design of the Beelzebub, but in practice this was not the case and could have been compensated for by the thermal coupling of the IP-VAS BJTs.
Good.hbtaudio said:
Without resistors, the sensitivity will be 10C temperature difference -> 2.5x bias current. The thermal resistance between the dies, packages, and heatsink can easily cause that.
Ed
That would be too boring for me. But tell me, or better yet, give a tangible example: what is your daily design with which you enjoy music?
The no-fault amplifier would be the famous (amplifying) wire (with infinite power output) ..!
HBt.
My amplifier design is Top Secret 😉, but I have seen the same techniques in use in various amplifier designs on this board. I should clarify "no audible faults". An expensive analyzer could detect its faults.
Ed
Ed
👍 "excellent",
Let me take a guess:
Hellraiser with /and Spooky ..! Wolverine, the Amp with no audible faults ... right?
regards,
HBt.
Dear Marcel,
you're making me feel insecure with your short one-liner. Would you please enter into an analysis and explain in a little more detail for the readers of this forum why "old school /Alte Schule" should cause thermal problems - perhaps in the style of causal chains.
If I have expressed myself incomprehensibly, I apologize generously - a sleepless night lies behind me.
thx,
HBt.
I don't see what purpose R1 has. If you want to measure the voltage across a resistor to measure the quiescent current, you can measure across R2. If the emitter of Q2 were connected to its other side, it could improve thermal stability by limiting the transconductance at large current levels, but that is not applicable.
I simply have the same concern as Ed. Suppose you play loud signals for some time, causing substantial power dissipation in Q6. Due to the thermal resistance from junction to heatsink, Q6 will then become hotter than Q1, Q2 and Q4. As a result, its Is increases, or in other words, its collector current at a given base-emitter voltage increases.
If Q6, Q1, Q2 and Q4 are mounted on the same heatsink, the temperature-dependence of Q6 is only partly compensated for by Q4 due to the temperature drop across the thermal resistance from junction to heatsink of Q6, that is, due to Q6 being hotter than the rest. If this causes thermal runaway, the amplifier will self-destruct. If the self-heating due to the increase of the quiescent current after playing loud is much smaller than the self-heating that caused it, it is not a big problem, although it still worsens the dependence of the distortion on the programme dynamics.
It might have its advantages to put Q1, Q2 or both on a separate heatsink. Due to the gain of the VBE multiplier, the self-heating of Q6 may then even be overcompensated. I don't really know what effect the thermal capacitance of the main heatsink will have, though.
I simply have the same concern as Ed. Suppose you play loud signals for some time, causing substantial power dissipation in Q6. Due to the thermal resistance from junction to heatsink, Q6 will then become hotter than Q1, Q2 and Q4. As a result, its Is increases, or in other words, its collector current at a given base-emitter voltage increases.
If Q6, Q1, Q2 and Q4 are mounted on the same heatsink, the temperature-dependence of Q6 is only partly compensated for by Q4 due to the temperature drop across the thermal resistance from junction to heatsink of Q6, that is, due to Q6 being hotter than the rest. If this causes thermal runaway, the amplifier will self-destruct. If the self-heating due to the increase of the quiescent current after playing loud is much smaller than the self-heating that caused it, it is not a big problem, although it still worsens the dependence of the distortion on the programme dynamics.
It might have its advantages to put Q1, Q2 or both on a separate heatsink. Due to the gain of the VBE multiplier, the self-heating of Q6 may then even be overcompensated. I don't really know what effect the thermal capacitance of the main heatsink will have, though.
It's a neat simple design but I guess Marcel is looking at what happens if the output stage current drifts up because Q6 and Q7 heat up. This means the voltage across R1 goes up and that turns Q2 on harder which will exacerbate the problem. Better top take the emitter of Q2 to the output rail. You could run a DC analysis to see what the spread is over temperature.
🙂
🙂
As long as all transistors heat up equally, like they normally do in SPICE, there is no problem. If there is an issue, you will have to take temperature differences between the transistors into account, as well as the effect of large signals, to see it.
Regarding the effect of large signals, the loop gain of the positive feedback via the self-heating increases at larger currents. The emitter resistors that are usually used to improve thermal stability are of the order of 1/gm or even smaller in the quiescent point, but >> 1/gm at higher current levels.
Regarding the effect of large signals, the loop gain of the positive feedback via the self-heating increases at larger currents. The emitter resistors that are usually used to improve thermal stability are of the order of 1/gm or even smaller in the quiescent point, but >> 1/gm at higher current levels.
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I wonder if you could make an Even Older School version with only NPNs. The old PNPs were rather bad and even worse on ICs.
Tom
Tom