I just don't like junctions turning on and off at arbitrary points on the transfer curve. It will certainly cause high order harmonics, and induce spikes into the rest of the amplifier. And probably will not improve distortion at all, compared to an ordinary EF which has more current gain and where the drivers never turn off.
If you still want to use it, it is your call, I will simply relate my experience with this sort of thing. I am not telling you not to.
If you still want to use it, it is your call, I will simply relate my experience with this sort of thing. I am not telling you not to.
I can totally see your point of view on this, but in practice it seems to work well. The little circuit on my bench right now has no visible crossover glitch in the error signal . It's obviously still there somewhere, it's apparently quite small. I have not yet connected the circuit to any FFT analysis.
Edit: I have been scoping the error signal with a MUSIC source, I will try a sine later, that will probably reveal any less obvious crossover artefacts.
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Two of these in a mint tin?
What is the output bias current through the 1R resistors?
What is THD at 9V into 6R, at 1KHz and 20KHz?
I feel this circuit is more complex than it needs to be. Q3 and Q6 will not perform as well as a simple bootstrap I think (reference VAS bootstrap to Q12 emitter).
Use BC5xx for Q9/10 for high CM output impedance. Use BC5xxC for Q11 for more VAS gain. Use BC5xxC for Q1/Q2 for lower distortion and more linear input impedance.
Your CFP output will show Gm doubling. While both outputs are conducting Ro will be .5R, at excursions 1R. In order to bring Ro back to normal at crossover the CFPs each must have an Ro of 1R at the crossover point. The Ro of a CFP is the Re of the driver divided by the current gain of the output. At 13mA the output has a current gain of no more than 26, and the driver an Re of 4.5, so output impedance is about 170mR. If you don't have serious Gm doubling, then something is broken, or you are simply not drawing more than 26mA from your load.
What is THD at 9V into 6R, at 1KHz and 20KHz?
I feel this circuit is more complex than it needs to be. Q3 and Q6 will not perform as well as a simple bootstrap I think (reference VAS bootstrap to Q12 emitter).
Use BC5xx for Q9/10 for high CM output impedance. Use BC5xxC for Q11 for more VAS gain. Use BC5xxC for Q1/Q2 for lower distortion and more linear input impedance.
Your CFP output will show Gm doubling. While both outputs are conducting Ro will be .5R, at excursions 1R. In order to bring Ro back to normal at crossover the CFPs each must have an Ro of 1R at the crossover point. The Ro of a CFP is the Re of the driver divided by the current gain of the output. At 13mA the output has a current gain of no more than 26, and the driver an Re of 4.5, so output impedance is about 170mR. If you don't have serious Gm doubling, then something is broken, or you are simply not drawing more than 26mA from your load.
By tweaking the bias while scoping the error signal, optimal bias appears to be 9mA. Above that results in gm doubling, below that results in crossover distortion. Wouldn't you get gm doubling with any other over-biased output stage?
The usage of the 2n390x was just due to me having lots on hand, this was never intended to be a super-refined design - it's on a breadboard - I can't physically fit any niceties on there without it getting more haywire than it is already!
The usage of the 2n390x was just due to me having lots on hand, this was never intended to be a super-refined design - it's on a breadboard - I can't physically fit any niceties on there without it getting more haywire than it is already!
Also, regarding the issue with the emitter resistors, while admittedly high in value, (but what I had nearby), wouldn't the gain reduction around the crossover point counteract any gm doubling effect those resistors would have? There will be a curved turn-on slope which you attempt to geometrically mirror with the other half.
If I had lower values at hand I would have used them, of course.
If I had lower values at hand I would have used them, of course.
What about a TDA1521
http://www.datasheetcatalog.org/datasheet/philips/TDA1521Q.pdf
Very few components. Room for a bottom fan with ventilation holes trough the bottom (Allowed?) each side.
External symetrical supply.
http://www.datasheetcatalog.org/datasheet/philips/TDA1521Q.pdf
Very few components. Room for a bottom fan with ventilation holes trough the bottom (Allowed?) each side.
External symetrical supply.
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That makes sense. With 9mA bias you get 410mR-600mR Ro for the CFP's, much closer to 1R.
Emitter resistors MUST be relatively high in this application to achieve low bias current without excess distortion. 1R is a good compromise for 9W. The dynamic emitter resistances are such that they can never perfectly cancel. However the best crossover occurs in an EF when the emitter resistors have a voltage drop of 26mV or so. We bias the current of the output stage so that at crossover, the emitter resistances will equal the degeneration resistances and this way, when both outputs are conducting and the degens are seen in parallel, the added emitter resistances make up for the halved degen resistance.
This design is interesting from a design angle perspective, but what are you comparing it to? What is it supposed to be better than? If it is not supposed to be a competing design then maybe you should make your own thread, after all this is a contest.
Your CFP does evidently have the advantage that it can use smaller emitter resistors than an equivalent EF. An equivalent EF would need degeneration of 2.66-8R, which is too much if you want to push 2A. However it has worse current gain and I suspect it will have more distortion overall than a simple EF biased at 29mA or so with .82R degeneration.
Emitter resistors MUST be relatively high in this application to achieve low bias current without excess distortion. 1R is a good compromise for 9W. The dynamic emitter resistances are such that they can never perfectly cancel. However the best crossover occurs in an EF when the emitter resistors have a voltage drop of 26mV or so. We bias the current of the output stage so that at crossover, the emitter resistances will equal the degeneration resistances and this way, when both outputs are conducting and the degens are seen in parallel, the added emitter resistances make up for the halved degen resistance.
This design is interesting from a design angle perspective, but what are you comparing it to? What is it supposed to be better than? If it is not supposed to be a competing design then maybe you should make your own thread, after all this is a contest.
Your CFP does evidently have the advantage that it can use smaller emitter resistors than an equivalent EF. An equivalent EF would need degeneration of 2.66-8R, which is too much if you want to push 2A. However it has worse current gain and I suspect it will have more distortion overall than a simple EF biased at 29mA or so with .82R degeneration.
I wonder what is the discrete parts version of DMOSfet outputs?
Yes, a non-switching amp will do nicely.qusp said:
That size stereo amp will fit on a credit card size PCB or can be made point to point even smaller.bobodioulasso said:
There's also the available option of NuVistor and subminature-8 tubes/valves boosted with a solid state output.
The contest is not about distortion numbers. Well, partly.
I take "useful efficient linear" to mean, it can get loud enough for computer monitors or a bedroom, it has low enough distortion to be considered enjoyably transparent, and doesn't smoke, burn, or explode.
I think for a discrete design the transistor count is one of the most important qualities. For AB outputs we will need at least 4 transistors for two channels. 8 if we use drivers. 10 with a VAS. At least 12 with all this and a singleton input.
I take "useful efficient linear" to mean, it can get loud enough for computer monitors or a bedroom, it has low enough distortion to be considered enjoyably transparent, and doesn't smoke, burn, or explode.
I think for a discrete design the transistor count is one of the most important qualities. For AB outputs we will need at least 4 transistors for two channels. 8 if we use drivers. 10 with a VAS. At least 12 with all this and a singleton input.
keantoken,
These are oscillograms taken from the amp with no emitter resistors.
Here is the error signal with zero bias:
Here is the error signal with what appears to be optimal bias.
It's amazing how the artefacts aren't visible in the error signal. Note that the funny notches on the peaks of the sine wave are from the signal generator.
The picture was far less pretty with 1R emitter resistors, sadly, though it was possible to bias to an optimum. Sadly some resistance is needed in a real amp.
These are oscillograms taken from the amp with no emitter resistors.
Here is the error signal with zero bias:
Here is the error signal with what appears to be optimal bias.
It's amazing how the artefacts aren't visible in the error signal. Note that the funny notches on the peaks of the sine wave are from the signal generator.
The picture was far less pretty with 1R emitter resistors, sadly, though it was possible to bias to an optimum. Sadly some resistance is needed in a real amp.
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