Re: Mike
You are absolutely correct on that last point...many amps. with correctly designed, three-resistor single slope schemes work just fine...the difficulty with Selfs' so-called 'load invariant' design for example, is if you scale the power supplies to +/-40v, (note that he recommends usage up to these values), the protection network will be activated almost constantly....
In general, i would suggest folks give the simple single slope scheme with two resistors a wide berth........
ALW said:
That may be one of the reasons, and you obviously have a vested interest there
I can think of many designs though that use similarly simple, single-slope, SOA protection and sound MUCH better, so I feel, with the greatest of respect, that's not the most important factor.
Andy.
You are absolutely correct on that last point...many amps. with correctly designed, three-resistor single slope schemes work just fine...the difficulty with Selfs' so-called 'load invariant' design for example, is if you scale the power supplies to +/-40v, (note that he recommends usage up to these values), the protection network will be activated almost constantly....
In general, i would suggest folks give the simple single slope scheme with two resistors a wide berth........
"i think the most significant short coming of Selfs' designs as published, is the extremly poor single slope S.O.A protection network"
Mikek, have you ever built one of Self's designs? Why do you conclude that the SOA network is the biggest shortcoming? If this were fixed then what do you think the next biggest shortcoming would be?
Mikek, have you ever built one of Self's designs? Why do you conclude that the SOA network is the biggest shortcoming? If this were fixed then what do you think the next biggest shortcoming would be?
charge storage and removal
quote
"However, i think you'll find that i specifically refered to the double complementary emitter follower, (modified Darlington if you like), with cross-coupled, (shared) emitter resistors for driver(s), as used, (for instance), in the Leach 'low-TIM'.
The shared emitter resistor in the complementary driver stage facilitates the reverse bias of the base-emitter junction of the output BJT being turned off..(see D. Selfs' book), which improves the rate of charge extraction significantly...."
I looked up the Leach circuit at
http://users.ece.gatech.edu/~mleach/lowtim/
I presume the complementary emitter with shared resistors
refer to Q14-Q17. Can someone explain
1. how this configuration helps to reduce (eliminate?)
charge storage phenomenon in the output devices
and thereby improve performance?
2. Is there a reason why Leach uses 2 pairs of
complementary devices instead of one? I presume it is
to quicken the removal of stored charge?
3. What amp performance parameter does the stored
charge degrade?
Thanks in advance, your help is much appreciated.
Yv
quote
"However, i think you'll find that i specifically refered to the double complementary emitter follower, (modified Darlington if you like), with cross-coupled, (shared) emitter resistors for driver(s), as used, (for instance), in the Leach 'low-TIM'.
The shared emitter resistor in the complementary driver stage facilitates the reverse bias of the base-emitter junction of the output BJT being turned off..(see D. Selfs' book), which improves the rate of charge extraction significantly...."
I looked up the Leach circuit at
http://users.ece.gatech.edu/~mleach/lowtim/
I presume the complementary emitter with shared resistors
refer to Q14-Q17. Can someone explain
1. how this configuration helps to reduce (eliminate?)
charge storage phenomenon in the output devices
and thereby improve performance?
2. Is there a reason why Leach uses 2 pairs of
complementary devices instead of one? I presume it is
to quicken the removal of stored charge?
3. What amp performance parameter does the stored
charge degrade?
Thanks in advance, your help is much appreciated.
Yv
I'll take a stab - maybe someone else can then explain where I am wrong so I can learn where I am mis-thinking things.
1. how this configuration helps to reduce (eliminate?) charge storage phenomenon in the output devices and thereby improve performance?
The way I look at it is to consider a basic, plain old common emitter, single stage amplifier. To simplify things, assume a digital mode of operation. If we apply a voltage to the base, the transistor will turn on and saturate, quickly causing curent to flow in the collector resistor and develop an output voltage. If we remove the base voltage, the transistor will turn off, thus removing collector current and the collector voltage will return to Vcc. Note, however, that while the collector voltage will immediately drop towards Vsat when turned on, when turned off, the collector voltage will rise relatively slowly back towards Vcc as there is nothing to drive this voltage to where it should be. The charges that have developed have no discharge path, in other words, other than though the collector load and the reversed biased transistor junction. If the load has a high impedance, the discharge will take longer. This is why high speed digital circuits need to be actively driven both high and low.
We have the same basic situation in an audio power amp. The NPN output transistors will be in cutoff nearly half the time, and when they are first cut off, unless the base is driven, the residual charge on the transistor will have no low impedance path to discharge through. This is the purpose of the Class A driver stage - it will drive the base voltage negative and force the stored charge to discharge rapidly rather than slowly discharge through reversed biased PN junctions.
2. Is there a reason why Leach uses 2 pairs of
complementary devices instead of one? I presume it is
to quicken the removal of stored charge?
If you are referring to the 2 pair of driver transistors, I think it has more to do with getting sufficient current gain. We do not want to load down the voltage amplifier (VA) stages. For a 120 watt output at 8 Ohms, you need nearly 4 Amps of current. The MJ15003 has a minimum beta of 25 (per Leach), thus the drivers need to be able to put out 160 mA (I am ignoring bias currents). The MJE15030 driver has a minimum gain of 20, so it needs 8 mA of current. If the amp had just a common Darlington output, the voltage amplifier would have to provide 8 mA of drive. While this is not a lot of current you might say, at the 31 volt signal level that would be needed, the load to the VA is 3.9 KOhms (per R = V/I), and this would significantly degrade the voltage gain. By adding a predriver (which Leach says has a gain of 40), the current is reduced to .2 mA, and this sufficiently unloads the VA such that gain variations due to load current are negligable.
Note that for lower power amplifiers, a triple Darlington may not needed.
3. What amp performance parameter does the stored
charge degrade?
I suppose harmonic distortion.
While on the subject of the Leach Amp, and on other designs that utilize base resistors in the output drivers, the above calculations show a problem. Note that under worst case conditions the driver transistors need 160 mA of base current. If one uses the 10 Ohm base resistors Leach has been showing on all Version 4 amplifiers, there would be a 16 mV drop across them. While feedback can help correct for this drop, I am nervous about this. Leach claims he added the base resistors due to some problems one person had with the Super Amp with oscillations. The base resistors cured the problem, so he started adding base resistors to all his designs. To me, adding resistors in the signal path to prevent a problem that has apparantly only occured once, and in a different design, is not a good enough reason for me, as a DIY'er, to use them in my own amplifier. Therefore, I recommend omitting these resistors. If one happens to have problems with oscillations, they can always be added back in. Especially in a low feedback design, the voltage error will not be fully corrected. I have no idea if it is audible or not, but if people can detect sonic differences in a single resistor swap, then I have to belive that even one unneeded resistor has to have some impact.
1. how this configuration helps to reduce (eliminate?) charge storage phenomenon in the output devices and thereby improve performance?
The way I look at it is to consider a basic, plain old common emitter, single stage amplifier. To simplify things, assume a digital mode of operation. If we apply a voltage to the base, the transistor will turn on and saturate, quickly causing curent to flow in the collector resistor and develop an output voltage. If we remove the base voltage, the transistor will turn off, thus removing collector current and the collector voltage will return to Vcc. Note, however, that while the collector voltage will immediately drop towards Vsat when turned on, when turned off, the collector voltage will rise relatively slowly back towards Vcc as there is nothing to drive this voltage to where it should be. The charges that have developed have no discharge path, in other words, other than though the collector load and the reversed biased transistor junction. If the load has a high impedance, the discharge will take longer. This is why high speed digital circuits need to be actively driven both high and low.
We have the same basic situation in an audio power amp. The NPN output transistors will be in cutoff nearly half the time, and when they are first cut off, unless the base is driven, the residual charge on the transistor will have no low impedance path to discharge through. This is the purpose of the Class A driver stage - it will drive the base voltage negative and force the stored charge to discharge rapidly rather than slowly discharge through reversed biased PN junctions.
2. Is there a reason why Leach uses 2 pairs of
complementary devices instead of one? I presume it is
to quicken the removal of stored charge?
If you are referring to the 2 pair of driver transistors, I think it has more to do with getting sufficient current gain. We do not want to load down the voltage amplifier (VA) stages. For a 120 watt output at 8 Ohms, you need nearly 4 Amps of current. The MJ15003 has a minimum beta of 25 (per Leach), thus the drivers need to be able to put out 160 mA (I am ignoring bias currents). The MJE15030 driver has a minimum gain of 20, so it needs 8 mA of current. If the amp had just a common Darlington output, the voltage amplifier would have to provide 8 mA of drive. While this is not a lot of current you might say, at the 31 volt signal level that would be needed, the load to the VA is 3.9 KOhms (per R = V/I), and this would significantly degrade the voltage gain. By adding a predriver (which Leach says has a gain of 40), the current is reduced to .2 mA, and this sufficiently unloads the VA such that gain variations due to load current are negligable.
Note that for lower power amplifiers, a triple Darlington may not needed.
3. What amp performance parameter does the stored
charge degrade?
I suppose harmonic distortion.
While on the subject of the Leach Amp, and on other designs that utilize base resistors in the output drivers, the above calculations show a problem. Note that under worst case conditions the driver transistors need 160 mA of base current. If one uses the 10 Ohm base resistors Leach has been showing on all Version 4 amplifiers, there would be a 16 mV drop across them. While feedback can help correct for this drop, I am nervous about this. Leach claims he added the base resistors due to some problems one person had with the Super Amp with oscillations. The base resistors cured the problem, so he started adding base resistors to all his designs. To me, adding resistors in the signal path to prevent a problem that has apparantly only occured once, and in a different design, is not a good enough reason for me, as a DIY'er, to use them in my own amplifier. Therefore, I recommend omitting these resistors. If one happens to have problems with oscillations, they can always be added back in. Especially in a low feedback design, the voltage error will not be fully corrected. I have no idea if it is audible or not, but if people can detect sonic differences in a single resistor swap, then I have to belive that even one unneeded resistor has to have some impact.
Yves:
The common driver resistor referred to in Leach's amplifier is R36. Self and Sloan (and others) recommend paralleling a 1uF film capacitor across this resistor to facilitate charge removal. It
improves high frequency distortion (as measured with THD meter; IM distortion might be more revealing).
I have done this on two different versions of the amplfier with
no problems; I can't say that I heard any difference in the sound.
I left in the 10 ohm base resistors, which may be a factor; as Jeff
points out, these resistors could be removed (and replaced with jumpers, of course). If there are no stability problems without the resistors, fine.
Paralleling output transistors helps to deal with 'beta droop' at higher currents; three or even four pairs of output devices would
not be a bad idea. Some output transistors exhibit less loss,
such as the 2SA1302/2SC3261 (best) or MJ21193/94. I have
used the latter because they are in TO-3 metal packages, which
are compatible with the only heatsinks I have.
Self thinks the problem ultimately lies in the driver transistors, if I read his articles correctly, but I don't know of any recommended
parts. The common practice as noted above is parallel output devices (often necessary to improve SOA anyway) and to use better devices.
Neither here nor there: I wonder why virtually no one uses intermodulation distortion as a benchmark? I have to use a (Heathkit) THD analyzer because it's the only working analyzer
I have, but if I could ever repair this much-abused Crown IMD (both meters are blown, among other things), I'd use that instead.
The common driver resistor referred to in Leach's amplifier is R36. Self and Sloan (and others) recommend paralleling a 1uF film capacitor across this resistor to facilitate charge removal. It
improves high frequency distortion (as measured with THD meter; IM distortion might be more revealing).
I have done this on two different versions of the amplfier with
no problems; I can't say that I heard any difference in the sound.
I left in the 10 ohm base resistors, which may be a factor; as Jeff
points out, these resistors could be removed (and replaced with jumpers, of course). If there are no stability problems without the resistors, fine.
Paralleling output transistors helps to deal with 'beta droop' at higher currents; three or even four pairs of output devices would
not be a bad idea. Some output transistors exhibit less loss,
such as the 2SA1302/2SC3261 (best) or MJ21193/94. I have
used the latter because they are in TO-3 metal packages, which
are compatible with the only heatsinks I have.
Self thinks the problem ultimately lies in the driver transistors, if I read his articles correctly, but I don't know of any recommended
parts. The common practice as noted above is parallel output devices (often necessary to improve SOA anyway) and to use better devices.
Neither here nor there: I wonder why virtually no one uses intermodulation distortion as a benchmark? I have to use a (Heathkit) THD analyzer because it's the only working analyzer
I have, but if I could ever repair this much-abused Crown IMD (both meters are blown, among other things), I'd use that instead.
"If one uses the 10 Ohm base resistors Leach has been showing on all Version 4 amplifiers, there would be a 16 mV drop across them. While feedback can help correct for this drop, I am nervous about this. "
This is a why bother? I would be more concerned about the extra 2,000mV base drive required on the outputs because of the 0R33 emitter resistors when driving full output at 4R from 50V supplies.
This is a why bother? I would be more concerned about the extra 2,000mV base drive required on the outputs because of the 0R33 emitter resistors when driving full output at 4R from 50V supplies.
leach ckt
JeffR, Damon,
Many thanks, I am beginning to understand it better.
The drivers of the output devices need to be in Class A.
If that is the only reason, then R36 could be split into two
and their common point
connected to the output too (R45 and R46 common node).
It seems that not connecting R36 and not splitting it up
and having a pnp (Q17) are required to speed up
base discharge. Is it because the pnp is required to actively
pull charge away that a single-ended Q16 will not be able
to do? This is where I am stumped.
cheers
Yv
JeffR, Damon,
Many thanks, I am beginning to understand it better.
The drivers of the output devices need to be in Class A.
If that is the only reason, then R36 could be split into two
and their common point
connected to the output too (R45 and R46 common node).
It seems that not connecting R36 and not splitting it up
and having a pnp (Q17) are required to speed up
base discharge. Is it because the pnp is required to actively
pull charge away that a single-ended Q16 will not be able
to do? This is where I am stumped.
cheers
Yv
If R36 is split in two and the common points tied to the output, I think the stage will no longer be Class A, but AB. For example, if the output goes negative, with the top end of R36 being tied to the load, Q16 can now be reversed biased and cut off. I highly advise not splitting up R36!
You need Q17 to drive Q19 and Q21. Q16 is needed, also, to pull charge from Q19/21, just as Q17 is also used to pull charge from Q18/20. I don't really know how audible this charge pulling is, but it is real and something the CE stage gives for free.
You need Q17 to drive Q19 and Q21. Q16 is needed, also, to pull charge from Q19/21, just as Q17 is also used to pull charge from Q18/20. I don't really know how audible this charge pulling is, but it is real and something the CE stage gives for free.
Jeff R said:
I know of at least two:
- Efficiency: if the charge is not removed quickly enough, both output transistors are conducting simultaneously (one driving the load, the other trying to switch off). The current of the one trying to switch off has to be absorbed by the other one. The feedback will (hopefully) take care of keeping the output intact, but especially at higher frequencies this can lead to much higher dissipation. That is why testing a slow output stage at full power at 10kHz is Not A Good Idea.
- Distortion: As mentioned above, the feedback will try to channel the excess current to the other output device, but this will unavoidably lead to higher THD, as the feedback cannot compensate 100%, especially at higher freqs where this happens most, as the loop gain is lower and consequently feedback is weaker there.
Jan Didden
"The emitter resistors are a necessity - the base resistors are not. However, I think the value of the emitter resistors could be reduced to maybe .1 Ohms (per Self)."
You're still being silly. Even if you reduce the emitter resistors to 0R1 from 0R33 there will still be 600mV drop compared to a small fraction of that in the base resistor. Your sense of perspective is skewed.
As an absolute error, it is very real. That's what the feedback is there for. If you want to run without global feedback you run it class A, or use a CE output stage like Nelson Pass did in the Stasis designs for both Threshold and Nakamichi.
You're still being silly. Even if you reduce the emitter resistors to 0R1 from 0R33 there will still be 600mV drop compared to a small fraction of that in the base resistor. Your sense of perspective is skewed.
As an absolute error, it is very real. That's what the feedback is there for. If you want to run without global feedback you run it class A, or use a CE output stage like Nelson Pass did in the Stasis designs for both Threshold and Nakamichi.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.