No. It is because the CFP has a much higher transconductance at low currents than EF or Darlington. This is because the frst transistor in a CFP provides voltage gain to the base of the second transistor.
I analysed this for a three-part article in Wireless World in Sept, Oct and Nov 2000; my simple mathematical model gave very similar results to those found by Self and thus explains his results. Unfortunately it is not available online.
I analysed this for a three-part article in Wireless World in Sept, Oct and Nov 2000; my simple mathematical model gave very similar results to those found by Self and thus explains his results. Unfortunately it is not available online.
I see, so instead of just gm of the output stage, its gm(driver)*ro*gm(output transistor), so far less bias is required to get it out of the crossover dead zone?
Yes, near enough. If you assume that the output transistor starts to turn on when its Vbe is about 0.6V then you find that the voltage gain of the first (driver) transistor is about 24 (working from memory here!) because its current has to vary inversely with its collector resistor in order to maintain the 0.6V. The gm of the output varies with current in the usual way, but near the crossover point much of the current actually comes from the driver.
Every time this issue comes up I say that one day I will do an online write-up of my findings, but I never get around to it. I suppose if I wait long enough the magazine article copyright will expire and then I can just post a scan of it.
Every time this issue comes up I say that one day I will do an online write-up of my findings, but I never get around to it. I suppose if I wait long enough the magazine article copyright will expire and then I can just post a scan of it.
BTW, note that when looking at distortion over level, the onset of crossover distortion occurs at a higher level in the optimally biased EF2 vs. the optimally biased CFP. (That would be fig. 6.45 vs. 6.46 in the 5th edition.) That's the A to AB transition, so the higher quiescent current is good for something at least. Also note how the crossover region a few pages on is bigger but not as nasty-looking.
CFPs make brilliant Class A (headphone) output stages though.
CFPs make brilliant Class A (headphone) output stages though.
I too read that with great interest. In simulations, with the lower bias I get current glitches in the drivers and higher simulated distortion. Based solely on the simulation, I come up with about 120mA where these symptoms recede. Of course simulations lie and I don't have a good explanation for the glitches. The drivers have enough standing current I think.
DF, if it ever becomes available, we will all be grateful. Possibly they would give permission as it has been 14 years?
I did model my MX-50 ( improved) as a class A for a learning experience. It could do 6W easy. Tempting, really tempting as it looked very good. But then, my Grado's seem happy with four op-amps.
DF, if it ever becomes available, we will all be grateful. Possibly they would give permission as it has been 14 years?
I did model my MX-50 ( improved) as a class A for a learning experience. It could do 6W easy. Tempting, really tempting as it looked very good. But then, my Grado's seem happy with four op-amps.
Hi,
The transition region for CFP is much narrower in voltage
than EF. Consequently with the same emitter resistors
the EF needs a lot more current than the CFP.
Actually D.Self goes quite thoroughly in how to set
optimum bias and the myth of it being related to
current, when in fact it is very much a voltage.
It has not much to do with the transconductance,
other than optimum bias is a choice of the best
gm variation "wingspread" of the output stage.
Its simply due to the effect of the emitter resistors
on the the 2 Vbe CFP and 4 Vbe EF bias loops
and the way the Sziklai pair works re Vbe drive.
As Self takes a lot of time to point out, the optimum
bias is the voltage across the emitter resistors, not
the standing current, and essentially changing the
emitter resistors does not change the voltage,
it changes the optimum current for voltage.
rgds, sreten.
From Self's perspective its probably as far as he's concerned
self evident EF needs much more current for the same values
of emitter resistors and the currents need no further explanation.
The transition region for CFP is much narrower in voltage
than EF. Consequently with the same emitter resistors
the EF needs a lot more current than the CFP.
Actually D.Self goes quite thoroughly in how to set
optimum bias and the myth of it being related to
current, when in fact it is very much a voltage.
It has not much to do with the transconductance,
other than optimum bias is a choice of the best
gm variation "wingspread" of the output stage.
Its simply due to the effect of the emitter resistors
on the the 2 Vbe CFP and 4 Vbe EF bias loops
and the way the Sziklai pair works re Vbe drive.
As Self takes a lot of time to point out, the optimum
bias is the voltage across the emitter resistors, not
the standing current, and essentially changing the
emitter resistors does not change the voltage,
it changes the optimum current for voltage.
rgds, sreten.
From Self's perspective its probably as far as he's concerned
self evident EF needs much more current for the same values
of emitter resistors and the currents need no further explanation.
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No, the correct bias is a current because that is what determines transconductance in a BJT. It only looks like a voltage because the current (which depends on the 'emitter' resistor, among other things) also passes through that same resistor. It has everything to do with transconductance, because the aim is to get the quiescent transconductance of each half roughly equal to the inverse of the emitter resistor so that quiescent and Class B conditions (one half cut off) give the same output stage total transconductance. The net result is that someone who hasn't thought about exactly what is going on may get the impression that voltage is what matters.
The formula I arrived at was Iq(mA) = 575/R + 1/r, where R is the resistor between base and emitter of the output BJT and r is the 'emitter' resistor. This then would give Vqr(mV) = 1 + 575r/R. For typical values of R=100 and r=0.22 I calculated Iq=10.3mA; Self found 11.5mA.
I haven't seen the later versions of Self's book, so I don't know to what extent (if any) he includes my theory. In my article I include a short table comparing his experimental/simulated findings with my calculated results - the two agreed surprisingly well.
The formula I arrived at was Iq(mA) = 575/R + 1/r, where R is the resistor between base and emitter of the output BJT and r is the 'emitter' resistor. This then would give Vqr(mV) = 1 + 575r/R. For typical values of R=100 and r=0.22 I calculated Iq=10.3mA; Self found 11.5mA.
I haven't seen the later versions of Self's book, so I don't know to what extent (if any) he includes my theory. In my article I include a short table comparing his experimental/simulated findings with my calculated results - the two agreed surprisingly well.
As this question often crops up, here is a brief summary of how I arrived at the above formula.
A CFP is essentially a follower, so like all followers an important parameter is transconductance. A CFP output stage has two contributions to transconductance; from the driver and the output. The driver contribution is just Id/25 - where Id is the driver quiescent current. The output contribution (for lowish currents, where the output base does not unduly load R) is 24*Io/25 - where Io is the output BJT quiescent current. The factor of 24 comes from the voltage gain of the driver. The total transconductance is thus (Id+24*Io)/25.
For good crossover performance this has to be equal to 1/r - where r is the 'emitter' resistor. Then quiescent transconductance (two CFP and two 'r' in play) is equal to the big signal transconductance of 1/r. So we want (Id+24*Io)/25 = 1/r. However, we also have a constraint that the output BJT must have started turning on. For this we assume that it has Vbe about 0.6V, hence Id(mA) = 600/R.
Putting this together we have
(600/R + 24*Io)/25 = 1/r so Io = (25/24)[1/r - 24/R] - note that this means that r should be less than R/24 if we want the output BJT to be turned on at the correct quiescent current - this condition is not difficult to satisfy. Then we have
Iq(mA) = Id + Io = 600/R + (25/24)[1/r - 24/R] = 575/R + 1.04/r
As r usually is not a precise component then 1.04 can be treated as 1. Actually, I could probably treat 575 as 600 too!
The assumptions here are:
1. BJTs behave like BJTs should (Ebers-Moll etc.) so transconductance is proportional to current - this is a good assumption, although the details depend on temperature.
2. A BJT will turn on when Vbe gets to 600mV - a rough guess, which will not be too far out, and the value chosen here ends up in the Iq formula so not too difficult to modify if necessary.
3. Class AB - for Class A you just want the current which matches the supply voltage and load impedance.
4. Quiescent transconductance should equal big signal transconductance - this is an approximation because there will be dips and peaks in the 'wingspread' diagram; perhaps slightly more current than the formula indicates may be best, so there is no chance of a sudden dip in transconductance.
What happens if a much higher current is chosen - as often seems to be the case? Then CFP transconductance becomes very high, so for small signals the overall output stage transconductance becomes 2/r and for large signals 1/r - classic gm-doubling. For an 8ohm load and large signal this gives 3rd-order distortion around r/16 (i.e. about 6r%). For small signals there may be little distortion. The transition region will be around 2*Iq*8 in voltage - 2V for 75mA quiescent, or about 250mW.
If the correct quiescent current is chosen then we can assume that variations in transconductance are unlikely to exceed half that above and could be much better. Distortion will thus be more like r/32 or 3r%. The transition region for 10mA will be around 3mW. The intriguing thing about this is that a correctly biased CFP output might, just, satisfy undemanding requirements without any negative feedback!
A CFP is essentially a follower, so like all followers an important parameter is transconductance. A CFP output stage has two contributions to transconductance; from the driver and the output. The driver contribution is just Id/25 - where Id is the driver quiescent current. The output contribution (for lowish currents, where the output base does not unduly load R) is 24*Io/25 - where Io is the output BJT quiescent current. The factor of 24 comes from the voltage gain of the driver. The total transconductance is thus (Id+24*Io)/25.
For good crossover performance this has to be equal to 1/r - where r is the 'emitter' resistor. Then quiescent transconductance (two CFP and two 'r' in play) is equal to the big signal transconductance of 1/r. So we want (Id+24*Io)/25 = 1/r. However, we also have a constraint that the output BJT must have started turning on. For this we assume that it has Vbe about 0.6V, hence Id(mA) = 600/R.
Putting this together we have
(600/R + 24*Io)/25 = 1/r so Io = (25/24)[1/r - 24/R] - note that this means that r should be less than R/24 if we want the output BJT to be turned on at the correct quiescent current - this condition is not difficult to satisfy. Then we have
Iq(mA) = Id + Io = 600/R + (25/24)[1/r - 24/R] = 575/R + 1.04/r
As r usually is not a precise component then 1.04 can be treated as 1. Actually, I could probably treat 575 as 600 too!
The assumptions here are:
1. BJTs behave like BJTs should (Ebers-Moll etc.) so transconductance is proportional to current - this is a good assumption, although the details depend on temperature.
2. A BJT will turn on when Vbe gets to 600mV - a rough guess, which will not be too far out, and the value chosen here ends up in the Iq formula so not too difficult to modify if necessary.
3. Class AB - for Class A you just want the current which matches the supply voltage and load impedance.
4. Quiescent transconductance should equal big signal transconductance - this is an approximation because there will be dips and peaks in the 'wingspread' diagram; perhaps slightly more current than the formula indicates may be best, so there is no chance of a sudden dip in transconductance.
What happens if a much higher current is chosen - as often seems to be the case? Then CFP transconductance becomes very high, so for small signals the overall output stage transconductance becomes 2/r and for large signals 1/r - classic gm-doubling. For an 8ohm load and large signal this gives 3rd-order distortion around r/16 (i.e. about 6r%). For small signals there may be little distortion. The transition region will be around 2*Iq*8 in voltage - 2V for 75mA quiescent, or about 250mW.
If the correct quiescent current is chosen then we can assume that variations in transconductance are unlikely to exceed half that above and could be much better. Distortion will thus be more like r/32 or 3r%. The transition region for 10mA will be around 3mW. The intriguing thing about this is that a correctly biased CFP output might, just, satisfy undemanding requirements without any negative feedback!
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I found the same, in fact the bias that Spice recommends is not far off different between EF and CFP.
They can, but I've found them to be remarkably accurate for voltages, currents, shoot-through in CFPs, compensation requirements for NFB loops, thermal control of q-currents etc. and tend to trust them more often than not.
I have never verified the match between distortion results and simulated distortion but in practice I do bias my CFP outputs closer to the level you would for an EF and have not found the results wanting.
Hi,
Anyone who arrives at the conclusion that CFP and EF have similar
bias current requirements for similar emitter resistor values is
simply way off base. They simply don't. See D. Self. *
FWIW to the OP :
it is shown by example, you are left to work out the explanation.
rgds, sreten.
* You can call his methods dubious, lowest distortion at full power
is not that practically useful. However biasing CFP to EF levels
will inevitably lead to a region of GM doubling for the CFP.
(D. Selfs methodology proves this is bad, at full power.)
You could argue for the first watt or so its better than the
gymnastics the CFP at "optimum bias" does at low powers.
Anyone who arrives at the conclusion that CFP and EF have similar
bias current requirements for similar emitter resistor values is
simply way off base. They simply don't. See D. Self. *
FWIW to the OP :
it is shown by example, you are left to work out the explanation.
rgds, sreten.
* You can call his methods dubious, lowest distortion at full power
is not that practically useful. However biasing CFP to EF levels
will inevitably lead to a region of GM doubling for the CFP.
(D. Selfs methodology proves this is bad, at full power.)
You could argue for the first watt or so its better than the
gymnastics the CFP at "optimum bias" does at low powers.
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Joined 2009
Paid Member
I'm a big fan of D. Self, his writings are very useful and informative. He noted that CFP has worse performance than EF at very low powers (he didn't explain why) but is better at higher powers. I'm sure his analysis of the optimal bias is well reasoned and accurate.
I didn't pick a slightly higher bias because I wanted better performance at low power - although now that you point it out this would be my preferred tradeoff. But the amplifier I made sounds sooooo darn goooood there's no way I'm tinkering with it. It has beaten out every other amplifier I've ever heard, SS, tube, DIY or commercial. And others who have built it have had similar feelings. If it works, it works.
I didn't pick a slightly higher bias because I wanted better performance at low power - although now that you point it out this would be my preferred tradeoff. But the amplifier I made sounds sooooo darn goooood there's no way I'm tinkering with it. It has beaten out every other amplifier I've ever heard, SS, tube, DIY or commercial. And others who have built it have had similar feelings. If it works, it works.
Hi,
Leach has a paper that claims Gm doubling doesn't exist. I'm not going
to work through DF's theory and assumptions that suggest the same.
Simple fact is D. Self has simulated it and measured it in practise. Theory
that doesn't match simulation and measurement is always flawed, and it
is that theory that needs critical examination, not the facts.
Both are easy to explain and he alludes to both. The EF at low power is much more
linear than the CPF due to the much wider central section compared to CPF, hence
lowerdistortion, however at near full power the CPF shenanigans at low powers
don't matter so much and the wider variation of the EF causes more distortion.
FWIW chasing lowest THD+N near full power is a utterly pointless exercise IMO,
which Self does. I'm far more interested in below 1W and say around 10W, than
a low number for THD+N that is only achievable very near full power output.
rgds, sreten.
Leach has a paper that claims Gm doubling doesn't exist. I'm not going
to work through DF's theory and assumptions that suggest the same.
Simple fact is D. Self has simulated it and measured it in practise. Theory
that doesn't match simulation and measurement is always flawed, and it
is that theory that needs critical examination, not the facts.
Both are easy to explain and he alludes to both. The EF at low power is much more
linear than the CPF due to the much wider central section compared to CPF, hence
lowerdistortion, however at near full power the CPF shenanigans at low powers
don't matter so much and the wider variation of the EF causes more distortion.
FWIW chasing lowest THD+N near full power is a utterly pointless exercise IMO,
which Self does. I'm far more interested in below 1W and say around 10W, than
a low number for THD+N that is only achievable very near full power output.
rgds, sreten.
Hi,
i just have read Arto Kolinummis book "Audio Power Amplifiers" and he also says
that EF have lower distortion at lower levels while CFP might be better in the high
power region. As most music is played in low to medium level regions, EF might
be the better choice for hifi. He also writes that at higher levels our ears produce
more distortion itself and higher distortion levels with EF will not mind that much -
while at lower levels distortion is more noticeable.
I have no experience with with comparing EF or CFP amps in practice but Artos
arguments sound reasonable to me.
regards, Dirk
i just have read Arto Kolinummis book "Audio Power Amplifiers" and he also says
that EF have lower distortion at lower levels while CFP might be better in the high
power region. As most music is played in low to medium level regions, EF might
be the better choice for hifi. He also writes that at higher levels our ears produce
more distortion itself and higher distortion levels with EF will not mind that much -
while at lower levels distortion is more noticeable.
I have no experience with with comparing EF or CFP amps in practice but Artos
arguments sound reasonable to me.
regards, Dirk
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