AndrewT said:
Sorry about that
I should have mentioned what topology.I wanted to make CFP output channels, and since A CFP stage only have 2 VBE voltage drops, instead of 4 like EF, I wanted to see if a single NPN/PNP pair with the 2 thermal diodes would do the trick. I'm going to test a set. If it works, it means you can design simple amps with just a single pair output with Thermaltraks. I'll keep you posted.
You are right. The diodes would be between the bases of the drivers/
I know the output stage still gains current when the outputs heat up. I was thinking that when outputs get hot, the diodes could reduce the bias of the drivers, which are more sensitive, therefore lowering the amp bias as a whole. I'll try and see. I know it's not the normal way, but that's why I want to try it. If it don't work, we'll know for sure.
I imagine if it works, maybe we could let Rod Elliott know, and he could make them work with the P3A amp and make a new PCB for the Thermaltraks.
I know the output stage still gains current when the outputs heat up. I was thinking that when outputs get hot, the diodes could reduce the bias of the drivers, which are more sensitive, therefore lowering the amp bias as a whole. I'll try and see. I know it's not the normal way, but that's why I want to try it. If it don't work, we'll know for sure.
I imagine if it works, maybe we could let Rod Elliott know, and he could make them work with the P3A amp and make a new PCB for the Thermaltraks.
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
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
This is only back of the envelop calculation, but if it is correct then we can design a very nice Vbe multiplier with the Thermaltraks transistors.
If we use an ouput stage with as push pull class A drivers made of two thermaltraks we have 4 diodes to play with: two from the final EF and two from the driver.
We can then use a classical Vbe multiplier like in Leach amp and put the four diodes in the upper leg of base biasing ( perhaps two diodes in the bases of the drivers to compensate for the two Vbe drops)
If we parallel the driver thermaltrak diodes with the correct resistor we can adjust the drift of the Vbe of the driver thermaltrak diodes to be the same as the Vbe drift of the drivers
because the drift of a diode increases by 0.2mV/°C per decade of decrease in current. This resistor bleeds some current from the diodes
The same can be done for the two other diodes.
We can then adjust the base potentiometer for minimum distortion at room temperature and ajust the // diodes for minimum distortion at high temperaturs. There will be some interraction but it shoud ( hopefully converge).
In such a situation, thermal transient crossover distortion should be attenuated.
This is again back of the envelope ideas.
JPV
If we use an ouput stage with as push pull class A drivers made of two thermaltraks we have 4 diodes to play with: two from the final EF and two from the driver.
We can then use a classical Vbe multiplier like in Leach amp and put the four diodes in the upper leg of base biasing ( perhaps two diodes in the bases of the drivers to compensate for the two Vbe drops)
If we parallel the driver thermaltrak diodes with the correct resistor we can adjust the drift of the Vbe of the driver thermaltrak diodes to be the same as the Vbe drift of the drivers
because the drift of a diode increases by 0.2mV/°C per decade of decrease in current. This resistor bleeds some current from the diodes
The same can be done for the two other diodes.
We can then adjust the base potentiometer for minimum distortion at room temperature and ajust the // diodes for minimum distortion at high temperaturs. There will be some interraction but it shoud ( hopefully converge).
In such a situation, thermal transient crossover distortion should be attenuated.
This is again back of the envelope ideas.
JPV
Re: do we need stable bias current with temp in the output stage? NO
I think I agree with you completely.
The only caveats are that deviations from the ideal must be kept in perspective.
The real-world situation with typical non-ThermalTrak transistors is far worse than being off by the difference of 26 mV and 31 mV. This only corresponds to a couple of degrees C of thermal mis-tracking.
Self grossly under-estimates the problem of thermal mis-tracking in regard to the creation of crossover distortion. He especially under-rates the problem of transient under-biasing of the output stage during thermal changes.
Another thing to bear in mind is that the 26 mV number only applies to ideal transistors with no ohmic base or emitter resistance. In the real world, the "right" value of voltage across the physical emitter resistor is somewhat less, because part of the actual effective total emitter resistance is buried inside the power transistor in the form of internal emitter resistance or base resistance divided by beta.
Finally, in this connection, people must be mindful of the effect of base stopper resistors when they are used. A 5-ohm base stopper resistor divided by a nominal beta of 50 is 0.1 ohm, quite significant in comparison to typical emitter resistors. Moreover, that value of beta will vary with current.
In this connection, one must also bear in mind that the internal resistances and betas of the PNP and NPN output transistors may not be the same, in some cases possibly calling for slightly different optimal values of emitter resistors top and bottom.
Cheers,
Bob
JPV said:
I think I agree with you completely.
The only caveats are that deviations from the ideal must be kept in perspective.
The real-world situation with typical non-ThermalTrak transistors is far worse than being off by the difference of 26 mV and 31 mV. This only corresponds to a couple of degrees C of thermal mis-tracking.
Self grossly under-estimates the problem of thermal mis-tracking in regard to the creation of crossover distortion. He especially under-rates the problem of transient under-biasing of the output stage during thermal changes.
Another thing to bear in mind is that the 26 mV number only applies to ideal transistors with no ohmic base or emitter resistance. In the real world, the "right" value of voltage across the physical emitter resistor is somewhat less, because part of the actual effective total emitter resistance is buried inside the power transistor in the form of internal emitter resistance or base resistance divided by beta.
Finally, in this connection, people must be mindful of the effect of base stopper resistors when they are used. A 5-ohm base stopper resistor divided by a nominal beta of 50 is 0.1 ohm, quite significant in comparison to typical emitter resistors. Moreover, that value of beta will vary with current.
In this connection, one must also bear in mind that the internal resistances and betas of the PNP and NPN output transistors may not be the same, in some cases possibly calling for slightly different optimal values of emitter resistors top and bottom.
Cheers,
Bob
Re: Re: do we need stable bias current with temp in the output stage? NO
I agree with you. The resistance seen in the emitter can be substantial compared with Re. This is perhaps an explanation of the optimal voltage at room temperature that can be significantely different from the theoretical one.
But if we tune the Vbe multiplier to generate at room temperature the optimal voltage drop on Re using a spectrum analyzer, then we can tune the drift of the EF diodes ( tweaking the // resistor) to minimize the distortion at full temperature. We can also make the tracking of the drivers accurate. This seams to me to take into account more non ideal aspects. But perhaps using the thermaltraks as such is enough. My points is that they are wonderfull components and they give you more degrees of freedom to fight the transient transient crossover distortion.
JPV
Bob Cordell said:
I agree with you. The resistance seen in the emitter can be substantial compared with Re. This is perhaps an explanation of the optimal voltage at room temperature that can be significantely different from the theoretical one.
But if we tune the Vbe multiplier to generate at room temperature the optimal voltage drop on Re using a spectrum analyzer, then we can tune the drift of the EF diodes ( tweaking the // resistor) to minimize the distortion at full temperature. We can also make the tracking of the drivers accurate. This seams to me to take into account more non ideal aspects. But perhaps using the thermaltraks as such is enough. My points is that they are wonderfull components and they give you more degrees of freedom to fight the transient transient crossover distortion.
JPV
Re: Re: Re: do we need stable bias current with temp in the output stage? NO
Yes, you are exactly right. Many do not realize that the Vbe temperature coefficient depends on the current density. Indeed, this is the basis of operation of the bandgap voltage references invented by Bob Widlar.
Cheers,
Bob
JPV said:
Yes, you are exactly right. Many do not realize that the Vbe temperature coefficient depends on the current density. Indeed, this is the basis of operation of the bandgap voltage references invented by Bob Widlar.
Cheers,
Bob
Re: Re: Re: Re: do we need stable bias current with temp in the output stage? NO
I found a nice formula on the Vbe drift in my old book from P.E. Gray and C.L. Searle.
From there and for low Kelvin Temp variation I derived the following very simple expression:
The Vbe drift variation with current density is:
k/q * ln(I2/I1)
where k/q = 0.086 mV /°K
For a ratio of 10 in the currents it gives the rule of thumb
0.086 * ln (10) = 0.2 mV/°C per decade
So the tempco of the Vbe of a diode increases by 0.2 per decade of current . This is for temperatures in the 300 375.
It is therefore possible to tune the tempco of the Thermaltrak diode to minimise the transient distortion in temperature. This would be if it works a major achievement in audio class B power amplifiers
This is another view in front of people criticizing this component.
I have designed a very complete amplifier block around triple T topology with thermaltracks and LME 49810 front end. The PCB layout is terminated on paper but I need one of my friend to put it in the cad/cam software. I hope to be able to order the PCB very soon and then make these interesting experiments.
JPV
Bob Cordell said:
I found a nice formula on the Vbe drift in my old book from P.E. Gray and C.L. Searle.
From there and for low Kelvin Temp variation I derived the following very simple expression:
The Vbe drift variation with current density is:
k/q * ln(I2/I1)
where k/q = 0.086 mV /°K
For a ratio of 10 in the currents it gives the rule of thumb
0.086 * ln (10) = 0.2 mV/°C per decade
So the tempco of the Vbe of a diode increases by 0.2 per decade of current . This is for temperatures in the 300 375.
It is therefore possible to tune the tempco of the Thermaltrak diode to minimise the transient distortion in temperature. This would be if it works a major achievement in audio class B power amplifiers
This is another view in front of people criticizing this component.
I have designed a very complete amplifier block around triple T topology with thermaltracks and LME 49810 front end. The PCB layout is terminated on paper but I need one of my friend to put it in the cad/cam software. I hope to be able to order the PCB very soon and then make these interesting experiments.
JPV
I've come very late to this thread, but I would like to say how much I've enjoyed it. Hats off to all those who have done measurements and got the facts. I look forward to doing some experiments with these little beauties when I've got a few other projects out of the way.
Now, calling Bob Cordell- right at the start of this thread you showed us a very interesting configuration you called a triple-Darlington T-circuit. I've never heard that term before, and I wondered if the "T-circuit" bit referred simply to the fact that a triple Darlington is used, or if it referred to the feedback control system. The latter is full of interesting possibilities, and I'd be interested to know more about it.
Now, calling Bob Cordell- right at the start of this thread you showed us a very interesting configuration you called a triple-Darlington T-circuit. I've never heard that term before, and I wondered if the "T-circuit" bit referred simply to the fact that a triple Darlington is used, or if it referred to the feedback control system. The latter is full of interesting possibilities, and I'd be interested to know more about it.
Doug,
In your article(s) on thermal compensation, your models, as I understood them, indicated that there is an optimal delay in thermal compensation as in your use of thermal insulators in your model. Won't the ThermalTraks then act too fast in the compensation scheme? Or, perhaps, is the unexplanably large diode molded into the package, instead of being formed on the same die, an actual advantage according to your model?
In your article(s) on thermal compensation, your models, as I understood them, indicated that there is an optimal delay in thermal compensation as in your use of thermal insulators in your model. Won't the ThermalTraks then act too fast in the compensation scheme? Or, perhaps, is the unexplanably large diode molded into the package, instead of being formed on the same die, an actual advantage according to your model?
DouglasSelf said:
Hi Doug,
Nice to have you aboard. As Andy mentioned, this circuit is essentially the Locanthi "T" circuit adapted for use with the ThermalTrak transistors. What I referred to as the triple Darlington T circuit referred only to the conventional Locanthi "T" circuit aspect of the design, and did not refer to the biasing feedback control system. Obviously, some elements of the feedback control approach can be applied to non-ThermalTrak output stages as well. That feedback control bias speader shares some similarities to the way in which bias control was achieved in my MOSFET power amplifier with error correction.
Cheers,
Bob
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