driver emitter resistors.
- Thread starter Dan Moos
- Start date
If you are using a resistor in each emitter tied to the output then 1xVBE/Iq=R (VBE being the base emitter on voltage of the output transistor being driven and Iq the desired idle current of the driver transistor). If your using a single resistor between the driver emitters then it is 2xVBE/Iq=R . Iq for the VAS is determined by its load.
Lots of packaged amps on here have 20 to 40 ma per TO3 output transistor as the quiescent current; there must be a reason. Using djoffe's circuit to increase the cold idle current to 20 ma on my ST120 made it sound better, IMHO.
One further consideration for multiple output transistor amps, is that the higher the value of the emitter resistor, the better the current sharing between mismatched parallel output transistors. Or to put it another way, the less time and the fewer temperatures you have to match Vbe on the various transistors installed in parallel. The highest I have seen is 0.5 ohm on Peavey amps where the factory states that transistors can be replaced one at a time with factory selected parts. The lowest value I have seen on here is 0.1 ohm on designs tested with simulations but not sold commercially.
One further consideration for multiple output transistor amps, is that the higher the value of the emitter resistor, the better the current sharing between mismatched parallel output transistors. Or to put it another way, the less time and the fewer temperatures you have to match Vbe on the various transistors installed in parallel. The highest I have seen is 0.5 ohm on Peavey amps where the factory states that transistors can be replaced one at a time with factory selected parts. The lowest value I have seen on here is 0.1 ohm on designs tested with simulations but not sold commercially.
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You probably have checked a few amplifier schematics as they appear on the forum. Most "basic" and conventional BJT amps settle for 0R22 emitter resistors and many calculations you see assume this. There are some reasons like efficiency and device mismatch already mentioned, to go lower or higher in value but given the better matching possible and likely in purchasing modern output transistors from authorised sources, there is not much reason to change from 0R22 - a good compromise.
It's not easy to decide. Mainly depending by the type of the driver.
The Toshiba and Sanken types such as 4793/1837 and 5171/1930 works properly with 5-6mA, but the Motorola/Onsemi types such as MJE150xx like the higher bias up to 50-100mA.
Sajti
The correct current depends on topology; low for complementary pair, high for Darlington. The current and 'emitter' resistor value are related, as the aim is to ensure that the effective transconductance of the output pair at quiescent current is roughly equal to the transconductance of one side when the other side is switched off.
Alternatively, follow the 'first watt' brigade and use a lot of current, thus enlarging any 'crossover' distortion but also moving it away from the crossover point.
Alternatively, follow the 'first watt' brigade and use a lot of current, thus enlarging any 'crossover' distortion but also moving it away from the crossover point.
Maybe my misunderstanding, but the thread starter asked about the driver bias, not the output.
Sajti
The drivers emitters resistors should be low enough to allow
fast sinking of the charge accumulated in the output devices
base/emitter/collector parasistic capitance , particularly
when clipping occur.
An usual value , as can be seen in tons of power amps , is about
100R , wich is in line with Satji s 5-6mA for fast drivers.
IIRC Bob Cordell prefer higher currents/lower resistors ,
perhaps due to the use of MJEs wich as pointed by Satji
need a higher current to achieve a said Ft.
As for the VAS , let say that the current drained by the output
stage must be no more than 10-20% of its current.
Just to be clear to a few that misunderstood, it was the drivers that I was asking about. I'd hoped the thread title was enough to avoid confusion 
.
If I'm understanding what some have said, I need to look at the transistor curves to see at what current the transistor is most happy at as far as gain and Ft.
.If I'm understanding what some have said, I need to look at the transistor curves to see at what current the transistor is most happy at as far as gain and Ft.
ChristianThomas
Member
No, I'm not sure that's the case, Dan. There are quite a number of things that come into play. One really needs to see the topology you are planning to work with and what exactly you are calling a driver. Are we talking about picking up from the output from the VAS? Or, at the other extreme, would you call the first transistor in a CFP a driver? And is it just emitter resistors, or are we looking at the whole current going down that arm?
Perhaps the most basic criterion must be that your output devices are able to deliver the current you expect, especially when their beta has dropped off to 20, or perhaps even 5. A proportion of that has to be available at the base. So that, taking into account wide tolerances, defines your maximum.
If one looks at that last as a simple example (the CFP), Cordell puts 100R in there as a 'reasonable' value, but then complains (he is not a CFP fan, though Doug Self is) that there is only 5mA of switch-off current. Quite why he doesn't halve or quarter the value, I don't know - perhaps he has an answer to that. But this is another factor to take into account.
A third, which is covered by a few people above, is the output impedance on each half. In some designs you can even see asymmetrical choices of resistors for just this reason. And on quasi-complementary designs, if you don't hate them as Self and others do, it can get really quite complicated.
You really do need bench measurements for this, as ultimately it cannot be done on paper alone. But to see people who put a lot of effort into their output stages have a look at the Bryston amps, where the schematics are kindly published on line.
Perhaps the most basic criterion must be that your output devices are able to deliver the current you expect, especially when their beta has dropped off to 20, or perhaps even 5. A proportion of that has to be available at the base. So that, taking into account wide tolerances, defines your maximum.
If one looks at that last as a simple example (the CFP), Cordell puts 100R in there as a 'reasonable' value, but then complains (he is not a CFP fan, though Doug Self is) that there is only 5mA of switch-off current. Quite why he doesn't halve or quarter the value, I don't know - perhaps he has an answer to that. But this is another factor to take into account.
A third, which is covered by a few people above, is the output impedance on each half. In some designs you can even see asymmetrical choices of resistors for just this reason. And on quasi-complementary designs, if you don't hate them as Self and others do, it can get really quite complicated.
You really do need bench measurements for this, as ultimately it cannot be done on paper alone. But to see people who put a lot of effort into their output stages have a look at the Bryston amps, where the schematics are kindly published on line.
@Christian Thomas: that was the sort of info I was looking for.
So as an example to see if I am on the right track, lets say I have a simple output section where one driver emitter follower drives one output emitter follower (plus the PNP counterparts respectively).
Would I:
first figure out the current I require from the output transistors.
then figure out the worst case scenario for beta for the output transistors, and make sure the drivers can deliver the required current to make that worst case comfortably sufficient.
do the same with the VAS current with respect to what the drivers will require.
After all of that, take turn of currents, Ft, and the curves into account to tweak.
Spice model and prototype the thing to make sure.
Is that the process in a nutshell? I basically want to be at the point where I can explain every component choice and value rather than just building circuits "out of a cookbook"
So as an example to see if I am on the right track, lets say I have a simple output section where one driver emitter follower drives one output emitter follower (plus the PNP counterparts respectively).
Would I:
first figure out the current I require from the output transistors.
then figure out the worst case scenario for beta for the output transistors, and make sure the drivers can deliver the required current to make that worst case comfortably sufficient.
do the same with the VAS current with respect to what the drivers will require.
After all of that, take turn of currents, Ft, and the curves into account to tweak.
Spice model and prototype the thing to make sure.
Is that the process in a nutshell? I basically want to be at the point where I can explain every component choice and value rather than just building circuits "out of a cookbook"
Being a real old
timer, I can explain why output transistor 100r BE resistors came to be preferred .
Nothing at all related to getting best Ft, linearity, low crossover distortion or whatever but because all power transistor datasheets offered 2 VCE ratings.
The "raw" one , base open, was called simply VCE.
The other one, called VCER was measured with a resistor from Base to Emitter.
VCER was typically 10V higher than VCE, an important consideration at that time.
Some datasheets offered a graph, VCER vs RBE ; typically 1K offered a few volts improvement, and best bang for the buck came from 100r .
Any lower was seen as abusing the driver for very little advantage.
As a side note, sometimes there was another rating shown: VCEX , some extra 20V above raw VCE, using -1.5V (reverse bias) applied BE.
Interesting for transformer driven amps and self oscillating power converters.
As a practical example, the workhorse of the Industry was the 2N3055, with the following ratings:
VCE=60V
VCER=70V (using 100 ohms)
VCEX=80 to 90V
timer, I can explain why output transistor 100r BE resistors came to be preferred .Nothing at all related to getting best Ft, linearity, low crossover distortion or whatever but because all power transistor datasheets offered 2 VCE ratings.
The "raw" one , base open, was called simply VCE.
The other one, called VCER was measured with a resistor from Base to Emitter.
VCER was typically 10V higher than VCE, an important consideration at that time.
Some datasheets offered a graph, VCER vs RBE ; typically 1K offered a few volts improvement, and best bang for the buck came from 100r .
Any lower was seen as abusing the driver for very little advantage.
As a side note, sometimes there was another rating shown: VCEX , some extra 20V above raw VCE, using -1.5V (reverse bias) applied BE.
Interesting for transformer driven amps and self oscillating power converters.
As a practical example, the workhorse of the Industry was the 2N3055, with the following ratings:
VCE=60V
VCER=70V
(using 100 ohms)VCEX=80 to 90V
using peak currents to determine the suitability of devices in each stage is critical to getting the amplifier to meet the current demand of the speaker.
That is a quite different design method compared to determining a suitable quiescent current for each stage.
I always start with peak output current and work back through the stages selecting suitable devices.
After a suitable device has been chaosen I then try to identify a suitable bias current. For me that is much less defined.
This Thread may well give me a better understanding of what is required for driver bias.
Certainl, the practice of adopting 5 to 6mA for driver bias in a CFP seems to me quite wrong.
But where the driver is a CFP and the predriver (two stages inboard of the output EF) then a 5 to 6mA bias would seem far more appropriate. Here the common emtter driver could have 50mA to 150mA of bias current.
That is a quite different design method compared to determining a suitable quiescent current for each stage.
I always start with peak output current and work back through the stages selecting suitable devices.
After a suitable device has been chaosen I then try to identify a suitable bias current. For me that is much less defined.
This Thread may well give me a better understanding of what is required for driver bias.
Certainl, the practice of adopting 5 to 6mA for driver bias in a CFP seems to me quite wrong.
But where the driver is a CFP and the predriver (two stages inboard of the output EF) then a 5 to 6mA bias would seem far more appropriate. Here the common emtter driver could have 50mA to 150mA of bias current.