Biasing Load invariant amplifier complemtary feedback CFP output stage

Hi i have built an amplifier Load invariant design by Douglas Self that uses 2 ouput transistor in parralel ie (complemtary feedback CFP output stage) and want to bias the output correctly. The recomended voltage drop across the two output resistors for a single output CFP transistor design is 10 mV. Do i double this for 2 ouput transistors in paralel ie 20 mV

I have attached the circuit diagram.
Thanks
Heath

Hmmmm.  Interesting one.  Did the diagram not come with any instructions?  Off the cuff, I would say yes, because you'll have the same quiescent current flowing through 2 transistors instead of one. which will give double the quiescent current, which will give double the voltage across R36 and R37.  I presume you set this with PR1.   But I would hate to second-guess the renowned Mr Self, who knows far more than I do.  So visit his site at http://www.douglas-self.com and use the Email link on the right about half way down,  and ask him yourself!

Hi,In theory, yes, double the number of transistors means doubling the bias current through the resistors, and hence double the voltage drop.
I would use a separate resistor for each transistor emitter, keeping the voltage drop consistent, to allow for any variation in transistor hfe.
Regards,Richard.

Very good thinking Richard.  Hopefully this would keep the currents around the same in each transistor, but you might get 11mV in one and 9 mV in the other.  But if you're really being finicky with things like this, you should hfe-match your transistors, both the pair of NPNs and the pair of PNPs, and between the NPNs and PNPs, so they are all as near as possible the same hfe.

Hi David,Yes, matching transistors is a key aspect to reduce distortion in the output stage where possible. I think the optimal approach is separate resistors, one per transistor, to ensure reduced power dissipation for the resistors, at the quiescent current operation, and reduce the power rating of the resistors too.
Regards,Richard.

Hi David i did email Douglas Self and his reply email is below. So i took from his email that 10mv is more or less optimal across resistors R36, R37 for multiple output transistors for a (cfp) complementary feedback output stage.
I emailed Douglas back and said i would set the voltage across R36, R37, he did not say not to.There were building instructions which included the circuit diagram I posted, The instructions did say set the voltage across R36, R37 to 10mV but when i started to read Doug sels amplifier handbook i noticed that the 10vm was used on the trimoidal amp with asingle (cfp) ouput stage which is why i asked the question.I still dont know why the voltage drop across R36, R37 for multiple cfp output transistors does not change much from 10mV. 

RegardsHeath
"
Hello

 

I would ignore Sloan’s book.

 

I think you may be a bit confused , as p320 refers not to the
Load Invariant but the Trimodal amplifier design.

 

For the Load Invariant the relevant resistors are R36 and R37
and these are emitter resistors. (p169)

The bias voltage to be set up across them does not vary
significantly with the number of power devices.

 

Douglas Self"

Heath....OK, so it would seem that having multiple output transistors is more for spreading the load and keeping the transistors well within their limits than increasing the power capability for lower resistance loads.
R36/37 are used for short circuit protection.  If the voltage on either of them reaches around 600 mV, they will drive transistors Q12/Q13 on, which in turn will short circuit the drive on transistors Q14/Q15 which in turn will remove drive from the main output transistors Q16/17/18/19.And while looking at this something hit me.  The resistors R36/37 are in the COLLECTORS of the output transistors, not the emitters.  So they will play no part in equalising the currents.  This is why Mr Self does not recommend using separate resistors.   There are in fact no emitter resistors used at all.  Which, IMHO, makes it even more necessary to match the transistor gains as closely as possible.If you were using low impedance speakers, e.g. 4 ohms, you might want to reduce R36/37 to avoid the amplifier going into short circuit protection mode (which will massively increase distortion) with the increased currents in low loads.  But otherwise Mr Self is right - using separate resistors is not necessary.I'm sorry for my misinterpretation and oversight here.  This shows why Mr Self has written books about amplifiers and I haven't

David I don't know why the 10mv Voltage Drop Across R36, R37 when Biasing The Output Does Not Increase For 2 Transistors In Paralell. Have You Any Idea.
RegardsHeath

Heath...ok, the quiescent current is set by the circuitry around Q8 which generates a small voltage (about 1.2 volts or so) to bias Q14 and Q15.  These transistors DO have those resistors R36 and R37 in their emitter leads so will act as constant current sources, supplying the bases of the 2 or 4 output transistors.  When you put 2 output transistors in place of one like this, you don't increase the gain, just the current carrying capability.  The currents through the output transistors also flow through R36/R37 but if the voltage across R36/R37 rises, it will tend to cutoff the driver transistors Q 14/15.  So it's a self regulating system (pardon the pun Mr Self, I should have said negative feedback .  The bias current for the output transistors will be divided by 2 for 2 transistors, it's a current drive not a voltage drive.  it will only be exactly half-half is the output transistors are matched.
If because of using 2 output transistors in each leg you wanted to double the output (current, not voltage) of the amplifier too, you would halve the value of R36/37, for example by putting two 0R1 resistors in parallel.  Because they drive the biasing circuitry (and the short circuit protection) you wouldn't have separate resistors for each transistor.  But Mr Self is trying to spread the load between the transistors, not increase the output.Low distortion amplifier design is quite an art and Mr Self is very much a master of it.BTW what voltage power rails  are used for this amp?
Thanks // David

Hi David thanks for your reply. I understand the points about increasing current cababilty, spreading the load between two transistor instead of one. I did not realise Q14, Q15 are constant current sourcces keeping the current constant. So these transistors keep the voltage drop across R32 constant which intern keeps the voltage drop across the ouput base/emitor power transistors Q16, Q18 constant without increasing v drop and current through R36 and R37.  If i have interepreted your email correctly. Not sure.
I am running the amp from +/- 50v DC which is +/- 5V DC higher than the recomended +/- 45v DC. The amp seems to o.k on this voltage I have tested it into 5 ohms resistive load at 26V output ie 135 watts.
RegardsHeath

I only had a torroidal with 35v AC secondarys giving +/- 50V DC.
RegardHeath

Heath - thanks, yes I think your understanding is correct (and hope mine is too .  135W is a lot of power but the output transistors are rated at 200W - so you'd probably get away with one instead of two - but good audio design is always conservatively rated.
Bear in mind that your rail voltages willl effectively reduce somewhat under load.  You're only 10% out which is not a lot.  Strangely enough there was a question just posted on just this issue...Amplifier design has come a long way since the days of my youth.  I used to use a 60W design with 2 of the venerable 2N3055s and 3 other transistors, but it gave surprisingly good results.  Purists today would probably look down their noses at it because it had 0.5% distortion instead of 0.05%.....Cheers // David

Hi,Apologies - quick reply - i missed the fact that the amplifier is a common emitter and not a common collector/emitter follower output stage. The design in the book - has two devices per output stage, so any setup instructions that came with the kit, should be followed.
Regards,Richard.