What is this diode for?

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From Onsemi's thermaltrak audio transistor's app note (AND8196/D) there is this schematic.

What is the purpose of D4-D5? Should it be connected to supply rails (for not-exceeding rail voltages), not to base of Q9-Q10?

If it is connected like that, it will bypass the VBE drop of Q7-Q8 plus drop on R18-R19?
 

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GK

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Joined 2006
It looks like a very basic current limit. The diode (D4, D5) will conduct current from the base of the pre-driver transistor (Q7, Q8) when the current through the driver transistor’s series base resistor (R18, R19) gets too large. I’ve seen this done before, but with two diodes connected in series, as a single diode drop is too close to the base-emitter drop of the pre-driver transistor.
D4 will start seriously stealing base current from Q7 at only a couple of mA through R18, but when that happens, VAS transistor Q4 will just turn on harder to compensate (due to the action of nfb) and provide direct drive to the base of driver transistor Q9 through D4!
It looks to me like the designer of this circuit may have been aware of this deficiency, as the VAS transistors (Q4, Q6) are biased with hardly any quiescent current, thus limiting their drive capability.
R8 sets the LTP tail current to approximately 1.4mA, giving ~700uA through each LTP collector. That gives a drop of 0.7V across R2 - only just enough to keep VAS transistor Q4 turned on! Q3 mirrors the collector current of Q2 to give the same balanced drive to the other VAS transistor, Q6. R11 and R12 therefore limit the collector current of each VAS transistor to approximately 7mA.

This really is a horrendous circuit – you only need to look at the THD plots for confirmation. To call it a “standard industry design” is a bit of a joke. The basic topology may be common, but the way it has been mutilated for this implementation definitely isn’t.
It’s also a bit ridiculous that, in an application note outlining an alternative, non-adjustable biasing scheme for a bipolar power amplifier, no mention at all is made of the quiescent current of the output transistors. Just look at the enormously high THD for low signal levels, where a class AB amplifier should be operating in class A.
 
IMHO, D-4 conducts and bypasses the base drive of Q-7, shutting off the drive of Q-7 when the voltage swing exceeds the +45 Vol;t rail of the power transistors.

The driver and CVoltage gain stages are powered via a higher ... 55 V Volt rail, and you want to drive the Power Transistor base more positive than its collector.

D-4 does that ... Clamps the Drive Voltage to approx the 45V rail.
 

GK

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Joined 2006
forr said:
As the associated transistor Vbe is higher than at least 0.5 V, the diode would certainly be conducting all the time. As said by Lumanauw, maybe an error in the drawing.


I agree that the diode will be conducting, but I don't think there is an error in the drawing. Looking at the rest of the circuit, I just think the designer didn't know what he was doing. Why on earth would anyone bias the VAS transistors so low?

The pre-driver transistors (Q7, Q8) would essentially do nothing, with the under biased VAS transistors (Q4, Q6) providing drive current to the driver transistors (Q9, Q10) directly through the diodes (D4, D5).
This somewhat less than ideal scenario would go a long way in explaining the amplifiers hopeless THD performance.
 
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