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I am trying to understand a simple class-AB output stage. The bias voltage for the transistors is made by using two diodes. The following picture show the circuits. The currents drawn are how, i believe, the currents will flow.

enter image description here

I want to understand the currents, to be able to find a value for the resistor, R.

Especially i am thinking of the following:

  • Is the transistor Q2 solely driven by the input? I think this as there is no other path for the base current.
  • Will the current through R be used for driving the transistor Q1 and the diodes?

Then, if this is the case, how can a reasonable value for the resistor R be found?

I know that the lowest possible voltage across R is 5.5 V, and i know that I_C for Q1 is at most 260 mA. Given this i believe that the minimum current for driving the transistor is: $$R = \frac{5.5}{(\frac{0,260}{\beta})}$$

This i enough current for driving the transistor, but how is it possible to determine the needed current through the diodes? I suspect that it is a quite low current. Will the easiest way to just undersize the resistor, and forget about the diodes?

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  • \$\begingroup\$ .What is your load impedance? What peak volt swing do you want without undue distortion?R can get rather low .Have you considered using a current source instead of R? \$\endgroup\$ – Autistic Aug 30 '18 at 11:17
  • \$\begingroup\$ My load impedans is 32 ohm. My peakvoltage will be 8.3 V, and the system is suppled with +-15 V. A current source could be used, if the value of R is too low. \$\endgroup\$ – keffe Aug 30 '18 at 11:19
  • \$\begingroup\$ forum.allaboutcircuits.com/threads/… and here electronics.stackexchange.com/questions/309936/… \$\endgroup\$ – G36 Aug 30 '18 at 13:05
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Is the transistor Q2 solely driven by the input? I think this as there is no other path for the base current.

You are only thinking about the DC (biasing currents) and yes, then the currents could (I am not saying that they always do) flow like you have drawn.

However the DC biasing is only one aspect of how this circuit works. The signals are small variations of this DC biasing current. So let's say that 10 mA is flowing out at Vi, if the signal causes that current to vary between 9 mA and 11 mA then on average there is still flowing 10 mA out of Vi. But the signal current is +/- 1 mA !!!

The current through R is your choice. What happens if you increase the DC current through R, what happens if you lower it? Think about the collector currents of the transistors as well.

Class AB means that for large signals part of the output stage will have no current (Ic = 0 A), what does that mean in relation to having a large or a small DC biasing current flowing through the transistors?

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  • \$\begingroup\$ If i increase R too much i suspect that not enough current can flow, and Q1 will go into saturation, not delivering enough collector current. If i lower the value much, unnecessary current will flow. If the bias current is too large, the outout of the transistors will be on for longer. Please correct me if im wrong. \$\endgroup\$ – keffe Aug 30 '18 at 11:37
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    \$\begingroup\$ Q1 cannot go into saturation, for sat. to occur the voltage at the collector would have to go below the voltage at the base. That cannot happen in this circuit. What happens when R has a high value is that the biasing current will become zero. Then it would be a class B stage! For low value R indeed the transistors will be on "longer", you mean they will keep Ic > 0 even for large signals. That's class A operation. \$\endgroup\$ – Bimpelrekkie Aug 30 '18 at 11:47
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You do not need to get too close to the rail in your application .This means that a resistor could suffice .Your minimum expected volts across R is 6 .If R was 330 ohms your circuit will work even with low gain transistors.A say 18mA current source would also work .Emitter resistors would make idle current more predictable and reduce thermal runaway problems .I have used old school BD140 transistors as diodes and bolted them to the common heatsink with simple m3 bolts and mica washers.When you run the current source the idle current is more constant if supply volts changes .

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