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enter image description here

enter image description here

The goal of this question I am having trouble with is finding the minimum output voltage, in which there are two cases: one in which Q1 turns off, and the second in which Q2 turns goes into saturation. I cannot seem to figure out how to find Q2 collector current, I, in this class A output stage. We were told in class that the diode, Q3, keeps the base-emitter junction at 0.7 V, but I cannot help think there is more to it than that. The one solution I could find online said that I is found as such:

enter image description here

I'm not sure how it is possible to find with the given information. How can the base current equal to the collector current with such a high beta? What am I missing here? Is that solution even correct?

first two images are from Sedra/Smith Microelectronics, 7th edition, chapter 12. last image is a solution from Slader by SSoliman (https://www.slader.com/textbook/9780199339136-microelectronic-circuits-7th-edition/983/problems/1/)

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  • \$\begingroup\$ Both the drawing and question image seem to be someone else's so you need to credit them. (This is site policy.) Hit the edit link ... \$\endgroup\$ Commented Nov 14, 2020 at 22:44
  • \$\begingroup\$ Wait, I think I may see it. Is it because the base-emitter junction is in parallel with the diode, making the collector and current through R equal? \$\endgroup\$ Commented Nov 14, 2020 at 22:45
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    \$\begingroup\$ @ModularMan Q2 and Q3 are a current mirror. R sets the collector current of both at 9.3 mA. So Q2 always attempts to sink 9.3 mA. Q1 can source as little as 0 mA (but cannot sink any.) So if Q1 isn't sourcing any current and Q2 is sinking 9.3 mA, then it must be the case that RL is sourcing 9.3 mA. That happens when the output is -9.3 V. \$\endgroup\$ Commented Nov 15, 2020 at 7:52
  • \$\begingroup\$ Jonk, thank you for your response. I now see exactly what you mean. I would accept your answer, but it is a comment. \$\endgroup\$ Commented Nov 15, 2020 at 19:04

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