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enter image description here

Im currently trying to solve this problem for total gain, though I'm stuck right off the bat with DC analysis. I was told the emitter current for Q1 and Q2 will be 0.9 mA though I have no idea why.

Why is the emitter current for Q1 and 2 0.9mA?

Would Ic at Q1 be Ic1=(9.3--10)/22k = 0.872mA?

edit: I understand the AC portion so don't bother with that, just need help with DC.

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  • \$\begingroup\$ How you get AC portion whitout DC? \$\endgroup\$ – Martin Petrei Dec 10 '15 at 19:47
  • \$\begingroup\$ @MartinPetrei phrasing, I meant I understand how to go about an AC analysis, but I need to do the DC first. \$\endgroup\$ – Hatandboots Dec 10 '15 at 19:53
  • \$\begingroup\$ You know that \$V_{BE}=0.7V\$ so what is the current mirror current? SUPERHINT: you only need the left vertical of this schematic for this answer. \$\endgroup\$ – Daniel Dec 10 '15 at 20:09
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    \$\begingroup\$ Q1 is a diode, assume Vbe = 0.7 V: (20V - 0.7) = 19.4 V / 22k = 881uA. Q1 and Q2 form a mirror so assuming Vin does not force Q2 in saturation, Ic(Q2) is also 881uA. Now calculate volatage across 10k resistor, -0.7 V gives voltage across emitter resistor of Q4 etc etc \$\endgroup\$ – Bimpelrekkie Dec 10 '15 at 20:23
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    \$\begingroup\$ @FakeMoustache Oh you're right. I didn't see -10V down at the bottom... 877uA. \$\endgroup\$ – Daniel Dec 10 '15 at 20:38
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The 0.9mA is probably an approximation (and a crude one at that) to the 0.877mA figure you reach by assuming Vbe = 0.7V. Knowing the collector current of Q2, which is equal to the emitter current of Q3, knowing the DC current gain of the transistors (which should have been supplied, but you might also assume that you're working with transistors with high enough DC current gain that the base currents are effectively zero and collector current equals emitter current) you can calculate the collector current of Q3. Following, you have a known voltage drop over the 10K resistor, which you can use to calculate the voltage over the 10K emitter degeneration resistor of Q4, to determine its emitter and collector currents. The process for Q5 will be identical, and you'll reach a complete DC solution.

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  • \$\begingroup\$ Ok I was hoping that was being rounded up. thanks. \$\endgroup\$ – Hatandboots Dec 11 '15 at 2:56

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