# Common emitter bjt amp Q point with and without current mirror

What is the real benefit of voltage divider CE BJT amplifier schematic like that? If I use different BJT's in real life with various betta I'll give different voltage drops XMM1. Different betta of real BJT's willn't allow to center the Q-point voltage XMM1 on output curve. Or it's not critical to allow Q-point some drift on output curve?

I'm also tryring to use current source instead R3 to fix Q1 collector current but I can't understand is that schematic usable in real life? And I can't understand how to select R3, R4 values if I know Q1 collector current (that equal to I_ref tuned by R1)?

• I thought I saw a comment from you asking for a derivation of an equation in electronics.stackexchange.com/questions/260979/… . Was that question resolved? – jonk Oct 19 '17 at 18:50
• Also, just as an aside, your attempt (as I see it) of a high gain of 100 at a quiescent current near $1\:\textrm{mA}$ is miscalculated. But that's a different story for later. – jonk Oct 19 '17 at 18:58
• @jonk The question was about VB = Rth * (Vcc/R1 - Ib) equation. I don't fully understand how to derive it? – MaxMil Oct 20 '17 at 10:09
• @jonk If I use the current mirror like in second schematic what I need to set Q1 collector Q-point voltage to half of 10V? – MaxMil Oct 20 '17 at 10:19
• It's easy. The base voltage is equal to the Thevenin divider voltage minus the Thevenin divider resistance times the base current (which loads the divider.) Imagine the BJT base isn't connected. Then it's obvious there is a Thevenin voltage and resistance, correct? Now, if you draw a small current from this source, there will be a voltage drop. Correct? Do you see this? – jonk Oct 20 '17 at 15:22

• In the simple CE configuration you provide as the first diagram, it is common to design it without over-dependence on the $\beta$ of the BJT. As a result of that approach, it usually turns out that variations of $\beta$ between parts in the same family (at worst, on the order of $\le 50\%$) has only a relatively small impact on the quiescent point, so long as you've made your voltage divider "stiff" enough. What impact there is, is due to the variations of base current as a loading on the voltage divider's Thevenin resistance. So this is quite predictable, too.
• In the simple CE configuration you provide as the first diagram, it is also common to design it without over-dependence on the $V_{BE}$ of the BJT. ($V_{BE}$ variations are due to saturation current variations, which themselves can vary by a factor of as much as 3 between parts. But since this is part of a logarithm function, the impact is less than you may imagine at first.) This is achieved usually for a different reason -- temperature stability. But it also has another impact, which is to reduce dependence of the quiescent point on $V_{BE}$ variations. This is often the larger impact on the quiescent point, though, than variations of $\beta$ -- though of course it all depends upon the design goals and the actual design choices that are made.