Help Plotting Q-Point

WBahn

Joined Mar 31, 2012
29,976
So the key point here -- and it is one that you've used before but perhaps you weren't too clear on the why -- is that whenever we say Ic = Ie, it is because we are making the assumption that the beta is very large (and, in the limit, infinite). Since Ic = βIb and since Ie = Ic + Ib, we have:

Ie = Ic + Ic/β = Ic(1 + 1/β)

If β is very large (or infinite), the this becomes

Ie = Ic

This is to say that the base current becomes so small that it can be neglected. Now, having said that, the base current may have a negligible impact on the output side of the circuitry but still have a very significant effect on the input side. This is because the other currents (such as in the biasing network) may be so small that even a tiny base current is significant by comparison.

But, for this analysis, we assume that β is inifinite meaning that Ib is zero.
 

WBahn

Joined Mar 31, 2012
29,976
The question regarding if the transistor is local to the transistor, meaning that it doesn't involve Vcc or Ic max or things like that. Basically, the transistor is active if Ic > 0 (if it ended up being greater than Ic max, it could still be active, it just means the circuit has changed) AND the Vce>0 (or, technically, greater than Vcesat, but we are assuming that Vcesat is 0V) AND that the base-emitter junction is forward biased.

The big thing to note is that the load line has the collector voltage going all the way down to 0V, but this line is dictated by the collector resistor. The transistor imposes a tighter constraint because it goes into saturation before Vc drop anywhere near that low.
 

WBahn

Joined Mar 31, 2012
29,976
Gonna take a break, need to let this next question sink in a bit.
Be sure to reflect on the comments I've made on your answers to the prior questions (and the stuff that MrChips has been saying). That should make things quite a bit easier.
 

MrChips

Joined Oct 2, 2009
30,707
One thing that was ignored is the effect of Ib on the base bias provided by the base bias resistors R1 and R2.

At the design stage, one would select R1 and R2 to draw a current that is at least 10 times Ib.

In this circuit, R1 at 1MΩ is too large and R1 and R2 together will draw 50μA, which is about the same value of Ib. Hence Vb is going to be much less than what was assumed.

Edit:

We assumed Ib to be small compared with Ic so that we can make the approximation that

Ic ≈ Ie

This is still a valid assumption.

But if Ic = 5mA and beta = 300, Ib = 17μA which will bring down the base bias voltage.
 
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