Hello again,

Again we are faced with what is allowed to be assumed and what is not allowed.
In this case, we might ask if we can keep rpi constant which would be close if we assumed very small AC changes at the input. If we can not keep rpi constant, then we must include the way it changes as the input AC current changes.

Another simplification we might notice is that because rpi is defined as:
rpi=VT/ib

then that means that according to the original equations the voltage drop between base and emitter is only 0.026v and that may or may not be allowed. Probably a better assumption is 0.026+0.7 or something like that, where rpi is the dynamic resistance of the base emitter diode and the static resistance is 0.7/ib. The combined resistance would then be used to solve for the operating point.

When i use the 0.026v alone i get a gain of -4.115 and that is by considering that the input voltage is some DC voltage plus some AC voltage, and using the peak AC voltage to calculate the voltage gain, not just the DC voltage. The AC gain is different because rpi changes with base current.
I have not done it with the extra 0.7v drop, but since that is in series with the base and it is constant, then it should be able to be added to the DC input bias voltage (i assume is Vbb).

Some questions that have to be answered in order to get the right results, and some of these can only come from knowing how the course work progressed in the past or what the instructor is looking for.
This happens a lot when questions that are not simple come up because we dont know the list of assumptions and presumptions that we would have known had we been sitting in the class since day 1 ourselves.
Of course if the OP can provide a little more detail that would help. The best way is probably to show a detailed example from the past that has already been solved. The assumptions then become more transparent to any readers of the first post.

Assuming we can either use just 0.026 or just later add 0.7 to the input bias DC voltage, a way to solve this is to come up with the solution for Vc after considering all variables, then use:
Av=(Vc1-Vc0)/(Vin1-Vin0)

which is simply the difference in the collector voltages with the starting input DC bias voltage calculated with Vc=4v, and the collector voltage with an increment in DC bias of say 0.1v, and divide that by the increment in DC bias voltage.

Since calculating Vc with all variables considered is key to the solution, you can use this formula to check your results, but dont rely only on this formula rather work it out for yourself and then compare:
Vc=(B*R3*VT)/(B*R2+R2)-(B*Ein*R3)/(B*R2+R2)+(B*E2*R2)/(B*R2+R2)+(E2*R2)/(B*R2+R2)

Factored a little:
Vc=E2-B/(B+1)*(Ein-VT)*R3/R2

Note in the above that the only variables are Vc and Ein, where Ein is the bias voltage plus the AC peak voltage: Ein=Edc+Vpk.
To calculate the initial input DC bias alone (Edc), first set Vc=4 and Vpk=0 (known values of those two) then solve for Edc. Once you have Edc then you can either calculate the first derivative or just use a small increment such as 0.1v as Vpk to calculate the gain, which will then be (Vc1-Vc0)/Vpk.
Sometimes it is good to go both ways, calculate the gains with a positive excursion and then with a negative excursion, then calculate the average gain.

Isn't all this somewhat beyond the pale, given the first sentence in post #1: "I am trying to solve this small signal problem".

 

Isn't all this somewhat beyond the pale, given the first sentence in post #1: "I am trying to solve this small signal problem".

Hi,

I dont think so, because there are many levels of perfection that might be sought after when doing circuits like this. For example, for some circuits with transistors the instructor would give Beta as 100 and constant, while others would want the student to look it up on the data sheet, and others might want to allow it to vary with collector current. Also, some would say that a constant 0.7v base emitter drop is ok to use, while others would want them to use the exponential diode model. It all depends on the course level. Usually the more advanced grade level the more detail is asked for. So it is good to query at least a little to find out what is going to be acceptable.
There are other variations here too, like sometimes they want the student to use a matrix rather than just a couple equations, or use a couple equations and not a matrix, etc. I could give numerous examples, but one of the most common is with the diode because there are so many model levels like, "zero ohms when on, infinite ohms when off", then there is "0.7 volts when on, open circuit when off".