Ic(max) is the Vce = 0 point on the load line.

Determine Ic at Q-point and calculate voltage across Rc, then Vc and Vce.

 

Is this a correct circuit analysis? Also is Ic at the Q-point found using beta?

 

This is the maximum current you can possibly get down the Rc + Re chain.
This fixes the Vce = 0V point of the load line. The other end of the load line is Ic = 0, Vce = Vcc.

You don't know beta.

You have already calculated Ve. Now calculate Ie and Ic.

 

I do know beta because he gave the capacitor MPSA09. Average beta is 300. I am unsure if It = Ib

 

You do not need to know beta. You have already calculated an approximate value for Ve. Use that.

 

ok so. 5.8mA x 4.7k and 5.8mA x 8.2K . I am completely lost on what these voltages are, and what you are asking me to calculate.

 

so you want me to calculate the voltage at Q-point. How do I go about figuring that out? I need the current at Q-Point. T_T

 

You are totally lost.

I have to repeat. The 5.8mA you calculated is the point on the graph where
x = Vce = 0V
y = Ic = 5.8mA

Put a dot here.

Next put a dot at
x = Vcc = 75V
y = Ic = 0mA

Draw a straight line connecting these two points. This line is your Load Line. The Q-point must lie somewhere on this line.
Now forget about that 5.8mA. It will never be used in your calculations again.

Now calculate Ie = Ve/Re = 23.3V/4.7k

 

Haha thanks for the clarification . I knew what you meant by the 5.8mA. I was just confused why you were having me calculate Ie again when I had already did before. I thought you were implying I had done something wrong. The Vce and Ic should be correct on my last picture from page 1. It was the Ic(max) I had wrong. Now 4.9mA is the operating current correct?

 

Ok, so first I calculated Vb using the formula (R2/R1+R2)75V. I then used the formula Vbe=Vb-Ve to get Ve=23.9V-.7V . Finally I used the formula Vce=Vc-Ve to get Vce=53v-23.9V. The only problem is that the voltage Vce and Ic Dont line up correctly on the load line so I think i am sitll doing it wrong.

Look at the axis labels of your graph. There is NO mention of Vce there! The vertical axis is Ic, the collector current. The horizontal axis is Vc, the voltage at the collector (relative to ground). It is NOT Vce. As I pointed out before, you are GIVEN Vc in the problem statement. Is it 53V. All you have to do is find the collector current that corresponds to a collector voltage of 53V.

The fact that your load line is a straight line means that you can establish a REALLY good estimate for it. Note that if Vc were 50V, you would be exactly 2/3 of the way along the bottom. This means that you would also be exactly 2/3 of the way down the vertical from 9.1mA toward 0mA, so you would be at ~3mA. Since your 53V is a bit further to the right, your actual operating point will be a bit further down. So you should expect a collector current of slightly under 3mA.

Now, part of the problem is the problem itself. Is this the entire problem? Because it is not self-consistent. If you aren't given that the collector voltage is 53V, then just the circuit itself gives you enough information to find the collector voltage and it comes out to be significantly less than 53V -- in fact it is more like 35V.

Now, that is assuming the beta is infinite, which you have assumed when you said that Ic = Ie. But, for this circuit to have Vc be 53V, the beta has to be pretty low. Does this problem come from a set, or a point in the course, where you aren't supposed to automatically assume high beta?

 

Ok I am trying to be as clear as possible for you, I am new to this stuff and it is not easy and I also know it is frustrating trying to teach a dummy.

 

And I also missed you post where you mentioned that the instructor told you that the problem wasn't supposed to include any mention of Vc being measured to be 53V. So that makes the problem much simpler since now it is reasonable to assume beta is very large and Ic ~= Ie.

I'll post a list of questions that should help you in a couple of hours. Right now I have to get out the door.

 

Okay. You've actually gotten most of the peices at one point or another, but I don't get a warm fuzzy that you really have a handle on what that you've done is correct and what isn't. This all seems more like "a happening", as one of my former profs used to say. So let's look at a coherent approach to the problem and, again, most of these steps you have done at one point or another.

Q1) If we assume infinite beta, then what can be said about the relationships between the base current, Ib, the emitter current, Ie, and collector current, Ic, in the transistor?

Q2) Given the answer above, what is the base voltage, Vb, on the transistor?

Q3) Given the answer above, and assuming that the transistor is "on" (a point that must be verified later), what is the emitter voltage, Ve?

Q4) Given the answer above, what is the emitter current?

Q5) Given the answer above and the answer to Q1, what is the collector current?

Q6) Given the answer above, what is the collector voltage?

Q7) Given the answer above and the answer to Q3, what is the collector-emitter voltage, Vce?

Q8) Given the answers up to this point, is the transistor in the active region? Justify you answer.

Plot the point, (Vc, Ic) on your load line. That is the Q-point.

Now imagine that the base voltage is varied such that the transistor just barely reaches cutoff.

Q9) What information do you know about Ic and Ie at this point?

Q10) What is the collector voltage, Vc, at this point?

Plot this (Vc, Ic) point on your load line and label it "cutoff".

Now imagine that the base voltage is varied such that the transistor just barely reaches saturation.

Q11) What information do you know about Vce at this point?

Q12) What is the collector voltage, Vc, at this point?

Plot this (Vc, Ic) point on your load line and label it "saturation".

Now, the line that you have been showing on the loadline plot is the region of operation that the presence of the collector resistor places on the relationship between the collector voltage and the collector current. The smaller region between the "cutoff" point and the "saturation" point is the region of operation imposed by the transistor (and emitter resistor) and the desire to keep the transistor in the active region.

Bonus Question (to see if you are understanding things): What happens (to Vc and Ic) as the base voltage is adjusted further to move the transistor deeper and deeper into saturation?
Connect the "cutoff" point and the "saturation" point by a straight line (

 

 

 

 

You do not need to calculate Ib.

If beta = 100, Ib will be 1% of Ie, hence we can ignore Ib in next step:

Ic = Ie

Vc = Vcc - Rc x Ic

 

 

 

Sorry forgot to put "a" and "b" into the description.

As long as "a" does not reach 0 or exceed Ic(max) it is in the active region.

As long as "b" does not reach 0 or exceed Vce(max) it is in the active region.