1) (2 pts) Given two voltages, v1(t) and v2(t), and given a differential voltage defines as vd(t) = v2(t) - v1(t), write v1(t) and v2(t) in terms of the common-mode and differential-mode voltages.
The first and foremost thing that you need to look for when reading a question is what you are being asked for. You were asked to "write v1(t) and v2(t)". That means that your answer needs to be " v1(t) = 'something'" and "v2(t) = 'something else'". Many people never has any expression anywhere that did this. The next part says to write these two voltages "in terms of the common-mode and differential-mode voltages". That means that the "something" and the "something else" had better involve vcm and vd.
Several people did regurgitate the formulas v1(t) = vcm + (vd / 2) and v1(t) = vcm - (vd / 2) (and got half credit for this answer) but this shows that you are merely memorizing a bunch of formulas and not understanding them - at the very least it shows a degree of carelessness that needs to be corrected. It was stated in class that you had the freedom to define vd two different ways and that it didn't matter which you used as long as you were consistent. Merely spitting back the above formulas indicates either a lack of understanding this point or, again, carelessness, since no effort was made to be sure that your answers were consistent with the definition of vd as given in the problem statement. Get in the habit of asking if the answer makes sense.
2) (2 pts) In order to keep a BJT operating such that the linearized small signal models are reasonably accurate, the small signal component of the base-emitter voltage needs to be on the order of 10mV. Yet for a differential pair, the magnitude of the differential input signal can be about twice that. Why?
Again, the key is to understand what you are being asked for. A claim is being made that the differential input voltage to a BJT-DP can be about twice as large as the allowed small signal vbe for a BJT. You are asked to explain why this claim is true - or, if you think it isn't true, to justify why the claim isn't true.
Many people were on the right track but merely stated that because you had two BJT's in the circuit that you could have twice the voltage. That doesn't cut it since I can easily design a circuit having two BJT's for which this claim would be false. The key is to relate the input differential mode voltage to the to the small signal base-emitter voltage each transistor. If you are asked something about the relationship between x and y, your answer needs to discuss the relationship between x and y.
3) (3 pts) Draw the basic PNP-based BJT Differential Pair.
Quite a few people just replaced the NPN transistors with PNP transistors. If it were this simple, why would we have two types of transistors? Doing this also reflects a basic lack of understanding of how a transistor works and/or how a differential pair works since doing this mans that that bias current supply has no way to establish the quiescent base-emitter voltage. Several people also blindly took the output voltage across the collectors after having done this straight swap without recognizing that in doing so the two collectors are shorted together and hence there can be no output voltage under any conditions. Again, you have to get in the habit of looking at your work and asking if it makes sense.
4) (3 pts) We have found that the differential voltage gain of the basic BJT differential pair is given by Ad = gm*RC. How would you respond to the claim that, without changing anything else in the circuit, the gain can be made arbitrarily large by either increasing the transconductance or collector resistance. In other words, what mechanism prevents this from being true?
No one got this, even though it was explicitly covered in the lecture and the implications it has for your design project were explicitly pointed out. A few people danced around the issue in a way that perhaps reflected some comprehension of the key issue - but I couldn't tell. Most people tried to claim that gm and RC are interdependent since gm is IC/Vt and made the claim that IC is determined by RC such that if you increase RC you decrease IC. This isn't true - as long as the transistors are operating in the active region, IC is completely independent of RC - the purpose of the bias current source is to establish an average current in each transistor and the purpose of the transistors are to steer that current between the two collector resistors. The purpose of the collector transistors is to convert the collector current into a voltage drop. The output voltage is then the difference between the two voltage drops. However, the average voltage drop across RC is equal to IC*RC and, at some point, this voltage drop becomes too great and the transistors leave the active region and enter saturation.
A few people made statements along the lines of RC = VC/IC. This generally reflects a fundamental lack of understanding of the basic electronics principles and, instead, often reflects a "throw equations at the problem" approach. In this case, you are throwing Ohm's Law at a problem and claiming that if I take any voltage and divide it by any current that I get a resistance. Not true. Ohm's Law states that the resistance is the ration of the voltage across that resistance to the current through that resistance. In this case, VC is NOT the voltage across the collector resistor, it is the voltage at the collector (relative to ground) and unless the other side of the resistor is tied to ground, the voltage across the resistor is going to be something different (in this case, it is VCC - VC).
GENERAL OBSERVATIONS
The fact that no one really touched on the key concept in #4 even though we covered it in class and even though it was pointed out that it directly impacts your Design Project concerns me more than just a bit because many of you will, as a result, spend hours struggling to get your design to work because you are not paying attention and/or comprehending the practical significance of what we are discussing in lecture. I am trying to encourage you to do so by explicitly pointing out things that you will have to deal with on your design projects - but I can't force you to make the effort to understand. That's fine - it just means that you will have to pay for the knowledge you gain through lots of hours spent fighting your design Certainly this is how I paid for much of what I know.
What it comes down to is that the profession of engineering is both an art and a science. The art is learning from your own mistakes and the science is learning from the mistakes of others. In lecture I am giving you the opportunity to practice the science of engineering - to learn from my mistakes and the mistakes of countless others that have gone before us and to learn what we have learned from those mistakes. You can either do that or you can choose to learn the same things by making those same mistakes. Making your own mistakes and learning from them is by far the best teacher - but it is very time inefficient - consider that the lessons to be found in the lecture represent the combined experience of thousands of people gained over hundreds of years (electronics, per se, may not go back that far but the mathematical tools we use most certainly do). You simply don't have time to make that many mistakes and re-invent the wheel. So you need to get as much knowledge as you can from our mistakes so that you can make quality mistakes that really advance your understanding.