I have been getting really good help with understanding EX 9.3. Understanding EX 9.4 will be paramount as well. I have a lab where I have to design a Level Shifter similar to Example 9.4, plus a couple of the homework problems are similar to this example. One of the biggest crimes is that the book doesn't ever draw the full circuit. I will paste the Book's small explanation of Level Shifters and the one example problem. I tried to get help from my instructor on this one, but it sound like maybe they are mishmashing small signal models and large signal models and not being clear which parts of the solution correspond to which of the 2 models. I will also bring my attempt to draw the circuit. I think I have the basic idea down, but some of the specifics.







This is my attempt to draw the CE Amplifier-LevelShifter Sandwich, but I somehow doubt its going to be as simple as CTRL-C then CTRL-V the pieces together so to speak.



I have plenty of questions about the solution. But perhaps some of them might be cleared up by getting the darn circuit drawn & labeled properly.

• The Common Emitter Aplifiers in this text each have a voltage divider on the base. What should the resistor values be on the voltage dividers? (4 unknown resistor values here)
• What should the collector and emitter resistor values be on each CE amplifier? (4 more unknown resistors here)
• Are the 2 CE Amplifers and level shifter connected to the same V_CC value, or will each of these 3 items have their own separate value V_CC.
• Are the Emitter Resistors connected to some V_EE value that needs to be determined, or simply to ground?

 

I won't be able to proceed with circuit analysis until I get the particulars of this drawing cleared up mentioned in the bullet points. Does anyone have any insights on any of the 4 bullet points?

 

Firstly, the transistor in 9.7c should be NPN, but is drawn as PNP.
Whoever published the book scores zero out of ten for proof reading.
”op amps have level shifters in them to compensate for offsets” - no they don’t - they have offset null connections which compensate for offsets by changing the balance of the long tailed pair.

 

Firstly, the transistor in 9.7c should be NPN, but is drawn as PNP.
Whoever published the book scores zero out of ten for proof reading.
”op amps have level shifters in them to compensate for offsets” - no they don’t - they have offset null connections which compensate for offsets by changing the balance of the long tailed pair.

Thank you for the tips Ian. I really appreciate the reply!

So on my drawing I should redraw as NPN. A big problem I am having is its impossible for me to analyze the circuit until I know how the rest of it is hooked up. The text did not provide a drawing of the full circuit, which leaves allot of unknowns for me.

• What about the possible presence of unknown voltage divider resistors on my drawing? (all the CE amplifiers have a voltage divider on the base in the text, so I threw that in on both CE amplifiers, but I don't know what values of resistors to use on the drawing).
• Also, what about the voltage connections for my collectors and emitters?

 

Let’s put the correct NPN transistor in, and to make it easy to understand, let’s replace R1 by a zener diode or some other voltage reference.
Now you should see that you have made a constant current source with the new transistor, zener and Re.
Let‘s also assume, although it doesn’t say, that Vee is a negative voltage. It doesn‘t matter how negative, but not too small.
Rb also isn’t too helpful in understanding what is going on, so short it out.

Now do you see how it works?

 

Okay, so I will replace R1 with zener diode for both the left CE amplifier and , will I leave R2 as a pullup to "V_CC" for the zener. Speaking of Vcc, one uncertainty I have is whether both CE ampliers AND the level shifter share the same value V_CC. I don't want to jump to that conclusion unless someone with more knowledge and intuition than me would think that is a reasonable idea.

V_EE, conversely appears to be simply "ground" in the CE amplifier entry in the table. I'm uncertain whether to connect the emitters to ground or to V_EE of the level shifter. I once again don't want to jump to possible errant conclusions.

The following is what I thought I was supposed to do, is cut and paste the following circuit onto both sides of the level shifter. Is this the right idea, or am I totally off base here? The circuit for the CE amplifier in this text is as follows:


Some assumptions I made (I hope their correct) is that i can ignor R_load and replace with an open. For the capacitors, do I just simply treat them as open circuits (caps are open to low frequencies, especially DC0 during Large Signal Model analysis, but then short them for small signal analysis (caps at high frequencies are short circuits)

 

Okay, so I will replace R1 with zener diode for both the left CE amplifier and , will I leave R2 as a pullup to "V_CC" for the zener. Speaking of Vcc, one uncertainty I have is whether both CE ampliers AND the level shifter share the same value V_CC. I don't want to jump to that conclusion unless someone with more knowledge and intuitiion than me would think that is a reasonable idea.

V_EE, conversely appeears to be simply "ground" in the CE amplifier entry in the table. I'm uncertain whether to connect the emitters to ground or to V_EE of the level shifter.

The following is what I thought I was supposed to do, is cut and paste the following circuit onto both sides of the level shifter. The circuit for the CE amplifier in this text is as follows:
View attachment 320548

Some assumptions I made (I hope their correct) is that i can ignor R_load and replace with an open. For the capacitors, do I just simply treat them as open circuits (caps are open to low frequencies, especially DC0 during Large Signal Model analysis, but then short them for small signal analysis (caps at high frequencies are short circuits)

That CE amplifier is not what is in figure 9.7b!
The lower portion of 9.7b is not a common emitter amplifier at all, it’s a constant current source
(and the top half is a common collector amplifier)

 

That CE amplifier is not what is in figure 9.7b!
The lower portion of 9.7b is not a common emitter amplifier at all, it’s a constant current source
(and the top half is a common collector amplifier)

The problem statement seems to suggest (to me at least) that it requires 2 CE amplifiers need to be attached to each side of Figure 9.7b. In my understanding, 9.7b is only the level shifter, but does not include the (seemingly necessary to me at least) CE amplifiers. The problem statement says the 2 CE amplifiers are in "series" with the level shifter so I tried to do my best to construct the full circuit. Should I omit the 2 CE amplifiers? Perhaps the problem statement just giving me extra information about the topology that I don't really need to solve the problem.

 

Your task is to design the level shifter. Imagine that someone else designed the two CE amplifiers and then person that build them discovered the need to level shift the signal from the output of one to the input of the other and they've come to you to do that for them, under the constraint that, for whatever reason, you can't change either of the CE amplifiers.

 

Your task is to design the level shifter. Imagine that someone else designed the two CE amplifiers and then person that build them discovered the need to level shift the signal from the output of one to the input of the other and they've come to you to do that for them, under the constraint that, for whatever reason, you can't change either of the CE amplifiers.

Perhaps I have been overcomplicating the problem the whole time. Does this mean I basically wouldn't need to figure out how the person(s) who designed the CE amplifiers configured their circuits, and thus can omit the CE amplifiers entirely from my drawing of the equivalent circuit for the purposes of this problem?

 

Perhaps I have been overcomplicating the problem the whole time. Does this mean I basically wouldn't need to figure out how the person(s) who designed the CE amplifiers configured their circuits, and thus can omit the CE amplifiers entirely from my drawing of the equivalent circuit for the purposes of this problem?

It doesn't necessary mean that you can ignore the CE amplifiers entirely, but whatever aspects of them you need is provided. For instance, they tell you what the quiescent output voltage of the first amplifier is, as well as it's collector resistor. You may or may not need those pieces of information, but you don't need any other information.

 

If you ever come across this problem in real life, it’s a pretty safe bet that you will be dealing with AC signals, and all you will need is a capacitor!

 

If you ever come across this problem in real life, it’s a pretty safe bet that you will be dealing with AC signals, and all you will need is a capacitor!

So... people don't need amplifiers that can amplify DC signals?

Almost everyone has devices in their home that need to do this, such as a bathroom or food scale.

I think if would be a shame if an EE program ignored such circuits and only expected students to be able to work with AC signals.

 

Almost everyone has devices in their home that need to do this, such as a bathroom or food scale.

Wouldn‘t a strain gauge be amplified in a single stage, with a single op-amp, or an instrumentation amplifier?
Level shifting is a potential way of introducing errors, so it would be avoided if at all possible.

 

Wouldn‘t a strain gauge be amplified in a single stage, with a single op-amp, or an instrumentation amplifier?
Level shifting is a potential way of introducing errors, so it would be avoided if at all possible.

The strain gage signals are pretty low level signals.

Besides, if we only expect students to learn about AC-coupled stages, who's going to design those opamps and instrumentation amps?

 

who's going to design those opamps and instrumentation amps?

Hopefully, someone who has learned a way of doing it that doesn’t drift with temperature and doesn’t depend on the Vbe voltage of a transistor.

 

Hopefully, someone who has learned a way of doing it that doesn’t drift with temperature and doesn’t depend on the Vbe voltage of a transistor.

Oh, I will grant you that there are plenty of second order effects that aren't dealt with, or probably even mentioned, in an academic environment. That's true across the board -- and it pretty much has to be if we want degree programs to take less than a couple decades to complete. Even if they never end up using a level shifter in a circuit, the exposure to the concept and particularly how a topology is constructed to address a problem and how the math goes into it and comes out of it, is useful to help them develop problem solving skills, both related to transistor circuits and in general. Sadly, those opportunities are getting watered down and pushed aside in favor of memorizing a bunch of formulas for a test and relying on software tools to make their decisions for them.

 

I think I have figured out a couple of very important pieces of this after consulting a couple other texts!

Things I have figured out so far:

• The Resistor R_B shown in Figure 9.3is the Thevenin equivalent Resistance of the first CE. This value is 1 k-ohm in EX 9.4.
• V_C of CE Amp 1 @DC is V_BB, which is 4V.
• V_C(t) of CE Amp 1 @ AC is "v_in"

I have managed to figure a few things out after dozens of hours spent trying to understand this one example. But I am still enormously overwhelmed by the weight and quantity of things that puzzle me about this problem. These are disappointingly meager gains after spending an absolutely monstrous quantity of time trying to understand this subject.


1. One thing that puzzles me in the textbook is that they get v_out = r_o2*ic?






2. How does the author get 2.29k for R_E?
It seems as if the author defies the impossible by cranking out R_E=55k-ohm for EQ. 9.36 with just this one equation, but both Vout and R_E are unknowns. I plugged their value of 55k in, and they would have somehow needed to know that Vout would be 3.3V (yet that value is not given?)?



3. I DO understand why R'B is R1||R2, but I don't understand how why they are multiplying Beta,R'E, & 0.1 together. I especially don't understand what the 0.1 is where it came from, an why it's being used here.



4. I See that the problem says the level shifter is connected to +/- 10V supplies. Is Vcc=10V and Vee=-10V?

5. Sort of a followup to the last question, What is being averaged here exactly? I see 10V and 1V being averaged together does this have something to do with +/-10V supplies? And, if so, wouldn't we need to factor in more like 20V since there is a positive and negative rail of the same magnitude? Also, I presume the 1V comes form the 1V bias of the 2nd stage from the problem statement.

 

on example 9.4, As far as solving for RE, I get that we are supposed to use Equation 9.36. But not all the data is seems to be present.

I have the following data:
Vbb =4V
Vbe =0.7v
RC = 1k-ohm
Ic = 1 mA
Beta = 100

Equation 9.7 has 7 inputs, of which I can only account for the above 5.

If I had these and Vout I could solve for Re. Or, if I had these 5 and Re I could instead solve for Vout, but the problem doesn't seem to have either.

 

The problem says to provide a 1 V bias to the second stage.

What do you interpret this to mean?