Book Example for Level Shifter CE Sandwich Also Doesn't Sense

Jony130

Joined Feb 17, 2009
5,539
This exercise is confusing. They are talking about CE amplifier and voltage at the collector. But then they use T1 as an emitter follower and set DC voltage at the output to 1V. Strange.

1. One thing that puzzles me in the textbook is that they get v_out = r_o2*ic?
Notice that for the AC signal, Ib2 is 0A. No AC source signal is present at the T2 base, therefore ib2 * β2 must also be 0. Wich in the case of a current source means an open circuit.
And this is why the Author wrote vout = r_o2*ic?

2. How does the author get 2.29k for R_E?
Well, KVL. Vbb = Ie*Rb/(β1 + 1) + Vbe + Ie*RE + Vout. Thus, for Vout = 1V, Vbe = 0.7V and β1 = 100 we have

Re = (Vbb - Vbe - Vout)/Ie - Rb/(β1 + 1) ≈ (4V - 0.7V - 1V)/1mA - 10Ω = 2.3kΩ - 10Ω = 2.29kΩ

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.
It is another rule of thumb. In short, we want to have a stiff voltage divider (R1, R2). To achieve this, we usually choose a voltage divider current much larger than the base current, 10 x times usually.
Another way of achieving a stiff voltage divider is to assume that R'B is 0.1*β*R'E.

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

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.
We want to have Vout = +1V and this is the collector voltage of a current source (T2). The "bottom end" of a current source is connected to -10V (Vee).
Thus, the available voltage for T2 is 11V. T2 should be NPN type, of course, not PNP type, as was shown on the schematic.
The T2 will work as a current source only if T2 is in active region Vce > Vce(sat). So the Author decided to pic Vce2 in the middle of an available voltage 11V/2 = 5.5V.
And this is why we have R'E = 5.5V/1mA = 5.5kΩ

WARDEVIL_UFO

Joined Nov 16, 2010
47
The problem says to provide a 1 V bias to the second stage.

What do you interpret this to mean?
Thanks for clearing that up. I definitely had interepreted the word "bias" to mean something else entirely.

At work , Our circuits have several housekeeping voltages called "Bias Supplies" on our circuit boards that supply voltage to IC's as well as a pullup voltages for optocoupler outputs and op-amp Open Drain & Open-collector style outputs, so that they can operate. We use the word "Bias" to something much more similar to V_EE, V_CC or V_DD, or V_SS. So that part must have been throwing me off since we use the terminology very differently. For instance, all op-amps and other IC's would have the + and - rails connected to a 12V BIAS supply, the output for the open-collector would also be pulled up to the 12V. To me, it seems very unfamiliar and against what I am accustomed to to have the DC component of a transistor output referred to as a "bias voltage". To be honest, I had actually thought at first that the biases referred to a possible V_CC of the two CE amplifiers.

Anyways, it looks like I am in business as far as finding R_E goes:
Equation 9.36:
$V_{out} = V_{BB} - \frac{R_B*I_C }{ \beta}-R_C *I_C-V_{BE}$

Solving EQ 9.36 for R_E:
$R_E = \frac{\beta*(V_{BB} - V_{BE} - V_{out}) -R_C *I_C}{\beta*I_C}$

Substituting known Values:
$R_E = \frac{100*(4V - 0.7V - 1V) - (1 k\Omega) *(1mA)}{100*1mA}=2290\Omega$

WARDEVIL_UFO

Joined Nov 16, 2010
47
$11V = V_{R'E} + V_{CE(T2)} =V_{R'E}+V_{R'E}= 2*V_{R'E} \implies V_{R'E}=5.5V$
We want to have Vout = +1V and this is the collector voltage of a current source (T2). The "bottom end" of a current source is connected to -10V (Vee).
Thus, the available voltage for T2 is 11V. T2 should be NPN type, of course, not PNP type, as was shown on the schematic.
The T2 will work as a current source only if T2 is in active region Vce > Vce(sat). So the Author decided to pic Vce2 in the middle of an available voltage 11V/2 = 5.5V.
And this is why we have R'E = 5.5V/1mA = 5.5kΩ
It makes so much sense when you put it the way you did. Before, it looked like the were averaging 11V and 1V together. But this is not really what is being done.

If I am following correctly, the full explanation of this step is as follows:

So basically, they implicitly did KVL at the bottom transistor. The voltage drop from the (should-be) collector to VEE is:
$V_{out} -V_{EE} = V_{R'E} + V_{CE(T2)}$

Substituting known values:
$1V - (-10V) = V_{R'E} + V_{CE(T2)}$

Which simplifies to:
$V_{R'E} + V_{CE(T2)} = 11V$

We know that the voltage across the sum of the V_R'E and V_CE must add to be 11V total. On the previous page, the books said that they were "setting the current-source transistor operating point in the middle of the dc load line". I didn't grasp before, but I think I grasp now that this verbiage essentially implies we much choose a resistor R'E such that:
$V_{RE} = V_{CE}.$

With the assumption that R'E is chosen such that V'RE and V_CE(T2) are equal in mind, we can further simplify:
$11V = V_{R'E} + V_{CE(T2)} =V_{R'E}+V_{R'E}= 2*V_{R'E} \implies V_{R'E}=5.5V$

Is this thought process correct?

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Jony130

Joined Feb 17, 2009
5,539

WARDEVIL_UFO

Joined Nov 16, 2010
47
Looks good.
Thank you. For my own learning, I am going to build the level shifter and current source this example with NPN transistors and Resistors in my lab. Will I need to physically build the 2 CE amplifiers (one on the input, and one on the output) Or could I just replace them in the lab with their Thevenin equivalent circuits (Vth in series w/ Rth) for each of the 2 CE amplifiers?

WBahn

Joined Mar 31, 2012
30,290
Thank you. For my own learning, I am going to build the level shifter and current source this example with NPN transistors and Resistors in my lab. Will I need to physically build the 2 CE amplifiers (one on the input, and one on the output) Or could I just replace them in the lab with their Thevenin equivalent circuits (Vth in series w/ Rth) for each of the 2 CE amplifiers?
You can replace them with their equivalents, but keep in mind that it's two different equivalents -- the first is the equivalent as seen at the output and the other is as seen at the input.

Also, the equivalent circuit is only valid at DC. It's a different circuit at AC. But that's probably good enough for your purposes (if I understand them correctly).

Since the level shifter you are implementing doesn't include current mirrors, you shouldn't run into transistor matching issues. But if you ever get tempted to implement a mirror with discrete transistors, you will need to take matching into account, which can usually be dealt with using ballast resistors in the emitter paths.

WARDEVIL_UFO

Joined Nov 16, 2010
47
Thank you. For my own learning, I am going to build the level shifter and current source this example with NPN transistors and Resistors in my lab. Will I need to physically build the 2 CE amplifiers (one on the input, and one on the output) Or could I just replace them in the lab with their Thevenin equivalent circuits (Vth in series w/ Rth) for each of the 2 CE amplifiers?
Also, the equivalent circuit is only valid at DC. It's a different circuit at AC. But that's probably good enough for your purposes (if I understand them correctly).
I had a feeling that might be the case. I am definitely willing to settle for seeing the DC level shifted empirically.

If I am slightly more ambitious, I think including AN AC source (chosen with a specific amplitude and frequency baked in) into the thevenin equivalent might be able to shift that particular signal (and only that particular signal) to the new 1V operating point, with Rth solved for that particular configuration.

I'm getting closer to the end now

Like the textbook, I solved for V_R’2, and found V_R2 = 3.8V.
Transferring all solved values to the schematic, I have:

With V_R2 found, I would think we would be able to find V_R1 directly through KVL with the loop that has R1, R2 and VEE.I have:
$V_{R1}+V_{R2}+V_{EE}=0$

Solving for V_R1:
$V_{R1} = -V_{EE} + V{R_2} = -(-10V) + 3.8V = 6.2V$

Another way of obtaining of V_R1 = 6.2V would seem to be recognizing that V_R1 is in parallel with the series combination of the base-emitter drop and the drop across R’E. In that case:
$V_{R1} = V_{BE}+V_{R’E} = 0.7V + 5.5V = 6.2V.$

V_R1 = V_BE+V_R’E = 0.7V + 5.5V = 6.2V.

So no matter what way I work this, I seem to get 6.2V

The text on the other hand gets a different result:

I found that if I replace the 0.001 mA with 1 mA, and zero out the first term in the other bracket, (55k/100), I get 6.2V. Are they making a mistake by throwing in R’E/Beta, or am I missing something?

Also, I don’t understand why they are equating V_R1 to V_BB which was previously specified to be 4V.

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WARDEVIL_UFO

Joined Nov 16, 2010
47

Above appears to attempt an application of Ohm's Law inside the Product-over-Sum formula for Parallel Resistance. But it looks like they are using the voltage across 5.5V voltage of R'E instead of the voltage found for R'1. The 3.8/I makes some sense to me, since that is the voltage we found for V_R2. But using 5.5V for VR1 makes no sense at all to me. They solved for V_R1 to be 6.75V. I found it to be 6.2V. It seems like they are disregarding their own result.

Jony130

Joined Feb 17, 2009
5,539
The text on the other hand gets a different result:
Here the Author tries to include the voltage divider loading effect (Ib2). And solves for VBB (Thevenin's voltage for T2 base).

I found that if I replace the 0.001 mA with 1 mA, and zero out the first term in the other bracket, (55k/100), I get 6.2V.
It is a mistake, it should be 1mA.
Are they making a mistake by throwing in R’E/Beta, or am I missing something?
55kΩ is R'B, not R'E. So the Author "transforms" RB into the emitter side. To include the voltage divider loading effect.
The better way is to "transform" R'E to the base side and then
VB'B = 1mA/100 * (55kΩ + 100*5.5kΩ) + 0.7V = 10μA*605kΩ + 0.7V = 6.75V

Also, I don’t understand why they are equating V_R1 to V_BB which was previously specified to be 4V.
Mistake. It should be VB'B alone.

But it looks like they are using the voltage across 5.5V voltage of R'E instead of the voltage found for R'1. The 3.8/I makes some sense to me, since that is the voltage we found for V_R2. But using 5.5V for VR1 makes no sense at all to me.
Yes, another mistake. It should be 6.2V.

They solved for V_R1 to be 6.75V. I found it to be 6.2V. It seems like they are disregarding their own result.
Yes, they solved for VB'B. And in the end, the Author decided to skip the loading effect.
This really must be a poor book.

One of the ways to do it correctly is:
6.75V = 10V* R1/(R1 + R2) (1)
55kΩ = (R1 * R2)/(R1 + R2) (2)

The solution is:
R1 = 170kΩ
R2 = 82kΩ

Also, what does this example have to do with CE amplifier? Is beyond my mind.

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WARDEVIL_UFO

Joined Nov 16, 2010
47

Here is an updated understanding of what I calculate the voltages to be.

Jony130

Joined Feb 17, 2009
5,539
Here is an updated understanding of what I calculate the voltages to be.
Looks fine to me. Good job.

WARDEVIL_UFO

Joined Nov 16, 2010
47
Looks fine to me. Good job.
So, the book is wrong then? They seemed to get 6.75V for V_R2, yet appear to use 5.5V for V_R2 in the computation below. Should I just stick my value of 6.2V (instead of 5.5V) in, and then solve for I?

I am almost finished with this! Woohoo!!

Jony130

Joined Feb 17, 2009
5,539
So, the book is wrong then?
This is what I said in my answer. You didn't read it? Or you don't understand it?

The book answer is R1 = 134kΩ and R2 = 93kΩ thus, if we assume that β = 100 and Vbe = 0.7V.
We have:

VB'B = Vee* R1/(R1 + R2) = 10V* 134kΩ/( 134kΩ + 93kΩ) ≈ 5.9V
Rth = R1 || R2 ≈ 55kΩ

So the emitter current will be equal to Ie = (VB'B - Vbe)/(Rth/(β+1) + R'E) ≈ ( 5.9V - 0.7V)/(550Ω + 5.5kΩ) ≈ 0.86mA
As you can see, the book is off by 14% from wanted 1mA.

"My solution"
R1 = 170kΩ
R2 = 82kΩ

VB'B = 10V* 170kΩ/( 170kΩ + 82kΩ) ≈ 6.746V
Rth = R1 || R2 ≈ 55kΩ
Ie = ( 6.746V - 0.7V)/(550Ω + 5.5kΩ) ≈ 1mA

Should I just stick my value of 6.2V (instead of 5.5V) in, and then solve for I?
Yes, but you don't have to strictly follow the design procedure shown in the book.

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WARDEVIL_UFO

Joined Nov 16, 2010
47
This is what I said in my answer. You didn't read it? Or you don't understand it?
My apologies. So, what I think must have happened is when I started drafting a reply, one of my posts was the most recent at the time. I had some delays submitting it because of stuff that came up, and then by the time that I finally did submit, I had missed your post this morning! I should have checked to see if someone snuck in a reply while I was drafting. Very sorry about. This was not my intention at all. It looks like we both posted at 8:22 at nearly the same time, with me slightly later.

I'm going through your 8:22 am post as well as the one from just a little bit ago now and going to run the calculations as I go along. I appreciate all your help so much Jony. you explain things very clearly. Between you, Ian, and Wbahn, I am finally grasping this example!

WARDEVIL_UFO

Joined Nov 16, 2010
47
"My solution"
R1 = 170kΩ
R2 = 82kΩ

VB'B = 10V* 170kΩ/( 170kΩ + 82kΩ) ≈ 6.746V
Rth = R1 || R2 ≈ 55kΩ
Ie = ( 6.746V - 0.7V)/(550Ω + 5.5kΩ) ≈ 1mA
I went ahead and simulated your result with LTSpice. I'm impressed your results are nearly perfect! I also simulated my own results and the book's as well. I found that as far as accuracy goes, ranking the results Jony > WARDEVIL_UFO > Book, with the book being the least accurate of all. I am gong to post your's, but omit the others. Your's produces an IE' of -987uA, which is very close to -1mA and a DC output of 977 mV which is very close to 1V.

For convenience, i tried my best to keep the Reference Designators the same as the schematic in the text.

After reviewing your solution, it is clear you are doing some sort of impedance transformation, and it sounds like there are a couple of transforms that could be attempted. I'm still trying to wrap my head around how it might be visualized on a schematic (especially the before and after). The way you mention that the base side stuff can be moved to the emitter side, or conversely the emitter side stuff moved to the base side, feels kind of remiscent of a different transformation. In particular the feel of it sounds similar to the idea of using a transformer turns ratio to reflect primary side impedances to the secondary, and vice versa. In this case, it sounds like Beta is taking on a similar role that the turns ratio would in an impedance transform, as an analogy.

The other thing I think you might be doing is finding is placing a box around V_EE, V_R1,V_R2 and base-emitter V_BE portion of the transistor and producing a thevenin equivalent:

Jony... everyone.. I massively appreciate your patience.

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Jony130

Joined Feb 17, 2009
5,539
β
The other thing I think you might be doing is finding is placing a box around V_EE, V_R1,V_R2 and base-emitter V_BE portion of the transistor and producing a thevenin equivalent:
Yes, the Thevenin is very useful in this case

We can replace the voltage divider (this gray rectangle) with Thevenin's equivalent circuit

$V_{TH} = V_{CC} \times \frac{R_2}{R_1+R_2}$

$R_{TH} = R_1||R_2 =\frac{R_1 \times R_2}{R_1+R_2}$

As for the transformation. Based on the Thevenins equivalent circuit we can write the KVL equation and solve for IB current this way:

Vth = Ib*Rth + Vbe + Ie*Re

Ie = Ib + Ic = Ib + Ib*β = (β +1)*Ib

Therefore

Vth = Ib * Rth + Vbe + Ib*(β +1)*Re

Now we can solve for the base current:

Ib = (Vth - Vbe)/(Rth + (β +1)*Re)

Notice that now the Re resistor "appears" on the "base side" as a beat+1 large resistor.
Therefore the equivalent circuit is:

This is the equivalent from the base point of view.

But we can write KVL equations slightly differently and solve for Ie current directly.

Vth = Ib*Rth + Vbe + Ie*Re

We knows that Ib = Ie/(β+1)

Therefore

Vth = Ie/(β+1)*Rth + Vbe + Ie*Re

And we solve for Ie current:

Ie = (Vth - Vbe)/( Rth/( (β+1) + Re )

As you can see this time Rth "appears" on the "emitter side" as a beta + 1 smaller resistance.
And all of this because Ib = Ie/(β +1) and Ie = (β +1)*Ib
And the equivalent circuit:

I hope that this appendix helps.

WARDEVIL_UFO

Joined Nov 16, 2010
47
I hope that this appendix helps.
This appendix is EXTRAORDINARILY helpful! This was perfect! We can consider this thread solved!! You outdid yourself Jony! I am extraordinarily grateful to you and everyone who helped me refine my understanding to the point where I am now at! I appreciate you guys!

I did the following step-by-step manipulations of the given circuit to make it closely resemble the form for what you have in the grey rectangle in the diagram, and was able to reason out the final missing links on my own, thanks to this appendix.

Initially, the negative voltage sources screwed super majorly with my head on how to redraw the circuit, and the question of whether Vth should be referenced to V_EE or ground was a concern The only difference is that that What we called R1 and R2 were swapped, and instead of V_CC, I have -V_EE = -(-10V) = 10V, since R1 & R2 are biased by a negative voltage source V_EE connected to ground, and not V_CC in this particular problem. After redrawing and then simulating, I have my answers to those questions, and Vth should be referenced to V_EE!! Which for anyone else who read this I will paste my work to help others get there who might read this.

Also, your explanation for the results obtained on the Transformation was absolutely flawless! Ordinarily I thought of Beta as more of an agent for current transformation, which it is in its most standard definition. In fact you proceeded from the definitions, applied them to this problem, and then with some algebraic manipulations, it becomes evident that we can reinterpret Beta as resistance transformation as well!

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