Differential Amplifier circuit

Discussion in 'Homework Help' started by PsySc0rpi0n, Dec 5, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hello...

    I'm studying the attached circuit at the moment and I need help to at least confirm my calcs!

    The first problem is to find the transistors Q-point.
    I've started by replacing the signal power supplies by their internal impedance.
    Then I have evaluated the Iee current by:

    Iee = ( V1 - (-Vee) ) / Ree
    Iee = ( -0.7 - (-10) ) / 2.2kΩ = 4.227mA

    With this value I can also evaluate Ie1 and Ie2, Ie1 = Ie2 = Iee/2, Ie1 = Ie2 = 2.114mA, because the 'left' side of the circuit is theoretically equal to the 'right' side of the circuit.

    Can anyone confirm this steps?
     
  2. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Well, I think I'm not correct because voltage at the node '1' is not 0.7V from the B-E junction as there is one resistor, Re1, in the middle!

    So, I took another approach but I would like to confirm this!

    I tried to write an equation for the net including B-E1 juntion, Re1, Ree and -Vcc like this:

    Vbe + Re1*Ie1 + Iee*Ree + (-Vcc) =0

    Knowing that Iee = 2*Ie1

    Vbe + Ie1 * (Re + 2*Ree) - Vcc = 0
    Ie1 = (Vcc - Vbe) / (Re + 2*Ree)
    Ie1 = (10-0.7) / (0.1 + 2*2.2) = 2.378 mA

    But now, as I don't know the transistor gain, I'm not sure if I'm supposed to assume a gain of 100 or 200, for instance, or if is there any other way of evaluating Ic and Vce to find the Q-point of the transistors!
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    If ever there was a case for a .DC sweep analysis, this is it...

    2.1.gif
     
  4. PsySc0rpi0n

    Thread Starter Well-Known Member

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    Ok, but I need to find the transistors Q-point. So I need to confirm my calcs...
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your equations looks good but the result is wrong. According to Google the correct result is 2.0666mA. So Ic = Hfe/(Hfe+1)*Ie ≈ Ie
     
  6. PsySc0rpi0n

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    Yeah, I already got that too... I did 10.7/4.5 instead of 9.3/4.5
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Now, to find Vce, can I build a net equation including the following:

    -Vcc + Rc1*Ic1 + Vce + Re1*Ie1 + Ree*Iee - Vcc = 0
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For sure you can do that. But I prefer this way:
    Ve = -Vbe = -0.7V
    Vc = Vcc - Ic*Rc = 10V - 2.07mA*2.2kΩ = 5.446V and Vce = (5.446V - (-0.7V)) = 6.14V
     
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  9. PsySc0rpi0n

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    Well, I have considered the small Ib and evaluated Vce but the result is not very close to your's:

    Vce = 2*10 - 2.2*2.059 - 0.1*2.067 - 2.2*4.23 = 5.9575V...

    Is this result acceptable, or is it wrong somehow?? I have considered the gain from LTSpice equal to 255.

    But even if I consider Ic1 = Ie1 = 2.07mA, I get Vce = 5.933V
     
  10. Jony130

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    But Iee = 2*Ie = 2*2.0666mA = 4.1332mA

    So Vce = 2*10 - 2.2*2.059 - 0.1*2.067 - 2.2*4.133 = 6.17V
     
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  11. PsySc0rpi0n

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    Ok, I got where I went wrong... I assumed a wrong value of Iee that I have previously wrongly calculated!

    Thanks, for now! Going to write this calcs on the paper!
     
  12. hp1729

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    Nov 23, 2015
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    Gain or 255 out of a 2N2222?
     
  13. PsySc0rpi0n

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    Isn't 'Bf' the gain on LTSpice model's parameters?
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    You are getting better at spotting things, which is very good. You are correct that the voltage at Node 1 is not 0.7. But this is not only due to the emitter resistors, but also to the fact that, even without them, the voltage at Node 1 would be -0.7 V. You also got careless with your original equation when you said that the current would be related to (V1 - (-Vee)). Vee is the emitter power supply (which in this case happens to be -Vcc). So if you plugged this in you would have (V1 + Vcc). Don't be complacent about minus signs, they are THE most significant digit in a value (with rare exception)!

    Your equation is correct, but you messed up in plugging the numbers into a calculator. Redo the math and you will get 2.067 mA.

    And you STILL refuse to track your units! Your last line should be

    Ie1 = (10 V - 0.7 V) / (0.1 kΩ + 2*2.2 kΩ) = 2.067 mA

    You have enough problems making mistakes in your work, why do you refuse to use the single most effective error detection tool available to the engineer?

    You can assume a reasonable transistor gain (for a 2N2222 that would be closer to 100 than 200, but either should give results that are good enough). For most DC calculations such as this, you can often assume infinite beta and get results that are good enough. Just run the numbers. If you assume infinite beta, then the current in each collector resistor is 2.067 mA and the voltage at each collector is 5.45 V. If you assume a beta of 100, you get a collector current of 1.023 mA and a voltage at each collector of 5.50 V. Does that 0.05 V difference matter very much. It's a 1% error and your resistors are probably on that order and your components are almost certainly not matched that well.
     
    Last edited: Dec 5, 2015
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  15. WBahn

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    You know that the Ve of the transistors is about -0.7 V. You know the emitter current in each transistor and, from that, the collector current (assume a reasonable beta or use infinite beta). From that you can get Vc for the transistors. Well, Vce = Vc - Ve = 5.45 V - -0.7 V = 6.15 V. The error due to the finite beta will be smaller than the error due to assuming a fixed 0.7 Vbe.
     
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  16. PsySc0rpi0n

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    WBahn, you're right... But it's not a question of refusing to use the tool... I just forget to use it! I'll try to make more use of the tool in the future!

    About considering beta as infinite, and the error being greater because of considering Vbe = 0.7, of course you're also right, but for now, because I already write down the math into the powerpoint, I'm going to leave it as it is. But in the future, I might consider your suggestion!

    After Jony130 explain his approach, I also think it is a simpler approach, but I still haven't my intuition as sharp as yours, so I went by the harder way, as it is normal in students! :p

    I hope I can get more experience so that I can make things simpler, as you guys do!

    Next, I'm going to calculate gains in Differential and Common Mode! Hope I'm using the correct terms... I've been looking in H&H book for these amplifiers, but I couldn't find them! :(
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    Not to worry -- much of learning is doing it the hard way and then seeing it done the easy way. That is a much more effective way of learning the easy way then just having it shown to you first. It's a shame that that's so often the case, but it is.
     
  18. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, now I have the following schematic to analyse! It's the Signal equivalent circuit we use to evaluate the Differential Mode Gain!

    https://awwapp.com/b/upnoo4gtg/

    I just have 2 questions:

    In a similar schematic I have from our teacher, he says that Ie1 gets into node marked in red, Ie2 gets out of the same node towards TR2 and Iee = 0 A, because there is no current flowing through Ree, and therefore, we can consider node 1 grounded! Why is Ie2 getting out of the node 1 towards TR2 and not the other way around?

    The other question is about the polarity of the 2 input signal voltage supplies. One has the greater potential grounded and the other has the lower potential grounded, right?
     
  19. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I don't see any diagram. As for Iee = 0A. Do not forget that now you are talking about small-signal analysis (AC current not DC). And this means that the AC component is 0A for Iee. But the DC current will flow. Why is that ?
    The differential amplifier is a cascade connection of the two "basic" BJT amplifiers. The first one is CE (Common emitter) and the second stage is CB (Common base) stage.
    [​IMG]
    Also notice that Vdi = Vbe1 + Veb2 (from KVL loop). And this means that as Vbe1 increase Vbe2 must decrease by the same amount. For example if Vbe1 changes from 0.60V to 0.61V the Vbe2 will drop by 0.01V from 0.6V to 0.59V. And this will increase Ie1 current from 2.07mA to let as say 2.17mA (due to Vbe1 increase) and Ie2 will decrease by the same amount form 2.07mA to 1.97mA

    ΔIe1 = 2.17mA - 2.07mA = 0.1mA
    ΔIe2 = 1.97mA - 2.07mA = - 0.1mA


    So the AC current sum of the emitters current Iee = ΔIe1+ΔIe2 = will be equal to 0A.
    Because the AC component of a Ie1 and Ie2 are equal in magnitude but 180° out of phase.

    This is not entirely true for your circuit with Ree resistor instead of a current source because with Ree resistor the common mode voltage will slightly change Iee current. But you don't have to worry about it right now.
     
  20. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, the diagram that I have is this:
    aac-1.png

    Ok, I understood the question about the Iee being equal to 0A. What I don't understand is why Ie2 flows OUT of the node marked as 1 and not the other way around. Or better, in a point of view I understand, but from another point of view I don't.

    Let's see:

    In one hand, I understand because if we know that Iee = 0A, so the Ie1 can only flow towards TR2 emitter. But in this case, another question rises: TR2 is supposed to be an NPN tyoe, as is TR1, right? So how does the Ie2 current flows into TR2 emitter?

    In other hand, if I don't notice at the first time that Iee = 0 A and I try to draw first the directions of Ie1 and Ie2, I would draw Ie2 flowing into node 1 and not flowing out of the node.

    Can you clear me about why am I still having this question?
     
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