Need help with BJT current source

Discussion in 'Homework Help' started by doze, Feb 10, 2013.

  1. doze

    Thread Starter New Member

    Feb 10, 2013
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    I don't know how to approach this circuit. I wish to get a an 400uA<Ic<2ma if I can but the way its configured is wierd. I don't know much about these things. what I do know is we can't use current mirrors and that this current source is to be used to power a Differential amplifier for a Adm=40 and CMRR greater than 70dB

    Circuit is attached

    I guess another question is, how much Ic for the current source do I actually need to meet that specification. I know Ib3<<IRb2. Other specs:

    Vic = 0 V Isupply ≤ 3 mA Zero-to-peak un-clipped swing at Vo1 ≥ 2.5 V VCC =
    VEE = 5 V |Adm| = 40 Operating frequency: 1 kHz
    Rid ≥ 20 kΩ CMRR ≥ 70 dB
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    How did you calculate the resistor values that are shown in the schematic?

    Here is the formula for the current sink with voltage divider:

    \mathsf{I=\frac{Vb-0.6}{R_E}

    Vb is decided by the 2 voltage divider resistors on the base (Rb1, Rb2 in your diagram)
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    (Vb-0.6V) is the voltage on the emitter of the current source transistor, it is NOT the voltage across the emitter resistor (unless the emitter resistor's other side is tied to ground, which it is not in this case).

    Try:

    I=\frac{(V_b-V_{be})-V_{EE}}{R_E}

    where R_E is Rb3 in your circuit.

    BTW: Your VEE is -5V, not 5V. It makes a HUGE difference.equation is

    But the above equation is an approximation based on the assumption that the base current is a negligible portion of the current in the bias circuit. That is not the case here unless you assume a beta that is greater than 1000 or so. Even with a beta of 300, the base current is half what the nominal current is in the bias resistors.
     
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  4. WBahn

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    Mar 31, 2012
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    I don''t see how you are making such an assertion. The 2N2222 has a β>100. Yet your emitter current in Q3 is 141 times the current in the bias resistors even assuming no base current at all.
     
  5. doze

    Thread Starter New Member

    Feb 10, 2013
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    The values in the Differential pair are random, I was testing how the current source reacts to it.

    For the current source:
    IeRe=nVbe with n= 4 and Vbe=.7V
    I2R2=(Beta(n+1)*Vbe)/m*Ic R1= Beta*(n+1)*Vbe/(m+1)*Ic After I couldn't get Ib<<Rb2, I just started tweaking various Rb values
     
  6. doze

    Thread Starter New Member

    Feb 10, 2013
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    Let me clarify. The R values in the DIfferential pair are arbitrary. I assumed a Beta of 200 because my lab measurments have given me a value close to it.

    I dont know much about this stuff, I just picked a design class to see if I want to do it. I like it but my lack of knowledge is killing me
     
  7. doze

    Thread Starter New Member

    Feb 10, 2013
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  8. doze

    Thread Starter New Member

    Feb 10, 2013
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    one more question, I am trying to pick values for the differential pair now. For Re and Rc.
    The lab manual says the Adm(small signal differential-mode gain)was : vod/vid~= -Rc/re1+Re

    My book says its = -gmRC. do I use the gmRC method to guess a reasonable Rc value?

    PS: going to bed now, I'll be back in 5 hours or so
     
  9. doze

    Thread Starter New Member

    Feb 10, 2013
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    any suggestions?
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    Don't be so impatient. People here have lives and need to sleep some time. It took 20 days before my first post received a response.

    What is gm, in terms of re and Re?

    If you are taking a design course without the circuit analysis prereqs, it's not surprising that things aren't going well. That's like taking a course in laproscopic surgery without having taken anatomy in order to see if you want to be a surgeon.
     
  11. doze

    Thread Starter New Member

    Feb 10, 2013
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    I took the circuit class but of its prereqs, in one the teacher didn't cover enough material and the other, it was a summer class that was sped up. All you had time to do was to study how the test is going to be. Either way yes I am struggling in it.

    gm~= Ic/Vt as far as I can tell. My lectures and books haven't told me otherwise. Anywho I was able to get the Diff gain i needed by adjusting Rc

    My question now is: How can I get a voltage swing for Vo1 to center around 0 and if that is not possible, how can I get a bigger Vo1 Voltage swing?
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    At the end of the day it doesn't matter if you slept in class, had a lousy teacher, had a pathetic text book, just weren't into the material, or had too much other stuff going on. I've pretty much had all of the above at one time or another. At the end of the day all that matters is that you have a hole in your knowledge and skill set and it is up to you to fill it in.

    You need to understand gm (transconductance) better than just some equation. What does it mean? What does it tell you? How is it defined?

    It is defined as the change in output current that results due to a change in input voltage. When talking about a transistor as a disembodied component, your output current is the collector current and the input voltage is the base-emitter voltage. Based on that and the Ebers-Moll model you should be able to derive the equation above in about five lines of math. But you can also talk about the gm of the circuit in which the transistor is a part. You need to be clear on which is being discussed/used and keep them straight.

    This isn't design, this is "a happening", in which you more or less randomly try things until you find something that works.

    Unless you are going to go back and learn circuit analysis, you pretty much will always be reduced to hoping for a happening, so just start changing values until you get something that works.
     
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