Help with project

Discussion in 'Homework Help' started by welkin87, Nov 18, 2010.

  1. welkin87

    Thread Starter Member

    Nov 18, 2010
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    Greetings! I am new to this forum and am a junior in Electrical Engineering. I am in need of some help on a project. This was a group project for school but I got screwed over by my 2 lab partners so I'm left doing this on my own and electronics aren't exactly my expertise compared to other areas I'm studying.

    I have to design a BJT Differential Amplifier in which part of the overall design involves designing a current source.

    Requirements:
    - ±10V voltage supplies
    - single ended output difference mode gain ≧ 100
    - common mode gain ≦ 0.1
    - CMRR ≧ 60 dB
    - output voltage swing (peak to peak) ≧ 6V (symmetric)
    - BJT array chip preferred for the input stages (i have no clue what this is)

    Any ideas for how to get started? I've been using google a LOT and reading the book but I'm still confused. Any help would be greatly appreciated. I've always had trouble with electronics and I have no interest in pursuing this field of electrical engineering when I graduate (nothing against electronics, i'm thankful people are good at it). I'm wanting to focus on electrical energy systems and power grid planning and design.
     
  2. bertus

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  3. PRS

    Well-Known Member

    Aug 24, 2008
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    You seem to have a bad attitude toward electronics. The first thing to do is perform a para-dine shift: "I love electronics!"

    You get to ...

    This is common practice with Diff. Amps.

    All of this is easily achieved.

    Look up the LM3046N Transistor Array data sheet and order one. You can't get it at Radio Shack. This chip has 5 BJTs on one chip and two of them are already set up emitter to emitter. The advantage of having all of the transistors on one chip is that they all have the same current gain, e.g. Beta. Thus they are "matched." Another advantage is that they have low Cu and Cpi values and thus perform at higher frequencies than discrete transistors. By the way, are you actually building this circuit or merely simulating it?

    Like Bertus, I'd like to see your proposed schematic and then we can work from there.
     
  4. welkin87

    Thread Starter Member

    Nov 18, 2010
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    Thanks for the quick replies! I'm still working on some hand calculations. I'll post a schematic and some values I'm computed tomorrow.

    I have chosen my transistor though (or transistors). There's a small shop in the basement of our Electrical Engineering building that sells everything one would need for their projects. I picked up a couple of MPQ3904's. They are an NPN quad BJT chip. Would they work out ok for what I have to do? I assumed they would since they will be matched.
     
  5. welkin87

    Thread Starter Member

    Nov 18, 2010
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    @ PRS:

    Sorry, just saw your question. I have to simulate and build it (I use LT Spice).

    I do like electronics, I just don't like it when I don't understand it and I'm having a hard time wrapping my head around the differential amplifier. It took me forever to grasp the single BJT amplifiers like the common-emitter, collector etc.

    When doing my hand calculations, is there any recommended starting point for differential amplifiers? The hardest part for me is finding a good starting place. It seems like there are so many assumptions one can make in rough hand calculations as well as certain values that can be ignored.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Take a look at the workings of a 741 Op Amp, which is a relatively simple and functional, though not "optimal" op-amp, it has been around forever and many advances have happened since to allow rail to rail output and less noise, but it is a good start.
     
  7. welkin87

    Thread Starter Member

    Nov 18, 2010
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    I can't use an op-amp for this project. I have to use BJT's. I do wish we could use op-amps though!:cool:

    I am confused about common-mode and differential-mode gain. I can't seem to find a good comparison of the two. Any thoughts? Thanks!
     
  8. thatoneguy

    AAC Fanatic!

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    Common mode is a signal that is on both inverting and non-inverting inputs, it should sum to zero (be rejected). Example: Twisted pair wire, noise appears on both wires of the same polarity and amplitude, this noise is rejected by a differential amplifier.

    Differential mode gain is amplifying the difference between the inverting and non-inverting inputs. This amplifies only the difference between the two wires on input, in the twisted pair analogy, the actual signal, since each wire has a different voltage on it to carry the signal, the noise being the same signal on both wires.

    The 741 link I gave you shows the internal construction of an Op amp (Differential Amplifier), and should give you ideas on starting, the stages are essentially input, gain, and output.
     
  9. bertus

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  10. welkin87

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    Nov 18, 2010
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    I have read through that page and it was pretty helpful. Could my design be as simple as that one and still meet the gain requirements? Is it even a possibility to just use 2 BJTs?

    I'll be posting my hand calculations shortly cause I'm having trouble meeting the common-mode gain requirement (Acm ≦ 0.1), but I'm exceeding the diff.-mode gain requirement by a lot (about 388).
     
  11. PRS

    Well-Known Member

    Aug 24, 2008
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    I'm glad to see you hung in there. You can't use that design because your specs said you need to use a constant current source and so this requires a third transistor. See my attachment. It's a schematic of a standard diff amp with a current source. The current source is biasing the diff amp. A good current for small transistors like these is 1 mA. So have your current source provide 2 mA and it will be split between Q1 and Q2 (the two upper amps) evenly so that 1 mA flows through each of those transistors. And with this in mind consider the need for your voltage swing of +/- 6 volts. The latter tells me that the Ve of your diff amps needs to be below -6 volts.

    We can trade that gain for other qualities. As I read it, you only needed a gain of 100 V/V.
     
  12. welkin87

    Thread Starter Member

    Nov 18, 2010
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    I can't thank you guys enough for helping me out! It means a lot. I was stressing to no end when I realized I was doing this by myself.

    @ PRS
    That makes sense. I was wondering about that current source. Correct me if I'm wrong, can that current source be treated like a common-emitter for my calculations? Or is it a common-collector? (it looked to me like a C-E)

    Here's some questions I wanted to ask before I get too far just to start over:

    1. If the single-ended diff.-mode gain has to be at least 100, then Add will need to be at least 200, right?

    2. If CMRR has to be greater than 60 dB, that's a value of 1000, correct?

    With respect to the current source, should I start my hand calculations with the differential amp stage or the current source?

    Sorry for all the questions, I'm trying to learn as much as I can from this, but I only have until this Tuesday evening at 5pm to get the actual circuit checked off by a TA. Will I be able to finish this by then?

    Thanks again!
     
    Last edited: Nov 20, 2010
  13. PRS

    Well-Known Member

    Aug 24, 2008
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    The best I can do is to show you a solved example similar to your problem. This example comes from a book by Lorne MacDonald. Refer to the included picture.

    R3 and R4 form a voltage divider with Vee.

    VR3 = (Vee X R3)/(R3 + R4) = 6 volts = Vr4 for the same reason.

    VR5 = 6 volts - Vbe(Q3) = 5.4 volts

    IE3 = VR5/R5 = 2mA

    The current in Q3 splits evenly through Q1 and Q2 so that

    IE1 = IE2 = 1mA

    VR6 = IC2 * R6 = 5 volts

    VR2 = IB1 * R2 = (IE1/Beta)*R2 = 1 volt assuming Beta = 100

    VR7 = VR2 for the same reason.

    VC1 = 12 volts

    VC2 = Vcc - VR6 = 7 volts

    VCE1 = Vcc - VE1 = 13/6 volts

    VCE2 = VC2 - VE2 = 8.6 volts

    VCE3 = VC3 - VE3 = 5 volts where VC3 = VE1 = VE2

    Av' = R6/(re1 + re2) where re = Vt/IE = 26mV/1mA = 52 ohms

    = 5k/52 = 96 volts per volt

    But there is voltage loss by the divider ratio of R1 in series with Zin.

    Zin = R1 + [R2 //( Beta*2*re2)] where Beta*2*re2 is the resistance at the emitter reflected to the base of Q1.

    Zin = 5943 ohms

    Therefore the voltage loss due to R1 and Zin = Zin/(R1 +Zin) = 0.83

    So the total gain, Av = 96*.83 =79.6 volts per volt

    And if there is a load resistor, there is divider action there, too.

    The signal swing is 2*Vce(Q2) or 2*IC2*RC, whichever is smaller.

    2*8.6 = 17.2 volts peak to peak but...

    2*(1mA*5k)= 10 volts peak to peak and therefore the output signal swing is 10 volts peak to peak.

    ******************************************

    I hope this helps. Designing is a similar process, but think about the problem in reverse. Start by getting your output swing of 12 volts peak to peak. To get a gain of 100 with 1mA collector current Rc must be at least 6.8k or so.
     
  14. welkin87

    Thread Starter Member

    Nov 18, 2010
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    Are the capacitors required? One of the major opening points of the chapter this project was pulled from was the use of direct coupling and not using capacitors (although my teacher didn't say that we couldn't use them). I didn't notice any other guys in the lab using caps.

    One other question I had about the two schematics you've posted for me. I was reading the check-off sheet for the project and it looks as though I'll be connecting two AC supplies to the amp (I'm assuming they'll connect to the base terminals of Q1 and Q2). Will this change much on the schematic?

    Here is the check-off sheet for the project:
    [​IMG]
     
    Last edited: Nov 21, 2010
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Caps are usually needed to keep the bias in working range by blocking a DC offset on the input, and to remove any from the output.

    --ETA: Looks like you need to have an inverting and non-inverting input, this isn't available in the schematic I looked at above.
     
  16. welkin87

    Thread Starter Member

    Nov 18, 2010
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    I was just looking at an example in my textbook and they're shooting for a gain of 100 (Add). "For Add=(gm)(Rc)=40(Ic)(Rc), a gain of 100 can be achieved with a voltage drop of 2.5V across the resistor Rc." (from the textbook)

    I can't figure out how they made that assumption. How did they figure out that a 2.5V voltage drop was needed? I'm needing a gain of 200.

    Update: I think I figured it out. Vc=IcRc and 100/40=2.5
     
    Last edited: Nov 21, 2010
  17. welkin87

    Thread Starter Member

    Nov 18, 2010
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    Ok, I've finished some hand calculations that meet all the requirements, but I haven't done the calculations for the current source yet. I found a Diff. Amp. Design example in the text that I somehow missed, so I used it as a guide and below is what I ended up with.

    If any of you wouldn't mind looking over the calculations I have so far, I would really appreciate it. (I'm using a MPQ3904 for my BJTs)
    [​IMG]
     
  18. welkin87

    Thread Starter Member

    Nov 18, 2010
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    Any thoughts at all?
     
  19. thatoneguy

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    Feb 19, 2009
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    Do you have a schematic to go with the values calculated?
     
  20. welkin87

    Thread Starter Member

    Nov 18, 2010
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    I meant to attach it to my last post.
    Assumptions: beta=100, VA=75V (early voltage)

    I'm having trouble figuring out the current source. The book has a current source design example but I can't figure out how to get r0. (didn't find it in the datasheet)

    [​IMG]
     
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