opamp design

Discussion in 'Homework Help' started by prescott2006, Oct 2, 2009.

  1. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    1. how can i modify the circuit to get a gain of 4000 and 9000? how should i start?
    2. how to replace the current source in practical, i.e in lab.?
    3. to get such a high gain the rc1 seem has to replaced with active load.but what is meant by active load and how to build it?
    thanks.:)
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    The controlled gain of an op amp is realized through feedback, and not in fiddling with the gain of the input differential amplifier. Take a look at out Ebook material on op amps and how to control their gain in follower and inverting configurations -http://www.allaboutcircuits.com/vol_3/chpt_8/3.html .
     
    Last edited: Oct 3, 2009
  3. Ron H

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    I think he's talking about open loop gain.
    To increase the open loop gain, I would do the following:
    1. Make RC2 an active load (current sink).
    2. Eliminate RE2.
    3. Eliminate the RE's (you'll need Q1 and Q2 to be matched to minimize input offset voltage).
    4. Make the tail current sink dependent on a Vbe, so it (mostly) cancels the effect of Vbe3.

    This is homework. Do some research on BJT current sources and sinks, and post a proposed solution, so we can critique it.
     
  4. beenthere

    Retired Moderator

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    Oops - I really have got op amps on my mind lately...
     
  5. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    thanks for your reply.not i do not want to provide a solution,but we haven't learn about the current source or active load.but the lecturer want us to modify the circuit to get a gain value of 4000 (i not sure he wants close loop or open loop gain). can u draw a circuit for my reference? because i even do not know where to start.
     
  6. Ron H

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    Can you do anything you want with the circuit?
     
  7. prescott2006

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    Nov 8, 2008
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    i think so, but need to modify base on the circuit provided.
     
  8. Ron H

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    The active load in the location you suggested will get the gain up into the range of 4000-10000, depending on transistor betas. I also removed gain degeneration resistors. See attachment.
    I simulated this, using a tail current sink of 200uA, and RC2=15k.
    Do you need to put in real parts in place of the current sink?
     
  9. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    ya,because i need to verify the circuit in lab so i need replace the current source with real circuit.by the way, what value should i use for capacitor CC and RL?
     
  10. Ron H

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    First, we need to know what kind of transistors are you going to use (part numbers).
    For open loop gain measurement, RL can be omitted.
     
  11. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    1.he does not state which transistor will be use,in case we are using 2n222 or 2n5120,how can we construct the current source circuit?
    2.what is the suitable value for capacitor CC?
    3.the open loop gain is measured by placing the probe at where?
    4.in case he mean modify the circuit to obtain closed loop gain,what step should i proceed?
    5.i had simulate the circuit u gave me,but the input at the base seem so big,what is going on?
     
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  12. Ron H

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    1. 2N222 is not a valid number. Did you mean 2N2222?
    2N5120 is a dual PNP. Are you sure you are using these?

    2. Cc almost has to be determined empirically (by simulation or in hardware). I am not interested in deriving the value analytically. Its purpose is frequency compensation, so the amplifier doesn't oscillate when feedback is applied. If you use a single value, it must be chosen for unity gain, noninverting, since this is the most feedback you can have. The optimum value is somewhat dependent on the types of transistors you use, which is why I asked.

    3. The way I measure open loop gain is with feedback applied. I'll show you the scheme after we get the transistor types nailed down.

    4. The gain spec is almost certainly open loop. This circuit is not suitable for a closed loop gain of 4000 or more.

    Have you studied the applications of op amps, and how to apply negative feedback?
     
  13. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    it is not written on the lab manual,so let us just assume we r using beta=100.ya,i have basic knowledge on feedback.so can you please show me how to measure open loop gain?
     
  14. Ron H

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    See this article.
    BTW, in my simulations, I found that a good compensation network consisted of 100pF in series with a 1k resistor.
    Beta is only one important spec when determining the compensation network. Collector-base capacitance is another.
     
  15. prescott2006

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    Nov 8, 2008
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    i have connect it base on your scheme,but i not sure the connection is correct or not?seem i am on the wrong path,can you please send me your connection diagram?
     
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  16. Ron H

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    Don't you need to add a battery for +15V to your schematic? You have one for -15V.
    Change Cc to 100pF in series with 1kΩ, as I indicated previously.
    Otherwise, your circuit looks OK.
    Be advised that the generic models for PNP and NPN may not have capacitances like a real transistor, so your AC and transient simulations may be very unrealistic.
     
  17. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    +15 is on top, -15 is at below (my lecturer say there not suppose to be grounded and should be -15)
    but i still confuse of how to measure the open loop gain.can you explain to me more detail?i want to ask whether bd139 and bd140 is suitable in my design or not.
     
  18. Ron H

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    You're missing the point about the battery. You have a battery for -15V. You don't have a battery for +15V. I have never seen a simulator that would interpret a node labeled "+15V" as a power supply. You need to add a +15V battery to your schematic.

    To measure gain, you need to use the AC analysis mode (not transient mode), and plot the Bode plot of Vout. You can set Vin to 1V, because the simulator linearizes the circuit at the quiescent point, so it can't be overdriven, no matter how much AC voltage you apply. This does not work this way in transient simulations, even if you apply a sine wave.
    Sweep the AC signal from 10Hz to 10MHz. measure the output voltage and Vx at low frequency (10Hz should be OK). Divide Vout by Vx, and multiply by 101. This is the gain.
    BD139 and BD140 will make a low-bandwidth op amp, but they are medium power devices, possibly suitable for the output emitter followers. They have no specs for capacitance and Ft (gain-bandwidth product), which makes it hard to predict how well (or poorly) they will work.
    You should use general purpose small signal transistors for all. I personally would use 2N3904 and 2N3906, and maybe PN2222 and PN2907 for the output emitter followers. Some of the BC series transistors would also be fine, but I am not familiar enough with them to make recommendations.
    The input differential pair and the PNP current mirror transistors need to be matched pairs. You get this automatically when you simulate, but not with hardware.
     
    Last edited: Oct 4, 2009
  19. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
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    OK,finally i arrive at this circuit,i want to replace the current source with real circuit.but i don't know how connect them,can you show me?
     
  20. Ron H

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    I edited my last post. You probably missed the edits.
    In your schematic, you have drawn a current source. You actually need a current sink.

    WHERE IS YOUR 15v BATTERY?

    See my attachment. This includes the gain measurement circuit.
     
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