High gain amplifier ideas?

Discussion in 'Homework Help' started by blackclothman, Apr 22, 2012.

  1. blackclothman

    Thread Starter New Member

    Apr 21, 2012
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    Hi,

    I was assigned in class to design a high gain (>100dB) MOS amplifier with just two gain stages (using CD 4007 CMOS chip). I've tried to read up on some literature to see what could be possible, but it just seems that I can't find a starting point to tackle this this problem. It would be a great help if anyone had any suggestions that would help me get going in the right direction (even places to find relevant info would be great)!

    There are other specific requirements as well, but I would much rather tackle this problem on my own for the most part. I would just like some help getting on track, since I am completely lost right now.

    Thank you!
     
  2. justtrying

    Active Member

    Mar 9, 2011
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  3. picbuck

    New Member

    Dec 13, 2010
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    I'm a TL072 guy, myself. But it might help your thinking to consider that 100 is not exactly "high" gain.

    A first stage (or preamp) with a gain of 10, which is piddling, a following stage also with a gain of 10, also piddling, and there ya go.

    Have you tried the obvious? Google for "CD4007 amplifier circuits" sans quotes.

    -a-
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,652
    2,348
    Hello,

    @picbuck, The OP is talking about 100dB and not 100 times.
    The 100dB is MUCH more gain.
    We do not know if he means POWER gain or VOLTAGE gain.

    Bertus
     
  5. blackclothman

    Thread Starter New Member

    Apr 21, 2012
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    I am sorry for the confusion. Yes, it is actually very high gain: 100 dB = 10^5 V/V gain. And it is voltage gain, not power gain.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,788
    4,808
    Perhaps I am misinterpretting what was meant, but while I know that some people use dB to be "voltage gain" (he shudders as he types), that does not appear to he the case here and the 100dB was properly indicating a power gain.

    1 bel, by definition, is the base ten logarithm of the ratio of two powers 10log(P1/P0). A decibel is, therefore, 1/10 of a bel and thus 1dB = 10log(P1/P0). All else being equal, power is proportional to the square of the voltage and, hence, 1dB=20log(P1/P0).

    The point is that if given a "gain of 1000" and asked to express it in dB, it is necessary to know whether that gain is voltage or power. But if told that the gain is 20dB, it
    "should be" unambiguous. We are talking about a power gain of 100W/W or, equivalently, a voltage gain of 10V/V.

    I've seen people (including a few college professors) that maintain that a decibel is ALWAYS 20log(A/B) where A and B could be anything. In fact, I've had one say that they don't even have to be the same units, that it could be the ratio of kWh to number refrigerators. As somewhat of a purist, I maintain that the argument to any transcendental function, of which log() is one, must be dimensionless.

    I also argue (sometimes successfully) that it might be the case that representing something using the concept of a decibel might be handy and useful, but do not pretend that it is, or has units of, dB. Call it something, anything, else.

    Now, other purists have argued for an even harder line stance than mine. They say that to truly be a dB, it must represent a true power ratio and not, as is commonly the case, a hypothetical ratio under the assumption of equal load impedances. For power budget applications, I can see their point. But that horse has long since fled the barn and so we have to work with it under those assumptions in most cases.
     
  7. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    So assuming a max output of 5VAC, input would be 50uV.
    At this level, noise will be significant--will be difficult to obtain better than about 30db signal to noise ratio (guess)

    One trick that greatly increases voltage gain is to use a constant current load 'resistor' --what this does is to force the load line horizontal so that any perturbation in gate voltage causes tremendous excursions in drain voltage--care must be taken to design a practical self-biasing arrangement--adjustment may be required.

    This could be accomplished via the use of a constant current diode that is essentially a two lead JFET device.
     
  8. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Here are the gain and frequency response curves of an ordinary Cmos logic inverter like in a CD4069. You can connect a CD4007 to do the same. The supply voltage affects the voltage gain and bandwidth.
     
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