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Amplifier design

  1. Jan 24, 2015 #1

    Zondrina

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    Homework Helper

    1. The problem statement, all variables and given/known data

    Given the availability of resistors of value 1 kΩ and 10 kΩ only, design a circuit based on the non-inverting configuration to realize a gain of +10 V/V.


    2. Relevant equations


    3. The attempt at a solution

    So I was doing some of those good old design problems, and I came across this one.

    I know the gain of the standard non-inverting configuration is given by ##\frac{v_o}{v_i} = 1 + \frac{R_2}{R_1}##. So I want to design the circuit such that:

    $$10 = 1 + \frac{R_2}{R_1} \Rightarrow \frac{R_2}{R_1} = 9$$

    I'm limited to only the ##1k## and ##10k## resistors provided, and this provides a design problem. Usually, we would like the input impedance to be a large as possible to retain as much of the signal in accordance with ##V = IR##. Similarly, we would like the output impedance to be as small as possible. Looking at the equation:

    $$\frac{R_2}{R_1} = 9$$

    It doesn't take long to realize I must choose ##R_1 << R_2##, which of course is the exact opposite of what a good amplifier should have. Ideally I want to choose ##R_1 >> R_2##, but I can't figure out the ideal resistor values.

    Is this just a bad amplifier design?
     
  2. jcsd
  3. Jan 24, 2015 #2

    gneill

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    Staff: Mentor

    If Vin is applied to the + input of the op amp then the input impedance it sees is that of the op amp, i.e. very large indeed. The feedback loop involves the op amp output and the - input of the op amp only.

    Don't be afraid to stick several of the available resistors in series to make up larger values, after all it is an academic not an economic exercise :)
     
  4. Jan 24, 2015 #3

    Zondrina

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    The inverting amplifier suffers from a low input impedance problem in general; it causes a problem because ##R_2## can potentially be unrealistically large (this is easy to see considering the inverting amplifier gain). The problem can be corrected by adding a few extra resistors around ##R_2##; Adding a resistor ##R_3## in parallel with ##R_2## using the virtual ground at the input terminals, and a resistor ##R_4## just before the output node.

    The gain is then (which we want to be 10V/V):

    $$\frac{v_o}{v_i} = - \frac{R_2}{R_1}\left(1 + \frac{R_4}{R_2} + \frac{R_4}{R_3} \right) = 10 \frac{V}{V}$$

    Choosing ##R_1 = 10k##, we would need to choose ##R_2 = 10k## so the first term is unity. This makes it easier to select ##R_4## and ##R_3## so the gain is 10. Choosing ##R_4 = 10k## to preserve the signal, and ##R_3 = 0.833k## to satisfy the equation, the circuit will realize a gain of 10.

    Now there's an issue, I can only use ##1k## and ##10k## resistors, so ##R_3## is going to have to be some combination of resistors.

    Sticking 5 ##1k## resistors in series, and then putting those in parallel with another ##1k## resistor, provides the desired resistance.
     
  5. Jan 25, 2015 #4

    rude man

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    How about one feedback resistor & one to gnd, then dividing the + input down to get Av = 10?
    Total no. of resistors = 4.
     
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