Please explain the gain function for given circuit

Discussion in 'The Projects Forum' started by pratapkollu, Jul 25, 2007.

  1. pratapkollu

    Thread Starter Active Member

    Dec 13, 2006

    Dear Members,
    For the circuit shown in figure above,
    input - 0.5V (p-p),100 kHz sine wave
    required output - 10 V (p-p), 100 kHz sine wave
    op amp used is TL082
    Result: I cannot get 10 V, I can only obtain 5 V (p-p) signal

    I changed C7 = 103, R15 = 420 Ohms, Near J3 = 10K(feed back resistor)
    Now I can get 12 V(p-p)signal, please explain what is the design equation thats is governing the gain here.

    Thanks ans Regards
  2. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    You need to be using +/-12V supplies or greater. Perhaps you should put a meter on these at the opamp and check?

    What is the load impedance? Is it driving too much of a load?

    The 1K input and 20K feedback should provide the required gain of Av = 20. You're schematic looks okay, provided the feedback isn't getting messed up because of the J3 arrangement.

    The output will be 180deg out (inverted) from the input. Although the Unity Gain Bandwidth is only 3MHz, it shouldn't be that bad at 100KHz. Are you sure the input frequency is 100KHz? Try it at 10KHz.

    Can the source adequately drive the 1K input? Try changing the input to 10K and the feedback to 200K (the 0.1uF input cap is fine, Fc = ~1.5KHz as shown). Do the actual componet values match those on the schematic?
  3. hgmjr

    Retired Moderator

    Jan 28, 2005
    \frac{e_o}{e_i}=\left(\frac{-R_J_3}{R_1_5}\right)\left(\frac{s}{s+\frac{1}{R_1_5C_7}}\f{ }{ }\right)

    Here is the expression for the gain in the s-domain
    where e_i
    - represents the input signal
    and e_o
    - represents the output signal.

    and  s = jw