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2018-10-08

2018-10-08

作者: carpediemmlf | 来源:发表于2018-10-09 23:32 被阅读0次
  • Modelling non-ideal operational amplifier
  1. gain
  2. Z_{in}
  3. Z_{out}
  • Non-idealness
  1. A is not \infty but 10^4 \sim10^6
  2. Z_{in} is not \infty, Z_{out} is not 0
  3. complex A(\omega)
  • Non-ideal non-inverting case
  1. Added r_{in} and R_{out}
  2. Conserve current and use Ohm's law
  3. V_{out} =V_{in}\cdot {(open\ loop\ gain)} - i_{out}\cdot {(output\ impedance)}
  • For large A, high r_{in} and low R_{out}
  1. gain = 1+\frac{R_2}{R_1}
  2. Z_{out} \rightarrow \frac{R_{out}}{A}(1+\frac{R_2}{R_1})
  3. Z_{in} \rightarrow \frac{r_{in}A}{1+\frac{R_2}{R_1}}, which reaches 10^{12} \Omega
  • gain = 1 when R_2 = 0, this is a "buffer" to send signal from system into another device without affecting the system

  • Inverting case

  1. gain = -\frac{R_2}{R_1}
  2. Z_{out} \rightarrow \frac{R_{out}}{A}f(\frac{R_2}{R_1})
  3. Z_{in} \rightarrow R_1 + low\ valued\ f(\frac{R_1, R_2, V_{in}}{A})
    use a non-inverting one to achieve high input impedance
  • Over a large frequency range, the phase shift of the feedback could become \pi s.t. negative feedback becomes positive feeback and leads to saturation/oscillation. This is countered by built in reduction of A.
  • Interlude on feedback
    Output = A(input + \beta input)

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