I amplify a signal which including AC and DC.Why is the output waveform changed when C1 is 0.1u?And how to decide the value of C1? enter image description here


1 Answer 1


When Capacitor Reactive Impedance, Xc(f) rises to affect impedance ratio with R such that they are equal, you can measure many responses;

  1. the voltage amplitude divider ratio (gain) reduces times \$\dfrac{1}{\sqrt(1+1)} = 0.707 = -3~dB\$
  2. the current phase shift changes by 45 deg, (=trig. angle with equal sides of impedance) which as already started to shift with 0.1uF that raises the impedance of Xc(f) to 26.5 kohm
  3. the -3dB breakpoint is \$ω_1=1/(R_1C_1)=2\pi f\$
  4. you can visualize all of this with a log impedance RLC nomograph and measure the impedance ratio of any XL(f) or C(f) vs V as the division ratio becomes a difference on a log scale. like 1000 is a difference of 3 decades
  5. R+ C Series Impedance = \$Zc(f)=R+jXc(f)\$

    1. Also with this you can "ballpark" estimate corner frequencies and LC resonant circuits and Q gain factors for LC intersections with R differences (Ratios). e.g. L//C//R is high impedance
      so $$Q_p=R/X_L(f)=R/X_C(f)$$
    2. at LC value intersection (=resonant f)
      then series resonant f (L+C+R) $$Q_s=\frac{X_L(f)}{R}=\frac{X_C(f)}{R}$$ (using absolute values or ignoring phase due to j

    Here is one example of the RLC Impedance nomograph.

enter image description here

  • \$\begingroup\$ sorry for the late reply.if i want to avoid attenuation and phase shift,the impendance of capacitor must be very lower than 47k.is that right? \$\endgroup\$
    – 甚麼甚
    Jan 24, 2020 at 4:17
  • \$\begingroup\$ Usually 1 or 2 decade below @ f falstad.com/afilter/… \$\endgroup\$ Jan 24, 2020 at 4:33
  • \$\begingroup\$ tinyurl.com/umlnwmo \$\endgroup\$ Jan 24, 2020 at 4:48

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