How do active and passive filter gains compare?

Active can exceed 1; passive is always <= 1

Which statement is correct?

Active limited by op-amp bandwidth; passive works up to RF

Active works to RF; passive only audio

Asked 5/26
2076 Asoj · 4+3 marks

Active vs passive; non-inverting (s+8)/(s+2)

Short answer

Active filters use op-amps, require power, can provide gain >1, and avoid inductors; passive filters use only R, L, C with gain ≤1. For T(s)=(s+8)/(s+2) using the non-inverting topology: R1=16.67 kΩ, R2=50 kΩ, C2=10 µF (DC gain 4, HF gain 1).

What are the differences between active and passive filter? Design an active filter using non-inverting op-amp configuration with following transfer function. T(s) = (s+8)/(s+2)

Differences between active and passive filters

Point	Active filter	Passive filter
Elements	R, C and op-amps/transistors	R, L, C only
Inductors	Not needed	Needed (bulky at low f)
Gain	Can be > 1	Always ≤ 1 (loss)
Power supply	Required	Not required
Loading	Op-amp buffers; easy cascading	Stages load each other
Frequency range	Limited by op-amp bandwidth (up to ~MHz)	Works up to very high (RF) frequencies
Signal level	Limited by supply (saturation)	Handles large voltage/current
Noise	Op-amp adds noise	Low noise (only R noise)
Size/IC	Small, can be integrated	Large; inductors cannot be integrated
Sensitivity	Often higher	Low (doubly terminated ladders)

Design of $T(s) = (s+8)/(s+2)$

Non-inverting configuration: $Z_1$ from the inverting input to ground, $Z_2$ from output to inverting input:

T(s) = 1 + \frac{Z_2}{Z_1}

Zero 8 > pole 2, high-frequency gain 1, DC gain 4. Use $Z_1 = R_1$ , $Z_2 = R_2 \parallel C_2$ :

T(s) = 1 + \frac{R_2}{R_1(1 + sR_2C_2)} = \frac{s + \frac{1}{R_2C_2}\left(1+\frac{R_2}{R_1}\right)}{s + \frac{1}{R_2C_2}}

          +---R2---+
          |        |
          +---||---+
          |   C2   |
   +------+-|-\    |
   |        |  >---+---- Vo
  R1  Vin---|+/
   |
  GND

Matching: $\dfrac{1}{R_2C_2} = 2$ and $1 + \dfrac{R_2}{R_1} = \dfrac{8}{2} = 4$ , so $R_2 = 3R_1$ .

Normalized: $C_2 = 1$ F, $R_2 = 0.5\ \Omega$ , $R_1 = 0.1667\ \Omega$ . With $C_2 = 10\ \mu$ F: $R_2 = 50\ \text{k}\Omega$ , $R_1 = 16.67\ \text{k}\Omega$ .

Answer: $R_1 = 16.67\ \text{k}\Omega$ , $R_2 = 50\ \text{k}\Omega$ , $C_2 = 10\ \mu$ F (DC gain 4, HF gain 1).

✓ Answer checked by Claude Sonnet 5.5

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