# 3rd Order High-pass Filter with 1 Op Amp

Tutorial

A 3rd order high-pass filter with 1 Op Amp can be used together with a 3rd order low-pass filter with 1 Op Amp to make a frequency splitter, for example. Design is similar to a 3rd order low-pass filter, so let’s do it.

## Basic equations for third order high-pass filters

The transfer function of a 3rd order high-pass filter is:

A  s/ω₁  s²/ω₂²
H(s) = —————————————————————————————————
(1 + s/ω₁)(1 + s/(Q ω₂) + s²/ω₂²)

where:
A – gain;
ω1 – radial frequency of the first stage, ω / K1;
ω2 – radial frequency of the second stage, ω / K2;
Q – quality factor of the second stage;
ω – pass-band radial frequency of the filter;
s – complex frequency.

Open brackets to have another form of the transfer function:

A  a s  c s²                       A  a c s³                         A  ps3 s³
H(s) = ————————————————————————— = ————————————————————————————————————— = ———————————————————————————
(1 + a s)(1 + b s + c s²)   1 + (a + b) s + (a b + c) s² + a c s³   1 + ps1 s + ps2 s² + ps3 s³

ps1 = 1 / ω₁ + 1 / (Q ω₂)
ps2 = 1 / (ω₁ Q ω₂) + 1/ω₂²
ps3 = 1 / (ω₁ ω₂²)

Knowing a −3 dB frequency and filter type, we can get K1 and K2 from tables in books [1] or compute them to get ω1 and ω2.

## Design Equations

The transfer function is:

A ps3 s³
H(s) = ———————————————————————————
1 + ps1 s + ps2 s² + ps3 s³

Assuming an ideal Op Amp, the factors are:

ps1 = (C1 + C2) R1 + (C2 + C3 + C4) R2
ps2 = ((C1 C2 + (C1 + C2) (C3 + C4)) R1 + C3 C4 R3) R2
ps3 = R1 R2 R3 C3 C4 (C1 + C2)

the gain is:

C1 C2
A = − ————————————
(C1 + C2) C3

## Example

Let’s compute the a third order Butterworth filter with 1 kHz corner frequency and unity gain.

Set C1 = C2 = 10 nF (E24).

C1 C2             10 nF × 10 nF
C3 = − ——————————— = − —————————————————————— = 5 nF ≈ 5.1 nF (E24)
A (C1 + C2)     (−1) × (10 nF + 10 nF)
1       1           1                 1
ps1 = ———— + —————— = ————————————— + ————————————————— ≈ 3.183e−4
ω/K1   Q ω/K2   2 π × 1 kHz/1   1 × 2 π × 1 kHz/1

1           1                        1                          1
ps2 = ——————————— + ——————— = ——————————————————————————————————— + ———————————————— ≈ 5.066e−8
ω/K1 Q ω/K2   (ω/K2)²   2 π × 1 kHz/K1 × 1 × 2 π × 1 kHz/K1   (2 π × 1 kHz/1)²

1                       1
ps3 = ———————————— = ———————————————————————————————— ≈ 4.0314e−12
ω/K1 (ω/K2)²   2 π × 1 kHz/1 × (2 π × 1 kHz/1)²

There are 3 equations and 4 unknown values: R1, R2, R3, C4.

To find a solution, set a C4 value and solve the equations to find R1, R2, R3. It is hard to find an analytical solution, so we will do it numerically.

Let’s set C4 = 75 nF (E24).
The solution is: R1 ≈ 7.52 kOhm ≈ 7.5 kOhm (E96), R2 ≈ 1.86 kOhm ≈ 1.87 kOhm (E96), R3 ≈ 37.6 kOhm ≈ 37.4 kOhm (E96).

Simulation confirms that our solution is correct.

Magnitude of the input impedance is varying from a very high value at low frequencies to about 300 Ohm at 100 kHz.

Let’s ensure that the circuit is stable and see step response of the circuit using parameters from the OPA2134 datasheet.

A Fully Differential Amplifier can also be used, but the R1 and R2 values must be multiplied by 2.

## Design Equations

The transfer function is:

A ps3 s³
H(s) = ———————————————————————————
1 + ps1 s + ps2 s² + ps3 s³

Assuming an ideal Op Amp, the factors are:

ps1 = (C1 + C2) R1 + (C2 + C3) R3 − C3 R2 R5 / R4
ps2 = C1 C2 R1 R3 + C3 [R3 (R1 (C1 + C2) + C2 R2) − (C1 + C2) R1 R2 R5 / R4]
ps3 = C1 C2 C3 R1 R2 R3

the gain is:

A = 1 + R5 / R4

## Example

Let’s compute the same third order Butterworth filter with 1 kHz corner frequency and unity gain.

For unity gain: R5 = 0, R4 is not installed.

The factor values are already known:

ps1 ≈ 3.183e−4
ps2 ≈ 5.066e−8
ps3 ≈ 4.0314e−12

There are 3 equations and 6 unknown values: C1, C2, C3, R1, R2, R3.

Solutions with C1=C2=C3 are the most interesting to optimize our bill of materials.

Set C1 = C2 = C3 = 24 nF (E24). Now we can find other values solving the equations numerically:

R1 ≈ 4.76 kOhm ≈ 4.75 kOhm (E96), R2 ≈ 32.76 kOhm ≈ 32.4 kOhm (E96), R3 ≈ 1.87 kOhm ≈ 1.87 kOhm (E96).

Simulation confirms that our solution is correct.

Magnitude of the input impedance is varying from a very high value at low frequencies to about 4 kOhm at 100 kHz.

Let’s ensure that the circuit is stable and see step response of the circuit using parameters from the OPA2134 datasheet.

The step response is similar to the ideal filter step response. There is no oscillation, so the circuit can be used.

## Conclusion

A 3rd order high-pass filter can be built using only one Op Amp. Multiple feedback and Sallen-Key topologies can be used.

An Op Amp’s Gain-Bandwidth Product (GBW) limits the working frequency range of the filter. Use this equation to estimate the GBW of the Op Amp and adjust it to your requirements:

GBW(Hz) = 100 Q G F3

where:
Q is the quality factor of the filter;
G is the specified gain;
F3 is the cutoff frequency at -3 dB;
100 is the gain margin.

## References

1. Analog Devices. “Linear Circuit Design Handbook”. Chapter 8, “Analog Filters”. Link.

2. Nuhertz Technologies, Active Filter Module. Link.

3. "idealCircuit", a simulator. Link.

4. "Filter Designer", a multistage analog active filter design tool for Android. Link.

5. "Circuit Calculator", an electronics design tool for Android. Link.

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