# Compensation for Error Caused by Limited Gain-Bandwidth of Operational Amplifiers in Low-pass Filters

• Tutorial

An operational amplifier has the internal compensation circuit for stability which limits its working bandwidth. Frequency response of the compensated Op Amp has slope of −6 dB/octave or −20 dB/decade. Unity gain frequency defines the bandwidth where the Op Amp is able to amplify a signal. If we multiply the gain and frequency at any point, the result is the same, allowing us to use this parameter to select the appropriate Op Amp. It is called Gain-Bandwidth Product, GBW or GBP. The limited open-loop gain introduces a closed-loop gain and phase error.

But we want to optimize our circuits, right?

 Closed-loop gain of the Non-Inverting and Inverting Amplifier

The closed-loop gain of the non-inverting amplifier is:

       A
G = ———————
1 + A β


and for the inverting amplifier:

      A (1 − β)
G = − —————————
1 + A β


where:

A is the open-loop gain;
β is the feedback fraction.

Let’s calculate, for example, an inverting amplifier with desired gain of 1 when A = 100, β = R1 / (R1+R2) = 0.5:

     A (1 − β)     100 × (1 − 0.5)     50
G= − ————————— = − ——————————————— = − —— ≈ −0.98
1 + A β       1 + 100 × 0.5      51


and A=1:

     A (1 − β)     1 × (1 − 0.5)
G= − ————————— = − ————————————— ≈ −0.333
1 + A β       1 + 1 × 0.5


But to make a real Op Amp stable, A is frequency dependent and there is a phase shift, how you can see in the picture below.
 Frequency response of the compensated Op Amp without a feedback network

Gain-Bandwidth Product (GBW) = A × F is a constant, and the greater the GBW is, the faster and expensive the Op Amp is. Of course, A cannot be infinity, so we see a shelf at low frequencies due to a finite gain.
 The inverting amplifier with ideal and compensated Op Amp.

The figure shows the difference between the ideal and compensated Op Amp with GBW = 1 MHz. You can see that the cyan line of the compensated Op Amp is always below the yellow line showing its open-loop gain and there must be some margin.

However, the simulator shows that gain at 1 МГц is not -9.55 dB, but about -7 dB due to a phase shift at the output.

The closed-loop gain error versus gain margin for the non-inverting amplifier looks likes that:
 Closed-loop gain error vs Gain margin of the non-inverting amplifier.

## Error compensation for 2-nd order low-pass filter

This equation is usually used to compute the GBW of an Op Amp:

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.

Why? A low-pass filter with Q > 0.707 has a peak and we need to multiply the gain with this peak value to account it. The peak value is:

              Q
β = ————————————————————— ≈ Q
sqrt(1 − 1 / (4 Q^2))


The most of used filters are low-pass filters for ADCs and DACs. So even for an LPF with 150 kHz an Op Amp with GBW of 15 MHz is desirable.

Texas Instruments’ engineers shared a method to reduce the requirement in [1][2].
Let’s try to figure out how to use their equations in practice.

### Multiple FeedBack LPF

Let’s compute a 2-nd order LPF with bandwidth of 150 kHz, gain of 1 and quality factor of 0.707.
 2-nd order MFB LPF

The required GBW is:

GBW(Hz) = 100 × Q × G × F3 = 100 × 0.707 × 1 × 150 kHz ≈ 10.5 MHz

Let’s try to use an Op Amp with GBW of 1 MHz.

Add R4 in series with C2:

            1                   1
R4 = ———————————————— = ——————————————————— = 2122 ≈ 2.1k (E96)
2 × π × GBW × C2   2 × π × 1MHz × 75pF


Change R3:

R3' = R3 − R4 = 4990 − 2100 = 2868 ≈ 2.87k (E96)

And finally enter the values into a simulator:
 Error compensation for 2-nd order MFB LPF

The top circuit and green color on the charts: the ideal amplifier.
The middle circuit and cyan color on the charts: the Op Amp with GBW of 1 MHz.
The bottom circuit and yellow color on the charts: the Op Amp with GBW of 1 MHz and with the error compensation.

When R3 is too low, the R3’ value may be negative. In such cases you should recalculate the filter with a greater R3 value.

### Sallen-Key LPF

Let’s compute a 2-nd order LPF with bandwidth of 150 kHz, gain of 1 and quality factor of 0.707.
 2-nd order Sallen-Key LPF

The required GBW is:

GBW(Hz) = 100 × Q × G × F3 = 100 × 0.707 × 1 × 150kHz ≈ 10.5 MHz

Let’s try to use an Op Amp with GBW of 1 MHz.

Add R5 in series with C1:

            R3 + R4                  ∞ + 0                      1
R5 = ————————————————————— = ———————————————————————— = ———————————————————— = 1061 ≈ 1.07k (E96)
2 × π × GBW × C1 × R3   2 × π × 1MHz × 150pF × ∞   2 × π × 1MHz × 150pF


Change R2:

R2' = R2 − R5 = 4990 − 1070 = 3920 = 3.92k (E96)

And finally enter the values into a simulator:
 Error compensation for 2-nd order Sallen-Key LPF

The top circuit and green color on the charts: the ideal amplifier.
The middle circuit and cyan color on the charts: the Op Amp with GBW of 1 MHz.
The bottom circuit and yellow color on the charts: the Op Amp with GBW of 1 MHz and with the error compensation.

## Error compensation for Type II compensation network with Op Amp

Let’s try to apply the same method to the Type II compensation network with Op Amp used in switching-mode power supplies.

Consider a Type II circuit with parameters: the zero at 2 kHz, the pole at 300 kHz, the middle gain is 0 dB.
 Type 2 compensation with Op Amp

Conservative calculation of the required GBW:

GBW(Hz) = M × Fpole × Gfp = 100 × 300kHz × 0.707 ≈ 21 MHz

where:

M=100 is the gain margin of the Op Amp at a frequency in times;
Fpole is the pole frequency in Hz;
Gfp is the gain at the pole frequency in times.

Christophe Basso in [4] gives another calculation:

GBW(Hz) = M × (20 Fcross) × Gfc = 20 × 20 × 5kHz × 1 ≈ 2 MHz

where:

M=20 is the gain margin of the Op Amp at a frequency in times;
20 Fcross is the frequency with maximum phase boost in Hz with the factor to account the pole position;
Gfc is the gain at the Fcross frequency in times.

The transfer function of the circuit is:

                         (C1 R1 s + 1)
H(s) = − ——————————————————————————————————————————————
(s Rfb1 (C1 + C2))(s C1 C2 R1 / (C1 + C2) + 1)


Add R2 in series with C2. The new transfer function is:

                      (1 + C1 R1 s) (1 + C2 R2 s)
H(s) = − ——————————————————————————————————————————————————————
(s Rfb1 (C1 + C2)) (s C2 C1 (R2 + R1) / (C1 + C2) + 1)


Change the C2 value and find the R2 value:

                  1
C2' = C2 − ————————————————
2 × π × GBW × R1


            1
R2 = —————————————————
2 × π × GBW × C2'


Now try to use an Op Amp with GBW = 1 MHz in this circuit.

Recalculate the circuit values using the equations above and see the result.

                     1
C2' = 56pF − —————————————————— = 40pF ≈ 39pF (E24)
2 × π × 1MHz × 10k


The R2 value:

             1
R2 = ——————————————————— = 4k ≈ 3.9k (E24)
2 × π × 1MHz × 39pF


And finally enter the values into a simulator:
 Error compensation for Type 2 compensation network with Op Amp

The top circuit and green color on the charts: the ideal amplifier.
The middle circuit and cyan color on the charts: the Op Amp with GBW of 1 MHz.
The bottom circuit and yellow color on the charts: the Op Amp with GBW of 1 MHz and with the error compensation.

## Error compensation for Type II compensation network with Optocoupler without Fast Lane

There are 2 variants of the circuit: with an Op Amp and a shunt regulator like TL431.
We will use the same parameters: the zero at 2 kHz, the pole at 300 kHz, the middle gain is 0 dB, the current transfer ratio of an optocoupler is 1.
 Type 2 compensation with Optocoupler without Fast Lane

The transfer function of both circuits is (C1=Cz, R1=Rz):

         CTR Rp       (1 + C1 R1) s
H(s) = − —————— —————————————————————————
Rd   (C1 Rfb1 s) (1 + Cp Rp s)


where CTR is the current transfer ratio of the optocoupler.

Add Rc in series with Cp. The new transfer function is:

         CTR Rp   (1 + C1 R1 s) (1 + Cp CRc s)
H(s) = − —————— ————————————————————————————————
Rd   (C1 Rfb1 s) (1 + Cp (Rp + Rc) s)


Change the C2 value and find the Rc value:

                  1
Cp' = Cp − ————————————————
2 × π × GBW × Rp


            1
Rc = —————————————————
2 × π × GBW × Cp'


Try to use an Op Amp with GBW = 1 MHz in this circuit.
Recalculate the circuit values using the equations above and see the result.

                     1
Cp' = 51pF − —————————————————— = 35.1pF ≈ 36pF (E24)
2 × π × 1MHz × 10k


The Rc value:

             1
Rc = ——————————————————— = 4.421k ≈ 4.42k (E96)
2 × π × 1MHz × 36pF


The extra DC gain is:

    CTR Rp   1 × 10k
A = —————— = ——————— = 1
Rd       10k


And finally enter the values into a simulator:
 Error compensation for Type 2 compensation network with Optocoupler without Fast Lane

The top circuit and green color on the charts: the ideal amplifier.
The middle circuit and yellow color on the charts: the Op Amp with GBW of 1 MHz.
The bottom circuit and cyan color on the charts: the Op Amp with GBW of 1 MHz and with the error compensation.
Do not forget that a real optocoupler has a parasitic capacitance and its value should be taken into account.

## Conclusion

The described methods help to reduce the requirement to the Gain-Bandwidth of a used Op Amp and cost of circuits.

For filters, they work well with commonly used quality factors below 1.

## Similar posts

AdBlock has stolen the banner, but banners are not teeth — they will be back