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?

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

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 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:

Now try to use an Op Amp withGBW = 1 MHz in this circuit. Recalculate the circuit values using the equations above and see the result. Adjust the C2 value:

Try to use an Op Amp withGBW = 1 MHz in this circuit. Recalculate the circuit values using the equations above and see the result. Adjust the Cp value:

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

A G = ——————— 1 + A βand for the inverting amplifier:

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

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

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:

## 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 × F3where: 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 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. The required GBW is:GBW(Hz) = 100 × Q × G × F3 = 100 × 0.707 × 1 × 150 kHz ≈ 10.5 MHzLet’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 × 75pFChange R3:

R3' = R3 − R4 = 4990 − 2100 = 2868 ≈ 2.87k (E96)And finally enter the values into a simulator: 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. The required GBW is:GBW(Hz) = 100 × Q × G × F3 = 100 × 0.707 × 1 × 150kHz ≈ 10.5 MHzLet’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 × 150pFChange R2:

R2' = R2 − R5 = 4990 − 1070 = 3920 = 3.92k (E96)And finally enter the values into a simulator: 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. Conservative calculation of the required GBW:GBW(Hz) = M × Fpole × Gfp = 100 × 300kHz × 0.707 ≈ 21 MHzwhere: 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 MHzwhere: 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

1 C2' = 56pF − —————————————————— = 40pF ≈ 39pF (E24) 2 × π × 1MHz × 10kThe R2 value:

1 R2 = ——————————————————— = 4k ≈ 3.9k (E24) 2 × π × 1MHz × 39pFAnd finally enter the values into a simulator: 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. 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

1 Cp' = 51pF − —————————————————— = 35.1pF ≈ 36pF (E24) 2 × π × 1MHz × 10kThe Rc value:

1 Rc = ——————————————————— = 4.421k ≈ 4.42k (E96) 2 × π × 1MHz × 36pFThe extra DC gain is:

CTR Rp 1 × 10k A = —————— = ——————— = 1 Rd 10kAnd finally enter the values into a simulator: 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.## References

- TI, Vito Shen, Thomas Kuehl, «Compensation Methodology for Error in Multiple-Feedback Low-Pass Filter, Caused by Limited Gain-Bandwidth of Operational Amplifiers»
- TI, Vito Shen, Thomas Kuehl, «Compensation Methodology for Error in Sallen-Key Low-Pass Filter, Caused by Limited Gain-Bandwidth of Operational Amplifiers»
- Christophe Basso, «Understanding Op Amp Dynamic Response In A Type-2 Compensator (Part 1): The Open-Loop Gain»
- Christophe Basso, «Understanding Op Amp Dynamic Response In A Type-2 Compensator (Part 2): The Two Poles»
- Christophe Basso, “The TL431 in Switch-Mode Power Supplies loops”
- «idealCircuit», simulator
- «Circuit Calculator», electronics circuit design tool