# A High-IIP3 Third-Order Elliptic Filter With Current-Efficient Feedforward-Compensated Opamps

TL;DR: A low-distortion active filter is realized using current-efficient feedforward-compensated operational amplifiers in the integrators and feedforward current injection in the summing amplifier to counter process variations and set the bandwidth accurately.

Abstract: A low-distortion active filter is realized using current-efficient feedforward-compensated operational amplifiers in the integrators and feedforward current injection in the summing amplifier. A third-order elliptic low-pass filter with two possible bandwidth settings of 17 and 8.5 MHz consumes 1.8 mW from a 1.8-V supply and occupies 0.17 mm2 in a 0.18- μm CMOS process. The measured maximum signal-to-noise and distortion ratios at the two bandwidth settings are 50.5 and 52.5 dB, respectively. The corresponding third-order intermodulation intercept points (IIP3) are +28.2 and +30.8 dBm. Automatic tuning is used at the startup to counter process variations and set the bandwidth accurately.

## Summary (2 min read)

### Introduction

- A switchable bandwidth is required for multistandard radios.
- Among integrated continuous-time filters, active RC filters provide the best distortion performance at medium frequencies because they consist of integrators and amplifiers using operational amplifiers in negative feedback loops, which suppress the distortion arising from active elements.
- The output current driven from the opamp can be reduced by injecting a replica of the current to the output through a separate path [6].

### II. THIRD-ORDER ELLIPTIC FILTER

- Fig. 1(a) shows the block diagram of the third-order elliptic filter.
- The integrator outputs are scaled for equal maximum transfer function magnitude.
- A 5-bit control word b〈4 : 0〉 switches the resistors and the capacitors and varies the RC product from 55% to 175% of the midcode value to compensate for process variations.
- The automatic tuning scheme is shown in Fig. 1(c).
- After five cycles of successive approximation, the bits converge to a value that sets the time constant of the replica integrator to 100 ns.

### III. ADVANTAGES OF FEEDFORWARD OPAMPS

- Fig. 2 shows the macromodels of the feedforward- and Miller-compensated opamps.
- The first stage consists of gm1 loaded by RL1 and C1, and the second stage consists of gm2 loaded by RL2 and CL.
- The distortion of the filter with feedforward-compensated opamps is significantly lower than that with Miller-compensated opamps.
- The input-referred noise voltage of the two opamps is the same because the same value of gm1 is used for both.

### IV. CURRENT-EFFICIENT FEEDFORWARD OPAMP

- Fig. 4 shows the feedforward-compensated opamp proposed in [5].
- Feedforward compensation is provided by using another nMOS differential pair M3.
- Separate common-mode feedback circuit (CMFB) stages are used to drive current sources M4 and M5 and to stabilize the common-mode output of each stage.
- The common-mode voltage is fed back through gm,cm (a differential pair with a current mirror load).
- The opamp consumes about 0.2 mA and is used for all integrators and the summing amplifier in Fig. 1.

### V. SUMMING AMPLIFIER

- Fig. 7(a) shows the conventional summing amplifier.
- Because multiple inputs are summed, the loop gain tends to be lower, and the distortion tends to be higher than in an amplifier with a single input.
- With a real opamp, the virtual ground experiences a signal swing proportional to (a1x1 + a2x2 + a3x3)/R, i.e., the current driven from the opamp.
- Conventionally, this is done by increasing the gain of the opamp (at the relevant signal frequencies).
- In their implementation [see Fig. 7(c)], only a3x3/R, which is the largest component of the output current, is injected to the output using a transconductor.

### VI. MEASURED RESULTS

- The third-order filter with automatic tuning is fabricated in a 0.18-μm CMOS process.
- Fig. 8 shows the layout, the test schematic, and the test board.
- In the small number of characterized samples, the 3-dB bandwidth after automatic tuning is 16.3–16.6 MHz in the high-bandwidth mode and 7.5–8.4.
- As expected, the worst cases occur near the band edges due to peaking in the magnitude response at the internal nodes.
- The worst case inband IIP3 values at the high- and low-bandwidth settings are +28.2 and +30.8 dBm, respectively.

### VII. CONCLUSION

- Power-efficient feedforward-compensated opamps in the integrators and feedforward current injection in the summing amplifier enable filters with an inherently high IIP3.
- The power dissipation depends on the signal level.
- The FOM of the filters presented here is better than those that have a comparably narrow transition band.
- 1At an input signal level equal to the IIP3, the IM3 (extrapolated from smallsignal levels) is also equal to the IIP3.
- 2Where the integrated noise is not reported, it is calculated by multiplying the inband spectral density by the square root of the bandwidth.

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### "A High-IIP3 Third-Order Elliptic Fi..." refers background in this paper

...The FOM of the filters in [2] and [11] are lower, but these filters have a much broader transition band....

[...]

121 citations

### "A High-IIP3 Third-Order Elliptic Fi..." refers background in this paper

...11 [wireless local-area network (WLAN)] has twice the signal bandwidth in the “n” mode as in the “a, b, and g” modes [4]....

[...]

119 citations

116 citations

### "A High-IIP3 Third-Order Elliptic Fi..." refers background in this paper

...The FOM of the filters in [2] and [11] are lower, but these filters have a much broader transition band....

[...]

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##### Frequently Asked Questions (13)

###### Q2. How long is the replica integrator used?

The replica integrator is initially reset, and an internally generated reference voltage Vref (approximately 400 mV) is applied to it for a period of 100 ns.

###### Q3. How long does the replica integrator take to realize the transmission zero?

With nominal resistor and capacitor values, the replica integrator’s time constant 1/ω0 is 100 ns when the digital bits are set to midcode.

###### Q4. Why is the loop gain lower than in an amplifier with a single input?

Because multiple inputs are summed, the loop gain tends to be lower, and the distortion tends to be higher than in an amplifier with a single input.

###### Q5. How can the transfer function of the filter be accurately measured?

By determining the transfer function of the filter path and the buffer pathfor both positive and negative settings of the buffer gain, the external feedthrough can be cancelled, and the transfer function of the filter can be accurately measured [9].

###### Q6. What is the RC product of the integrator?

A 5-bit control word b〈4 : 0〉 switches the resistors and the capacitors and varies the RC product from 55% to 175% of the midcode value to compensate for process variations.

###### Q7. What is the CMFB circuit used to drive the current sources?

5. Separate common-mode feedback circuit (CMFB) stages are used to drive current sources M4 and M5 and to stabilize the common-mode output of each stage.

###### Q8. What is the effect of the opamp on the output current?

Its transconductance, which depends on Rgm and the transconductance of Ma, is adjusted to be equal to a3/R in the nominal process corner.

###### Q9. What is the way to measure the noise and noise ratios in a passive RC?

Power-efficient feedforward-compensated opamps in the integrators and feedforward current injection in the summing amplifier enable filters with an inherently high IIP3.

###### Q10. How many cycles of successive approximation do the bits of the replica integrator converge?

After five cycles of successive approximation, the bits converge to a value that sets the time constant of the replica integrator to 100 ns.

###### Q11. How many MHz of the low-bandwidth mode are characterized?

In the small number of characterized samples, the 3-dB bandwidth after automatic tuning is 16.3–16.6 MHz in the high-bandwidth mode and 7.5–8.4

###### Q12. What is the largest component of the output current?

In their implementation [see Fig. 7(c)], only a3x3/R, which is the largest component of the output current, is injected to the output using a transconductor.

###### Q13. What is the difference between the feedback capacitor array and the receiver?

Doubling the feedback capacitor array, as shown in Fig. 1(b), halves the bandwidth to 8.5 MHz while maintaining the same passbandnoise spectral density as required in WLAN receivers.