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Design Ideas: September 12, 1996

Feedback network silences op-amp resistor noise

Michael Steffes,
Burr-Brown Corp, Tucson, AZ

Once you've selected an op amp that offers low-voltage and -current noise terms, further noise reductions are hard to come by. Broadband pulse amplifiers often require as low a noise as possible to accurately detect or convert the input signal.Figure 1a shows the full noise model and output noise (eo) equation for a noninverting op-amp configuration.

The last term in the output-noise equation includes the effect of both Rf and Rg noise. You can make the last two terms insignificant by simply making the values of Rf and Rg very low. However, the output drive of the amplifier limits this approach, because Rf+Rg appear in parallel with the load.

Figure 1b shows an approach that limits this resistor noise and, hence, provides a lower integrated noise, without loading the output of the amplifier at dc. This technique significantly reduces the last two terms in the output-noise equation. This example uses an OPA643 low-noise, voltage-feedback op amp set up for a signal gain of +5.

At dc, the op amp drives a 1-k[ohm] load due to the feedback network. As the capacitors in the parallel gain-setting networks short out, this load reduces to approximately 50[ohm]. The capacitor values in Figure 1b hold a flat frequency response and place Rf2 and Rg2 in parallel with Rf1 and Rg1 at 10 MHz. The following equation is the full transfer function for the circuit in Figure 1b.

To hold a flat frequency response while shunting the resistor noises, you need to set Rf1/Rg1=Rf2/Rg2=C2/C1. In this case, C1 was selected simply to set 1/(C1×Rf2)=2[pi]×10 MHz. The selection of C2 sets 1/(C2×Rg2) at the same frequency. The feedback and gain-setting impedances then begin decreasing together at a pole frequency of 1/(C1×(Rf1+Rf2))= 474 kHz.

For broadband applications, the noise above 10 MHz dominates the integrated noise, which allows an easy comparison of the output noise of the two circuits. Otherwise, you can perform an integral of the noise power over the roll-off region to get an exact result. Above 10 MHz (with Rs=0) and without the shunting gain network, eo=12.2 nV/[sqr. root]Hz. With the shunting gain network, eo=9.2 nV/[sqr. root]Hz. The circuit in Figure 1b reduces the equivalent input referred noise from 2.44 to 1.84 nV/[sqr. root]Hz, a noise-power reduction of 43%. Silencing this feedback-network resistor noise can make an even more significant change in noise at lower gains and low op-amp input-voltage noises.

Table 1--Input-referred noise without and with feedback network
Gain (AV) en=1 en=1.2 en=1.4 en=1.6 en=1.8
2 2.88/1.14 2.96/1.32 3.05/1.51 3.14/1.69 3.25/1.88
3 2.38/1.10 2.47/1.28 2.58/1.47 2.69/1.66 2.81/1.86
4 2.10/1.07 2.21/1.26 2.32/1.45 2.45/1.65 2.58/1.84
5 1.93/1.06 2.04/1.25 2.16/1.44 2.30/1.64 2.44/1.83
Notes:
All noise units are in nV[sqr.root]Hz.
Input-referred noise=output noise/signal gain=eo/AV.Without network, the actual Rf=800V.
With network, the equivalent Rf=38.1V.
en=op-amp voltage noise.

For example, Table 1 shows the improvement in noise using the same feedback network as in Figure 1b for lower gains and lower-op-amp input-noise voltages. The calculated entries are based on the equation for eo in Figure 1a with Rs=0, ibi=2.4 pA, ibn=2.4 pA, and 4kT=1.6e­20 J/°K. Moving from left to right, the table shows the total input-referred voltage noise for high- and low-equivalent feedback resistors (without and with the feedback network, respectively) as the op-amp input-noise voltage (en) increases. The lower right entry of Table 1 is the actual design point of Figure 1b in which you can see some improvement in equivalent input noise. As Table 1 shows, this improvement becomes even more pronounced at lower gains and lower op-amp input-noise voltages. (DI #1886)


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