The perils of input capacitance
By Ron Mancini -- EDN, December 11, 2003
It is impossible to build an op amp without including some input capacitance, and the op amp's pc board adds even more (Figure 1). All of the capacitors except for the feedback capacitor, CF, are stray capacitances, and they influence the circuit's stability. When you set the capacitors to zero, an artificial condition, Equation 1 gives the loop gain (Reference 1). Op-amp open-loop gain, a, has magnitude and phase components, so it introduces phase shift into the Bode (logarithmic-stability) plot. The critical point on a Bode plot is the point at which the gain magnitude equals zero (gain=1); the difference between 180° and the actual phase shift is the phase margin.
EQUATION 1
The external components are resistive, and making RG=RF decreases loop gain by 6 dB. This decrease further enhances stability; the vertical intercept on the Bode plot drops 6 dB, but the pole locations stay constant. Equation 2 gives the loop gain for an inverting amplifier with real input capacitances (CF=0), as Figure 1 shows.
EQUATION 2
The input capacitance adds a pole to loop gain, and when the parallel value of RG and RF is small, say 500Ω, the pole location is at f=16.76 MHz. The pole introduces essentially zero phase shift at one-tenth of its location frequency, so input capacitance does not affect op amps with a gain bandwidth of less than 1.676 MHz. As the op-amp gain bandwidth increases beyond 1.676 MHz, phase shift from the pole adds to the loop-gain phase shift, and, depending on its phase response, the op amp overshoots, rings, and then oscillates.
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Increasing the parallel value of RG and RF causes the pole to decrease in frequency (f=0.1676 MHz at RF||RG=5 kΩ); hence, the phase shift occurs sooner, exacerbating the instability problem. Resistors in high-frequency op-amp circuits traditionally have small values to minimize the effect of stray input capacitance. An alternative option for the input-capacitance problem is to add a feedback capacitor, CF. Equation 3 gives the loop gain when the circuit has input and feedback capacitors.
EQUATION 3
The zero in Equation 3 always precedes the pole; thus, its phase shift cancels some negative phase shift until the pole comes into play. The circuit can be independent of both capacitors by making RFCF=RGCG. Normally, this option is not the best for closed-loop bandwidth performance, so engineers use smaller values of CF. You can optimize the resistor values, capacitor values, and op-amp bandwidth for the best high-frequency performance, but 2CF=CG is an excellent starting place for lab experiments.
Stability is the same for inverting and noninverting op amps because it's independent of the input. The inverting op amp acts very much like theory predicts, but the noninverting op amp has lower common-mode capability because part of the input signal feeds through the differential capacitor (CD) to the inverting node. The reduction of common-mode performance is noticeable only at high frequencies.
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| Author Information |
 Ron Mancini is staff scientist at Texas Instruments. You can reach him at 1-352-569-9401, rmancini@ti.com. |
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I suspect that what he means is Cg = (Ccm+Cd).
Matt - 2004-20-2 09:09:00 PST -
Hi,
Can you please clarify what is Cg in your Figure 1?
Thanks
VT - 2003-19-12 05:44:00 PST -
Ron must know what "Cg" is but he's not letting us in on the secret. The recommended solution for the problem is: RF*CF=RG*CG but there is no mention of Cg anywhere else...
Mike Wexler - 2003-16-12 13:10:00 PST
