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Hi,
I'd need to drive the input of the OPA828 with a sharp square signal (from 10 K and up to 1 MHz); I need to add a capacitor in the feedback loop to compensate for the pole introduced by the feedback resistor.
So, what is the input capacitance of the OPA828 since it is not mentioned in the datasheet?
Also, I need to use it as a buffer for the same square input. So, documentation of the LM6172 is suggesting to place in the feedback a resistor of 1K and a capacitor too of 2pF.
What would be the case for the OPA828?
Tanks
Hi Jean,
Thanks for clarifying on the load capacitance.
I think I have an explanation for why such a large capacitor is required across the feedback resistor in the buffer case. To begin, I started from your TINA schematic, but replaced the fixed load capacitor with an additional OPA828 instance, as shown. In some cases (especially op amps with anti-parallel diodes between their inputs) this can help to catch unusual transient behaviors early in a design.
The feedback loop for U2 in this circuit can be redrawn as:
This gives rise to the following feedback equation:
With a zero set by the feedback resistor and feedback capacitor, and a pole set by the feedback resistor and combined feedback and input capacitance.
With no feedback capacitor, we get a first-order pole around 14MHz with no zero.
I built a stability test circuit for this in TINA:
Running an AC simulation gives the following curves:
One guideline we use for stability is rate of closure between 1/B (reciprocal of feedback gain) and the amplifier’s loaded open loop gain (Aol). A 20dB/decade rate of closure is typically stable while a 40dB/decade rate of closure is typically unstable.
In this case, Aol decreases at approximately -20dB per decade while 1/B begins to slope upwards at ~14MHz (eventually reaching to a slope of +20dB/decade). By inspection, this gives a rate of closure somewhere near 40dB/decade, which is likely to be marginally stable at best. This is also confirmed by the phase margin of ~13 degrees.
Adding increasing amounts of feedback capacitance begins to decrease the rate of closure by shifting the pole frequency lower and adding a zero above it, as can be seen in the plot for the same circuit with a 10pF feedback capacitor:
With 10pF, the rate of closure is now closer to 20dB/decade, with an increase in phase margin to 59.29 degrees.
As feedback capacitance becomes larger and larger, it increasingly dominates over the Cin term, and the equation for B to reduces to:
or approximately an all-pass filter characteristic.
This same effect can be created by simply decreasing Rf (In fact, a simulation of the AC stability of the pure buffer case showed no concerns).
I suspect that remaining overshoot in this circuit may be caused by the OPA828’s slew boost circuitry, or by other higher-order effects inside the amplifier itself, as the overshoot does not exhibit the typical ringing behavior expected from a second-order system. If slew boost is responsible, simply slowing down the input signal's rise time slightly may eliminate this artifact.
Looking at a noninverting amplifier case, we can re-draw the feedback network as
With feedback described as
This is, again, a pole-zero pair. In this case, however, the pole frequency is determined by the parallel combination of Rf and Rin interacting with the parallel combination of Cf and Cin.
Substituting Rin = Rf, we find:
If we set Cf = Cin, we get B=1/2 (independent of frequency), giving us an all-pass characteristic as above.
For the noninverting amplifier circuit, the input resistors R3 and R4 interact with the input capacitance of the OPA828 as well, forming a low pass filter which slows down the input slew rate, helping eliminate the overshoot seen in the buffer configurations.
