filteroperational-amplifierpower supply
Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? Can the AD8031 op amp be removed? Also, to save power, can the voltage divider be replaced with a buck converter?
The "Dual Decompensated" in the IC's title is biting you.
Two things come to mind. The description in the data sheet mentions that that IC is stable in gains of 10 and more, and you are low gain.
Also
From the data sheet: High Speed Operation When the feedback around the op amp is resistive (RF), a pole will be created with RF, the source resistance and capacitance (RS, CS), and the amplifier input capacitance (CIN ≈ 2pF). In low closed loop gain configurations and with RS and RF in the kilohm range, this pole can create excess phase shift and even oscillation. A small capacitor (CF) in parallel with RF eliminates this problem. With RS (CS + CIN) = RFCF, the effect of the feedback pole is completely removed.
Even though you have no resistor in your feedback path, maybe R2 is working its way in somehow. You might try a small cap in parallel with R2, but be careful of your frequency response. You could also try a different configuration filter.
12V AC reaches a peak voltage of about 17 volts, then via the rectifier it might produce about 15V DC across a suitable capacitor (rated at 25 volts). Maybe you had the capacitors wired back to front or maybe you misapplied the bridge rectifier. Maybe the bridge was broken in a previous experiment.
I'd still look for a buck regulator like this: -
Found here on ebay costing $2.86. It's a switching buck regulator that has an adjuster to set the output to be anywhere from 3V to 40V: -
Type / name:LM2596HVS DC-DC step-down module
Input Voltage:4.5V ~ 53V
Output Voltage:3V ~ 40V
Output Current:3A (max)
Conversion efficiency:92% (the highest)
Output Ripple:<30mV
Switching frequency:150KHz
Operating Temperature:-45 ℃ ~ +85 ℃
Size:43mm * 21mm * 14mm (L * W * H)
Best Answer
The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance.
Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages.
Any buck converter will have some output ripple. If you used it here, that ripple would be coupled directly into your filtered signal. I wouldn't do it just to save something like 150 uA. (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade)
If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current (the AD8031 has 800 uA, you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more.
Aside
The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. I'd remove them.