Faster settling time for near DC pole low pass filter

Question

I need to clean up the noise from a voltage reference and I wanted to use a simple RC low pass filter. In order to lower the pole position I have to use a big resistor and/or a big capacitor. As the voltage reference should not have much intrinsic noise, what are the drawbacks of speeding up the capacitor charging (and thus the settling time) using a switch to bypass the resistor? The main problem I see is that the voltage reference should not be loaded with too much current, so a buffering stage could be needed in the bypass branch. As the time constant of the circuit will still be high, the circuit will act like a sample-and-hold for the last value of the noisy reference before opening the bypass switch, slowly converging to the low noise DC value, right? A big capacitor will also reduce the thermal noise of the resistor, right?

Answer 1

I dislike the standard approach with diodes to cut out the precharger (posted and explained by Neil_UK) in the context of voltage reference buffering because the output will quick-settle only to within some region set by the opamp gain. The remaining difference will settle very slowly given by R1*C1. So for settling at the target accuracy of a voltage reference, this circuit actually does basically not help.

Here is an alternative that:

• is simpler
• settles quickly to the actual reference voltage
• retains low noise filtering once settled

Even though \$\tau = 100 s\$ from \$R1\cdot C1\$, the filter settles very quickly. The time constant \$R2\cdot C2\$ sets when the fast charging is cut out.

The output voltage basically follows the input rise initially, but then gradually transits into a noise filtering state once settled.

Answer 2

Speeding up the 'start up' of a long time constant filter, by switching a lower R, or putting a schottky diode across the R, is a well known tried and tested thing to do.

Your voltage reference only has to avoid damage if it's too heavily loaded during start up, the output voltage is going to be wrong during start up anyway, so check its safety into a sort circuit, not the maximum current for regulation. As the charge current drops, it will come into regulation.

OTOH, buffering the reference during start up will allow a higher charge current for your filter FBC (very large capacitor), and the buffer does not need to be a precision type, as the output voltage will be wrong during startup.

If you consider a buffer connected by two back to back diodes to your filter, then it automatically disconnects when the voltage is close enough.

^{simulate this circuit – Schematic created using CircuitLab}

Analysing the operation. The filter is R1, C1.

Let's assume C1 is the same voltage as the reference V1, that is, the filter has settled. The amp -ve input will be at the reference voltage, so no current flows through R3. Its output will be at the reference voltage, so no current flows through the diodes. With normal minimal amplifier voltage offsets, little current flows through R3.

Now during startup, there will be a large error voltage on C1. The opamp amplifies this by the ratio of R2/R3. If the output voltage >0.7v, so if the error voltage is >70mV, one diode will conduct to charge the capacitor quickly. Once within 70mV, the diode stops conducting, and the final filter settling occurs in the normal way through R1.

The size of window outside which fast charging occurs can be controlled by the gain of the opamp. With 100x gain, the fast charge window would be 7mV. The limit to how small this window can be is given by the input offset error of the opamp, you do not want it turning on once the filter has settled.

Answer 3

A simple approach is to make the bypass path turn itself off as the capacitor charges up close to the reference voltage. Just bypass the high resistance with a low resistance in series with a diode.