Sorry if this is a bit of a sophomoric question, but when should you buffer an ADC input, specifically from a potentiometer. I've seen circuits add a low-pass to the wiper/ ADC input.. is this only necessary if you have switching noise on the reference voltage bus?
Electronic – When to use buffer/ low-pass filter on ADC input
adcfilter
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The cap near the power pin is not to protect the part from noise, but to keep the part from generating noise as the logic switching causes rapid changes in supply current. Ideally the cap would supply instantaneous demands for more current without increasing current all the way back to the power source.
The sum of the impedances on the PSU side of the circuit - the internal impedance of the PSU plus the inductance, resistance, and capacitance of the traces or planes - is enough to give some low pass filtering on the input side of the cap. I think of the cap as a tiny a power supply that can respond to demands with a bandwidth in the multi-MHz range. The larger regulators that supply a full circuit react far too slowly and the cap is a temporary source of power that replaces or bypasses (or decouples) the PSU. Placing the cap close to the power pin on a chip minimizes resistance and inductance that would slow the response.
CMOS parts consume most of their power while switching state. For microprocessors this means on clock edges and the current draw is in little fast spikes. The size of the spikes varies as fast as the clock as every instruction uses different combinations of internal circuits. Imagine the circuitry used in checking a register for zero versus fetching data from RAM. The power needed fluctuates at the clock rate. The greater the current changes, the bigger the cap. Calculating the right size is a matter of estimating for most of us and the 0.1uF ceramic cap is so common that it is very low cost. Capacitor construction is also a concern as well as change with temperature. Some can respond quicker than others and some vary by 80% over the commercial temperature range.
They are also called bypass caps because: 1) They can "bypass" (short) high frequency PSU noise to ground. 2) They can "bypass" the PSU and respond to high frequency demands for power.
Also called "decoupling caps", a more accurate term for high frequencies as they "decouple" the power demand between the part and the PSU.
The browser doesn't like that link to the datasheet, so I couldn't read it. Provide the link to just the PDF file, not a page with all kinds of fluff around it.
In any case, the reason for the low pass filter is that the motor current can have short term spikes and other noise, but what you care about is more of a recent "average", or more precisely, you care only about the low frequencies of the current signal. Since you only want readings at 10 Hz (a reasonable rate for looking at motor current), you should filter out frequencies above 5 Hz at least.
A simple way to accomplish this is with a R-C low pass filter:
The rolloff frequency of such a filter is
F = 1 / 2πRC
When R is in Ohms and C in Farads, then F is in Hz. In this example, the rolloff frequency is 4.4 Hz. That's the frequency at which it roughly starts to attenuate, with the attenuation being 3 dB at that point. Much below that frequency, the amplitude is unchanged. Much above that frequency and the amplitude falls off 6 dB per octave above the rolloff frequency, which is also the ratio of the rolloff frequency to the frequency being passed. For example, 100 Hz is 23 times the rolloff frequency, so this filter will attenuate a 100 Hz signal by 23 in voltage. If you stick in a 100 Hz at 10 V, you will get out a 100 Hz at 440 mV.
You also have to consider loading of the current sensor output and what maximum impedance the A/D input requires. The above is fine if the current sensor can drive a 1.2 kΩ load, and if the A/D is OK with its signal having 1.2 kΩ impedance. You can adjust this by changing the resistor but keeping the R*C product the same. For example, R1 = 12 kΩ and C1 = 3 µF would give you the same frequency response, load the current sensor output less, but also present a higher impedance signal to the A/D.
Best Answer
The output impedance of the potentiometer (with both ends connected to low impedances) varies over the pot's revolution. The greatest is at halfway. The impedance is the parallel combination of half the pot's resistances, or one-fourth of the potentiometer's resistance.
Each ADC will give a specification for maximum driving impedance. If this is greater than the pot's output impedance, then accuracy will be affected. Some ADC's care more than others about input impedance. Some datasheets will give you an equivalent circuit to the ADC's input impedance. You can use this to analyze the error you are likely to see, either by hand or with a Spice simulator.
In particular, watch out for some PIC microcontrollers which have a surprisingly low input impedance and maximum driving impedance spec.
The low-pass is for one of several reasons:
1) There is a low-pass filter for removing frequencies above the Nyquist frequency. But a hand-operated potentiometer just won't generate enough high frequencies for aliasing to be a problem.
2) The capacitor is right at the ADC input pin and is intended to reduce the driving impedance, at least at high frequencies. I've had mixed success with this method. It doesn't work if there is DC leakage at the ADC input.
3) The capacitor is across the potentiometer's wiper. The common failure mode for potentiometers is that the track gets dirty and makes intermittent contact. You have experienced this problem if you've ever had an old radio that made loud crackling sounds when you touched the volume knob. With a capacitor across the wiper, the wiper voltage doesn't change if it loses contact with the track intermittently. (This trick only works for DC signals.)
Remember, op-amps are pretty cheap. I would recommend always buffering the signal before ADC unless the volumes are high enough that the cost savings are worth it.