Question about Sigma Delta ADC at lower frequencies
william chan
Posts: 1,326
Considering the Propeller Demo Board Mic circuit,
if I run the Propeller at 40Mhz instead of 80Mhz, should both the 1nF capacitors be doubled to 2nF capacitors?
Thanks.
if I run the Propeller at 40Mhz instead of 80Mhz, should both the 1nF capacitors be doubled to 2nF capacitors?
Thanks.
Comments
The 1nf (0.001uF)capacitors are fine, their purpose is to provide a balanced Vref that is equal to the I/O threshold voltage of about 1.4V. Increasing their value would not be directly related to the speed of the Propeller, however an increased value might lower the noise floor that the I/O is sensitive to. In other words, by increasing the value, you will attenuate any of the small voltage changes from the input.
The most critical part in the ADC is the inputs hysteresis. To understand the ADC: the two resistors from input and compensation (100kOhm) are working as current sources. This is possible, as the capacitor represents a voltage close to the inputs threshold voltage. A resistor connected to (virtual) ground runs always a current proportional to applied voltage, so is a voltage controlled current source.
Best way to understand this ADC: remove the input and apply a 50% square wave to the compensation resistor. The capacitor will show a sawtooth voltage. Lets make a simple example: Vcc = 2 V. C = 1 nF, charged to 1V, R = 100K. Clock = 10 ns, so 10 ns (pulse) of 2V on the resistor, 10ns (pause) of 0V on the resistor. So the resistor will see a 1V voltage and there will be a current of 10µA. In 10 ns this current will charge the C with 10^-13 Cb. So the caps voltage will change by 10^-13 / 10^-9 = 10^-4 V. So the voltage at C is still very close to 1V and R can be really seen as a current source. We do not see a R/C exponential voltage behavior, we are very well locked to the linear region of the charging curve. With the next (de-)charging (pause) the voltage will go back. Let now have the input an hysteresis of 1 mV, so we see, that 10 consecutive pulses charge the capacitor by 1 mV, the input will flip output to generate pauses and after 10 of those the input will flip back. Over the time we count have of the possible compensation pulses as "pauses", so there is no current flowing to the cap beside the compensations resistor. If there is a current from the input resistor, the relation pulse to pause will change between 0 and 1, giving a value for the incoming current. As he is linearly related to input voltage, we so can determine the input voltage.
What is critical to the precision: Switching level of the input must be constant. The pulses and pauses must be properly formed (rise/fall time). The output voltages must be constant. Not critical: value of Cap and R if only constant over time (and both R equal), because the measurement is relative.
What is critical to noise: No fluctuations in threshold voltages. No fluctuation in compensation pulses duration and amplitude.
How to determine the values: R should not load the output to much to allow high peak/peak voltage. C should be such small, that not to many compensation pulses should be used to swing hysteresis.
Therefore: place components as close as possible to the chip. Clean supply voltage. If there is an influence of threshold by Vcc, distribute C between Vss and Vcc