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The capacitor does not "short out", it has charged up to a constant voltage by storing energy as electrical charge, and if something external tries to change the voltage over the capacitor, it means that more or less charge is needed to change the capacitor voltage up or down, and moving charges means is current flowing. So in short, a capacitor wants to ...

8

The fundamental nature of a capacitor is that stored charge = capacitance x voltage: - $$Q = CV$$ We also know that current is the rate of change of charge with time hence, if the formula is differentiated with respect to time we get this: - $$I = C\cdot\dfrac{dV}{dt}$$ The impact of this formula is that if a current is injected, the capacitor's terminal ...

8

The capacitor does not become an actual short circuit. When we say that a capacitor is a short circuit at high frequencies, we are talking about the impedance of a capacitor when a sine wave voltage is applied to it. The impedance is the voltage divided by the current (similar to resistance). $$Z_C = \frac{V}I = \frac{1}{2\pi jfC}$$ It's an imaginary ...

6

if a random noise spike is introduced into the system from the power source or from an external source through capacitive coupling, what happens to the charge already stored on the capacitor? It becomes easier to understand if you model the noisy power supply as a DC + AC noise voltage source having an output impedance "Z": This makes a simple voltage ...

4

why does noise need to be reduced Noise needs to be tackled because it creates problems. One example below: Assume the output signal is a pulse from a circuit which counts number of people entering a door. If the noise is not addressed, it will count wrong number of people. Connecting the orange pulse directly to a pulse counter will lead to wrong ...

2

Johnson noise is the white noise of resistors. 1) You can measure the integrated noise voltage: sample the voltage with a digital multimeter and process the statistics. The standard deviation will be the integrated voltage noise. You might need to know the bandwidth of the multimeter and its own noise if the noise levels are very low. 1) b) use an ...

2

No closed form expression exists for the capacities. The best you can do is numerical approximation. The capacity is the mutual information between Y and X, where Y=X+N, X is the constellation point (in your case a PAM), and N is Gaussian. Then, the mutual information is I(Y;x) = H(Y) - H(Y|X). The second term is easy (working it you yourself if you want). ...

2

The ADCs will have a specific maximum input impedance mentioned in the datasheet (for most of the MCUs). Even the recommend series resistor and the capacitor can be found. Since the signal is from a potentiometer, it is a slow changing signal. A capacitor of value 10nF or even 100nF at the input of the capacitor will help reduce the sampling glitch. The ...

2

Try adding decoupling capacitors. Decoupling capacitors are like sugar and your primary power supply is like fat. The first is fast and on-demand, the second is high energy but slow. If you try to do strenuous activity and your blood sugar is low, you pass out. Same thing with circuits, more or less. All connections have inductance so sudden changes in ...

1

In this situation, my analog input on an arduino is modulated by a potentiometer and I am seeing the values bouncing around a lot. I would like to implement a low pass filter Given the schematic you posted, it seems unlikely that the potentiometer output is so noisy - unless you have some troubles with the power supply. You could anyway put a capacitor at ...

1

Carriers crossing junctions exhibit SHOT NOISE, proportional to $\sqrt { 2 q I_B}$, where $I_B$ is the current (number of carriers per time unit) flowing though the junction.

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I am not familiar with the use of the term heteroscedastic , but we are aware of many sources of noise that increase such as thermal noise increases with temperature and Partial Discharge Noise increases with voltage getting closer to the breakdown voltage. We also know that Zener diode noise increases with current.

1

The fan current goes through the same ground trace as the PSU output. This will produce a voltage across the ground trace. The fan draws pulses of current during operation, so the voltage it produces will look like AC noise. As you haven't shown us the whole PCB layout I can't tell how long this trace is, but the part shown looks quite long and thin so it ...

1

I'll try to explain simply... WS2812B draws pulsed currents from the supply. When you have lots of these LEDs, the high amount of varying current will create voltage drop across the supply wires, which have both resistance and inductance. This causes the supply voltage to ripple, which will crash your microcontroller, and also makes the controllers inside ...

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