I have learned that the digital ICs would generate the power noise in the power supply line,however, I am not so clear about how.
In my opinion, two contributors: the change of the current would lead to output voltage deviation of the regulator due to the load transient response. Also, the load regulation would result in another deviation.
For example, a lm1117 regulator supplies power to a MCU. A 0.5A step current change causes more than 100mV deviation(figure 1). And 15mV max deviation if load current changes from 0 to 800mA(figure 2).
Is my opinion correct? Any other opinion? Thank you very much!
simulate this circuit – Schematic created using CircuitLab
figure 1.
fugure 2.
Best Answer
Your opinion is correct. The LM1117 (like virtually all other linear voltage regulators) cannot instantly accommodate rapid/instantaneous changes in load current - inside there is a feedback system that attempts to keep the output voltage stable at a set point and this mechanism targets accuracy mainly and this means it has a high loop-gain so that slow changes in load current do not produce significant changes in the long term output dc voltage level.
Whenever you have high loop-gain there is a possibility of high frequency instability and this is "cured" by reducing the loop gain (in an appropriate way) at higher frequencies. This keeps the device "stable" and accurate at DC but makes it vulnerable to rapid load current changes.
This is (usually) why linear voltage regulator data sheets specify that input and output capacitors must be fitted. It is also why you, as a designer, will realize the limitations and, when there is a possibility of sudden current changes, you apply local bulk capacitance at the point of the load change in order to restrict the sudden current change from having such a marked effect on the power rails of the regulator. To this end, digital chips require power decoupling capacitors (as do most other chips).
Linear voltage regulators are not perfect!