What is the significance of the input capacitance (680 uF) (uF = microFarad) at the input of the LM 2596 ? LM 2596 is step down voltage regulator.
Please refer: http://www.ti.com/lit/ds/symlink/lm2596.pdf
The 680 uF (microfarad) capacitor is the leftmost capacitor on the circuit on page 1. Usually to 'kill' the high frequency noise from supply voltage 0.1 uF is used. I am sure 680 uF serves different purpose. What is that ?
Any intuitive explanation would be great.
According to datasheet section 9.1.1 about input capacitor, it prevents large voltage transients at the input, and provides instantaneous current for the switching.
The "typical" 0.1µF capacitor you refer to is used to 'kill' very short pulses (in the oreder of nanoseconds) of low current (in the order of tens of milliamperes), as they occur when a logic IC switches its output state.
The LM2596 is a step-down converter. All it does is generating one current pulse after the other, but the operating frequency is around 150kHz, so the pulses are 6 microseconds long, that is 1000 longer than the current spikes generated by logic ICs. At the same time, the pulses can have up to 3 Amperes, so 100 times the current of a pulse created by a logic IC. This means the pulses have 100.000 times more charge than the standard pulses to be filtered by the 0.1µF capacitor. This explains, why the capacitor needed to filter the switching pulses of a switch-mode converter needs to be much bigger than a capacitor needed to filter the switching pulses of a logic IC.
