a question about buck circuit

[wei9321]

as the picture:
we know the RL circuit current curve discharging and charging is exponentional fuction rising and downing,but in buck ,when swich on and off ,the current curve is ramp fuction
why?
thanks !

 

[VSMVDD]

becouse of laws of conservation of energy...

lcr is not nulled by a diode just half dampened

capacitance is all around 1

and 1/ internal resistance of the cell dictates

infact 1/ (1/.404)

is the reactance... of this cell circuit

so missing 0,596 is your missing ? or .6 of 1 - loss in atomic friction as heat in a perfect system / latent heat

((in+i)/(out-o)) = ((input-o) / (output+i)) in a perfect system

but you must * .96 for a start to get rid of loss from latency

than / .707 to factor a perfect diode in

so 1 v all over can at max/min only equal .596 input
so 0.596 volts drop at 1ma

 

[keith1200rs]

Ignore the reply above - it is complete b0ll0cks.

The current rise is exponential - you just aren't waiting long enough so you are seeing the relatively straight part of the exponential curve. di/dt=V/L. If you have a resistor in there then the voltage drops as the current rises. That is what causes the exponential current rise. If you only look for a short time, before the volts drop across the resistor is significant, the current rise will almost be a straight line.

I can do a couple of simulations later to illustrate it if it will help.

Keith

 

[leo_o2]

A capacitor should be added at the output node of the Buck. In a steady state, the output voltage is constant if voltage ripple is supposed small enough. Then when the switch is turned on, the voltage drop on the inductor is VIN-VO. Here, VIN is the input voltage, and VO is the output voltage. According to di/dt=(VIN-VO)/L, so di/dt is close to a constant. When the switch is off, di/dt=(0-VO)/L=-VO/L, the slope is close to a constant too if the voltage drop on diode is neglectable.

 

[wei9321]

i have a another question
we know the fuction of L is to store energy
if we add a capacitor at output,the coverter output is like a L-c filter
can we see the L as a filter ?