Re: Buck circuit question

"The diode "clamps" the switch node from swinging
> negative."

?

So I'm a bit confused. So am I correct in thinking that thorught out the on/off phase, the current always flows from the inductor to the capacitor. The back e.m.f would go to the switch when turned off if it were not for the diode?

I thought that the diode would conduct as it were revese biased by the back e.m.f

I'm all condused again!



John Larkin wrote:
On Fri, 30 Nov 2007 18:24:23 +0000, Ben <nebATwalrus81DOTplusDOTcom>
wrote:

Hello,

Been working with a buck circuit, today and think I've got in a bit of mix.

Sorry for the bad ascii art!




Looks fine to me.


cct:

Switch Inductor
---------- \------|----LLLLLLL-----|---------o +ve output 5V DC
| |
| |
/ \ ------
Diode --- ------ Capacitor
| |
| |
| |
---------------------------------------------o ground 0V DC


Diode is a low power shotky switch is actually transistor.

Ok m so when the switch is closed the current flows through the inductor, creating a magnetic field. The cap charges and the output voltage of 5V is reached. So the p.d at the output is 5V say at 500mA
to keep things simple.

Right. The diode is now reverse biased, off.

When the switch is opened, there is a back e.m.f generated by the inductutor. Which basically means that the p.d at the didoe and inductior is -5V. The diode is now reverse biased.


No. When the transistor is off, the diode is forward biased. It keeps
the voltage from swinging very much negative, which it would do if the
diode weren't there. The diode "clamps" the switch node from swinging
negative.


So current flows
through the inductutor to the cap keeping the output voltage a 5V.

Well, it keeps pumping current into the capacitor and the load.



What about the current does it still stay at 500mA? I thought that caps could not store energy?

Caps do store energy. So do inductors.

The current through the inductor builds up when the transistor is on,
and decays when it's off. Those phases correspond to adding energy to
the inductor, and taking it out. If the L is big enough and the
frequency is high enough, the current through the inductor is *almost*
constant. In real life, there's a pretty-much triangular ripple
current in the inductor, because nobody wants to buy huge inductors.

The ripple current in the inductor in turn causes a bit of ripple
voltage across the capacitor, and the load.

John


.

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