On Sep 11, 1:44=A0pm, cabraha...@[EMAIL PROTECTED]
wrote:
> On Sep 1, 12:41=A0pm, exxos...@[EMAIL PROTECTED]
wrote:
>
>
>
>
>
> > Hi all,
>
> > While messing with buck/boost circuits I had some thoughts which don't
> > hold up in real tests....
>
> > A simple example (aside from losses) is that if transfer energy from
> > say 10V 10uF into 1uF the voltage will increase to preserve the
> > charge.
>
> > Now, in my circuit, I charge 25V 1,000uF capacitor and charge a 10uH
> > inductor. When the switch turns off, all the energy should be in the
> > 10uH inductance. So if I have a 100pF capacitor, then the voltage
> > should be like 100,000volts according to my workings out.
>
> > Now I ran a computer simulation on this, at best I can only obtain
> > 12KV on the 100pF. So most of the energy is lost in switching losses I
> > assume.
>
> > In realworld tests, I end up with =A0less voltage than I started out
> > with, So I am trying to find out why ?
>
> > I know charging 22uH inductor at 100khz can be used as a buck/boost
> > supply, I built a simple 12V to 30V inverter, can switch 10amps
> > easily. Though I am not running at 100khz, only 100hz. Though the
> > current pulse rises to something like 500amps over 500uS.
>
> > I am not sure I follow all this exactly, Or even if it will work ?
> > AFAIK, The longer a inductor has current pumped across it the more
> > charge it obtains over time. So at turn off, all the energy given to a
> > coil is recovered. It works well, even with my simple buck/boost
> > circuit.
>
> > So I am slightly confused as to why pu****ng 500A into a coil has no
> > effect. I can only assume I have a huge loss somewhere, Or I do not
> > follow the idea correctly ?
>
> > Cheers,
> > Chris
>
> You need to aquaint yourself with the buck-boost topology and theory
> of its operation. =A0If you go to TI's web site and download their buck-
> boost app note no. SLVA059, studying this thoroughly, you will
> understand why.
>
> In a nutshell, the 1000 uf input capacitor did not transfer ALL of its
> energy to the inductor. =A0The output voltage is related to the input
> voltage and the duty cycle as follows:
>
> Vout =3D Vin * (D/(1-D)) not accounting for losses.
>
> The energy transferred to the inductor is half the inductance times
> the difference between the square of the final current and square of
> the original current. =A0This is determined by the voltage across said
> inductor multiplied by the on time. =A0During the power switch off time,
> the volt-seconds must equal that of the on time. =A0The app note covers
> this. =A0The value of the input capacitor does not determine the output
> voltage. =A0The same goes for the output cap value. =A0The values of
both
> caps, however, are im****tant regarding ripple. noise, and transient
> response.
>
> Did I help? =A0BR.- Hide quoted text -
>
> - Show quoted text -
i haven't gotten the whole detail, but you seem to start with a bit of
a misunderstanding which i will now endeavor to correct:
you charge a capacitor by loading a voltage into it, then open
circuiting it; but, an inductor being essentially the opposite, you
charge it by loading a current into it and shortcircuiting it; whereas
a charged capacitor holds its energy in the electric field with zero
current, a charged inductor holds its energy in the magnetic field,
with zero voltage. If you short circuit a capacitor it loses its
charge, if you open circuit an inductor it loses its current. In real
life, the series resistance in the inductor windings decays the
charged current a lot more quickly than the leakage resistance in the
capacitor decays the charged voltage.
|
