by Eeyore <rabbitsfriendsandrelations@[EMAIL PROTECTED]
>
Apr 22, 2008 at 08:09 PM
Iain Churches wrote:
> "Eeyore" <rabbitsfriendsandrelations@[EMAIL PROTECTED]
> wrote
> > Ian Thompson-Bell wrote:
> >> Iain Churches wrote:
> >> > "Eeyore" <rabbitsfriendsandrelations@[EMAIL PROTECTED]
> wrote
> >> >> Iain Churches wrote:
> >> >>
> >> >>> The standard 6V3 winding would give 8.8V DC.
> >> >> Not with rectifier voltage drops it won't. More like 6.8V.
> >> >
> >> > I did not include the rectifier losses.
> >> >
> >> > The AC is at about 235V here this afternoon. I have a
> >> > Welter mains transformer on the bench in front
> >> > of me. The heater winding is 6.3V at 3A reg 5%. With
> >> > a BR156 rectifier and 10 000µF and drawing 3A I measure
> >> > 7.6V across the electrolytic. What does a SS regulator
> >> > require?
> >>
> >> Depends what you measure it with - a regular ac meter will not show
the
> >> ripple.
> >>
> >> Given c.dV = i.dt then dV(pp ripple) = i.dt/C and using your values I
get
> >>
> >> dV = 3Ax10mS/10,000uF = 3V pp ripple!! assuming 50Hz mains and full
wave
> >> rectification.
> >
> > The conduction period will be about 2.5 - 3 ms reducing that 10ms to
about
> > 7 ms. That gives 2.1 V pk-pk so about 750mV rms by my calculation.
>
> Better and better.
>
> But have you thought about (calculated) the ripple current
> in a 22 000µF cap in a heater supply drawing 3A? Herra Gud!
See my other post. You count BOTH the 10,000 and 22,000 uF. The resistance
of
the VR makes very little difference, indeed it serves mainly to extend the
conduction period.
Graham
