by Eeyore <rabbitsfriendsandrelations@[EMAIL PROTECTED]
>
Apr 22, 2008 at 07:21 PM
Iain Churches wrote:
> "Ian Thompson-Bell" <ruffrecords@[EMAIL PROTECTED]
> 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?
> >>
> >> Iain
> >
> > 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.
> >
> Sorry, Ian. I was writing a post while doing ten other things at once.
> Wives and secretaries can manage that admirably, I fail miserably:-(
>
> The heater supply chain has two caps with a low value WW pot between
> them. C1 is 22 000µF and C2 is 10 000µF.
That makes a BIG difference. The total capacitance is 32,000 uf, so
scaling my
previous calculations that gives ~ 250mV rms of 100Hz ripple.
Graham
