by "Iain Churches" <IainNG@[EMAIL PROTECTED]
>
Apr 22, 2008 at 06:41 PM
"Ian Thompson-Bell" <ruffrecords@[EMAIL PROTECTED]
> wrote in message
news:fukur1$1coj$1@[EMAIL PROTECTED]
> Iain Churches wrote:
>> "Eeyore" <rabbitsfriendsandrelations@[EMAIL PROTECTED]
> wrote in message
>> news:480DE286.CB70EB3C@[EMAIL PROTECTED]
>>>
>>> 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.
Regards
Iain
