Iain Churches wrote:
> "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:-(
ROFL
>
> The heater supply chain has two caps with a low value WW pot between
> them. C1 is 22 000µF and C2 is 10 000µF.
>
Ah, that's better, just over a volt ripple across the first cap and
thereafter depends on the WW pot value, but since 3A is flowing it
cannot be more than a couple of ohms I guess.
Cheers
Ian
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