by Ian Thompson-Bell <ruffrecords@[EMAIL PROTECTED]
>
Apr 22, 2008 at 11:28 PM
Eeyore 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.
>
> Graham
>
Yep, it is only an approximation. I would have thought the conduction
period would have been shorter than that.
Cheers
Ian
