On 8/30/2021 2:55 PM, Bob Engelhardt wrote:
> On 8/30/2021 2:31 AM, Phil Allison wrote:
>> bobenge...@gmail.com wrote:
>>>
>>>> ** Yep two:
>>>>
>>>> 1. The drop out voltage of a standard 12v reg IC is not 12.8 - more 
>>>> like 13.8.
>>>> 2. The AC supply varies and may be be 10% low on occasion.
>>>>
>>>> So, you need a 15VAC tranny.
>>>>
>>>> OR you can make a "voltage doubler" supply with two diodes and two 
>>>> electros.
>>>>
>>>> That will give about 33V DC - minus the same deductions.
>>>>
>>>
>>> Thank you - that is helpful in 2 ways: pointing out what I missed &
>>> was wrong about, and even more helpful is not having to correct my 
>>> analysis.
>>
>> **   If your 12v tranny is really small, like say 100mA  -  it will 
>> output about 15V to 16v  AC off load.
>>
>>       So would do at a pinch with a puny 10mA DC load .
>>
>>
>> .......  Phil
>>
>>
>>
>>
>>
>   Thanks again - I will take some measurements on the tranny.

Thanks again - that's good to know.  The OC voltage is 15v & it doesn't 
drop below 12v until loaded to 200ma.  That gives me a lot of working room!

On 8/30/2021 2:31 AM, Phil Allison wrote:
> bobenge...@gmail.com wrote:
>>
>>> ** Yep two:
>>>
>>> 1. The drop out voltage of a standard 12v reg IC is not 12.8 - more like 13.8.
>>> 2. The AC supply varies and may be be 10% low on occasion.
>>>
>>> So, you need a 15VAC tranny.
>>>
>>> OR you can make a "voltage doubler" supply with two diodes and two electros.
>>>
>>> That will give about 33V DC - minus the same deductions.
>>>
>>
>> Thank you - that is helpful in 2 ways: pointing out what I missed &
>> was wrong about, and even more helpful is not having to correct my analysis.
> 
> **   If your 12v tranny is really small, like say 100mA  -  it will output about 15V to 16v  AC off load.
> 
>       So would do at a pinch with a puny 10mA DC load .
> 
> 
> .......  Phil
> 
> 
> 
> 
> 
  Thanks again - I will take some measurements on the tranny.

bobenge...@gmail.com wrote:
>
> > ** Yep two: 
> > 
> > 1. The drop out voltage of a standard 12v reg IC is not 12.8 - more like 13.8. 
> > 2. The AC supply varies and may be be 10% low on occasion. 
> > 
> > So, you need a 15VAC tranny. 
> > 
> > OR you can make a "voltage doubler" supply with two diodes and two electros. 
> > 
> > That will give about 33V DC - minus the same deductions. 
> > 
>
> Thank you - that is helpful in 2 ways: pointing out what I missed & 
> was wrong about, and even more helpful is not having to correct my analysis.

**   If your 12v tranny is really small, like say 100mA  -  it will output about 15V to 16v  AC off load. 

     So would do at a pinch with a puny 10mA DC load . 


.......  Phil 

On 8/29/2021 8:48 PM, Phil Allison wrote:
>   bobenge...@gmail.com wrote:
> =======================
>>
>> I'm going to make a 12v DC power supply with a linear reg. I will
>> half-wave rectify 12v AC and have a smoothing cap. The load is only 10ma.
>>
>> So: 12v RMS is 17v peak, minus the diode drop of 0.7 is 16.3v peak.
>> Using a 47uF cap, the ripple will be 3.5v p-p. So the min voltage into
>> the regulator will be 16.3-3.5 = 12.8. Ripple calculated from Vpp =
>> i/fC (.010/(60*47e-6).
>>
> 
> ** Yep two:
> 
> 1.   The drop out voltage of a standard 12v reg IC is not 12.8 -  more like 13.8.
> 2.   The AC supply varies and may be be 10% low on occasion.
> 
> 
> So, you need a 15VAC tranny.
> 
> OR you can make a "voltage doubler" supply with two diodes and two electros.
> 
> That will give about 33V DC  -   minus the same deductions.
> 
> 
> 
> 
> .....  Phil
> 
> 
  Thank you - that is helpful in 2 ways: pointing out what I missed & 
was wrong about, and even more helpful is not having to correct my analysis.

 bobenge...@gmail.com wrote:
=======================
>
> I'm going to make a 12v DC power supply with a linear reg. I will 
> half-wave rectify 12v AC and have a smoothing cap. The load is only 10ma. 
> 
> So: 12v RMS is 17v peak, minus the diode drop of 0.7 is 16.3v peak. 
> Using a 47uF cap, the ripple will be 3.5v p-p. So the min voltage into 
> the regulator will be 16.3-3.5 = 12.8. Ripple calculated from Vpp = 
> i/fC (.010/(60*47e-6). 
> 

** Yep two: 

1.   The drop out voltage of a standard 12v reg IC is not 12.8 -  more like 13.8. 
2.   The AC supply varies and may be be 10% low on occasion. 


So, you need a 15VAC tranny. 

OR you can make a "voltage doubler" supply with two diodes and two electros. 

That will give about 33V DC  -   minus the same deductions. 




.....  Phil

I'm going to make a 12v DC power supply with a linear reg.  I will 
half-wave rectify 12v AC and have a smoothing cap. The load is only 10ma.

So: 12v RMS is 17v peak, minus the diode drop of 0.7 is 16.3v peak. 
Using a 47uF cap, the ripple will be 3.5v p-p.  So the min voltage into 
the regulator will be 16.3-3.5 = 12.8.  Ripple calculated from Vpp = 
i/fC (.010/(60*47e-6).

Am I missing anything?

Thanks, Bob