[Milsurplus] Attention Smart People:

[J. Forster]

Hey guys, I screwed up on my fist post on the topic. I glanced at it and
did not analyze it carefully.

Starting again: First label the parts:

+   D1|     D2|  C1   D3|

-   D4|     D5|  C2   D6|


The + & - are the reference polarities for the transformer. The diodes are
as in the original diagram.

Suppose everything is discharged, then the transformer is turned on the
half-cycle so the voltages are + and - as shown.

Start analyzing at the Xfmr - terminal. It is below ground, so D4 conducts
and D1 is reverse biased, hence OC. So the junction of D1, C1, and D2 goes
positive. D2 conducts, charging C2 to near peak. D6 conducts and charges
Cout.

Next analyzing what happens when the Xfmr is on the other half cycle and +
& - are reversed. The + terminal is below ground so D1 conducts and D4 is
reverse biased, hence OC. So the junction of D4, C2, and D5 goes posttive.
D5 conducts, charging C1 to near peak. D3 conducts and charges Cout.

So, at the end of the first cycle C1, C2, and Cout are all charged.

On the third half cycle, D4 again conducts and D1 is back biased so OC. C2
is again charges via D3, BUT C1 is still charged from the previous 1/2
cycle, so the D5, C1, D3 junction goes to Vc1 + Vpeak of Xformer. This
forward biases D3 and charges Cout to near twice Vpeak.

On the fourth 1/2 cycle, the same thing happens via the mirror image path.

Best,

-John

================




> AGREED, the input/transformer-secondary must be floating lest one of the
> left-hand diodes be directly across the input.  My real problem is that
> I fail to see how this is going to double any voltage.  The negative
> sides of the caps in the bridge are always going to be one diode drop
> above ground, and therir positive ends are essentially in parallel, one
> diode drop above the output..
>
> Al
>
> On 8/3/2010 9:03 AM, arc5 at ix.netcom.com wrote:
>> Please take a look at this full-wave doubler circuit:
>>
>> http://www.kwarc.org/bulletin/99-04/tech_corner.htm
>>
>> At the end of the page is this warning:
>>
>>      This circuit can be operated off the AC line 120 or 230 V).
>>      BUT the negative ground must be left floating since
>>      one side of the line Is grounded and YOU CANNOT ground
>>      both the Input and the NEG (or POS) DC or else B O O M!
>>
>> I guess I'm stupid (I've only had one cup of coffee, so there's my
>> excuse),
>> because I don't see a current path from the DC B+ side
>> to the AC line, other than ground which is supposed to be -0- potential
>> on both sides.  Now, if your B- ground came loose on the power supply
>> and you didn't have a grounded line cord installed
>> (which would be even stupider than me with this kind of voltage)
>> you'd have full B- on the chassis to ground and your wife
>> would have to pay for a funeral.
>>
>> OK... what am I missing here?
>>
>>
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>
> --
> Al Klase - N3FRQ
> Jersey City, NJ
> http://www.skywaves.ar88.net/
>
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