>Hi,
>
>I'm not so experianced this 3 phase transformers and have the following
>questions:
>
>I have a 3 phase transformer which has on the secundar 3 coils of about
>810VAC. If I take a normal diode bridge I will have Us=810V x squrt 2 x sqrt
>3 = 1980V. Actually I'm looking for a voltage in the area close to 4000VDC
>or at least closer to 2500VDC. Is there a methode to reach this goal?
>
>I saw a drawing for a 1 phase transformer with 2 diodes and 2 C's to double
>the voltage. One coil was conected between the C's and the other side of the
>coil was connected to the diodes. The diodes were connectet in such a way
>that the positive wave went to the upper C and the negative wave went to the
>lower C. So you can use the formular Us= 810V x sqrt 2 x 2 = 2290V.
>
>Does this methode work with 3 phases in the same way?
>
? As I see it, in order to make a three-phase transformer work in FWD
service, the common connection between the 3-windings needs to be
disconnected, and 3 isolated FWD circuits utilized. The D.C. outputs of
the 3 can then be combined. Each 810V-rms winding should produce c.
2300VDC no-load.
>Thanks for your help in advance and hope to catch you on the band
>
>Oliver, DL1EJA
>
>
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>
- R. L. Measures, a.k.a. Rich..., 805.386.3734, AG6K,
www.vcnet.com/measures.
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