Re: Diode Strings

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "robert heidlebaugh" <rheidlebaugh-at-desertgate-dot-com> 

Rick W: Thank you for your input; 1.414 x RMS (15 Kv is 22Kv) .In a full
wave rectifier the diode sees 2 x Epk. In a bridge the two diodes are in
series in conduction and in series when cut off ( reverse bias) with or
without a capacitor so the total voltage across the cut off diodes is Pk not
Pk/pk. That is the advantage of using a bridge. The disadvantage of a bridge
is it requires twice as many diodes. That is 4 strings not 2 strings. The
other advantage is a bridge produces full Pk voltage dc out not 1/2 Pk
voltage out.  If someone uses a NST  of 15 Kv rms with a full wave rectifier
having  one diode string to each leg  Ct  to ground, The output is 7 Kv x
1.414 ,about 11Kv, but the diodes see 2x PIV across the diodes . A bridge
would produce 15 Kv x 1.414 (22 Kv) but would see the same Piv across each
diode string as a full wave rectifier. Two diodes in series sees 2 x Pk / 2
diodes =  Pk each.
       Robert     H
-- 


 > From: "Tesla list" <tesla-at-pupman-dot-com>
 > Date: Sat, 21 Aug 2004 17:53:53 -0600
 > To: tesla-at-pupman-dot-com
 > Subject: Re: Diode Strings
 > Resent-From: tesla-at-pupman-dot-com
 > Resent-Date: Sat, 21 Aug 2004 18:11:39 -0600
 >
 > Original poster: "Richard W." <potluckutk-at-comcast-dot-net>
 >
 >
 > ----- Original Message -----
 > From: Tesla list <tesla-at-pupman-dot-com>
 > To: <tesla-at-pupman-dot-com>
 > Sent: Saturday, August 21, 2004 7:57 AM
 > Subject: RE: Diode Strings
 >
 >
 >> Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz>
 >>
 >> Perhaps I might be permitted to expand on this slightly:
 >>
 >> On 20 Aug 2004, at 7:42, Tesla list wrote:
 >>
 >>> Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com>
 >>>
 >>>> Yes, the PIV is 40 KV.  But it will be distributed across
 >>>> two legs of a bridge
 >>>
 >>> I have seen quite a few people mention this now so I couldn't let it
 >>> pass. it's wrong!
 >>>
 >>> The reverse voltage seen by any diode (or diode stack) in a bridge
 >>> rectifier is equal to the DC output voltage, ie 1.4 times the RMS AC
 >>> input voltage. The reason is that when one diode is conducting, the
 >>> voltage across it is "negligible" therefore the whole DC output
 >>> voltage must appear across the other diode on that side of the bridge.
 >>> (If you draw the diagram you can see that the two diodes are in series
 >>> across the DC bus.)
 >>>
 >>> So if you're making a rectifier to have 40kV DC output, your diodes
 >>> must be rated at least 40kV PIV each. Preferably 1.5 to 2 times more
 >>> for safety.
 >>>
 >>> Steve C.
 >>
 >> It actually depends on whether the bridge is feeding a capacitor or
 >> not which, if it is, the above arguments rigidly apply since the
 >> reversed diodes have to hold off the voltage across the capacitor
 >> plus the peak voltage of the transformer.
 >>
 >> Malcolm
 >>
 >>
 >
 > I setup a low voltage bridge using a filter cap  and using an O-scope looked
 > at the voltage across each diode. There was only 1/2 of the peak-to-peak
 > voltage across any one diode. In other words the diode in any leg needs to
 > be 1.414 x RMS even with a filter cap.
 >
 > Of course add some voltage rating for safety.
 >
 > Rick W.
 >
 >