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#1 |
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diyAudio Member
Join Date: Jan 2003
Location: Asia
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I have two 530VA 2X40V toroid lying around, can I use this for the gainclone project.
1. IMHO the output voltage is too high, is there any way to reduce the output voltage (like using a half wave rectifier)? 2. Also is the VA rating too much? 3. why no one use EI transformers for this application? many thanks DM
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#2 |
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diyAudio Member
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voltage is way too high
there's no such thing as too high VA rating ![]() some people do use EI for their gc |
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#3 |
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diyAudio Member
Join Date: Jan 2003
Location: Asia
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Thanks.
Could someone also kindly help me to understand the DC output voltage of the attached configuration (220V version)? As I understand, it is a full wave rectifier with a capacitive load (also take into account the 1000uF caps, shall I?), some the DC voltage should be: Vdc = 0.71 x Vac and the Vac is 96V (48V x 2) in this case ... how can it be? way too high? Where did I made a mistake? |
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#4 |
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diyAudio Member
Join Date: Jan 2003
Location: Asia
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Anyone?
BTW, if this configuation is used, what VA rating of the transformer shall I use? is 100VA good enough, or 160VA better? does this also mean I need 4 x 100VA (or 160VA) per channel? Many thanks. |
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#5 |
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diyAudio Member
Join Date: Dec 2001
Location: SIUE, Illinois, USA
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out of curiosity, what are the caps and diodes between primaries for? some kind of low-level DC removal? and 1uF is pretty small for a filter cap.
and why not just draw it as 4 diodes in a row down in parallel with 4 in a row up? also, the unloaded Vdc will be about 1/0.707 or 1.414 Vac, this is because the caps will chrage to the peak voltage, and without any load they will never discharge. adding a load will create a ripple voltage which is pretty high with 1000uF. an approximation is I = C dv/dt = C Vrip 2 pi f so Vrip = I/ (2 pi C f) where f is either 50 or 60 for half-wave, or 100 or 120 for full-wave. i assume your looking for voltage rails of about 24-33V per rail, +-24 to +-33V ?
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