Veryveryfamous
Member
- Joined
- Feb 23, 2011
- Messages
- 7
Dear GroupDIY,
I'm looking to modify a British (240V/50hz) Korg Trident to run on US (120V/60hz) power by replacing its power transformer. Typically I would look up a data sheet for the transformer or use a step up transformer and take measurements but without either of those available I'm hoping to find a suitable replacement by making a few educated guesses.
The calculations should be simple but moving between Watts and Volt-amps (VA) psychs me out and I'd love to hear any insights. Hopefully I'm not overlooking something!
First, I wanted to figure out what kind of secondary voltage I need to look for. The keyboard has a linear power supply with custom regulators that supplies +/-15V and +/-5V (see attached). I disconnected the original transformer and ran a 24VAC sine wave into the primary and measured a 4VAC wave on the (unloaded) secondary to so with that ratio remaining constant I assume I'm looking for a transformer with at 36VCT or 40VCT secondary coil which would give each rail ~18-20VDC of headroom to regulate down to +/-15V.
The next step (and my main question!) is to figure out how big of a transformer I'd need. The service manual says that this keyboard consumes 53W of power so I know I'm looking for a transformer with a Volt-Amps rating of at least (and probably more than) 53VA, but how big is big enough?
I could base a conservative guess on the main line fuse but the fuse is rated at 2A so 2A*240V=480VA which seems way too big for the instrument. Alternately, this website suggests a way to calculate the VA for a similar power supply which comes down to: VA = 1.4 x ( WATTS + 2 IDC ) where IDC is the DC current output. This power supply has a 3.15A fuse for each rail so it seems my calculation would be 1.4 x (53W + 2(6.3A)) = 91.84VA which looks more reasonable although still pretty big. Is that a fair estimation?
Whats the best way to make this calculation so I don't end up with a transformer that's too big to fit in the instrument or so small it burns out?
I'm looking to modify a British (240V/50hz) Korg Trident to run on US (120V/60hz) power by replacing its power transformer. Typically I would look up a data sheet for the transformer or use a step up transformer and take measurements but without either of those available I'm hoping to find a suitable replacement by making a few educated guesses.
The calculations should be simple but moving between Watts and Volt-amps (VA) psychs me out and I'd love to hear any insights. Hopefully I'm not overlooking something!
First, I wanted to figure out what kind of secondary voltage I need to look for. The keyboard has a linear power supply with custom regulators that supplies +/-15V and +/-5V (see attached). I disconnected the original transformer and ran a 24VAC sine wave into the primary and measured a 4VAC wave on the (unloaded) secondary to so with that ratio remaining constant I assume I'm looking for a transformer with at 36VCT or 40VCT secondary coil which would give each rail ~18-20VDC of headroom to regulate down to +/-15V.
The next step (and my main question!) is to figure out how big of a transformer I'd need. The service manual says that this keyboard consumes 53W of power so I know I'm looking for a transformer with a Volt-Amps rating of at least (and probably more than) 53VA, but how big is big enough?
I could base a conservative guess on the main line fuse but the fuse is rated at 2A so 2A*240V=480VA which seems way too big for the instrument. Alternately, this website suggests a way to calculate the VA for a similar power supply which comes down to: VA = 1.4 x ( WATTS + 2 IDC ) where IDC is the DC current output. This power supply has a 3.15A fuse for each rail so it seems my calculation would be 1.4 x (53W + 2(6.3A)) = 91.84VA which looks more reasonable although still pretty big. Is that a fair estimation?
Whats the best way to make this calculation so I don't end up with a transformer that's too big to fit in the instrument or so small it burns out?