I will be using a 13.5 volt transformer with a 20 amp capacity. I will draw approx. 6 amps current DC- 60000µF filtering, capacitive input. Can someone tell me what I should expect my rail voltages to be? I thought it would be theoretically 1.4 times the 13.5= approx. 18.9V. I read in the pass labs A75 article that the transformer will load down, so that it will be about 1.2 times the 13.5= 16.2V. Is there any way to predict?
Steve
Steve
Steve,
Check out this site for Duncan's PSU
Designer.http://www.duncanamps.com/psud2/
It's pretty easy to use.
Rodd Yamashita
Check out this site for Duncan's PSU
Designer.http://www.duncanamps.com/psud2/
It's pretty easy to use.
Rodd Yamashita
sam9
Having bought a transformer or two, I can appreciate the problem. Some like Avel-Lindberg include a state called " regulation", expressed in percent. For instance, if you buy a 13.5V transformer with 6% regulation from them and measure the secondary voltage without a load it will be 13.5V x 1.06 =14.3V. This accounts for what they expect the drop under load to be. On top of that, manufactures want those specs to always be met so they tend to de-rate them a little, maybe another 5%. Transforners I've bought from Plitron and Avel-Lindberg both were this way.
This sounds nice unless you are in a situation where you have an upper limit on the rail voltage due to possible damage to components. I ran into this once and had to do a quickie fix by adding a pair of + & - voltage regulators to what should have been an unregulated supply. Eventually, I replaced the transformer.
So anyway here is the modified calculation I apply to the spec sheets:
(Stated Voltage) x 1.1414 x (1+stated "regulation" %) x (1.05 for under-rating).
Example (13.5V x 1.1414 x 1.06 x 1.05)-1 = 16.1V expected unloaded reading at the rsulting rails . The "-1" is what I expect a 25 amp bridge rectifier to drop, other rectification methods my differ.
Having bought a transformer or two, I can appreciate the problem. Some like Avel-Lindberg include a state called " regulation", expressed in percent. For instance, if you buy a 13.5V transformer with 6% regulation from them and measure the secondary voltage without a load it will be 13.5V x 1.06 =14.3V. This accounts for what they expect the drop under load to be. On top of that, manufactures want those specs to always be met so they tend to de-rate them a little, maybe another 5%. Transforners I've bought from Plitron and Avel-Lindberg both were this way.
This sounds nice unless you are in a situation where you have an upper limit on the rail voltage due to possible damage to components. I ran into this once and had to do a quickie fix by adding a pair of + & - voltage regulators to what should have been an unregulated supply. Eventually, I replaced the transformer.
So anyway here is the modified calculation I apply to the spec sheets:
(Stated Voltage) x 1.1414 x (1+stated "regulation" %) x (1.05 for under-rating).
Example (13.5V x 1.1414 x 1.06 x 1.05)-1 = 16.1V expected unloaded reading at the rsulting rails . The "-1" is what I expect a 25 amp bridge rectifier to drop, other rectification methods my differ.
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