you might appreciate , the cost saving of 2 diodes, supposing the load is 10 to 40 amps.What can the center tap transformer with two diodes at the end taps with ground on the center do for me as opposed to using a bridge between the center tap and one outer tap?
Generally, CT makes sense for high current (>10A) because of the power savings of only one diode conducting at time.What can the center tap transformer with two diodes at the end taps with ground on the center do for me as opposed to using a bridge between the center tap and one outer tap?
In practice, the kind power losses and ripple amplitude we're talking about here won't have a significant effect on the output voltage regulation stability of a linear supply. If anything, these higher losses have the potential of slightly reducing the maximum regulated output voltage. By 'slightly' I mean no more than 2 V.Will this set up have any impact on the voltage stability in a linear power supply once a load is applied.
This is why it's rated for √2 times the normal maximum current rating. For example, a 1A 12V centre tapped transformer can happily supply 1.414A when used in this configuration.2-diode runs cooler.
fwr 1/2 'copper losses', because all secondary is used.
...and 1/2 the voltage loss in diodes
Just to be clear, by "12 V C/T", you mean "6-0-6 1A" ? Or 12-0-12 ?a 1A 12V centre tapped transformer can happily supply 1.414A . <..> Another advantage is lower diode losses.
<..>output will only be 1/√2 the maximum rating, meaning a 10VA transformer should be derated to 7.07VA.
If you are comparing a 6-0-6 transformer used either2-diode runs cooler.
fwr 1/2 'copper losses', because all secondary is used.
Yes you're right, the copper losses are the same but because there's two coils the loss in each coil is half so you can safely increase the current by a factor of √2.B)both used, each with a single diode
the copper losses are the same
The copper losses are quartered aren't they?If the transformer is a 0 - 6, 0 - 6 type, and the windings are paralleled and a full bridge used, the current is split between the two secondary windings in each half cycle, so the copper losses are halved.
That depends on how the heat can get away from the secondary windings, and what the primary losses are. The two secondary windings are not like two separate resistors, each of which can loose heat independently. They are much closer thermally than that. I guess you can dissipate more heat if both windings are used, but I am sure that it won't be twice as much heat.Yes you're right, the copper losses are the same but because there's two coils the loss in each coil is half so you can safely increase the current by a factor of √2.
Yes, but there are two windings contributing to the loss, so the power isThe copper losses are quartered aren't they?
The current through each coil is halved and P = I²R so P goes down by a factor of four.