Continue to Site

Center tap transformer and 2 diode question?

Status
Not open for further replies.

xtrmi

New Member
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?

2-diode runs cooler.

fwr 1/2 'copper losses', because all secondary is used.

...and 1/2 the voltage loss in diodes

Last edited:
Will this set up have any impact on the voltage stability in a linear power supply once a load is applied.

This relates to my other issue; but I thought I'd make a new thread so people can find answers easier. On the other problem Marc I think I'm going to try a new regulator because no matter what I do the voltages drop drastically when the load is applied.

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?
you might appreciate , the cost saving of 2 diodes, supposing the load is 10 to 40 amps.
We see saving of energy being wasted across the additional 2 diodes in case of bridge.
In full wave,the Transformer windings take partial load only during half cycle for each winding. In bridge, the winding takes load during both half cycles, thus high power transformer tends to work too hot.

like wise there are disadvantages too.
the two windings may not be perfectly balanced, unless well designed, thereby causing little extra ripple. Also it would be more bulky, compared to single winding secondary.

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.

FWB makes more sense at less current because it better utilizes the Xformer secondary. The CT only uses half of the secondary at a time, hence you basically need two duplicate secondary windings (bigger size and more \$ on the transformer for a given power rating).

Last edited:
Will this set up have any impact on the voltage stability in a linear power supply once a load is applied.

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.

There are other likely causes more possible that cause drastic drop in voltage after a load is applied: i.e. not enough uF in the reservoir capacitor to hold up the voltage between line current cycles.

2-diode runs cooler.

fwr 1/2 'copper losses', because all secondary is used.

...and 1/2 the voltage loss in diodes

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.

Another advantage is lower diode losses.

Unfortunately because the output voltage is halved, the maximum power output will only be 1/√2 the maximum rating, meaning a 10VA transformer should be derated to 7.07VA.

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.

Just to be clear, by "12 V C/T", you mean "6-0-6 1A" ? Or 12-0-12 ?

Sorry, I didn't mean to imply a 1A 12V transformer will be 10VA, it'll be 12VA a of course.

Change that to a 5-0-5V 1A transformer.

Thank you for the help

I know biphase configuration has its benefits but I don't think it gets the most out of the transformer. You're better off using a centre tapped transformer for a circuit that needs both a positive and negative rail. Transformers without a centre tap are cheaper anyway.

2-diode runs cooler.

fwr 1/2 'copper losses', because all secondary is used.
If you are comparing a 6-0-6 transformer used either
A) with a full-wave-rectifier on one of the 6 V taps
or
B)both used, each with a single diode
the copper losses are the same.

With either configuration, the current in each half cycle is taken by one of the 0 - 6 windings. It really doesn't matter that whether the other half cycle is taken by the same winding or a different winding.

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.

B)both used, each with a single diode
the copper losses are the same
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.

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.
The 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.

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.
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.

I suppose it also depend on how long you want the power for. Transformers take a long time to heat up, so for a short term load, where all the heat goes into increasing the temperature of the windings, then using both will double the rating. For a continuous load it will be much less than double, and maybe only a marginal increase on using one winding.

The 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.

Yes, but there are two windings contributing to the loss, so the power is
2x (I/2)²R which is half.

Last edited:
Status
Not open for further replies.

Replies
1
Views
241
Replies
1
Views
3K
Replies
0
Views
402
Replies
14
Views
5K
Replies
16
Views
4K