bucking and boosting and current capacity of transformers

[cornytheclown]

What determines the current capacity of a transformer that is hooked
up in a boost or buck confiquration ??? Like say a 120 volt pri 28
volt secondary xfrmr that is designed to deliver 28 volts and is rated
at 40 va.... you are now using both windings to get an output of 150
volts roughly but how do you determine the current available thru the
"winding" now ??? Would the same va rating exist though now it would
be higher voltage and less current handling capacity ???

What happens to the va rating of the transformer when hooked up in a
bucking confiquration??? would the same rules as in boost apply ???

thanks

 

[John Popelish]

cornytheclown wrote:

What determines the current capacity of a transformer that is hooked
up in a boost or buck confiquration ??? Like say a 120 volt pri 28
volt secondary xfrmr that is designed to deliver 28 volts and is rated
at 40 va.... you are now using both windings to get an output of 150
volts roughly but how do you determine the current available thru the
"winding" now ??? Would the same va rating exist though now it would
be higher voltage and less current handling capacity ???

What happens to the va rating of the transformer when hooked up in a
bucking confiquration??? would the same rules as in boost apply ???

thanks

In pure buck or boost configuration (line voltage across the line
rated winding, and the load separated from the line by the low voltage
winding) the output current rating is just the current rating of the
low voltage winding.
So your example of a 40 VA, 28 V transformer, the secondary is rated
for 40 VA/ 28 V = 1.4 A.

If you connected the 28 V winding to add to the 120 volt primary, the
output would be rated for 120 + 28V = 148 V @ 1.4 A for a total VA of
149V * 1.4 A = 207 VA.

If you connected the 28 volt to buck the primary voltage you would
have a buck transformer rated for 120 - 28 V = 92 V @ 1.4 A for a
total VA of 92 V * 1.4 A = 129 VA.

However, is you hook the secondary in series as in the boost
configuration, but connect the line to the two ends of the string, and
the load across the 120 volt primary, you have both windings feeding
current into the load, so the total load current is higher. It can
then go as high as 1.4 A (through the secondary and 40 VA / 120 V =
..33 A (through the primary) for a total of 1.4 +.33 = 1.73 amps. The
voltage out is 120 V * (120/(120+28)) = 97 V. So the VA rating in
this configuration is 97 V * 1.73 A = 168 VA.

Actually you can go just a bit higher than that, because the core
losses go down a little with the lower volts per turn this
configuration applies to the core. So you can afford a little more
than rated copper losses before the thing exceeds its design
temperature.
--
John Popelish

 

[John Popelish]

phil-news-nospam@ipal.net wrote:

On Sat, 27 Mar 2004 11:40:23 -0500 John Popelish <jpopelish@rica.net> wrote:

| In pure buck or boost configuration (line voltage across the line
| rated winding, and the load separated from the line by the low voltage
| winding) the output current rating is just the current rating of the
| low voltage winding.
| So your example of a 40 VA, 28 V transformer, the secondary is rated
| for 40 VA/ 28 V = 1.4 A.
|
| If you connected the 28 V winding to add to the 120 volt primary, the
| output would be rated for 120 + 28V = 148 V @ 1.4 A for a total VA of
| 149V * 1.4 A = 207 VA.

You mean 148V ?

Yes. Spell checker doesn't catch numerical typos.

| If you connected the 28 volt to buck the primary voltage you would
| have a buck transformer rated for 120 - 28 V = 92 V @ 1.4 A for a
| total VA of 92 V * 1.4 A = 129 VA.
|
| However, is you hook the secondary in series as in the boost
| configuration, but connect the line to the two ends of the string, and
| the load across the 120 volt primary, you have both windings feeding
| current into the load, so the total load current is higher. It can
| then go as high as 1.4 A (through the secondary and 40 VA / 120 V =
| .33 A (through the primary) for a total of 1.4 +.33 = 1.73 amps. The
| voltage out is 120 V * (120/(120+28)) = 97 V. So the VA rating in
| this configuration is 97 V * 1.73 A = 168 VA.

I'm not 100% sure I understand your wording here. Are you just taking
the boost configuration and "reversing" the input and output?

Yes.

If this
is so, then it would seem to me that the way to build these would be to
connect them internally in the boost configuration and bring the leads
out marked for the 2 different voltages, basically an autotransformer
with 3 leads. (common, 120, and 148, in the above example).

That would be my first example.

But in this third configuration, you can not get more volt than the
line, only less. It is essentially a transformer form of a voltage
divider.

--
John Popelish