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I have a transformer-type PT 13/2/6 as the following page (Versions 34 to 37 from 68). Image like this:

PT 13/2/6

The voltage between points A and B is V AB (primary)= 230 vac

The voltage between points C and D are V CD (secondary 2) = 6 Vac

The voltage between points E and F are V EF (secondary 1)= 6 Vac

I have 2 loads, first load 5VDC/2.5A put on EF and second load 12VDC/1A put on CF(DE shorted).

My question: How much total power do I need? Is power rate 13VA enough?

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No - you cannot reasonably expect to do what you are trying to do.
You are attempting to draw more power from the transformer than it is designed for and MUCH more power from one winding than it is designed for.

I have 2 loads, first load 5VDC/2.5A put on EF and second load 12VDC/1A put on CF(DE shorted)

I'll take the VA rating as the same as DC Watts drawn - close enough in this case.

5V x 2.5A = 12.5 Watt
12V x 1 A = 12 Watt.

The 5V load is all on winding EF .
The 12V load is 50/50 on CD and EF.
So CD load = 12W/2 = 6W.
The EF load is 6W + 12.5W = 18.5W.

A transformer with two identical secondaries is usually designed to allow a maximum of about 50% of total power to each. You can unbalance them slightly but if you take all power from one winding you'll usually get extra losses and possibly transformer failure.
Here each winding is rate at about 13 VA/2 = 6.5 VA each.
You are trying to draw 6W from one winding (= OK) and 18.5W from the other (about 3 x overload).
Overall you are trying to draw 18.5 W from a 13 VA device ~= 40% overload.

You need a larger power rated transformer or smaller loads.

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As I read the datasheets, the total rated power output of the transformer is 13 VA.

All transformers in the PT13 series are rated at 13 VA, regardless of the output configuration.

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  • \$\begingroup\$ Hee hee - the obvious +1 \$\endgroup\$
    – Andy aka
    Dec 18 '13 at 21:34
  • \$\begingroup\$ A key part of the question is/was "How much total power do I need?" Your comment is useful, but does not address the majority of what was asked. \$\endgroup\$
    – Russell McMahon
    Dec 19 '13 at 9:18
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At points E and F the power for a given load is one quarter of what it will be for the same load placed at C and F. It's one-quarter because the voltage is half and power is proportional to voltage squared.

This is a slightly simplified explanation that doesn't consider the transformer's inability to proportionally regulate the output voltage when more power is being taken.

So, with this simplified explanation and assuming the same value load is connected to EF and EC, for a transformer that is 13VA (assuming this translates to watts exactly), the power at EF is one-fifth of 13 watts = 2.6 watts whilst the power at CE is four times more at 10.4 watts. Both add to 13 watts.

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  • \$\begingroup\$ Your answer does not make much sense in the context. I understand the principle that you are trying to convey, but he will not, and it will confuse him utterly if he tries to understand what you say. He specifically notes the voltages and loads he wishes to obtain, and your answert talks about loads which are essentially unrelated. \$\endgroup\$
    – Russell McMahon
    Dec 19 '13 at 9:20
  • \$\begingroup\$ @RussellMcMahon that's because he's edited his question and put some facts in!!! All the values for load current and voltage were not present in the original answer so I could only generalize an answer. \$\endgroup\$
    – Andy aka
    Dec 19 '13 at 10:05

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