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I have step down 12-0-12 1Amp transformer input volts 220V and output 12V. As it is step up transformer Np > Ns. I was wondering if I can change it to step up transformer by reversing the input to secondary coil and output at primary coil.

Problem :

There are 3 wire at secondary coil. how do I connect those wire so that I can use 2 wire as input?(or 12V to both end)

My questions:

1.step down can change into step up transformer?

2.is there any power loss?(compare to step up transformer,actually I see there is some thin and thick copper wire binding in the transformer. Will this affect the power loss?

update: I have 12V(Two 6V acid lead battery connected in series) I=16A(8A each battery) and i will add some resistor to get I=1A still working on the inverter part for AC current. The load is 9W CFL 100-300VAC.

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  • \$\begingroup\$ What did you intend to purchase? Please edit your question to give details of your application, the power source and the load as these will be important in giving the correct answer. \$\endgroup\$
    – Transistor
    Commented Jan 9, 2016 at 12:56
  • \$\begingroup\$ you have some mumbo-jumbo in your update. What is generating your 6+6V. What does 'I have I=16A' mean? Do you mean that your battery can supply up to 16 A? What are you planning to do with the resistor 'to get 1A'? What is your inverter design? CFL is compact fluorescent lamp; LED is light emmitting diode; I've never heard of a lamp that is both. It would be far more efficient and cost effective to build or use a 12 V DC LED lamp. \$\endgroup\$
    – Transistor
    Commented Jan 9, 2016 at 14:57
  • \$\begingroup\$ Too much to learn. DC is nice for LED. but i never worked with AC and now getting hands on. Q. Do CFL works in DC? \$\endgroup\$
    – chandrayaan1
    Commented Jan 9, 2016 at 15:19
  • \$\begingroup\$ Note that if you have a 12-0-12 winding that you plan to use as a primary you will either have to use it as a 24V input (windings in series) or use just one winding at half the rated current and VA of the transformer or arrange a Push-Pull circuit on the center tap or two split phase drives on the alternate legs. A 12-0,12-0 type of winding would allow you more flexibility letting you place the windings in parallel for 12V full power drive. \$\endgroup\$
    – KalleMP
    Commented Jan 9, 2016 at 15:24
  • \$\begingroup\$ @editint. No. CFLs require high voltage to 'strike' the lamp. You appear to be out of your depth with this project. If you want help then edit your post to tell us (1) What you are trying to build. (2) Schematics that you're using. (3) How far you've got. (4) Then ask for comments on the design - a design check. If you then decide to proceed break down the project into sections and ask a new question on each of those. I notice that instead of answering my questions in my last comment that you asked a new question instead. You'll need to do better than that. \$\endgroup\$
    – Transistor
    Commented Jan 9, 2016 at 15:35

2 Answers 2

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is there any power loss?(compare to step up transformer,actually I see there is some thin and thick copper wire binding in the transformer. Will this affect the power loss?

Primary and secondary share the same core so, the main thing to consider is whether the core saturates more when driven backwards. Here's a sort of worked example: -

If the primary winding is 10 henries inductance, on 240V AC supply the current will be: -

\$\dfrac{240V}{2\pi f L}\$ = 76 mA.

Let's say the transformer had 1000 turns on the primary so ampere turns are 76.

The 12V secondary has one twentieth of the primary turns i.e. 50 and its inductance will be \$20^2\$ times smaller at 25 mH. Now, if you applied 12V AC to the secondary you'd get a magnetization current of: -

\$\dfrac{12V}{2\pi f L}\$ = 1528 mA.

Ampere turns are 50 x 1.528 = 76 (i.e. just the same)

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Transformers are passive time invariant linear components. This means amongst other things you can use them backwards. In fact many old square wave inverters that are now being remanufactured and rebranded as modified sine use a 50Hz transformer with a LV CT winding in pushpull of the battery. Many solid state audio amplifiers use an output transformer backwards to drive a 100V line. Losses are approx the same forward or backward.

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    \$\begingroup\$ A competent transformer winder will allow for IR losses so ratios may not be identical - but close enough for most uses. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jan 9, 2016 at 7:30
  • \$\begingroup\$ My gut feel also tells me that the VA rating of the primary side would benefit from being designed slightly higher to account for the output load and the core and copper losses. So an optimised transformer might be calculated for 105 VA on the primary and 100 VA on the secondary to get the last bit of value from the copper and iron in large volume products. \$\endgroup\$
    – KalleMP
    Commented Jan 9, 2016 at 14:59
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    \$\begingroup\$ @Kalle you are correct - which may mean you can only drive the secondary with 100VA and collect 95VA from the primary. \$\endgroup\$
    – user16324
    Commented Jan 9, 2016 at 16:54

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