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Physical Reason For Reduced Current In Step Down Transformer

  1. Mar 26, 2014 #1
    Hi everyone,

    I've been reading transformers and this question came in my mind. I'm unable to find a satisfactory answer.

    Why does the current increase in a step down transformer? I am aware of V1I1 should be equal to V2I2, but I was looking for a physical reason for the current to increase.

    Current is flow of electrons along a conductors, the electrons vibrate and transfer energy to the neighboring electron and thus current flows.

    So in Step Down Transformer, what is the physical reason for the current to increase?

    You have less number of turns.

    A fellow student says, since number of turns is less, you have each turn of more cross section area to occupy full height of core. So since turns have more cross section area, the resistance is less and current flow increases by ohm's law.

    I read many books, but all gave reason of VI should be same on both sides. But this is mathematical reason, can anyone help me with a physical reason why current should increase if you have less number of turns on secondary.

    Thanks in advance.

    I am a student of electrical engineering.
  2. jcsd
  3. Mar 26, 2014 #2


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    I just think of the step down secondary as a voltage divider. If you have 2:1 coils then all of the secondary coil sits against 1/2 of the primary coil. So you step down from say 240 to 120 volts. From there the voltage "senses" the load and then the simple V=IR kicks in.

    Then there is the change in flux formula....someone else can explain this lovely phenomina.
    But you can see it dicates the voltage.

    V=-N* dΘ/dt

    N=number of turns in coil
    Last edited: Mar 26, 2014
  4. Mar 26, 2014 #3

    jim hardy

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    That same question puzzled scientists of the late 1800's.

    Basically it's a push-push balance between opposing magnetomotive forces....

    See if this textbook chapter from 1896 is any help.

    [ whole book available here: http://books.google.com/books/about/Dynamo_electric_Machinery.html?id=lhNDAAAAIAAJ ]

    I have the 1901 edition of the same book, and used it a lot in my career to help "figure out" things electrical and explain them in physical terms to my mechanical engineering peers.. in exchange for similar discourses in their areas of expertise, of course. we had a lot of fun educating one another.

    old jim
  5. Mar 26, 2014 #4


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    I look at the problem differently. It's easy for me to see intuitively why the current increases, because there are twice as many wires for the same flux to interact with. But, why does the voltage decrease? (the answer "because there is 2X the current" doesn't help). So I just accept IVin = IVout.
  6. Mar 27, 2014 #5
    At meBigGuy,

    The voltage induced per turn is same. E=N* dΘ/dt.

    Assumining no leakage, flux on both sides is same, so E per turn is same. Now since number of turns on secondary side is less(in case of step down transformer), you have less voltage that is less(emf per turn * no. of turns).

    But why does current increase?
    "It's easy for me to see intuitively why the current increases, because there are twice as many wires for the same flux to interact with."

    Compared to primary, there are half the number wires in secondary in case of step down transformer.

    My class mate says it could be because the secondary coil is of more cross section and less length. So less resistance.

    Now if resistance is less, the ohm's law can be used and current would reduce since R has reduces, so
    V = IR. R reduces and V increases.
    Last edited: Mar 27, 2014
  7. Mar 27, 2014 #6


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    Resistance inside a transformer is negligible, otherwise they'd be really inefficient and produce a ton of heat.

    Have you read the wiki description?

    The secondary side current is determined by the load. The primary side current is determined by the load side impedance, which also goes to infinity as the current drops to zero.
    Last edited: Mar 27, 2014
  8. Mar 27, 2014 #7

    jim hardy

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    Maybe the question is why does any primary current flow at all?

    Thought experiment: Ideal transformer, requires no magnetizing current. Well, to avoid division by zero let it require only a picoamp of magnetizing current at rated primary voltage.

    Now apply primary voltage:
    Flux appears in the core in amount necessary to counter applied primary voltage, say X volts per turn. (volts per turn at any given frequency defines flux). Primary current is 0.000000000001 amps, one picoamp for this really great almost ideal core.

    Primary volts is X volts per turn, so is secondary, so voltages in both windings are in proportion to their respective number of turns. Let's assume secondary has half the number of turns as primary.

    Now connect a load to secondary. Adjust that load so it draws exactly one amp.
    Now you have 1 X (number of secondary turns) ampere-turns opposing the flux from the primary winding.
    So flux will collapse, unless more current flows into the primary winding to maintain flux at the level of X volts per turn. Else the transformer cannot counter the applied voltage.

    So, what is primary current now?
    It's that 1 picoamp to make X volts per turn,, plus whatever is required to cancel out the secondary amp-turns.
    I get a half amp in primary to cancel the 1 amp in secondary, + that picoamp ,
    which is 0.50000000001 amps.
    Aha - allowing current to flow in secondary is what allows current to flow in primary ! (above and beyond our miniscule magnetizing current) !

    The amp-turns from primary and secondary are equal and opposite . It really is that simple.

    In the words of Silvanus Thompson's 1884 book, "It is a beautifully self regulating system."
    And that's because amp-turn is the basic unit of magnetomotive force. It goes by other names - oersteds, gilberts, but the amp-turns are what count.

    So since the MMF's (AMPS X TURNS) must be equal and opposite , the amp ratio will be inversely proportion to the turns ratio.
    In a decent transformer the magnetizing current is small compared to rated load , but not as small as in my example.

    Does that help?

    It's important to get that concept of balanced mmf's. Else you will have trouble with current transformers.
  9. Mar 27, 2014 #8
    Thanks. That helps tremendously. If a secondary load requires 1A, that 1A into number of secondary turns will oppose the flux, so the primary will draw current from source till the flux is opposed and at 0.5A the flux will be balanced, so it will stop extracting current.

    Thus amp-turns ratio is maintained.

    Conclusion: The amp-turns ratio maintenance is a result of the behavior and not the other way around.
  10. Mar 27, 2014 #9

    jim hardy

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    Conclusion - Chicken or egg?

    It's a result of those 1800's experimenters figuring out that if they arranged their windings on a closed-loop core so that all the flux went through both windings, it'd be well behaved and predictable. Earliest attempts were just two coils wound on an iron rod, like the electromagnet you made with a nail as a kid.
    They minimized leakage flux by careful arrangement of the windings, and minimized magnetizing current by closing the iron loop and laminating it.

    That old book i linked is fun - the author speaks of Steinmetz and Tesla 's work in present tense - they were contemporaries.

    old jim
  11. Mar 27, 2014 #10


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    Wish I could make my post go away. Not sure what I was thinking (other than wrong)

  12. Mar 27, 2014 #11
    Thanks for the book old jim. Unfortunately, I was unable to open it. A google page came and book title, but I couldn't find the button that would say, 'Read book now,' nor any option for buying or downloading the book.
  13. Mar 27, 2014 #12

    jim hardy

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  14. Mar 27, 2014 #13

    jim hardy

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  15. Apr 2, 2014 #14
    Hi guys, I have a question related to step down transformers that hopefully can be answered in layman's terms. I think I understood some of the responses above but I just need confirmation. Anyway, I have a transformer that I am trying to use to power a custom made heater element. I need approximately 40 amps to power it so I rewound a transformer to a 10:1 ratio (10 turns of 10awg wire on the secondary). I powered it up and made a couple measurements. I have 115 VAC on the primary side and 11.8 volts VAC on the secondary side but the amperage on the secondary side is negligible and averages around 3-4 amps. The resistance of the secondary side load is 2.75 ohms. I was under the impression that I would be able to significantly boost the output current but I can't seem to figure out how it works. I understand that IVin has to equal IVout but I don't understand how this can work without violating Ohms law (which it hasn't in my case). On the flip side, Since Ohm's law hasn't been violated here I don't understand how removing turns in the secondary side is supposed to effect Iout. How do I set up this transformer to give me the current needed to power this heater? I'm clearly missing something in the transformer theory department.

    On a side note, I apologize if I'm posting on the wrong forum. I realize my question is more of a novice DIY question and not really up to par with the rest of the discussion I see here. I would appreciate any advice nonetheless.

  16. Apr 2, 2014 #15

    jim hardy

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    You cited Ohm's law,,,,,
    so - what is the voltage that you must apply to a 2.75 ohm resistor in order to push 40 amps through it ?

    How much current will flow in your secondary with your 11.8 volts and your 2.75 ohm resistor?

    I think we can agree - Ohm's law has not been violated.

    take another look at post 6
  17. Apr 3, 2014 #16
    Without having a calculator handy it's something like 110V, which is pretty much outlet power... Unfortunately The circuit breakers don't like 40A. So I'm assuming the videos of these guys pulling 800A from a microwave transformer is because they are shorting the output so the resistance is essentially 0?

    Tell me if I'm on the right track here; Say I have 120VAC @ 5A on the primary side (120*5=600W) and I wound the secondary side so that it output 15VAC @ 40A (15*40=600W) I would be conserving the power across the transformer UNTIL I add the load resistance of 2.75 ohms. Now the current drops down significantly (15/2.75=5.45A). Do I assume that the change in values will be reflected on the primary side to maintain IVin=IVout? How does the secondary side affect primary side? From what I'm reading here it has something to do with the magnetic fields created by the windings? I guess the point of all this is that I won't be able to power this heater from the 120V outlet, correct?
  18. Apr 3, 2014 #17


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    No, for that kind of power in a residence you would need a dedicated 220 30A 10 AWG circuit and a 2:1 step-down 5KVA transformer to deliver the 40A at ~110 volts using 8-6 AWG (6 for safety factor) wire. Using a 220 20A heater is a much better idea at those power levels.
  19. Apr 3, 2014 #18
    ok, well i have a 220V 30A outlet in my work area for the dryer. I would use a typical heater except that this is a custom one that i need for vacuum forming some acrylic. thanks for the clarification. I'm not too far from my goal, it will just take a little rethinking.
  20. Apr 3, 2014 #19
    How is Ohm's law being satisfied? I=11.5/2.75=4A. But you're getting 40A. I'm confused.
  21. Apr 3, 2014 #20


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    That's his problem, he's NOT getting 40A @ 11.5V and even if the heater load resistance was adjusted downward to flow 40A @ 11.5v he still would not have the desired power level in his heater. The fundamental energy factors when dealing with the initial possible requirement of a power transformer is not current, it's desired power ,the resistance/impedance of the load and voltage level desired to power that load. The power and the resistance of the load determine what voltage is needed at that power level (the resistance can then be adjusted to the correct power level at that voltage). Once you have the voltage you calculate the current using the load resistance. From these basic facts combined with the supply power voltage and current capabilities you can begin to design a power transformer to meet your needs. In his case where the required voltage (115) is equal to the utility voltage (115) you don't need a transformer if the utility voltage circuit can also deliver the required amount of current.

    Because we often use standard voltages (110/220) on utility circuits we commonly use current flow as a proxy for power as the voltage is assumed constant. This view of current flow (in isolation from other factors) as power is a common source of confusion when looking at circuits for beginners.
    Last edited: Apr 3, 2014
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