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)
This answer explains that for an unloaded secondary, the natural phase relationship between primary voltage and secondary voltage is is zero degrees.
It therefore follows that if there is a secondary load current (due to a resistive load), the current in the primary due to that secondary resistive load must be 180 degrees out of phase with the secondary load current i.e. as current flows into the primary, current flows out from the secondary.
This of course is for an ideal transformer and a resistive load.
If you ignore the leakages and magnetization inductance of the transformer, and the load is reactive, then there will be a 90 degrees phase shift.
Bringing in leakage inductance and DC coil resistance will/can muddy the waters. Bringing in magnetization inductance muddies the water a bit more.
The low frequency transformer equivalent circuit is this: -
As you should be able to see, if you considered all the leakages, magnetization inductance and losses and then added a semi-reactive load, the phase angle is quite complex to calculate.
However I do-not know, in which-way Lenz's law acts in transformer;
because the law states the induced current will try to hinder the
cause.
Strictly speaking, it is voltage that is induced and any current that flows is subject to the that voltage, the load and the leakage inductance.
but when the secondary circuit of a step-up transformer turned On
(closed) , so-far I've know , the current in primary-coil goes-up
In normal usage, for a voltage transformer it is non-ideal to consider the secondary being short circuited. However, it makes no difference to the phase angle providing you obey the rules inherent to the model.
Best Answer
Good.
You have remembered wrongly, or misread.
You need at least a minimum number of turns on the primary in order to keep the core field below its maximum.
Core materials, iron or ferrite, have a maximum magnetic field. With iron, the limit tends to be saturation, with ferrite, heating from losses tends to bite before saturation. The maximum fields are typically in the order of 1.5T for mains iron transformers, and in the range 0.1T to 0.3T for ferrite depending on frequency and grade.
The core has an area \$A\$, and the core material has a maximum field \$B_{max}\$. The transformer has an operating frequency \$f\$. We can calculate the maximum voltage induced in one turn via the maximum rate of change of flux, which is \$V_{max} = 2\pi f B_{max} A\$ for a sinusoidal input, with all terms in SI units. This gives us the peak voltage. Divide by sqrt(2) to give the RMS voltage. We need enough turns to equal our input voltage.
To give an example, I have an iron toroid with cross section 13x25mm, operating at 50Hz, and I am going to assume a \$B_{max}\$ of 1.5T.
\$V_{max} = 2 \pi \times 50 \times 0.013 \times 0.025 \times1.5 = 0.153V \$
0.153V peak is about 0.108V rms. At 240v, we would need 2222 turns on the primary, minimum. A few more wouldn't hurt.
then you get a very low secondary voltage produced.
well, yes, if you shorted the secondary. But we don't do that. The resistance of the load tends to exceed that of the winding, and so it's the load that determines the current.
You do several things to design a transformer. Note that a full design could get much more detailed, but this gives you a good start.
1) Start with a big enough core for your power. Go to a transformer catalogue, and find a transformer with a similar power to what you want to use. Halve its weight, the copper is about the same weight as the iron. Start with a core this weight.
2) Calculate the minimum number of turns needed on the primary, as above
3) Caluclate the number of turns needed on the secondary, to give your design voltage, using the same volts per turn figure.
4) Choose a the maximum thickness of wire that will half-fill the winding window. 50% is a good fill factor, you will be doing well to get a better factor.
5) Check that the current density doesn't exceed about 3A/mm2 (a rule of thumb valid for small to medium transformers) at your load power.