Below there is an impedance matching aimed step-down transformer circuit where the primary to secondary turn ratio N1/N2=10. To actualize such turn ratio and voltage ratio, I set the inductor ratio L1/L2=100. I also set the magnetic coupling coefficient K1=1 to neglect any magnetic leakage. And I also set inductors' series or parallel resistances and capacitances to zero. But I guess this still does not represent an ideal transformer.

I have come to this conclusion because of the following observation. When the secondary is open V1/V2=10 where V1 and V2 are rms values of the primary and the secondary windings. This is expected since we set the ratio to 10. Below circuit and the plots shows this:

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But when the secondary is loaded with 8 Ohm load as in the below circuit, both V2 and V1 starts decreasing as we increase the loading by decreasing R2 as shown in below plots:

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And the the winding currents does not have ratio of 10:

enter image description here

My questions are:

  1. Why does the secondary voltage decrease when loaded? And most importantly why does the primary change at all? I think I have some fundamental wrong understanding of a transformer. I thought primary's action is independent from secondary. So I thought the process is unidirectional like in an amplifier i.e. output port is dependent on the input port but not the other way around. How to explain this?

  2. When the secondary is loaded the voltage decreases but voltage ratio is still 10 but why does the current ratio is very different and varies with loading? Isn't V1/V2 = I2/I1 always?

  • \$\begingroup\$ Why is there 800 ohms in the primary, is that the referred secondary resistance? \$\endgroup\$ – Chu Mar 14 '19 at 12:44
  • \$\begingroup\$ Just an example output impedance to be matched to 8 Ohm load. Only a random value to ask the question. \$\endgroup\$ – cm64 Mar 14 '19 at 12:47
  • \$\begingroup\$ And is V1 in your 1st graph the supply voltage or the voltage across the primary winding? \$\endgroup\$ – Chu Mar 14 '19 at 12:50
  • \$\begingroup\$ V1 is the voltage across the primary winding \$\endgroup\$ – cm64 Mar 14 '19 at 12:51
  • 1
    \$\begingroup\$ @Chu When we talk about "ideal" transformer does that mean that the transformer should have infinitely large winding inductances? \$\endgroup\$ – cm64 Mar 14 '19 at 13:25

For an ideal transformer the relation $$ P_{in} = P_{out} \Rightarrow P_{prim} = P_{sec} \Rightarrow V_{prim} \cdot I_{prim} = V_{sec} \cdot I_{sec} $$ and hence $$ \frac{V_{prim}}{V_{sec}} = \frac{I_{sec}}{I_{prim}} $$ is always valid.

The power drawn by the secondary side influences the power that primary side (has to) provide(s).
The voltage of the primary side influences the voltage of the secondary side.
So the "primary's action is dependent from secondary".

The current drawn on the secondary side is 'provided' by the primary side. This current on the primary side causes a voltage drop in R1, reducing the voltage V1 at the primary side of the transformer. The secondary voltage changes proportionally (to the transformer ratio) with the reduced voltage V1.

You can refer the impedance at the secondary side of the transformer to the primary side of the transformer. For the load applies $$ R_2 = \frac{V_{2}}{I_{2}} $$ If the load would be on the primary side, it would be $$ R_1 = \frac{V_{1}}{I_{1}} $$.

Using $$ \frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}} $$ and $$ \frac{I_{1}}{I_{2}} = \frac{N_{2}}{N_{1}} $$ you can make define the equivilant load at the primary side being

$$ R_1 = \frac{V_{1}}{I_{1}} = \frac{ V_{2} \cdot \frac{N_{1}}{N_{2}} }{ I_{2} \cdot \frac{N_{2}}{N_{1}} } = \frac{ V_{2}}{I_{2}} \cdot \bigg( \frac{N_{1}}{N_{2}} \bigg)^2 = R2 \cdot \bigg( \frac{N_{1}}{N_{2}} \bigg)^2 $$

enter image description here

From the equivalent circuit it is easier to see how the load current causes a voltage drop in R1 which lowers the voltage V1.

  • \$\begingroup\$ But I1/I2 is not N2/N1 see my third plot i.stack.imgur.com/aaMMG.png (N2/N1 is is fixed to 0.1) \$\endgroup\$ – cm64 Mar 14 '19 at 12:11
  • \$\begingroup\$ That has to do with the inductance of L1 and L2. Note that with an open secondary circuit, the peak of the current in L1 is 1.23 mA instead of 1V/800 Ohm = 1.2mA due to the impendance of L1. I think you should also replace the transformer by an equivalent circuit, but I don't know how. \$\endgroup\$ – Huisman Mar 14 '19 at 12:24
  • \$\begingroup\$ If the inductances are very large then I1/I2 approaches to N2/N1. Does the ideal transformer definition include that it has to have infinitely big inductance? \$\endgroup\$ – cm64 Mar 14 '19 at 12:28
  • \$\begingroup\$ It would be really nice to tweak your model and include the effect of the inductances. \$\endgroup\$ – cm64 Mar 14 '19 at 12:29

why does the current ratio is very different and varies with loading?

There are two currents flowing into L1 when a secondary load is connected: -

  • The primary magnetization current - dependent only on applied primary voltage and the inductance of the primary (not load dependent)
  • The primary referred secondary load current - this is the secondary current referred (via the turns ratio) to the primary

Both currents are 90 degrees out of phase when the secondary load is resistive so you have to use Pythagoras's theorem to calculate an effective value.

This means that you cannot attribute a simple ratio for the secondary and primary currents based on the turns ratio.

Here's a full equivalent circuit of a low frequency transformer: -

enter image description here

Isn't V1/V2 = I2/I1 always?

Only if you ignore the primary magnetization current and. for small transformers this cannot be ignored.

Why does the secondary voltage decrease when loaded?

Because the primary voltage decreases. This is because of the 800 ohm resistor that series feeds the primary.

  • \$\begingroup\$ If the transformer were "ideal", would that mean the inductances L1 and L2 would be infinitely large so that we can neglect the magnetization current? Because when I set L1 and L2 to very large values like 10000H to 100H the current ratios are equal to winding turn ratios. See the other answer he has a very nice equivalent circuit but neglects the inductance effect. Would be great if it could be modified to add only the inductor effetcs to that circuit. \$\endgroup\$ – cm64 Mar 14 '19 at 13:23

Reason why the secondary voltage drops when its circuit is closed: before the secondary circuit is closed, its voltage is determined only by the magnetic field created by the primary current (mutual inductance). When the secondary circuit is closed, its voltage now is determined by both the magnetic field caused by the primary and also by the magnetic field of the secondary current (self inductance), both being in opposite direction. this is why you end up with a lower secondary voltage. The same thing happens in the primary winding: these magnetic fields are in opposite directions in the primary windings too when the secondary circuit is closed, and this is what causes the voltage across the primary windings to decrease. This can be shown mathematically if you write kirchhoff's voltage law for both sides of the transformer by taking into account the mutual inductance of both windings. Thanks so much for posting the results of your simulation! I found them very useful.


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