Electronic – load impedance

A constant power load varies it's impedance on change of input voltage to keep the power constant. A constant impedance load is simply a load that presents an unchanging impedance, like a resistor. An L-Pad is used to change speaker output level whilst maintaining a constant impedance load to the amplifier.

A good example of a constant power load is a switching regulator. Since this has to maintain it's power into it's load, it must draw the same power from it's source even if the source changes voltage.
This is also an example of a negative impedance because in order to maintain the output power, if the voltage in drops, the current must rise (opposite to a standard resistor where the current and voltage rise/fall with each other)

Here is an example circuit, made from a LT1377 boost switching regulator:

Here is the simulation:

The input voltage V(in) (blue trace) starts off at 4V, and gradually rises up to 10V. We can see the power (red trace) stays constant at ~1W over a change of 6V at the input (it's not perfect as it's meant to represent "real life" and not 100% efficient, but it's pretty close)
We can also see the dynamic negative resistance characteristic (green trace) which is due to the input current falling as the voltage rises. Tt falls from ~300mA to ~120mA over the voltage rise from 4V to 10V - don't be confused by the minus sign, that's just the direction of measurement in LTSpice.
The dynamic resistance slope can be roughly calculated by (4V - 10V) / (300mA - 120mA) = -33.3Ω. Looking it another way, 6V / -33.3Ω = -180mA.

You should possibly consider looking at the transformer in two ways; one without a load and one with a load on the secondary.

Without a load on the secondary, the transformer is just an inductor and if you have components (such as L1 and R1) in series with the primary, the voltage developed on the primary will not be the full AC amount from your generator. It's a simple case of calculating the impedances and volt-drops. This is with the secondary unconnected remember.

The primary has inductance like any other coil but, for a transformer to more effective, it is desirable for the primary's self inductance to be high in power applications. If you looked at how much current flowed into the primary (secondary open circuit) you would find that the current was small compared to when driving a load on the secondary and it may have an inductance of several henries.

With 10 henries inductance, at 50Hz the impedance is 3142 ohms and from 230VAC would take a current of 73mA - that current through R1 (10 ohm) hardly drops any voltage.

It's a different matter when there is a load on the secondary. If the turns ratio is 1:1 and you have 100 ohms on the secondary, it is reasonable to argue that the impedance presented to the primary circuit is also 100 ohms. This assumes power out is close to power in. In fact the impedance relationship between primary and secondary is related to turns ratio squared. For instance if it is a 10:1 step-down transformer with a load of 100 ohms, the equivalent impedance at the primary is 10k ohms i.e. 10 x 10 x 100.

In summary, for a power transformer, you'd like the primary inductance to be infinite but that is impractical so you live with something that doesn't take too much current when the secondary is open circuit. The off-load current that flows is real current taken from the AC power and if everyone had low-impedance transformers the electricity companies would be supplying a load of current that doesn't get them revenue. This is a slight exaggeration but not far off the truth. On industrial sites power factor correction is used to minimize this effect but that's a whole new story!

And if your transformer primary was 100 ohm impedance you'd be seeing something less than half your AC voltage applied. If R1 was zero then you'd see exactly half.

As regards saturation I've shown the equivalent circuit of a transformer below. Note that saturation is caused by the current flowing through the magnetizing inductor which is nothing to do with load current: -

Here is a good document from Elliott Sound Products and please note what it says about maximum flux density therefore saturation:

Why doesn't the core saturate more under load conditions? Imagine two coils sharing the same magnetic core. Ignore magnetization currents and losses. The primary is 100 turns and the secondary is 10 turns. If the secondary load current is 10A, the primary current must be 1A and the ampere-turns is therefore the same on both coils. Are these ampere-turns additive or subtractive? They are subtractive and this can easily be seen with dot notation....

If current is flowing into the dot on the primary, current is flowing out of the dot on the secondary and this produces opposing fluxes in the magnetic material. When you think about this you have to be consistent and use the right-hand rule to see that the two fluxes oppose and cancel.

Because the dots are at the top on both coils, they are wound in the same direction and the currents are flowing in (primary) and out (secondary) therefore due to the RH rule the fluxes (due to ampere-turns) are cancelled.