Electronic – Multiple transformers to generate high voltage

All it takes to make high voltage AC is low voltage AC and a transformer.

To make high voltage DC, you have to chop it into (what else) AC, run it through a transformer, and then rectify it back to DC. Quite a bit more hardware is necessary.

So, with mass produced products, there's a strong economic bias to use AC high voltage, so that's what you'll see, unless there's a compelling reason that the high voltage needs to be DC.

A transformer essentially converts between voltage and current using a magnetic field. Because it is a conversion, then if the process is 100% efficient, then the output power and input power must be equal:

$$P_{in} = P_{out}$$

If they are not equal, then either you are losing energy in the transformer (inefficiencies), or gaining energy (perpetual motion anyone?!). The former can happen, the latter can't.

So based on this, what can we say about the voltage and current? Well, we know that:

$$P = I V$$

So: $$I_{in}V_{in} = I_{out}V_{out}$$

Lets say that you have a step up transformer with 10 turns on the primary and 50 turns on the secondary. This means you have a turns ratio of:

$$n = \frac{50}{10} = 5$$

So that means that the voltage will be increased by a factor of 5 (\$V_{out} = 5\times V_{in}\$). So what happens to the current?

$$I_{in}V_{in} = 5V_{in}\times(I_{out})$$

In order for both sides of that to stay equal (can't get energy from nothing!), then the current must be divided by 5. Basically you can say that:

$$I_{out} = \frac{1}{n}I_{in}\space\space\space\space\space\space\space\space\space V_{out} = nV_{in}$$

So what happens if you have a fixed load and change the number of turns? Lets do an example. We shall say that the input voltage is \$10\mathrm{V}\$, the transformer initially steps up by a factor of \$n=1\$, and then later by a factor of \$n=5\$. In both cases the output load is a \$2\Omega\$ resistor.

In the first case, your calculations are correct.

$$V_{out} = n\times V_{in} = 10\mathrm{V}$$ $$I_{out} = \frac{V_{out}}{R_L} = \frac{10}{2} = 5\mathrm{A}$$ $$I_{in} = n\times I_{out} = 5\mathrm{A}$$

Now lets go for \$n=5\$.

$$V_{out} = n\times V_{in} = 5\times 10 = 50\mathrm{V}$$ $$I_{out} = \frac{V_{out}}{R_L} = \frac{50}{2} = 25\mathrm{A}$$

Great, these match what you are saying. But this is where everything changes. We do the last step of the calculation:

$$I_{in} = n\times I_{out} = 5\times 25 = 125\mathrm{A}$$

Ahh, there we go. Notice that the input current goes up, quite considerably. This makes the scales balance so to speak - power in goes up in order to cope with the large power requirements of the load.