# Why does a step up transformer draw more current than a step down?

I am trying to understand why a step up transformer draws more current than a step down transformer in the inverse configuration. The circuit below shows a 2:1 stepdown so I get 10V input, 5V output. Primary current is 2.5 mA and secondary is 5mA. I understand the concept where if voltage is halved, current is doubled. But let's say I inverse the circuit below and switch the ratio to a 1:2 step up transformer (1H to 4H). Now if I put in 10V I am getting out 20V. Primary current is 40 mA and secondary current is 20mA. Why is there a higher current draw with a step up transformer? I thought it might be because I'm increasing 10V instead of decreasing 5V so there's a higher difference but if I stepdown even more say 10V down to 1V I'm getting 100uA on primary 1mA on secondary so the current draw is even less with a higher turn ratio on a step down.

• So if your input (source) and output (load resistor) do not change, what impact would 20V have on the resistor as compared to 5V? Oct 28 at 20:51
• Because you kept R2 constant. Oct 28 at 21:35

A good way to understand physics issues is to look for a conservation law, in this case, Conservation of Energy.

When a system is lossless, or in the case of a transformer very nearly lossless so we can approximate it to lossless, then energy out equals energy in.

You are keeping the load on the secondary constant at 1 kΩ, and the input voltage to the primary constant at 10 V.

If a 2:1 transformer is configured to step up, you're supplying 20 V to the resistor, and it will consume 20^2/1k = 400 mW. The transformer primary will draw this from the 10 V supply, at 400m/10 = 40 mA.

If a 2:1 transformer is configured to step down, you're supplying 5 V to the resistor, and it will consume 5^2/1k = 25 mW. The transformer primary will draw this from the 10 V supply, at 25m/10 = 2.5 mA.

You have an impedance of 1k on the secondary at all times, and your current draw calculations are correct -- there is no inconsistency.

You may think of this situation as the transformer actually transforming the impedance. With an impedance of $$\R\$$ on the secondary, the transformer + its load "looks" like a $$\R/N^2\$$ impedance (where N is the step-up factor), assuming an ideal transformer.

In the step-down configuration the (transformer + resistor on secondary) effectively looks like a 10 V / 2.5 mA = 4k Ohm. This matches the above formula.

On the other hand, in step-up, the combination of transformer and load looks like a 250 ohm load (i.e. 1kohm * 0.5^2), again consistent with the above formula.

In fact, this mechanism is used practically to adjust unequal impedances to match them,

A transformer passes power (voltage times current), not just voltage or just current.

In a step-up transformer, if the secondary voltage is twice the primary voltage, the primary current must be twice the secondary current for the primary power to equal the secondary power. (In practice, the primary power will be a little greater than the secondary power due to losses in the transformer.)

It's because energy is conserved, energy in must equal energy out.

Power is just "energy per second". Energy in per second is the same as energy out per second, and consequently power in equals power out.

Power is calculated as the product of current and voltage, $$\P=I \times V\$$, which means that the product of current and voltage on the primary side must equal the product of currrent and voltage on the secondary side:

$$I_{PRIMARY} \times V_{PRIMARY} = I_{SECONDARY} \times V_{SECONDARY}$$

Physically, $$\I \times V\$$ energy delivered to the transformer from either side becomes a magnetic field in the transformer's core, but that field is shared by both coils, so circuitry on either side can then obtain energy from that field. However, neither side can draw more energy from the field than the field contains, and this is manifest in the above equation.

From that perspective, there really is no such thing as a primary or secondary coil, both sides can deliver energy to the shared magnetic field, and both can draw energy from it, like a joint bank account. The primary might deposit 10 1-dollar bills, and the secondary can then withdraw a single 10-dollar bill.

The accounting is perfect, and something is getting that energy when the field collapses. Usually we want that something to be the thing connected to the secondary, but the secondary can only receive whatever energy was delivered to the field in the first place.

Consequently $$\I \times V\$$ on both sides must be the same. If you are wondering what happens if there's nothing connected on the secondary side to receive the energy delivered via the primary side, think of the voltage/current source on the primary side as both depositing and withdrawing at the same time. Sure, it delivers energy, but it's immediately returned to the source as soon as the source stops pumping energy in.

• Not that it's wrong, but I feel as though most people asking this question are searching for a more satisfying answer. i.e. What happens in the transformer on the secondary to determine the current on the primary. Oct 29 at 2:40
• @DKNguyen, I think that's probably true, and OP will find nothing useful here! I also think that from a perspective of conservation of energy, which I find deeply satisfying, it's a little piece of the puzzle that somebody could benefit from. Oct 29 at 3:08

For an ideal transformer, power out equals power in. Do the math; compute the numbers.

Consider an ideal 1:1 transformer with a driving circuit and a fixed load. The voltage and current on the primary side of the transformer will have the same magnitudes as they have on the secondary side.

Now we replace the 1:1 transformer with an ideal 10:1 step down transformer, keeping the same fixed load. Both the output voltage and output current have dropped by a factor of 10 reducing the power in the load by a factor equal to the square of the turns ratio. To keep the input power equal to the output power the input current must also drop by a factor equal to the square of the turns ratio (100X) because the input voltage has remained unchanged. The impedance, as seen by the driving circuit, has gone up by a factor equal to the square of the turns ratio.

Next we replace the 10:1 step down transformer with an ideal 1:10 step up transformer, keeping the same fixed load. Both the output voltage and output current have increased by a factor of 10 (as compared to the outputs from the 1:1 transformer) increasing the power in the load by a factor equal to the square of the turns ratio (as compared to the case of the 1:1 transformer). To keep the input power equal to the output power the input current must also increase by a factor equal to the square of the turns ratio (100X) because the input voltage has remained unchanged. The impedance, as seen by the driving circuit, has gone down by a factor equal to the square of the turns ratio (as compared to the 1:1 transformer).

EDIT

If we want to drive an 8R load from an op amp which is supplying an 8V rms output signal then the op amp would be required to supply 8 Watts of power at a current of 1A rms which is obviously not within the capabilities of an ordinary op amp.

But, we could place a 12:1 step down transformer in between the op amp and the 8R load which, with the same amplitude of driving signal, would reduce the power in the load by a factor of 144 (turns ratio squared) to about 55mW and increase the impedance seen by the op amp also by a factor of 144 to 8*144= 1152R. The current drawn from the op amp would then be reduced by a factor of 144 to 6.9mA which is within the capabilities of a typical op amp.

By adding the transformer we are limiting the power in the load to a value which can reasonably supplied by a typical op amp. To increase the power above that which the op amp can supply we could remove the transformer and replace it with a class AB push-pull driver stage which can take the required extra power direct from the power rails.