why does a step up transformer cause lower currents?

Question

The usual answer I hear is because power needs to be conserved, but I dont find that answer satisfying. Normally if I were to double a supply voltage in a circuit, then I would expect twice the current (assuming its a linear circuit). But if i have a 2x transformer (which AFAIK is made of inductors which are linear elements), then the output of the transformer will have twice the voltage and half the current. What exactly is going on electromagnetically that allows a transformer to violate ohms law and give lower currents with higher voltage?

Answer 1

Ohm's law is not violated because a transformer is not a resistor. The transformer simply changes the source's view of the load impedance. So if you have a 1:2 transformer connected to a source and a load resistor, then the source side would look like one quarter of the load resistance.

Transformers don't work at DC and hence understanding what's going on requires a bit more math and physics. Let's take a simple example circuit:

^{simulate this circuit – Schematic created using CircuitLab}

In this case, we have a source (V1), a load (R1) and a 1:2 step-up transformer (N = 0.5). Some current will flow on the primary side of the transformer, \$i_p(t)\$. The current in the primary will induce a magnetic field in the transformer core that's proportional to the current and to the number of turns in the primary:

$$B_p(t) = \mu_0 k_1 N_p i_p(t)$$

Don't worry about the \$k_n\$ constants, they're dependent on the construction of the transformer and they'll cancel out at the end, assuming the transformer is ideal. This results in a magnetic flux on the secondary:

$$\phi_s(t) = k_2 B_p(t) = \mu_0 k_1 k_2 N_p i_p(t)$$

The oscillating magnetic flux will induce a voltage in the secondary of the transformer that's proportional to the derivative of the magnetic flux and to the number of turns in the secondary:

$$v_s(t) = N_s \frac{d}{dt} \phi_s(t) = \mu_0 k_1 k_2 N_p N_s \frac{d}{dt}i_p(t)$$

The same thing can be calculated in the other direction:

$$v_p(t) = N_p \frac{d}{dt} \phi_p(t) = \mu_0 k_1 k_2 N_p N_s \frac{d}{dt}i_s(t)$$

If we assume an ideal transformer with 100% coupling between the primary and secondary, then \$\phi_s(t) = \phi_p(t)\$. If \$\phi_s(t) = \phi_p(t)\$, then \$\frac{d}{dt}\phi_s(t) = \frac{d}{dt}\phi_p(t)\$. This means we can write:

$$\phi_s(t) = \phi_p(t) = \mu_0 k_1 k_2 N_p i_p(t) = \mu_0 k_1 k_2 N_s i_s(t)$$

and

$$\frac{d}{dt}\phi_s(t) = \frac{d}{dt}\phi_p(t) = \frac{v_p(t)}{N_p} = \frac{v_s(t)}{N_s}$$

from which you can write

$$i_s(t) = \frac{N_p}{N_s}i_p(t)$$

and

$$v_s(t) = \frac{N_s}{N_p}v_p(t)$$

If \$N_s = 2 N_p\$ as in the example circuit, then \$v_s(t) = 2v_p(t)\$ and \$i_s(t) = \frac{1}{2}i_p(t)\$. This makes sense from a standpoint of conservation of energy; \$P = v_s i_s = 2 v_p \frac{1}{2} i_p = v_p i_p\$.

Ohm's law on the load resistor indicates that \$v_s(t) = 100\ \Omega\cdot\ i_s(t)\$. By plugging in both current and voltage relations, we get \$v_p(t) = 25\ \Omega\cdot\ i_p(t)\$. In effect, the transformer makes the 100 Ω resistor 'look like' a 25 Ω resistor from the standpoint of the source. Again, this makes sense - half the voltage and double the current is 1/4 the resistance. In general a transformer will 'transform' the load impedance by a factor of \$N_p^2/N_s^2\$

Answer 2

The usual answer I hear is because power needs to be conserved, ...

This is correct.

Normally if I were to double a supply voltage in a circuit, then I would expect twice the current (assuming its a linear circuit).

That would be correct if you didn't change the load resistance.

But if i have a 2x transformer ... then the output of the transformer will have twice the voltage and half the current.

This is true for a given output power.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) Reference setup. (b) Doubling the voltage quadruples the power into the same resistance. (c) By quadrupling the load resistance the power is the same for double the voltage.

To maintain output power for a doubling of output voltage the load resistance would be increased by a factor of 4 (the step-up ratio squared).

What exactly is going on electromagnetically that allows a transformer to violate ohms law and give lower currents with higher voltage?

Nothing.

Answer 3

Normally if I were to double a supply voltage in a circuit, then I would expect twice the current (assuming its a linear circuit).

Yes, but only if the load resistance stays the same.

But if i have a 2x transformer (which AFAIK is made of inductors which are linear elements), then the output of the transformer will have twice the voltage and half the current.

Yes, but only if you increase the resistance by a factor of 4.

What exactly is going on electromagnetically that allows a transformer to violate ohms law and give lower currents with higher voltage?

Nothing, because it's not. That is, Ohm's law holds. It's just that, with a higher voltage, for the same power you increase the resistance so as to get less current. Once you've done that, the transformer is supplying the same power but at less current. For the same size wiring, with the load current halved and the load resistance being the dominant term in setting the system current, the power lost in the wiring (assuming it is unchanged) drops by a factor of 4 and the system overall becomes more efficient.

Answer 4

Nothing to do with Ohms Law: the secondary side of the transformer is a separate circuit, electrically unconnected to the first.

In this case the relevant equation is P=VI combined with conservation of energy. The output power of the transformer cannot exceed the input power.

Answer 5

The behavior of a transformer requires a negative-feedback behavior inside the core, where primary-flux is almost perfectly cancelled by the secondary-flux.

If secondary currents increase, the direction of secondary flux change causes a drop in total summed core flux, and the energy source driving the primary is allowed to provide more current into the primary and thus into the secondary.

KEY to this behavior is a very small resistance between the energy source and the ideal primary winding mathematics model.

With FLUX defining the transformer behavior, a doubling of turns in either primary or in secondary is all that is needed to permit a precise halving of the current in that winding.

Answer 6

We call this type of circuit a magnetic circuit. It's equation is similar to Ohms law.

F=Φ*R

Where F = magnetomotive force, Φ = magnetic flux, R = magnetic reluctance

Magnetic flux is analogous to current.

Reluctance is analogous to resistance.

Magnetomotive force is analogous to electromotive force (voltage).

From Ampere's law we know that an inductor will produce currents perpendicular to the magnetic field lines. The strength of the current will be proportional to the number of coils in the solenoid.

From Faraday's law we know that a conductor in a changing magnetic field will produce a current in it. This current will be in the direction to oppose the forces of the changing magnetic field (Lentz's law).

So in any transformer, we have a conductor wrapped around each side of a metal core. They are not electrically connected, only magnetically coupled.

An alternating current in the coil on the primary side, induces a flow of magnetic flux circulating in the transformer core. The magnetic flux is a changing magnetic field which flows in the transformer core.

There is a another conductor wrapped around the secondary side of the transformer core. Since this secondary side conductor will be stationary in a changing magnetic field (induced by the primary side) it will produce a current in the secondary conductor. This current opposes the magnetic field which created it.

The primary and secondary "turns ratio" or number of coils or loops in the windings will control the voltage and current outputs relative to the input values.

As you can see above, the primary and secondary turns are directly proportional to the primary and secondary voltage.

However, the primary and secondary voltage (and turns) ratio is indirectly proportional to the current ratio.