8.2.1. Idling behaviour Idling features

A transformer idles where mains voltage U₁ remains applied to the primary side whilst no consumer is connected to the secondary side (Z_a) (Figures 125/126).

Primary circuit	U1 applies
	I0 flows (idling current)
Secondary circuit	Za = ¥
	I2 = 0
	U2 = U20

Idling current

The applied voltage U drives the idling current I₀. This is needed to establish the magnetic field Iµ. This lags behind the voltage U₁.

Figure 127 - Indicator image for idling operation

1 Iron loss current I_Fe

The phase position of the idling current I₀ to voltage U₁ can be determined according to the circuitry of Figure 128.

Figure 128 - Circuitry to determine idling losses

1 Rated voltage

The value of idling current I₀ is between 2 and 5 per cent of idling current in big transformers and up to 15 per cent in smaller transformers.

No-load curve

The idling curve I = f (U₁) in Figure 129 indicates that no-load current I₀ increases proportionally to the input voltage U₁. No-load current increases markedly over and beyond the input rated speed U_1n. It can, moreover, even attain values greater than the rated current.

Figure 129 - Idling curve of a transformer I₀ = f (U₁)

Transformers shall not be driven by voltages greater than the rated voltage.

Idling losses (iron losses)

The active power derived from the circuit in Figure 128 can only be transformed into heat in the input winding and iron core as no current flows into the secondary winding during idling. The active power P₀, which is converted into heat in the iron core, is made up of eddy current and hyteresis loss.

The following example shows that the iron losses almost always arise during idling.

Example:

The following idling values were measured in a transformer:

U_1n = 220 V; I₀ = 0.5 A; P₀ = 40 W; R₁ = 3.

What percentage of winding losses are contained in idling power?

Solution:

P₀ = P_VFe + P_VW

P_VW = 0.75 W

P_VFe = P₀ - P_VW = 40 W - 0.75 W = 39.25 W

Thus, the power loss determined during idling is an iron loss.

Iron losses are determined during no-load operation and are independent of load.

8.2.2. Short-circuit behaviour

Short-circuit curves

Secondary current I₂ increases if load resistance is decreased. Where Z_a = 0 the transformer has been short-circuited.

Primary circuit	U1 is applied
	IK flows
Secondary circuit	Za = 0
	U2 = 0

Short-circuit voltage

The short-circuited transformer can be replaced by resistor Z₁ which corresponds to the transformer internal resistor.

Figure 130 - Short-circuited transformer

1 Short-circuit current I_K

Figure 131 depicts the commensurate duplicate circuit diagram.

Figure 131 - Duplicate circuit diagram for short circuit run

1 Ohmic winding resistance, 2 Scattered reactance (is made up of the scatter flow of the input and output coils), 3 Inner resistance of the transformer (impedance)

During a short-circuit attempt (Figure 132) the input voltage given a short-circuited output winding is increased until primary and secondary nominal currents flow. The voltage applied to the input side is then the short-circuit voltage U_K.

Figure 132 - Circuitry to determine short-circuit losses

1 Short circuit voltage

The short-circuit voltage is the overall voltage decrease of a transformer during rated loading.

The relative short-circuit voltage U_K in % is determined by the following equation:

The relative short-circuit voltage is, on average, 2 to 10% of input rated voltage (U_1n) in mains transformers.

Short-circuit losses (winding losses)

In the short-circuit experiment (Figure 132) a power meter indicates short-circuit losses as the primary and secondary rated currents generate winding losses. The iron core is only slightly magnetised by the applied short-circuit voltage (U_K U₁).

The winding losses can be metered during the short-circuit experiment. They are dependent on the load current (P_VW = I₂ R).

8.2.3. Loaded voltage behaviour

In contrast to operational idling, during loading the secondary circuit is closed through an external resistance Z_a (Figure 126). Secondary current I₂ flows. According to the energy conservation law the transformer must also take up commensurate primary power, thus a primary current I₁ also flows.

Primary circuit	U1 is applied
	I1 > I0
Secondary circuit	Za < ¥
	I2 > 0
	U2 ¹ U20

Voltage curve U₂ = f (I₂)

As the curve in Figure 133 shows, terminal voltage U₂ decreases during loading.

Figure 133 - Voltage behaviour during loaded operation U₂ = f (I)₂

1 U_K small, 2 U_K big

Figure 134 depicts the duplicate circuit diagram for the loaded transformer.

Figure 134 - Duplicate circuit for the loaded transformer with a transformation ratio = 1:1

The duplicate circuit diagram corresponds to a transformer with a transformation ratio

As rated current flows the short-circuit voltage U_K decreases at the internal transformer resistance Z_i as a result of which the terminal voltage U₂ declines by the power decrease of the short-circuit voltage U_K.

Transformers with considerable short-circuit voltage U_K have powerful internal resistors, that is to say pronounced voltage changes as load alters.

UK = 2...10%	minimal voltage losses
	voltage-rigid behaviour
UK = 20...50%	considerable voltage losses
	voltage-flexible behaviour

Example:

A 220/42 V transformer has a short-circuit voltage of 10%.

How great is the voltage change between idling and rated current loading?

Solution:

Output voltages at differing loads

Given differing loads with ohmic, inductive or capacitive external resistance gives rise to the dependence of output voltage on load current as shown in Figure 135.

Figure 135 - Secondary terminal voltage depending on the degree and nature of loading

1 Idling, 2 Rated load

Given capacitive load, the output voltage may even be greater than no-load voltage.

The output voltage of a transformer depends on the

- degree of load current I₂
- the magnitude of relative short-circuit voltage
- the nature of the load (ohmic, inductive or capacitive).

8.2.4. Efficiency

Efficiency represents the ratio of power output to power input.

Efficiency is determined by the separate loss indication.

P_in = P_out + P_VFe + P_VW

Pout	rated power of the transformer given
	cos j = 1 (resistive load)
PVW	metered winding losses in short-circuit trial
PVFe	metered iron losses in no-load trial

The following equation applies for determining efficiency for partially inductive or capacitive load:

Example:

Given:

S_n = 100 kVA
P_VFe = 570 W
P_VW = 2.1 kW
cos j = 0.85

Sought: h

Solution: