Technical Article

# An Overview of Buck-Boost Transformer Sizing

June 28, 2021 by Alex Roderick

## There are two common methods of sizing a transformer. The first method is to determine the load power and calculate the required size of the buck-boost transformer. The second method is to determine the load current and the buck or boost voltage and calculate the required size of the buck-boost transformer.

One method of sizing a transformer is to calculate the load power in kVA. This method can be used when the buck-boost transformer is wired as an isolation transformer, and the entire load goes through the transformer secondary. See Figure 1. In the case of a single-phase motor, the power is the product of the current and the line voltage. The power is calculated as follows:

$$P=\frac{I\times E}{1000}$$

where

P = power (in kVA)

I = current (in A)

E = voltage (in V)

Figure 1. Standard two-winding transformers can be sized by determining the load power.

A motor nameplate gives the current at full load. For example, a 5 HP, 240 V single-phase motor may draw a full load current of 28 A. The load power is calculated as follows:

$$P=\frac{I\times E}{1000}=\frac{28\times 240}{1000}=6.72\text{ }kVA$$

Note

Buck-boost transformers rated for 60 Hz should be derated to operate at 50 Hz.

The load power is 6.72 kVA, and the transformer secondary needs to be rated to carry that load. The user should select a transformer slightly larger than the expected maximum load. Manufacturers have selection tables to help determine the appropriate transformer size.

The alternative sizing method for buck-boost transformers is to calculate the power, in kVA, from the load current. This method can be used when the buck-boost transformer is wired as an autotransformer. See Figure 2. In the case of a single-phase motor, the power is the product of the current and the voltage buck or boost of the transformer. The power is calculated as follows:

$$P=\frac{I\times E}{1000}$$

where

P = power (in kVA)

I = current (in A)

E = voltage (in V)

#### Figure 2. Buck-boost transformers wired as autotransformers can be sized by determining the load current and multiplying by the buck or boost voltage.

A motor nameplate gives the current at full load. For example, a 5 HP, 240 V, single-phase motor may draw a full load current of 28 A. When the line voltage is 208 V, the buck-boost transformer can be used to raise the voltage to 236 V. This is a value that can be used by the 240 V motor. With the addition of the boost transformer on the line, the primary of the boost is across the line, and the secondary is in series with the load across the line. The circuit current is now the total of the load and the transformer primary. The load power is calculated as follows:

$$P=\frac{I\times E}{1000}=\frac{28\times 32}{1000}=0.896\text{ }kVA$$

This calculation indicates that a 1 kVA buck-boost transformer can be added to a 208 V system to boost the voltage to 236 V to supply 28 A to a 5 HP motor.

## Boosting at the End of a Line

Many loads at the end of a long line suffer from low voltage during peak load periods. The resistance of the lines drops the line voltage when the current flow is high at peak load.

A buck-boost transformer can be used to boost the voltage at the end of a power line where there is a voltage drop because of the length of the line. For example, a shelter (a small structure used for protection from the weather) is located a sizable distance from the service and contains a 120 V, 2 kW heater, and a 100 W lamp (see Figure 3). With the light and heater OFF, the measured voltage at the shelter is near the nominal 120 V service.

##### Figure 3. The voltage at the end of a long line drops because of the line resistance when a large load is connected.

There is a significant voltage drop as the current flows through the long line to the loads. With the light and heater ON, the measured voltage at the shelter drops considerably, perhaps to 95 V. This is a drop of 25 V across the power line resistance. The heater output and the candlepower of the light are significantly reduced. At 95 V, the voltage is only 79% of the nominal value. Therefore, the current is also only 79% nominal. The power dissipated is proportional to the square of the current. Therefore, the heater output is reduced to about 1250 W, greatly reducing its effectiveness on a cold winter day. The lamp load is reduced to about 63 W.

Transformer Sizing

A buck-boost transformer can be used to boost the voltage at the end of the line. In order to size the buck-boost transformer, the power required for the transformer can be calculated from the load current. The current can be calculated by dividing the power by the applied voltage as follows:

$$I=\frac{P}{E}$$

where

I = current (in A)

P = power (in W)

E = voltage (in V)

For the 100 W lamp, the current is calculated as follows:

$$I=\frac{P}{E}=\frac{100}{120}=0.833A$$

For the 2000 W heater, the current is calculated as follows:

$$I=\frac{P}{E}=\frac{2000}{120}=16.667A$$

The total current, 17.5 A, is the sum of the individual load currents. The power required in the secondary of the buck-boost transformer is calculated by multiplying the current by the voltage rise in the transformer as follows:

$$P=\frac{I\times E}{1000}$$

where

P = power (in kVA)

I = current (in A)

E = voltage boost or buck (in V)

If the buck-boost transformer has a 7.5:1 turns ratio, the transformer can be wired to boost the voltage by 26.7%. At 95 V, this is 25.3 V, and at 120 V, this is 32 V. This is accomplished by placing the primary coils in parallel and across the line and by placing the secondary 16 V coils in series with the load across the line. See Figure 4. The buck-boost transformer size can be calculated as follows:

$$P=\frac{I\times E}{1000}=\frac{17.5\times 32}{1000}=0.560\text{ }kVA$$

A buck-boost transformer larger than 0.560 kVA can be installed to boost the voltage to the required level. When both loads are turned on, the voltage at the loads is 120.3 V. The lamp and the heater operate normally.

##### Figure 4. Buck-boost transformers can be used to boost the voltage at the end of a long line.

After the buck-boost transformer is installed at the shelter, the voltage can be measured. With the loads turned off, there is no current flow and no voltage drop on the power line to the shelter. The voltage will be boosted by the same percentage as before.

The lamp is a small part of the total load at the shelter. If the lamp is turned on, it draws only about 5% of the current that the heater draws, so the voltage drop across the line resistance is only about 5% of the 25 V drop shown previously. See Figure 5. Therefore, the lamp will see about 150 V instead of the 120 V it needs to operate efficiently. The current rises from 0.833 A to just over 1 A, and the power dissipated by the lamp rises from 100 W to about 150 W. This increase in voltage results in a much shorter life for the lamp as the filament will burn out faster. This is why buck-boost transformers are not recommended for boosting when there is a varying load.

## Wiring Alternatives

There are two common wiring alternatives that can reduce the problems with shortened lamp life. First, the lamp can be connected ahead of the buck-boost transformer. When the heater is not operating, the lamp operates normally. When the heater is operating, the lamp operates with a lowered voltage. The shelter will not have normal illumination, but the lamp will have a normal life.

##### Figure 6. A buck-boost transformer can be wired with a center-tapped secondary to provide intermediate voltages for some of the loads.

A second alternative is to center tap the buck-boost transformer between the two 16 V coils and use a 130 V 100 W bulb. See Figure 6. The current drawn by this lamp on 130 V is 0.769 A. By tying the lamp circuit between the coils, the voltage available to the bulb is 107 V when the circuit is fully loaded, which provides 68 W. With the heater OFF, the lamp sees approximately 134 V and provides 109 W. This is within the design parameters of the lamp bulb.