Introduction to Industrial Brakes

Electromechanical Spring Brakes: A Crucial Role in Various Industries

Electromechanical spring brakes play an essential role in many industries, ranging from material handling and wind turbines to industrial carts and lifting applications.

Choosing the Right Brake

Selecting the right brake depends on several factors. It is crucial to understand the application parameters, such as the type of load, motor rotation speed, inertia, and the maximum permissible braking time. Additionally, the operating environment (temperature, humidity, presence of dust or corrosive atmosphere) and the specific task requirements can be decisive. For instance, rotation motor brakes for tower cranes must remain permanently released at the end of the work shift (“weathervane mode”). This requirement, unique to such applications, necessitates brakes compatible with appropriate release accessories.

This article will demonstrate how to properly size a brake system.

DC or AC Brakes?

Electromagnetic brakes can be powered by direct current (DC) or alternating current (AC). Both types have their own advantages and disadvantages.

AC Brakes

AC electromagnetic brakes can connect directly to the motor’s power supply and are appreciated for their high dynamics. This means they release forcefully and quickly. This characteristic arises because the electrical impedance is significantly lower when the brake is closed than when it is open. This difference results in a high initial current surge, creating a strong magnetic field and, consequently, a high initial attraction force.

However, this desirable characteristic comes with a drawback: if the armature does not close due to incorrect adjustment or excessive brake disc wear, the AC brake may burn out within moments.

DC Brakes

DC brakes, on the other hand, do not suffer from this issue. However, they require a dedicated power supply or a current rectifier. Modern rectifiers with initial overexcitation can enable DC brakes to approach, or even match, the performance of AC brakes.

How to Choose the Right Brake

To choose the most suitable brake, it is essential to know the following data:

• Initial motor speed (N1), in rpm.
• Final motor speed (N2), in rpm, which is typically zero if a full stop is desired.
• Total inertia (JL), in kg·m².
• Required braking time (t3,req), in seconds.
• Brake rise time (t12), in seconds, defined as the time for the brake torque to rise from 10% to 90% of its nominal value.
• Load torque (ML), in Nm, applied to the motor shaft.
• Number of braking operations per hour.

Using this information, the braking torque (MK) can be calculated as:

MK = (JL × (N1 – N2)) / (t3,req – (t12 / 2)) ± ML

Where:

• JL is in kg·m²,
• N1 and N2 are in rpm,
• t3,req and t12 are in seconds, and
• ML is in Nm.

Simplifying the Formula

Typically:

• N1 is the nominal motor speed, e.g., 3000 rpm for a 2-pole motor or 1500 rpm for a 4-pole motor.
• N2 is the desired final speed, usually 0 for a full stop.
• t3,req is chosen by the user.
• t12, intrinsic to the brake, may not be known. It is often approximated as 5% of t3,req.

Thus, the formula simplifies to:

MK = (JL × N1) / (9.55 × (t3,req × 0.95)) ± ML

The load torque (ML) is added if it aids movement (e.g., descending load) or subtracted if it opposes movement (e.g., lifting load).

Example Calculation

Let us calculate the braking torque for a motor with the following parameters:

• Initial speed (N1): 1500 rpm
• Inertia (JL): 0.1 kg·m²
• Braking time (t3,req): 0.5 seconds
• Load torque (ML): 10 Nm (aiding movement, e.g., descending load)

Case 1: With Known t12

Assuming t12 is 25 milliseconds (0.025 seconds): MK = (0.1 × 1500) / (9.55 × (0.5 – (0.025 / 2))) + 10
MK = 32.2 + 10 = 42.2 Nm

Case 2: Without Known t12

Using the simplified formula: MK = (0.1 × 1500) / (9.55 × (0.5 × 0.95)) + 10
MK = 33 + 10 = 43 Nm

Finally, apply a safety factor of 2: MK_final = 43 × 2 = 86 Nm.

Sizing with Motor Power

If application details are unavailable, you can estimate the braking torque using motor power (P):

MK = (9550 × P) / N1

Where:

• P is motor power in kW,
• N1 is motor speed in rpm.

For a motor with P = 5.5 kW and N1 = 1500 rpm: MK = (9550 × 5.5) / 1500 = 35 Nm

After applying a safety factor of 2: MK_final = 35 × 2 = 70 Nm.

This approach ensures accurate sizing of electromechanical spring brakes for various applications.