I want to calculate the weights that can be added to the dc motor for loading arrangement. So, how can I calculate this? I am using the motor with 24 volts DC, 6000 rpm No-Load current= 600 mA, Stall current= 30A max. What is the maximum weight that can be added?

• Is 6000rpm the no-load speed? – Bruce Abbott May 11 '17 at 9:48
• If you mean the mechanical load then this is a mechanical design question not an electronic one (though I'm sure some here will have a go at answering it :-) ) – TonyM May 11 '17 at 10:16
• If you are asking, how much weight can this motor lift, there is no answer. You have not specified the mechanical linkage to the weights. In general, a motor can lift almost any weight with a sufficient gear ratio, but the tradeoff is speed. – WhatRoughBeast May 11 '17 at 10:34
• The information you've given isn't sufficient without making assumptions as to what sort of motor it is, what the gearing is (if any), or what distance the load is attached from the centre of the shaft. – Simon B May 11 '17 at 12:30
• Mechanical Power = Torque x Rotation rate. Electric power = Mechanical power/efficiency. [Approx formula (2% wrong) which is VERY useful. Watts = (Newton.metre torque) x RPM. – Russell McMahon May 11 '17 at 13:04

Unless specifically identified as a brushless DC motor, a DC motor is usually assumed to be a a classical DC motor with a commutator. A DC motor of the size described is assumed to be a permanent-magnet DC (PMDC) motor unless information about a wound field is provided. The steady-state performance of a PMDC motor can be analyzed using the equivalent circuit shown below.

Va = Supply Voltage (24 volts) Ra = Armature resistance including brushes E - Back emf = N / Ks; N = Speed (RPM); Ks = Speed constant (RPM/Volt) Ia = Armature current (Amps)

V = Ia x Ra + E

At Stall: RPM = 0, E = 0, Va = Ia x Ra therefore Ra = 0.8 ohms

At No-Load: Va = Ia x Ra + E therefore E = 23.52 volts

N = 6000 RPM = Ea/Ks; Ks = 6000 / 23.52 = 255.1 RPM/volt

Power

Electrical Input Power = Mechanical Output Power + Losses

Electrical Input Power (Watts) = V x Ia

Mechanical Output Power = Load Torque (Newton Meters) X Speed (RPM) x 0.1047

At No-Load: Torque = 0 so Output Power = 0 Input power = 0.6 x 24 = 14.4 watts Total losses = 14.4 watts. Losses in Ra = 0.6^2 x 0.8 = 0.288 watts. Total losses minus losses in Ra = mechanical losses = 14.112 watts. Assume the mechanical losses are due to bearing and brush friction and do not change with speed. Total friction losses = 14.112 W = Loss friction torque x 6000 x 0.1047. Total friction loss torque Mf = 0.0225 N-M.

Armature Power and Torque: Mechanical power developed in the armature before mechanical losses is P = Ia x E = Ia x N / Ks. The mechanical power developed is also the armature torque Ma x N x 0.1047 therefore 0.1047 x Ma x N = Ia x N / Ks.

Ma = Ia / (0.1047 x Ks) or

Ma = Ia x Km where Km = 1 / (0.1047 x 255.1) = 0.0374 N-m/amp

Stall torque is Km x stall current = 30 x 0.0374 = 1.12 N-m

The shaft torque, Ms, available to drive a load is the armature torque Ma, minus friction torque Mf. Ms = Ma - Mf.

In order to "calculate the weights that can be added to the dc motor for loading arrangement," it is necessary to determine the torque required from the motor shaft to move the weight in the required manner at the required speed. The torque available from the motor at any speed can be calculated from the motor analysis shown above. The continuous torque that the motor can provide without overheating can only be determined by testing the motor with the knowledge of the maximum allowable motor temperature. That information is normally provided by the motor manufacturer.