# what is cut off current in dc motor?

This question is conceptual question. May it is easy for engineers to explain cut off current in DC Motor.

In our formula, we are finding motor torque,

$$T_{motor} = C_{motor} \times (I_{cut} - I_{ideal})$$ where,

$$T_{motor} [Nm] = motor \ torque$$ $$C_{motor} \ [Nm/A] = motor \ constant$$ $$I_{cut} \ [A] = Cut \ off \ current$$ $$I_{ideal} \ [A] = Ideal \ current$$

In this formula, We know motor constant, motor ideal current but, we don't know cutoff current.

$$\textbf{what is cut off current in dc motor?}$$

• Never heard of such a thing. Could you explain your concept a little more? – JRE May 31 at 11:17
• Which still doesn't explain the term "cutoff current." – JRE May 31 at 12:10
• I suspect that cut-off current is the current at which the torque is insufficient to overcome internal friction. That can be easily determine and may be published in the form of a current vs. output torque curve, but is unlikely to be listed in specifications. – Charles Cowie May 31 at 12:51

After 30 years working with motors, I've never heard that term.

For fixed field (permanent magnet or shunt-wound) brushed DC motors, and BLDC motors without speed control, the torque is fairly linearly related to the current, but is offset by the mechanical losses from ideal i.e. the line doesn't intersect the torque and current axes at zero. Normally the offset is expressed as a no-load current, or a friction torque - the intersects with the vertical and horizontal axes respectively. The cutoff current you mention looks to be that no-load current.

The graph (from here ) shows a small DC motor, with the torque line intersecting the axis at zero torque (that I've highlighted in red) at about 0.15A (table shows a higher value, probably a maximum value rather than a typical on the graph).

What doesn't make sense is the notation "ideal" - the effective current that generates useful torque is the armature current less the no-load current.