I have a motor that I'm turning with another motor. I can determine the speed of rotation with a tachometer. One motor is powered with a power supply and it is coupled with another motor. I need to find the I-V characteristics of the second motor but I'm not sure how to test for it? I put a resistor across the terminals of the second motor and tried measuring the current through it, but I'm not sure if this is the right way to do it?

  • \$\begingroup\$ Let me see if I understand. Power supply is causing motor 1 to turn in the normal way. Motor 1 is mechanically connected to motor 2, causing motor 2 to spin. There is no power supply connected to the terminals of motor 2, but you would like to understand the I-V characteristics of motor 2, is that right? I think you should specify what type of motor it is, or if you don't know, where it came from (what kind of equipment was it used in). \$\endgroup\$ – mkeith Feb 16 '15 at 6:56
  • \$\begingroup\$ Yes, and it's a dc motor with a gearhead on it. amazon.com/HOSSEN%C2%AE-30RPM-Electric-Motor-Torque/dp/… \$\endgroup\$ – Jucesanc Feb 16 '15 at 7:04
  • \$\begingroup\$ Yes, I think to a first approximation, the motor will produce a voltage dependent on the speed. If you add a load, but keep the speed the same, the voltage will change a little, but the resistance felt by motor number 1 will increase. Motors are not an area of expertise for me, so I am keeping to the comment section. \$\endgroup\$ – mkeith Feb 16 '15 at 7:37

DC motors have a linear I-V relationship, and the specifications give enough information to calculate the I-V slope.

\begin{align} m &= \frac{12V}{600mA} = 20 \Omega\\ V_{emf} &= 12 - m I \end{align}

There is technically no way to directly measure \$V_{emf}\$, however it can be deduced by simultaneously measuring the current flowing through the motor along with the rotational speed for various steady loads (you will need at least 2 different loads). A cheap way to apply a load is to use your fingers and vary how hard you hold the shaft (the key is how steady the load is, not what it's actual level is).

There are a few things to note when dealing with real components:

  1. Ammeters have a burden voltage due to how they measure current (they have a known small resistance inside which they measure the voltage across).
  2. Some motors don't like to be stalled for significant periods of time, and may be damaged if stalled for too long.
  3. DC motors age. Depending on the quality of the motor, this may or may not be significant. This is especially a problem with brushed DC motors, which it looks like is what you're using. I've tested DC motors which have lost significant (1/2, or more) performance running them for less than an hour. I've also used DC motors which are many years old and have lost little performance.
  4. Real components have physical friction (either contact friction or air resistance).

Problem 1 is easy to solve: use kelvin sensing and adjust the power supply such that the voltage across only the motor itself is always the same.

Problem 2 is also simple to handle: just don't apply too great a load that the motor stalls.

Problem 3 is unavoidable. The only way to reduce this is to limit the amount of testing you do.

Problem 4 is usually negligible. If you need to characterize this then you will need a way to accurately measure the test torque. I am going to ignore this.

Once you have a few data points, get a line fit of I vs. rotational speed using your favorite line fitting algorithm. From the computed stall speed you can calculate the winding resistance of the motor. Similarly, you can compute the no-load speed from this line fit (point at which I = 0). At this point, \$V_{emf}\$ is the applied motor test voltage. Using the stall point and the no load point you can calculate the linear relationship between \$V_{emf}\$ and I:

\begin{align} m &= \frac{V_{test}}{I_{stall}}\\ V_{emf} &= V_{test} - m I \end{align}

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  • \$\begingroup\$ I don't understand how you got the slope of the I-V curve? Where did you get the 600 mA from? \$\endgroup\$ – Jucesanc Feb 20 '15 at 8:51
  • \$\begingroup\$ It's in the specifications for the motor you linked. \$\endgroup\$ – helloworld922 Feb 20 '15 at 10:10

I understand you have a brushed DC motor with permanent magnets, with an integral gearbox down to 30rpm. This suggests a gearbox ratio of at least 100:1.

Helloworld's answer has the right formula for an idealized motor. It's basically just an emf and a resistance.

However, given that you have a way to spin the motor with no current applied, I disagree with his statment that "there is technically no way to directly measure Vemf."

One way to do it is to run the motor off load, but it will still be loaded by the (considerable) friction of the attached gearbox.

The way to get round this and measure the emf under no current conditions is to spin the motor. The impedance of the motor should be negligible compared with the impedance of a regular voltmeter, so provided have a way to spin the motor and you can disconnect all loads, you should be able to measure the Vemf.

In theory, it should be proportional to the speed of the motor.

The current through the motor should be proportional to the torque developed by the motor.

If the emf is lower than the terminal voltage, the current will run in one way, and the motor will generate torque: i.e it will function as a motor, consuming electricity and generating mechanical power.

If you apply an external torque and try to make run the motor faster, the emf will get higher than the terminal voltage, the current will run the opposite way and the motor will react against the applied torque. i,e it will function as a dynamo, consuming mechanical power and generating electricity.

The current flowing will be as follows:


A theoretical perfect, resistance free motor connected to a constant terminal voltage will therefore run at a constant speed no matter what torque is applied to it.

The resistance can be measured by applying a voltage to motor and stalling it.

In theory the curve between I=0 and stall should be a straight resistive line, but I expect there will be some deviation.

What is not clear to me is whether you want to use this motor as a motor or as a dynamo. It is a geared motor, and the gearboxes tend to be horribly inefficient (consuming 50% or more of the torque of the motor in some cases) and this will have a massive effect on current draw. So if you want to use your motor as a motor rather than a dynamo, be sure to collect data when the motor under test is trying to drive the load, and not the other way round.

Also, be careful stalling these motors at full rated voltage. With the high gearing, the torque may be sufficient to damage the gearbox before the current can damage the motor. That said, I have gotten away with it with similar motors, despite what it said on the datasheet. (The worst that happened was a damaged shaft!)

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