Modelling the voltage measured across a dc motor

[pkc111]

TL;DR Summary

Hi there,



I have done an experiment measuring the voltage across the terminals of a dc motor, whilst running and then by holding it with my fingers to stop it. Done as demo for my high school class.



The voltage reading was higher when the motor was running, which was against my expectations as I thought the back-emf would have created a lower net voltage (being measured on the voltmeter on the motor terminals?).



So now I am guessing that the internal resistance of the motor is varying a

Hi there,

I have done an experiment measuring the voltage across the terminals of a dc motor, whilst running and then by holding it with my fingers to stop it. Done as demo for my high school class.

The voltage reading was higher when the motor was running, which was against my expectations as I thought the back-emf would have created a lower net voltage (being measured on the voltmeter on the motor terminals?).

So now I am guessing that the internal resistance of the motor is varying a lot to create this result?

Many thanks

 

[Baluncore]

When you run a DC motor on a fixed voltage, it produces a back_EMF, proportional to speed. A DC current flows through the motor, providing power to overcome losses. That DC current drops a voltage across the motor internal resistance.
The supply voltage = back_EMF + resistive drop.

When you stop driving the motor with a fixed supply voltage, the back-EMF remains due to momentum, while the motor current becomes zero, so the resistive voltage drop disappears. That is the voltage step that you see when you disconnect the motor.

 

[DaveE]

This question makes no sense to me without a description of your motor drive circuit (eg. battery?). In many applications, the voltage you measure across a DC motor is the voltage you put there. So I guess you have some source (series) resistance, or something similar.

Anyway, there's a 99.9999% chance that @Baluncore is right. When you slow the motor the back-EMF is reduced which means more of the source voltage is applied across all of the circuit's series resistances. This means more current is drawn from the source, which may cause it to decrease because of the source impedance.

 

[pkc111]

Thanks,
So how would I model (in a spreadsheet )the predicted net voltage across a small 4 V DC motor hooked up to a laboratory power pack set at 2V when the motor is slowed down to different % max speeds by finger resistance?

 

[pkc111]

Thanks,
So how would I model (in a spreadsheet )the predicted net voltage across a small 4 V DC motor hooked up to a laboratory power pack set at 2V when the motor is slowed down to different % max speeds by finger resistance?

and would I be better off trying to quantify a motor's back emf with a battery rather than a transformer power supply?

 

[Baluncore]

Torque is proportional to motor current. Find that parameter coefficient.

Back_EMF is proportional to motor speed. Find that parameter coefficient.
Spin the motor at a known speed, with no load, measure the voltage generated.

Measure internal resistance with the rotor locked, so there is no back_EMF.

The difference between supply and back_EMF, is available to drive current through the motor internal resistance, which provides torque.

 

[DaveE]

The difference between supply and back_EMF, is available to drive current through the motor internal resistance, which provides torque.

Plus any other resistance in the entire circuit. Which you must have, if you saw the motor terminal voltage decrease. So, what @Baluncore said, plus characterize your source resistance. How does the source voltage change with load current? I think this is what you are really seeing.

 

[pkc111]

Thank you!
I think I may try a battery instead to remove some variables :)

 

[pkc111]

So my DC motor with a 6V battery attached gives a 5V PD reading across the motor terminals. When the rotor is held still with fingers it reads 3V...whyso? I thought the voltage would go up when held due to the absence of a back emf.

Also when I use the motor shaft to drive an identical motor shaft (to use it as a generator), the "generator" produces a 4V reading. I assume this is the same as the back emf value in the motor? But the numbers arent agreeing?

 

[Baluncore]

So my DC motor with a "6V" battery attached gives a 5V PD reading across the motor terminals. When the rotor is held still with fingers it reads 3V...whyso?

The voltage is being divided between the internal resistance of the battery and the resistance of the motor windings and connections, which are of similar magnitudes.

Also when I use the motor shaft to drive an identical motor shaft (to use it as a generator), it produces a 4V reading. I assume this is the same as the back emf in the motor?

Yes. The driving motor has coil current through the winding resistance. The driven generator has zero coil current, so the 4 volt reading, is simply the back_EMF at that speed. The identical driving motor can be expected to have the same 4 V back_EMF.

The driving motor is providing energy to support losses in both motors, so it will need to draw twice the current it would normally at that speed. 6V - 4V = 2V across the coil resistance. If the motor was idling alone, it would only need 1V across the coil, so back_EMF is then 5V, meaning the driving motor would run 5/4 times faster if not driving the voltage generator.

 

[pkc111]

Ok..I think its becoming a little bit clearer.

The 5V to 3V drop in the net motor supply voltage when the motor is stopped with fingers still puzzles me.

Are we saying that the rise in voltage expected by absence of a back-emf in rotor when stopped with fingers, is more than made up for by a decrease in resistance of the coil and battery hence producing an overall net voltage drop?

Thanks

 

[berkeman]

The 5V to 3V drop in the net motor supply voltage when the motor is stopped with fingers still puzzles me.

Are we saying that the rise in voltage expected by absence of a back-emf in rotor when stopped with fingers, is more than made up for by a decrease in resistance of the coil and battery hence producing an overall net voltage drop?

Sorry if I missed it, but what is the DC resistance of the motor winding? You should be able to measure it with a DVM. That DC resistance is what will be presented to the driving voltage source (battery or whatever) when the motor is stalled. That is typically the lowest resistance that the voltage source will see, and is the reason that stalled motors often overcurrent the voltage source driving the motor (resulting in fires, or more hopefully blown fuses).

If the motor is not stalled, the drive current will be less than the stall current.

https://www.pololu.com/docs/0J73/4.1

 

[DaveE]

a decrease in resistance of the coil and battery

What decrease? These should be (relatively) fixed values.

 

[pkc111]

Sorry if I missed it, but what is the DC resistance of the motor winding? You should be able to measure it with a DVM. That DC resistance is what will be presented to the driving voltage source (battery or whatever) when the motor is stalled. That is typically the lowest resistance that the voltage source will see, and is the reason that stalled motors often overcurrent the voltage source driving the motor (resulting in fires, or more hopefully blown fuses).

If the motor is not stalled, the drive current will be less than the stall current.

https://www.pololu.com/docs/0J73/4.1

10 ohms

 

[Jigga]

So 10 ohms matches the peak current seen through the coil immediately during startup (6V/10 ohms = 0.6A) So I am mot seeing why the same situation (6V rather than 3V) does not exist during stall?

 

[Tom.G]

So 10 ohms matches the peak current seen through the coil immediately during startup (6V/10 ohms = 0.6A) So I am mot seeing why the same situation (6V rather than 3V) does not exist during stall?

Because the battery also has some internal resistance, and so do the connecting wires.

Here is why the voltage from the battery is 6V with no (or very little) current being drawn,
AND when the rotor is held still, the voltage is 3V:

1) You are measuring 1/2 the battery voltage at the motor
2) That leaves 3V loss in the battery internal resistance
3) Since the same current is flowing thru both the motor and the battery

AND the voltage drop is the same across each​

The internal resistance of the battery is the same as the motor resistance 10 Ohm

Cheers,
Tom

p.s. Batteries are not perfect. They are built as two conductors made of different materials and separated somewhat with an electrolyte between them. The battery Voltage is created by an electro-chemical reaction between the electrolyte and the two different electrode materials. This electrolyte may be a fluid, a paste, or a gel, and has a somewhat high resistance to current flow. That electrolyte resistance is what causes the battery terminal voltage to drop when current is drawn.

If you wish to learn more about batteries, https://batteryuniversity.com/ is a good source of information.

 

[pkc111]

so a 6V battery can have an immediate jump to 10 ohm (3V) internal resistance drawing 0.4 amp?

 

[Tom.G]

The resistance doesn't "jump", it is an inherent, ever-present, characteristic that gradually increases as the batteries discharge and age.

The resistance depends on the physical size of the battery, the particular chemistry used, the state-of-charge, temperature, and usage history.

Lets take an extreme example:
8 pieces of size AAA battery in series supply the same voltage as a car battery.
But they sure won't start a car, they have too high an internal resistance, and they can not supply enough total energy.

 

[pkc111]

 

[sophiecentaur]

@pkc111
If you draw a schematic diagram of the circuit (if you don't know what I mean then hunt around for a web page about motor back emf) you can copy that and get used to circuit diagrams as second nature. There will be a battery emf, a battery internal resistance (r), a motor winding (with a back emf) and a motor resistance (R). All in series. And then add the meter readings. i.e. present yourself with the problem in a way that will help you. (Like you get in problem exercises)

The voltage drops all round the circuit when the motor is stalled can only be caused by the two series resistances. If you are measuring an unexpected value of voltage drop then you must have an unexpected value of effective resistance somewhere. Will your meter measure the motor stall current?
Do you have any handy low value resistors (anything between 1 and 10 Ohms) and see if your estimated battery internal resistance makes sense.

PS That table could have been much more use if you'd used higher definition and checked legibility before posting. It may only need a change of (scanner?) settings, somewhere.