Why DC motor draws more current with friction?

[NuLED]

Hi - question here:

2 x AA batteries with open circuit voltage of around 3 volts.

When connected to a 3V DC motor, voltage (measured at battery terminals) drops to around 2.8 volts.

The question I have is, when fingers are placed against the motor shaft so that there is increased friction and it turns slower, the voltage actually drops when measured both at the batteries and at the DC motor terminals.

Why is this?

thanks

 

[tehtehteh]

the current increases when you put strain on the motor, therefore greater voltage drop?

 

[NuLED]

the current increases when you put strain on the motor, therefore greater voltage drop?

Thanks but why? Is there less resistance (i.e., more conductance) at the motor because it turns slower and the copper bits are touching longer?

 

[tehtehteh]

I don't know why really, I always just thought of it as a free spinning motor is doing very little work to keep itself turning over, and a stalling motor is doing a lot more

 

[NuLED]

I don't know why really, I always just thought of it as a free spinning motor is doing very little work to keep itself turning over, and a stalling motor is doing a lot more

Yup, that's what I want to know.

Changed the thread title to clarify question.

 

[(*steve*)]

recall that if you spin the motor by hand a voltage is generated

also recall that the motor has a low resistance and should draw a high current due to this.

and have you wondered why a motor doesn't turn faster and faster and faster without limit?

or perhaps why a motor runs faster at higher voltages?

Simplistically speaking that first observation will help explain all the others.

When the motor is stalled (e.g
held stopped) the current drawn is exactly what ohms law would suggest, based on the DC voltage applied and the resistance. And this may be many times the normal current.

as the motor is allowed to spin it also starts to act like a generator. The generated voltage "pushes back" against the voltage driving the motor in such a way as to reduce the effective voltage across the motor. If this generated voltage (also called "back EMF") is 1/4 of the applied voltage, then the DC resistance of the motor "sees" only 3/4 of the applied voltage and hence the current is only 3/4 of what you would expect from the DC resistance.

so as the motor spins faster and faster, the amount of power available to spin it even faster falls. Thus, even if the motor was perfect with no resistance in the bearings, etc, it would have a maximum speed at a given voltage.

this also explains why a higher voltage causes the motor to spin faster. It must do so, because the slower speed does not cancel out the incoming voltage.

so finally, why does the current increase when you load the motor? Well, you are slowing the motor and reducing the back EMF. This means that the dc resistance "sees" a higher voltage and thus a higher current flows.

 

[NuLED]

Thanks Steve for the detailed answer.

I had forgotten about back EMF. It's not a very intuitive concept to get my head around.

I guess if I connected an oscilloscope, you'd see the voltage drop rapidly as the motor spins up right as you close the switch.

It is a bit confusing to me that the battery terminals themselves reflect the reduced voltage. So the back EMF is going all the way to cancel out the battery voltage? This back EMF is in direct opposite phase to the battery DC voltage? (It's weird because all this is DC and not AC...)

 

[Minder]

The motor generates a voltage the same polarity as the supply so it opposes the applied voltage, for example a 100vdc motor rated at 1000rpm, will generate a voltage of ~100vdc when rotated externally at 1000rpm.
M.

 

[(*steve*)]

I guess if I connected an oscilloscope, you'd see the voltage drop rapidly as the motor spins up right as you close the switch.

No, you wouldn't. All you can see is the reduced demand for current which implies either an increase in resistance or some other force acting against the current. It's not resistance

If you disconnected the source of voltage, you would see the voltage being generated by the motor. However without a suitable load, the value might be other than what you expect.

It is a bit confusing to me that the battery terminals themselves reflect the reduced voltage.

They don't. The battery voltage is reduced because the battery is not perfect and cannot supply the current required without the voltage dropping. Use a larger battery of the same voltage, or a power supply capable of high enough current and this effect would be diminished or non-existant.

 

[Ratch]

Hi - question here:

2 x AA batteries with open circuit voltage of around 3 volts.

When connected to a 3V DC motor, voltage (measured at battery terminals) drops to around 2.8 volts.

The question I have is, when fingers are placed against the motor shaft so that there is increased friction and it turns slower, the voltage actually drops when measured both at the batteries and at the DC motor terminals.

Why is this?

thanks

A heavier load on the motor causes it to demand more current. More current causes the battery to lower its output voltage. This is because increased current across the internal battery resistance causes a larger internal voltage drop, which takes more voltage away from the battery output voltage.

Ratch

 

[Minder]

Most has been covered already.
The basic resistance of a DC motor is whatever the commutator including brushes is equal to.
Which in most cases is just a an Ohm or two, this results in a very high current if full voltage is applied at switch on, due to this very high current the motor starts to turn immediately and once it does, it starts to generate a voltage dependent on its rpm.
This voltage opposes the applied voltage so less and less current flows as it come up to speed, the generated field can never equal the applied voltage as no current would flow and at this point and it requires some current to overcome bearing friction and windage etc.
If a load is applied, this demands more current to sustain this load, the rpm drops in order for a greater difference between the voltages and accordingly greater current.
Voltage is = to rpm.
Torque is equal to current.
I recently did an empirical test on a T.M. motor that was rated 90vdc @ 18amps.
I had a DC supply that I could gradually increase from 0-90v using an off loaded motor..
The current remained at a constant 2 amps throughout the test range.
M.

 

[NuLED]

Sorry what is a T.M. motor?

 

[NuLED]

The battery voltage is reduced because the battery is not perfect and cannot supply the current required without the voltage dropping. Use a larger battery of the same voltage, or a power supply capable of high enough current and this effect would be diminished or non-existant.

OK let me experiment later using a wall power supply at 3V instead of the dry cells.

 

[NuLED]

Thanks y'all for your replies.

 

[Minder]

T.M. Treadmill (motor).
M.

 

[NuLED]

So I did further experiments on this using a power supply instead of dry cell batteries. Indeed the voltage did not drop (as much) at the power supply end, although it still dropped a bit (from 3.2 v to around 3.0 v) and significantly dropped at the motor end.

I measured the current with a DMM in-line, and current went as high as 600 mA once friction is applied to the motor shaft.

So my further question is, why does voltage get reduced when friction is applied?

Is the unstable back EMF causing inaccurate readings on the DMM?

Because current increases with voltage, the relationship is not inverse. With higher current being drawn, and reduced back EMF to counter the positive voltage from the supply, shouldn't voltage increase instead of decrease?

Thanks.

 

[BobK]

Current increases with voltage in a resistor. Other components are more complicated.

The current of a running motor is determined by back EMF. The current in a stalled motor is determined by Ohm's law.

Bob

 

[NuLED]

Yes, but the motor isn't stalled. Just dragged a bit with friction to slow down the shaft.

That's the confusing bit, why voltage measures LESS when the current has increased dramatically, and conceivably the back EMF has also decreased (algebraically, the supply voltage - back EMF = net voltage).

 

[(*steve*)]

That's the confusing bit, why voltage measures LESS when the current has increased dramatically

that's nothing to do with the motor. It's the batteries.

 

[NuLED]

I already tried with the wall plug (not batteries) but I have a bench top power supply, will try that next then.