why the supply voltage of dc motor decreases when the motor load increases?

When I plot the supply voltage of a one-cell LiPo battery during the working period, I found that the supply voltage would drop dramatically when the motor is commanded to speed up.

I think this is somehow against the model of DC motor. If the discharging rate of battery increases in order to generate more torque in the DC motor side, the supply voltage should increase instead of decrease. Because the motor can be regarded as a inducter.

Following is the plot,

Y-axis is the supply voltage of the battery, X-axis is the time. There are five dropping peaks when the motor is commanded to speed up.

Thanks.

The battery also has internal resistance. As the motor draws more current to accelerate and run at a higher speed, the increased current draw will create a larger voltage drop across the battery's internal resistance. The result is voltage droop.

Discharge rate would be measured in units of watts anyway, which is volts times amps. The increase in current will more than offset the drop in voltage so that ultimately you are drawing more power, just at a lower voltage.

When I plot the supply voltage of a one-cell LiPo battery during the working period, I found that the supply voltage would drop dramatically when the motor is commanded to speed up.

This is normal and is called "droop". Imagine if it did not droop - there would be no limit to the amount of power you could draw from the battery. This is clearly unreasonable so that makes droop reasonable.

I think this is somehow against the model of DC motor. If the discharging rate of battery increases in order to generate more torque in the DC motor side, the supply voltage should increase instead of decrease.

How would the battery voltage increase? The battery voltage is determined by the chemistry. Maximum voltage is always in the unloaded condition.

Because the motor can be regarded as a inducer.

Inductor not inducer.

There are five dropping peaks when the motor is commanded to speed up.

Perfectly normal.

Battery model

simulate this circuit – Schematic created using CircuitLab

Figure 1. Battery model and load.

The standard method of predicting battery voltage drop is to model the battery as a perfect (constant-voltage) battery with an internal series resistance. If the battery terminals are open-circuit the voltage measured on them will be the battery voltage. Once loaded the terminal voltage will drop proportional to the current drawn. If the internal resistance, $R_{INT}$ is known this can be calculated as $V_{DROP} = I \cdot R_{INT}$.

Motor model

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For a direct battery-powered situation such as yours the DC motor can be modeled as a resistor whose resistance is equal to the motor winding resistance in series with a DC "back-emf" (electro-motive force or voltage) voltage source. We know that a DC motor acts as a generator and that the voltage generated is proportional to speed. Therefore the back-emf is proportional to speed.

• On power-up the speed is zero so the back-emf is zero. The current from the battery will be limited by the battery and motor resistances. $I = \frac {V_{B}}{R_B+R_M}$.
• Once the motor starts to turn the back-emf will start to increase, opposing the flow of current. The current is now given by $I = \frac {V_B - V_M}{R_B+R_M}$ wher $V_M$ is the back-emf. (You can see that the previous equation is this equation with $V_M$ = 0.)

The result is that when you load the motor the speed drops, the back-emf, $V_M$, falls and the current increases.

Inductance of the coils doesn't play much of a part in a typical motor as the resistance is dominant. It's much more significant in PWM controllers and switches where the inductance generates spikes and transients which have to be handled properly to avoid destroying semiconductors and contacts as well as electro-magnetic interference.

If you are able to obtain or measure the resistance and inductance of your motor it will be instructive for you to work out the expected voltage across the inductive element using $V = L \frac{dI}{dt}$ where $\frac {dI}{dt}$ is the rate of change in current in A/s.

• I understand the internal resistance part that could perfectly explain the drop of the supplied voltage. There is still one place I do not understand. If the supply current in DC motor increases, like from 5c to 10c, the voltage in the inductor of the motor increases as well, at least at a moment. However, I did not see the sudden increase of the supplied voltage when the supplied current increases. – facebook-1536745818 Mar 31 '16 at 16:46
• See the update. – Transistor Mar 31 '16 at 17:59
• Appreciate your help. I am using PWM to control a small coreless dc motor for a quadrotor project. As you said, $V_{inducter}=L\frac{dI}{dt}$. If ${I}$ increases largely in a moment, $V_{inducter}$ should increase as well. However, in the plot, it is not shown that there exists a voltage increasing moment. This is the place where I am confused about. And since I am using the dc motor for quadrotor, I think the motor speed increases as the motor load increases, then the back-emf,$V_{M}$ increases as well. – facebook-1536745818 Mar 31 '16 at 18:15

Motor back-emf is caused by it acting as a generator. The voltage produced is proportional to speed, but it will never produce more voltage than the supply unless it is over-driven by an external force.

Inductance has effect during PWM, but you should have a flyback diode which recirculates induced current through the motor. The only time you would see the battery voltage rise due to inductance is if using dynamic braking, which requires synchronous drive (both upper and lower FETs which are switched on/off alternately by PWM).

simulate this circuit – Schematic created using CircuitLab