Gelöste Aufgaben/JUMP/E-Motor and Drive-Train

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Scope

Diagram: E-Motor and Drive Train

The Drive-Train consists of a DC/DC-converter, a DC Motor and a gear-box.

• DC/DC-converter: is supplied with the battery voltage U_B, the output voltage is controlled by the driver via setpoint “p“.
• motor: is a standard DC brushed motor, the manufacturer provides only few information on its characteristics - we’ll need to improvise.
• gearbox: has a gear ratio of ratio of n_G=100, its shaft rotates at speed ω_W and delivers a torque M_W to the front wheels.

The task is: provide a mathematical model for the drive train that accounts for load-alterations imposed by the driver. And we assume losses in the two converters - DC/DC and gearbox - to be negligible.

Structure

Block diagram

The drive train receives a "gas"-pedal position "p" from the driver and a battery-voltage U_B.

It delivers a torque M_W on the wheel and creates an electric current I_M through the motor.

Drive-train components.

The sub-model consists of  DC/DC-converter, Motor and gear-box:

DC/DC Converter

Losses in the DC/DC converter shall be small - so for input port “1“ and output port “2“ we obtain

$U_1\cdot I_1=U_2\cdot I_2$ .

Let the “gas”-pedal-indicator “p“ control

$U_2=p\cdot U_1\text{; }0\le p\le 1.$

with

$0\le p\le 1\text{ and }{U}_1=U_B$

Motor

Brushed-DC motor

We use a common electric circuit representation for a series wound motor, the field coils are connected electrically in series with the armature coils, resistance R sums up all electrical losses in the motor.

Gearbox

Losses in the gearbox shall be small - so for input (ω_M, M_M) and output (ω_W, M_W) we obtain the fixed relation

${\omega }_M\cdot M_M={\omega }_W\cdot M_W$.

And we have only one differential equation for the electrical components:

$\frac{d{I}_B}{dt}=\frac{U_L}{L}$,

the remaining equations are algebraic.

Model

For the motor, we find with Kirchhoff's law that

$U_M=U_R+U_L+e$

with U_R, U_L being the differential voltage over resistance R and inductance L respectively. “e” is the back electromagnetic force with

$e=k_e\cdot {\omega }_M$

and the electromotive force constant k_e. Note the ω_M is the differential rotational velocity between rotor and stator, i.e.

${\omega }_M={\dot{\psi }}_W(t)+\dot{\varphi }(t).$

Employing

$U_R=R\cdot I_M,U_L=L\cdot \frac{d{I}_M}{dt}$

and using

$M_M=k_t\cdot I_M$

with the armature constant k_t, we have the complete set of equations.

From the above, we find

$L\cdot \frac{d}{dt}{I}_M(t)=U_B(t)\cdot p(t)-R{I}_M(t)-e$

and additionally the algebraic equations

$\begin{array}{ll}{I}_B& =I_M\cdot \mathrm{p}(t),\\ U_R& =R\cdot I_M(t),\\ U_L& =U_B\cdot \mathrm{p}(t)-R\cdot I_M(t)-e,\\ U_M& =U_B\cdot \mathrm{p}(t)\end{array}$.

tmp

Electrical Components

Text

1+1=2

tmp

Mechanical Components

Text

1+1=2

Parameter

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References

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