2
\$\begingroup\$

following a tutorial on the web I created in LTSpice a DC motor model, but I have a question about the results, the model is:

LTSpiceModel

and this is the spice netlist:

V_MOTOR N001 0 PULSE(0 10 0.05 1u 1u 1 10 1)
R_MOTOR N001 N002 {R_M}
H_TORQUE H_TORQUE_PROBE 0 V_SENSE_1 {Kt}
R_Friction H_TORQUE_PROBE V_SEEN_BY_L_INERTIA {R_F}
V_SENSE_1 N004 N003 0
V_SENSE_2 N005 0 0
H_BE N003 0 V_SENSE_2 {K_bemf}
L_MOTOR N002 N004 {L_M}
L_INERTIA V_SEEN_BY_L_INERTIA N005 {L_I}
.tran 2
.PARAM R_M 0.5
.PARAM L_M 1.5m
.PARAM L_I 0.25m
.PARAM Kt 0.05
.PARAM R_F 0.0001
.PARAM K_bemf 0.05
.backanno
.end

The results are:

LTSpiceGraph

My question is: why V(V_SEEN_BY_L_INERTIA), representing inertia of the motor is not 0 at steady rate? As far as I know voltage across an inductor is given by L*d(i)/d(t) so as you can see by the green line when current is (almost) constant there should be no voltage drop across inductor! Obviously this is a result we expect considering that at steady rate the motor has an inertia, but I cannot understand how it is possible being no current change.

Thanks, Vincenzo

\$\endgroup\$
1
\$\begingroup\$

I've understood the problem: at what it seems Spice, if no argument is given, set a series resistance of 1 mOhm with the inductor, if a value of 0 is set for inductors series resistence the voltage across the inductor drops to 0 as desired.

Source: http://ltwiki.org/LTspiceHelp/LTspiceHelp/L_Inductor.htm

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.