0
\$\begingroup\$

I'm currently trying to model a simplified circuit of an inverter + 3 phase induction motor. For now, I want to represent the DC link as an ideal independent voltage source. To keep this example simple, I also want to keep frequency and input voltage the same, though I expect to be changing these to simulate a VFD

I'm using https://myelectrical.com/notes/entryid/251/induction-motor-equivalent-circuit as reference

The output I want is motor torque and resulting current on the DC link.

So, given \$V_{DC}\$, what would \$I_{DC}\$ be in terms of inverter equations and induction motor equations?

I think I'm most of the way there, but I can't find a full set of equations for this, only separate equations for motors and inverters.

schematic

simulate this circuit – Schematic created using CircuitLab

For input voltage to the motor, I think this is correct $$V_{L} = \frac{3}{\sqrt{2}\pi}V_{DC}\\ V_{P} = \frac{V_{L}}{\sqrt{3}}$$ It's my understanding that torque is per phase $$T = \frac{1}{2\pi n_s}\frac{R_{2}}{s}\frac{E_{2}^2}{\frac{R_{2}}{s}^2+X_{2}^2}$$ So total motor torque $$T_{3\phi} = 3T = \frac{3}{2\pi n_s}\frac{R_{2}}{s}\frac{E_{2}^2}{\frac{R_{2}}{s}^2+X_{2}^2}$$ To calculate back emf \$E_2\$: $$E_{2} = V_{P}-\frac{I_{1}}{Z_{1}}$$ Impedance per phase: $$Z_1=R_{1} + jX_{1}\\ Z_m=jX_{m}\\ Z_2=\frac{R_{2}}{s} + jX_{2}\\ Z=Z_{1}+\frac{Z_{m}Z_{2}}{Z_{m}+Z_{2}}$$ Would the calculation back to \$I_{DC}\$ be as follows? $$I_{1} = \frac{V_{P}}{Z}$$ $$I_{L} = \frac{I_{1}}{\sqrt{3}}$$ $$I_{DC} = \frac{3}{\sqrt{2}\pi}I_{L}$$

EDIT: Specifically, I want to know if the equations I have above are correct for determining the average current on the DC link (neglecting most losses for now)

\$\endgroup\$
0
\$\begingroup\$

The average DC current will be determined by the total power delivered to the motor plus the losses in the inverter. The reactive component of the motor current will be seen as bus capacitor ripple current. I don't believe you have taken that into consideration.

If the losses are ignored, the power delivered by the DC link is equal to the mechanical power delivered by the motor. The average DC current is the mechanical power in watts delivered by the motor divided by the DC bus voltage.

\$\endgroup\$
  • \$\begingroup\$ Correct, I'm looking at average only. The assumption is that everything here is ideal - no losses and perfect 3 phase sine, 120 degree phase difference. I can consider losses separately once I've got the basic equations in place which is what I'm having trouble with. The issue is more about what the equivalent impedance would be as seen at the DC link to then be able to work out the power. \$\endgroup\$ – SteelIXB Jan 24 at 19:56
  • \$\begingroup\$ See paragraph added to answer. \$\endgroup\$ – Charles Cowie Jan 25 at 4:29

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.