Simulated bode plot in Simplis is different from the result of Mathcad

Question

I built an open-loop-controlled buck converter in Simplis. The parameters are listed below:

\$V_g=12V,V_{out}=5V\$

\$D=0.52\$

\$C=100\mu F,ESR=5m\Omega; L=30\mu H,ESR=5m\Omega\$

I use Simplis to simulate the bode plot of \$G_{v_d}\$. And I use the Mathcad to simulate the same circuit. But the result was different. During the low frequency, the peak gain margin in Simplis is less than 0db, while in the Mathcad, the gain was about 20DB.

Here are the results I got.

the circuit in Simplis

the Gain bode plot in Simplis

the code in Mathcad

the curve in Mathcad

I don't know what kind of mistake leads to this result. If someone knows the reason, would you please tell me? I am a freshman in Power electric and I appreciate your guidance .

Answer 1

The control-to-outout transfer function of the CCM buck operated in voltage-mode control is the following:

Several comments regarding your results:

• in Mathcad, you have missed the pulse-width modulator (PWM) gain which is \$G_{PWM}=\frac{1}{V_p}\$ with \$V_p\$ the peak of the modulation ramp. This is your source \$V_3\$ in the SIMPLIS file. It must obviously appear in the control-to-output transfer function that will use for analytical results.
• the quality factor includes many contributors as you can see. In the formulas, these are values of passive elements such as ohmic losses of the capacitor and inductor. In reality, other losses affect \$Q\$ such as the MOSFET \$r_{DS(on)}\$, the diode recovery losses and even magnetic losses. This is the reason why you may see differences between your Mathcad analysis and the SIMPLIS one or a bench prototype. To make sure you have very close results between simulations and Mathcad, chose perfect switches in SIMPLIS (and not a MOSFET), 0-\$V_f\$ diodes etc. Finally, you may be interested to import SIMPLIS simulation results directly in Mathcad to superimpose magnitude and phase curves. See here for a tutorial.

Finally, you can have a look at the 60+ simulation templates I posted here for my next book on small-signal modeling of switching converters (TOC is here). Most of the examples work on Elements, the demo version and you can explore many structures.

Answer 2

Additional info and important take-aways on ideal lossless open-loop Buck converters: ( THINGS can go horribly wrong with the wrong load)

The equation for Q for \$r_C=r_L=r_{SW}=0,~~~\$, \$Q=C\omega_o,~~~~ Q=C\sqrt{\dfrac{1}{LC}} =R_L\sqrt {\dfrac{C}{L}}~~~~~~~~, \$ \$\omega_o=\dfrac{1}{2\pi f_o}\$

This only works when the SMPS is loaded. Otherwise with no load Q rises to infinity with RL.

So for the critical damping load = \$Q=1/2\zeta_{(zeta_=0.707)}~~~= C\sqrt{\dfrac{1}{LC}} =R_L\sqrt {\dfrac{C}{L}}\$

• thus the critical value is \$R_L=\sqrt{\dfrac{L}{8C}}\$
  • If R is 10 times higher it will look like below.

Then for Vpp ripple voltage to be <=1% f must exceed the value for dV/dt=Ic/C where \$dt~ 1/2f\$ so \$f_{min}=\dfrac{Ic}{2C*dV}\$ for Vpp=dV

I'll let you figure out f(min) for Load=1.5A, L=100uH, C=100uF, R=3.3 and dV=dI*R 0.1% of Vout=5V = 5mV, assuming ideal rC=rL=rSW=0. but Ic is not load current.

(this is why IC's use more complex startup Imax, good stability and ripple compensation)

Here again ideal Open-Loop but with a critical damping load.

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