+1 vote
70 views
ago in Modeling by Gabriele Maria Lozito (20 points)

Hello!

I'm trying to implement a simple push-pull converter for photovoltaic applications. However, I'm not sure I correctly understand the inner behaviour of the Push-Pull block. I'm trying this test:

1. DC input voltage, 10V

2. Resistive load, 10 Ohms

3. Internal modulator, 10kHz, [0, 180] carrier offset, [0 1] reference signal boundaries

4. n1 = 1, n2 = 10 (thus, I assume, 10x transformer gain)

5. duty cycle sent to the "in" pin equal to 0.25

As far as I remember from literature, voltage gain is 2D*(N2/N1). Thus, I should get Mv = 0.25*2*10 = 5 V/V. However, in output, I get twice the expected voltage (100V). 

As a diagnostic test, I doubled n1 to 2 (keeping n2=10): Vout dropped to exactly half (50V), suggesting n1 and n2 might not be defined on the same basis (e.g. n1 = full primary winding turns, n2 = single secondary half-winding turns). Or maybe the modulator works in a different way from the one I'm assuming. 

Moreover, I'm getting a small efficiency loss that seems to be load dependent (lower resistive load leads to higher losses). I suppose those are conduction losses on some inner elements? In case, what is the default R_on?

Unfortunately on the manual there is very little detail concerning the model inner structure (which would also be important to assess the magnetic/inductive part, which cannot be specified in any way). 
Is there any additional documentation? Or perhaps a way to see the submodel? Or some implemented example that could point out any error from my side?
Thank you very much.

Gabriele

3 Answers

+1 vote
ago by Filip Bakic (58 points)
selected ago by Gabriele Maria Lozito
 
Best answer

on this figure 2 Push-Pull converters are shown upper that operate in DCM and bottom that have additional inductor to stay in CCM,

 

As the output waveforms show, the upper converter has a discontinuous output current, while the lower converter has a continuous output current. You can also see that these two converters produce different output voltages. The lower converter achieves a voltage gain close to 2D × (N2/N1), diode and switches are not ideal, whereas the upper converter does not.

These waveforms were obtained from a variable-step simulation in TyphoonSim.

+1 vote
ago by Filip Bakic (58 points)

Hello,

The voltage gain you are referring to is obtained when the converter operates in continuous conduction mode (CCM). To ensure that it remains in CCM, you should add an output inductor so that the output current does not fall to zero.

I assume you have connected only a resistor in parallel with the output capacitor. This causes the converter to operate in discontinuous conduction mode (DCM). In DCM, the voltage gain is no longer equal to 2D × (N2/N1). Instead, it becomes dependent on the output load.

Could you also let me know whether you are running your simulations only in TyphoonSim or on a HIL device?

+1 vote
ago by Filip Bakic (58 points)

Regarding your question about the internal modulator, if you set the duty cycle to D through the "in" port, the modulator will generate two gate pulses with a duty cycle of D, phase-shifted by 180°. For a switching frequency of 10 kHz and a duty cycle of 0.25, the resulting gate signals are shown in the figure below.

 

The transformer parameters N1 and N2 represent the total number of turns in the primary and secondary windings, respectively.

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