Esercizi svolti di Power Electronics

Nella dimostrazione associata alle perdite dovute alle capacità parassita, solo la corrente I_out del generatore va considerata.

i_D = I_out + i_DS + i_DRV

i_D ∝ (V_GS - V_TH)²

Le approssimazioni sono ragionevoli?

Exercise:

2 tipi: - switching losses - operating condition

Rimanere coerente alle domande!

Buck converter L=12μH       f_s=100kHz   T_s=10μs P_out=240W      η=100% V_GG=0÷6V R_DRV=2Ω R_G=2Ω R_ON,g=6Ω

Switching losses?ΔV,ΔI, Q_G,swbisogna guardare il grafico!

ΔV=48V, data dell'applicazione di un segnale quadro ai capi del MOSFET ΔI??

i_DRV,ON = _{V_GG - V_SP}⁄^{R_DRV + R_ON,g + R_G} = _{6 - 3}⁄¹⁰ = 0.3A i_DRV,OFF = _{- V_SP}⁄^{R_DRV + R_ON,g + R_G} = _-3⁄¹⁰ = 0.3A

Q_G,SW = la carica che bisogna iniettare per passare dalla tensione di soglia fino alla fine del Miller plateau.

Q_G,SW = 6nC. Può anche essere calcolata: Q_G,SW = Q_GD + Q_GS - Q_G,TH = (5 + 3 - 2) nC = 6nC Q_G,TH = Q_GS. _{V_TH×3×2}⁄³ = 2nC

il?

The current I is given by

I = V/R = 3,6 A

2. Sketch the inductor and capacitor current waveform.

About the inductor we can calculate the peak current i_{L, peak}, which is

i_{L, peak} = (V_g - V) DT_s/L = 48-18/10x10^-6 - 9 A

and since this quantity must be equal to

i_{L, peak} = V_g/L D₂T_s, it's possible to define D₂T_s = 9 x 10 x 10^-6/58 = 5µs

Given the f_sw = 100kHz, DT_s = 3µs, D₂T_s = 5µs and D'T_s = 7µs, you get the following scheme.

Having the mean value of the output current I and having i_{L, peak}, we can also define the capacitor current plot.

Additional exercise

Let's check if we are in CCM or in DCM.

K = _2L/^{R T_S} = _{2 · 12 · 10^-6}/^{10 · 100 · 10-6} = 0.24

Since K_crit(D') = D'² = 0.8², K < K_crit(D') → DCM

Let's calculate the output voltage

_V/^V_g = ₂/^{1+√(1+4K' D_²)} → V = _2·Vg/^{1+√(1+4 · 0.24 · (0.2)²)} = _{2 · 12}/^{1+√(1+4 · 0.24 · (0.2)²)} = 4V

Having the output voltage,

I = _V/^R = ₄/¹⁰ = 0.4 A

It's possible to look for the i_{L, peak} = _{Vg - V}/^L · D T_S = _{12 - 4}/^{12 · 10-6} · 0.2 = 1.3A

To correctly plot i_L we need to look for D₂ T_S. Since i_{L, peak} = _V/^L · D₂ T_S,

D₂ T_S = i_{L, peak} · _L/^V = 4 µs

The current on the capacitor is the one of the inductor lessened by I.

Let's consider the i_c current. We have that

i_{L peak} = V_g DT_S = 12 = 2 A

D₂ T_S = L i_{L peak} / |V| = 5 µs

• So the ripple is

S = ΔV / V = 2.3%

Let's calculate t_a.

2/5 = 1.5/t_a ⟹ t_a = 5/1.5/2 = 3.75 µs

So we have that

ΔQ = |(1/2)t_a1.5| = 2.8125 µC

ΔV = ΔQ/2C = 140.6 mV

Switching losses

• Boost V_IN = 300 V
• V_OUT = 400 V R_OUT = 500Ω
• L = 400µH
• f_s = 100kHz T_S = 10µs

• V_GS vs. Q_GG plot
• V_GG ∈ [8 ÷ 10] V
• R_GAV = 2Ω
• R_GA = 2Ω
• R_QMA = 6Ω

- Calculate the inductor current, equal to the input voltage generator current.

P_t=(E_sw,off+E_sw,on)⋅50⋅10³=2,253 W

i_L(t)

τ(μs)

V_ds(t)

τ(μs)

V_in

224

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M76