After Q is turned off, the capacitor current becomes equal to I₀

$i_c=I_0=C_s\frac{dv}{dt}$

Concept:

For Single-phase Simi converter

Average output voltage

$V_{o\left(av\right)}=\frac{V_m}{\pi }\left(1+\cos \alpha \right)$

Power loss in battery = EI₀

 Calculation:

Power loss in battery = 1200 watt

EI₀ = 1200

$I_0=\frac{1200}{60}=20A$

Average output power V₀ = E + I₀R

= 60 + 20 × 3

= 120 V

Putting the value of output voltage

$120=\frac{90\pi }{\pi }\left(1+\cos \alpha \right)$

⇒ cos α = 0.33

Voltage equation across inductance is given by

$v_L=L\frac{dI}{dt}$

Calculation:

Circuit diagram for the boost chopper and its equivalent circuit with conducting switch open are shown below.

Applying KVL in the above circuit

V_s – V_L – V₀ = 0

⇒ V_s – V₀ = V_L

Rate of change of inductive current  is 

$V_L=L\frac{d{I}_L}{dt}$

$V_s-V_0=L\frac{d{I}_L}{dt}$

$\frac{d{I}_L}{dt}=\frac{V_s-V_0}{L}$

$V_{0,rms}=V_s\sqrt{\frac{\beta }{\pi }}$

Average power $=\frac{V_{o,rms}^2}{R}$

$\Rightarrow \frac{{\left(V_s\sqrt{\frac{\beta }{\pi }}\right)}^2}{R}=2000$

$\Rightarrow \frac{V_s^2\times \beta }{R\times \pi }=2000$

$\Rightarrow \frac{{\left(200\right)}^2\times \beta }{10\times \pi }=2000$

$V_{dc}=\frac{240\sqrt{2}}{2\pi }(1+\cos 45{}^{\circ })=92.21V$

$I_a=\frac{V_{dc}}{R}=9.22A$

$I_{D,rms}=\sqrt{\frac{1}{2\pi }\underset{\alpha }{\stackrel{\pi }{\int }}{i}_D^{2}d\theta }$

$=I_a\sqrt{\frac{\alpha +\pi }{2\pi }}$

When the switch Q is ON,

$L\frac{d{i}_s}{dt}=V_s,0\le t\le \delta T_s$

$\Rightarrow i_s\left(t\right)=I_1+\frac{V_s}{L}t$

Where I₁ = i_s (0)

$I_2=I_1+\frac{V_s}{L}\delta T_s$

Where I₂ = i_s (δ T_s)

Power balance of the converter:

p_in = p_out

$\Rightarrow V_s{I}_s=\frac{V_{dc}^2}{R}$

$I_{s\left(av\right)}=\frac{1}{2}(I_1+I_2)$

$=\frac{1}{2}(2I_1+\frac{V_s}{L}\delta T_s)$

$=I_1+\frac{V_s}{2L}\delta T_s$

$V_s(I_1+\frac{V_s}{2L}\delta T_s)=\frac{V_s^2}{R{(1-\delta )}^2}$

$\Rightarrow I_1=\frac{V_s}{R{(1-\delta )}^2}=-\frac{V_s}{2L}\delta T_s$

For continuous source current,

I₁ > 0

$\Rightarrow \frac{V_s}{R{(1-\delta )}^2}>\frac{V_s}{2L}\delta T_s$

$\Rightarrow R<\frac{2f_{s}L}{\delta {(1-\delta )}^2}$

Maximum value of $R=\frac{2f_{s}L}{\delta {(1-\delta )}^2}$

$=\frac{2\times 20\times {10}^3\times 500\times {10}^{-6}}{\left(0.7\right){(1-0.7)}^2}$

= 317.46 Ω

For R load, voltage is continuous when α ≤ 30° and average output voltage is given by

$V_o=\frac{3V_{ml}}{2\pi }\cos \alpha$ 

Maximum possible output voltage $=\frac{3V_{ml}}{2\pi }$

Required average output voltage $=\frac{1}{2}\left(\frac{3V_{ml}}{2\pi }\right)$

$Now,\frac{3V_{ml}}{2\pi }\cos \alpha =\frac{1}{2}\left(\frac{3V_{ml}}{2\pi }\right)$ 

⇒ α = 60°

So, the required value can’t be obtained for α < 30°.

For α > 30°,

Average output voltage, $V_o=\frac{3V_{mp}}{2\pi }\left[1+\cos \left(\alpha +{30}^{\circ }\right)\right]$ 

$=\frac{3V_{ml}}{2\sqrt{3}\pi }\left[1+\cos \left(\alpha +{30}^{\circ }\right)\right]$ 

$\Rightarrow \frac{3V_{mL}}{2\sqrt{3}\pi }\left[1+\cos \left(\alpha +{30}^{\circ }\right)\right]=\left(\frac{3V_{ml}}{2\pi }\right)\left(\frac{1}{2}\right)$ 

⇒ α = 67.69°

From the given diagram we can say that two step down chopper is operating in parallel.

V_S = 100 V

δ = 0.3

For step down chopper

V₀ = δ V_S

δ = duty cycle = 0.3

Load current $I_0=\frac{V_0}{R}=\frac{0.3\times 100}{1}$

= 30 Amp

For source current I, adding currents i₁ & i₂

Frequency f = 1kHz

$T=\frac{1}{f}=1msec$

δ = 0.3 (T_on = δ T)

∴ T_on = 0.3 m sec

$\frac{T}{2}=0.5msec$

$i_{av}=\frac{I_0\times 0.3}{0.5}=\frac{30\times 0.3}{0.5}$

= 18 A

Power delivered to the load. P₀ = V₀₁ I_or cos ϕ

V₀₁ is rms value of fundamental output voltage

I_0r is rms value of load current

$V_{01}=\frac{4V_{dc}}{\sqrt{2}\pi }$

Given that, V_dc = 400 V

$V_{01}=\frac{4\times 400}{\sqrt{2}\times \pi }=360.12V$

Power delivered, P₀ = V₀₁ I_or cosϕ

$=360.12\times \frac{200}{\sqrt{2}}\times \cos {30}^{\circ }$

= 44.1 kW

The output voltage for a signal phase full bridge inverter is

$V_o=\underset{n=1,3,5}{\sum ^{\mathrm{\infty }}}\frac{4V_s}{n\pi }\sin n\omega t$

$V_{01}=\frac{4V_s}{\pi \sqrt{2}}=\frac{4\times 12}{\pi \times \sqrt{2}}=10.8V$

Primary voltage of transformer (E_p) = 10.8 V

Secondary voltage of transformer (E_s) = 230 V

Secondary current $\left(I_s\right)=\frac{230}{100}=2.3A$

$\frac{E_P}{E_s}=\frac{I_s}{I_P}$

⇒ Primary current $\left(I_p\right)=I_s\times \frac{E_s}{E_p}$

$=2.3\times \frac{230}{10.8}=48.98A$