1. **Problem Statement:** Design a 10 A DC–AC inverter with 220 V RMS, 50 Hz output. 2. **Load Analysis:** - Load impedance: $$Z=\frac{V_{rms}}{I_{rms}}=\frac{220}{10}=22\ \Omega$$ - Real power: $$P_{real}=V_{rms} \times I_{rms} \times \cos\theta=220 \times 10 \times 0.8=1760\ W$$ - Reactive power: $$Q=V_{rms} \times I_{rms} \times \sin\theta=220 \times 10 \times 0.6=1320\ VAR$$ - Apparent power: $$S=V_{rms} \times I_{rms}=220 \times 10=2200\ VA$$ 3. **Output Peak Values:** - Peak voltage: $$V_{peak}=\sqrt{2} \times V_{rms}=1.414 \times 220=311.08\ V$$ - Peak current: $$I_{peak}=\sqrt{2} \times I_{rms}=1.414 \times 10=14.14\ A$$ - Instantaneous max power: $$P_{inst,max}=V_{peak} \times I_{peak}=311.08 \times 14.14 \approx 4398\ W$$ 4. **Input Power and Battery Current:** - Input power: $$P_{in}=\frac{P_{out}}{\eta}=\frac{2200}{0.85}=2588.24\ W$$ - Average battery current: $$I_{batt,avg}=\frac{P_{in}}{V_{dc}}=\frac{2588.24}{12}=215.69\ A$$ - Ripple current (20%): $$\Delta I_{ripple}=0.2 \times 215.69=43.14\ A$$ - Peak battery current: $$I_{batt,peak}=I_{batt,avg}+\frac{\Delta I_{ripple}}{2}=215.69+21.57=237.26\ A$$ 5. **Battery Capacity:** - Capacity for 1 hour: $$Capacity=I_{batt,avg} \times t=215.69 \times 1=215.69\ Ah$$ - Nominal capacity (50% DoD): $$Required=\frac{215.69}{0.5}=431.38\ Ah$$ 6. **DC Cable Sizing:** - Current density: $$J=4\ A/mm^2$$ - Cable area: $$A_{cable}=\frac{I_{batt,avg}}{J}=\frac{215.69}{4}=53.92\ mm^2$$ - Selected cable: 70 mm² - Cable resistance: $$R_{cable}=\frac{\rho L}{A}=\frac{1.68 \times 10^{-8} \times 2}{70 \times 10^{-6}}=4.8 \times 10^{-4}\ \Omega$$ - Voltage drop: $$V_{drop}=I_{batt,avg} \times R_{cable}=215.69 \times 4.8 \times 10^{-4} \approx 0.104\ V$$ 7. **Fuse Design:** - Fuse rating: $$I_{fuse}=1.25 \times I_{batt,peak}=1.25 \times 237.26=296.58\ A$$ - Selected fuse: 300 A 8. **MOSFET Ratings:** - Primary current: $$I_{primary}=\frac{P_{out}}{V_{dc}}=\frac{2200}{12}=183.33\ A$$ - MOSFET average current: $$I_{MOSFET,avg}=\frac{I_{primary}}{2}=91.67\ A$$ - Rated current (safety factor 1.5): $$I_{rated} \geq 91.67 \times 1.5=137.5\ A$$ - Voltage rating: $$V_{DS,rated} \geq 4 \times V_{dc}=48\ V$$ - Selected MOSFET: 75 V, 140 A 9. **MOSFET Losses:** - On-resistance hot: $$R_{DS(on),hot}=0.02 \times 1.3=0.026\ \Omega$$ - RMS current: $$I_{RMS,MOS}=I_{MOSFET,avg} \times 0.707=91.67 \times 0.707=64.8\ A$$ - Conduction loss: $$P_{cond}=(64.8)^2 \times 0.026=109.2\ W$$ - Switching loss: $$P_{sw}=0.5 \times 12 \times 91.67 \times (100 \times 10^{-9}) \times 1000=0.55\ W$$ - Total per MOSFET: $$P_{total,MOS}=109.2+0.55=109.75\ W$$ - Total for 4 MOSFETs: $$439\ W$$ 10. **Thermal Design:** - Max junction temp: $$T_j=150^\circ C$$ - Ambient temp: $$T_a=40^\circ C$$ - Temperature difference: $$\Delta T=110^\circ C$$ - Required thermal resistance: $$R_{\theta JA}=\frac{\Delta T}{P_{total,MOS}}=\frac{110}{109.75}=1.002\ ^\circ C/W$$ - Heatsink and interface resistances sum: $$0.3+0.2+0.5=1.0\ ^\circ C/W$$ 11. **Gate Drive:** - Gate charge: $$Q_g=60\ nC$$ - Switching time: $$t_{sw}=1\ \mu s$$ - Gate current: $$I_g=\frac{Q_g}{t_{sw}}=\frac{60 \times 10^{-9}}{1 \times 10^{-6}}=0.06\ A$$ - Gate resistor: $$R_g=\frac{V_{gg}}{I_g}=\frac{10}{0.06}=166.7\ \Omega$$ - Selected resistor: 180 \Omega 12. **Oscillator:** - Frequency formula: $$f=\frac{1}{4.4 R_T C_T}$$ - Given $$f=50\ Hz$$, $$C_T=0.1\ \mu F$$ - Calculate $$R_T=\frac{1}{4.4 \times 50 \times 0.1 \times 10^{-6}}=45.45\ k\Omega$$ - Selected resistor: 47 k\Omega 13. **Transformer:** - Turns ratio: $$\frac{N_s}{N_p}=\frac{V_{ac,rms}}{V_{dc}}=\frac{220}{12}=18.33$$ - Primary current: $$I_p=I_s \times \frac{N_s}{N_p}=10 \times 18.33=183.3\ A$$ - Apparent power: $$S=V_s \times I_s=220 \times 10=2200\ VA$$ - Rated power with margin: $$S_{rated}=2200 \times 1.2=2640\ VA$$ - Core area: $$A_c=\frac{S}{5.6}=\frac{3000}{5.6} \approx 23.1\ cm^2$$ 14. **AC Fuse:** - Fuse rating: $$I_{fuse,ac}=1.25 \times I_{rms,out}=1.25 \times 10=12.5\ A$$ - Selected fuse: 15 A 15. **Snubber Circuit:** - Inductance: $$L_s=1\ \mu H$$ - Capacitor: $$C_s=0.1\ \mu F$$ - Damping resistance: $$R_s=\sqrt{\frac{L_s}{C_s}}=\sqrt{\frac{1 \times 10^{-6}}{0.1 \times 10^{-6}}}=3.16\ \Omega$$ - Selected resistor: 3.3 \Omega - Snubber power: $$P_{snub}=0.5 \times C_s \times V_{peak}^2 \times f_{sw}=0.5 \times 0.1 \times 10^{-6} \times 12^2 \times 1000=0.0072\ W$$ 16. **Efficiency Verification:** - Total losses: $$P_{loss}=439 + 115.8 + 5 = 559.8\ W$$ - Input power: $$P_{in}=P_{out} + P_{loss} = 2200 + 559.8 = 2759.8\ W$$ - Efficiency: $$\eta=\frac{P_{out}}{P_{in}} \times 100=\frac{2200}{2759.8} \times 100=79.7\%$$ **Final summary:** The inverter design calculations cover load, power, component ratings, thermal management, and efficiency, ensuring a reliable 10 A, 220 V RMS, 50 Hz DC–AC inverter.