Design goals: mains input, switching frequency 32KHz, 0~220V 50~100Hz 1200W output.
1. Determine the input voltage
After checking, my country's mains voltage standard, when using 220V single-phase power supply, is +7% and -10% of the rated value. Therefore, the input voltage range is 198VAC-235.4VAC.
2. Determine the input current
In the formula, Po is the output power of the switching power supply, eta is the power supply efficiency, umin is the minimum value of the AC input voltage, and cosφ is the power factor of the switching power supply.
Po: switching power supply design power 1200W
Umin: The minimum input of mains power is 198V
η: Normal switching power supply efficiency is higher than 80%, here we directly take the lowest value of 0.8
cosφ: single-phase uncorrected estimate 0.6, three-phase uncorrected estimate 0.85, corrected estimate 0.95
After calculation, Irms = 12.6 ≈ 13A.
3. Insurance tube
Rated voltage: Just be greater than the input voltage. Input voltage 220V, optional specifications 250V, 300V, 350V.
Minimum rated current: greater than the rated current, which is 12.6A.
Commonly used sizes: 5*20, 6*30, 10*38.
Final selection: 250V, 30A, 10*38mm (same model as fuse holder).
4.EMI circuit
X capacitor: On switching power supplies, X2 safety capacitors are generally used. Two-stage X capacitor, the first stage uses 0.47uF, the second stage uses 0.1uF; the single stage uses 0.47uF.
X capacitor bleed resistor: IEC60950 stipulates that the voltage needs to drop to 37% within 1s, and IEC60065 stipulates that the voltage needs to drop to 35V within 2s. If according to IEC60950, the time constant RC < 1, then the front-stage discharge resistance R < 2.1MΩ, and the rear-stage discharge resistance R < 10MΩ.
Y capacitor: From the perspective of withstand voltage, only Y1 capacitor can be selected. GJB151 stipulates that the capacity of the Y capacitor should not be greater than 0.1uF; and for machines working in subtropical zones, the ground leakage current is required not to exceed 0.7mA; for machines working in temperate zones, the ground leakage current is required not to exceed 0.35mA. I = U*(2πf * C) < 0.35mA, calculated C < 4.7nF.
Common mode inductor: rated current is 12.6A greater than the average input current. EMI test frequency, conduction 150KHz~30MHz, usually the cutoff frequency can be set around fo = 50KHz (generally not less than 10KHz); among them, the Y capacitor of C2 is assumed to be 2.2nF; calculated L1 = 4.6mH.
Final selection: X2 capacitor, 0.47uF for the first stage and 0.1uF for the second stage;
5. Precharge resistor
Withstand voltage: maximum input voltage, which is 235.4V.
Resistance value: After simulink simulation, when the capacitor has been selected, the charging time is 10~20S (error 5%) when it is about 1500Ω.
Instantaneous power: The instantaneous power of the cement resistor is approximately 10 times the rated power. That is, instantaneous power = (Umax ^ 2) / R, where Umax is the maximum peak voltage (235.4*1.414) and R is the precharge resistance (1500*0.95). After calculation, it is 77.75W, that is, the rated power is 7.775W.
Final selection: withstand voltage 750V, resistance 750Ω, power 5W, number of series connections 2.
6. Precharge relay
Rated voltage: maximum input voltage, which is 235.4V.
Rated current: input current 12.6A.
Final selection: rated voltage 250VAC, rated current 15A.
7. Determine bus parameters
Bus voltage: Output voltage * 1.414 / 0.9 / 0.9 (grid fluctuation) / 0.9 (bus fluctuation) = 427V.
AC transformer: bus voltage/1.414 ≈ 300V
Minimum equivalent resistance: (minimum bus voltage ^ 2) / (maximum output power / efficiency), calculated to be 97.1706456Ω.
Pulsation period: Mains power is 50Hz. After rectification by the rectifier bridge, the frequency is 100Hz. That is, the pulsation period is 0.01S.
Bus voltage fluctuation: The target value is set to, and the decrease shall not be less than 90%.
Precharge time: The target value is set to 10-20S, charge to 405V (95%).
8. Isolation transformer
Frequency: 50Hz
Primary: 220V 12.6A
Secondary: 300V 9.24A
Power: input current * input voltage = 2772W ≈ 3000W
9.Input rectifier bridge
Forward current: Peak current * 1.2 (margin), that is, 9.24 * 1.414 * 1.2 = 15.67 ≈ 16A
Reverse withstand voltage: Peak voltage * 1.2 (margin) * 2 (anti-leakage current coefficient), that is, 321 * 1.414 * 1.2 * 2 = 1089.3456V
Final selection: forward current 35A, reverse withstand voltage 1000V.
10. Bus capacitance
Withstand voltage: Peak voltage * 1.2 (margin) = 300 * 1.414 * 1.07 * 1.2 = 544.6728V
Capacity: e ^ ((-1 * t) / (R * C)) > Voltage fluctuation, where t is the pulsation period and R is the equivalent resistance. Calculated, C > 976.758μF.
Final selection: withstand voltage 400V, capacity 1.2mF, leakage current 1.5mA. 2 in series and 2 in parallel. (Capacity error ±20%, so C > 1221μF).
11. Bus capacitance balancing resistor
Resistance value: (bus peak voltage / number of series connections) / (leakage current * number of parallel connections * 20) = 3888.5Ω
Power: (bus peak voltage/number of series connections) ^ 2/resistance = 13.9986W
12.Full-bridge MOS
Drain-source rated voltage: Peak voltage * 1.5 (margin) = 321 * 1.414 * 1.5 = 680.841V
Rated current: Output current * 1.1 (margin) * 1.414 * 1.2 (ripple) = 10.18 ≈ 11A.
13.Output filter circuit
Cutoff frequency: f < switching frequency / 6 and f > maximum output frequency * 6, that is, [600,5333]
According to the formula f < 1 / (2 * π * sqrt(lc)), that is, lc ∈ [9e-10,7e-8]
Current ripple is less than 20%: (Uin - Uout) / L * (Uout / Uin) * T < Imax * 20%
Uin: transformer secondary voltage
Uout: output voltage
L: inductance
T: pulsation period
Imax: maximum output current
Transform the above formula L > ((Uin - Uout) * (Uout / Uin) * T) / (Imax * 20%)
At the beginning, when Uin = Uinmax, ((Uin - Uout) * (Uout / Uin) * T) / (Imax * 20%)
Secondly, when Uout = Uin / 2, ((Uin - Uout) * (Uout / Uin) * T) / (Imax * 20%) is the maximum
Therefore, L > ((Uinmax / 4) * T) / (Imax * 20%)
After calculation, L > 2.3mH
According to the filtering parameter LC ∈ [9e-10,7e-8]. If L = 3mH, then C is approximately equal to 0.3uF~23.3uF.
Ripple coefficient is less than 0.2%: capacitor ESR ripple + capacitor charge and discharge ripple < Vmax * 0.2% = 0.62216V
Among them, 1.Vesr = current ripple * ESR
2.Vc = Q / C = (1 / C) * ∫Icdt; In the inductor current, the area surrounded by the average current is charging, and the area is (1 / 2) * (current ripple / 2) * (T / 2) Therefore, ripple voltage = current ripple * ESR +
( Current ripple * T) / (8 * C) = Current ripple * (ESR + T / (8 * C))
When the capacitor N is connected in parallel, ESR = single ESR / N, C = single C * N. That is, ripple voltage = current ripple * (monomer ESR + T / (8 * C monomer)) / N
Under the condition that the ripple voltage is met, the capacitance value should be reduced as much as possible to meet the larger cut-off frequency.
If the false load loss does not exceed 5%, then Pdummyload < Poutmax * 0.05 = 60W, Umax ^ 2 / R < 60, R > 806.67Ω
Filter inductor: inductance * 1.3 (margin) = 2.3 * 1.3 = 3mH;
Current: (current + 1/2 ripple) * 1.3 (temperature rise margin) = (Imax + (Imax * 20% / 2)) * 1.3 = 7.8A ≈ 8A
Filter capacitor: Peak voltage * 1.2 (margin) = 220 * 1.414 * 1.2 = 373.296V
Capacity: N = 1, film capacitor ignores ESR. After calculation, C > 9.6856uF, and 0.3uF~23.3uF.
Final selection: inductor 3mH, 8A; capacitor 10uF, 900V; dummy load 10000Ω, 700V, rated power 10W, 2 in parallel.