# University of Nottingham H63PED Power Electronic Design Note

## Power stuff background

Phase Voltage, E.

Line (line-line) Voltage, $$V_{line} = \sqrt{3} E$$

RMS Line Voltage, $$V_{RMS} = \frac{V_{line}} {\sqrt{2}} = \frac{\sqrt{3} E} {\sqrt{2}}$$

$$P_{tot} = 3 \, V_{RMS} \, I_{RMS} \, \cos{\phi} = \sqrt{3} \, V_{LINE} \, I_{LINE} \, \cos{\phi}$$

Notation “415V, 50Hz”: $$V_{line-line,RMS} = 415 V$$ , $$V_{line-line,Peak} = V_{line-line,RMS} \cdot \sqrt{2} V$$

### Power Quality

Current harmonics

$$I_D$$ is current in DC side supply and I peak in AC side.

$$I_n = \frac{4}{n\pi} I_D \sin{(n\theta/2)}$$

where, n = 1,5,7,11,13 … , for fundamental frequency: n = 1

Distortion Factor $$\psi$$

$$\psi$$ = Fundamental RMS / Total RMS

$$\psi = \frac{ I_1/ \sqrt{2}}{ [\sqrt{I_1^2 +I_n^2}]/\sqrt{2}}$$

Total Harmonic Distortion (THD)

Harmonics RMS =  $$\sqrt{ (\text{Total RMS})^2 – (\text{Fundamental RMS})^2}$$

THD = harmonics RMS / Fundamental RMS

## Bridge Rectifier Calculations

Area under cosine
$$\frac {A}{\omega} (\cos{\alpha} – \cos{\beta} )$$

DC average, diode bridge
$$\text{Area in 1 Section} = \frac {\sqrt{3}E}{\omega} (\cos{60^{\circ}} – \cos{120^{\circ}} )$$

$$V_{XY} = \frac{Area}{time} = \frac{3 \sqrt{3} E}{\pi}$$

Mean voltage Loss due to overlap

$$\text{Area in 1 Section} = I_D h$$
$$\text{loss in mean} = \frac{\text{Area in 6 Section}}{\text{time for 1 cycle}} = \frac{6 \cdot I_D h}{2\pi / \omega} = \frac{3 I_D h \omega}{\pi}$$

DC average, diode bridge with overlap

$$V_{XY} = \frac{3 \sqrt{3} E}{\pi} – \frac{3 I_D h \omega}{\pi}$$

Overlap Duration, considering Vline Chart

$$2 \cdot I_D h = \frac {\sqrt{3}E}{\omega} ( \cos{0^{\circ}} – \cos{\mu} )$$

Thyristor Bridge Mean Voltage

$$\text{Loss in delay} = \text{Area under } V_{line-line} = \frac{\sqrt{3} E}{\omega} [\cos{0} – \cos(\alpha)] = \frac{\sqrt{3} E}{\omega} [1 – \cos(\alpha)]$$

$$\text{Mean loss in cycle} = \frac{ 6 \cdot \text{1 Area} }{2\pi / \omega} = \frac{3 \sqrt{3} E}{\pi} [1 – \cos(\alpha)]$$

$$\text{Mean Voltage, } V_{XY} = \frac{3 \sqrt{3} E}{\pi} – \frac{3 \sqrt{3} E}{\pi} [1 – \cos(\alpha)] = \frac{3 \sqrt{3} E}{\pi} \cos(\alpha)$$

Thyristor Bridge Mean Voltage with overlap

$$V_{XY} = \frac{3 \sqrt{3} E}{\pi} \cos(\alpha) – \frac{3 I_D h \omega}{\pi}$$

Ripple and smoothing

VTA = $$\Delta I_{p-p}$$ x $$L_{DC}$$

### ZVS Resonant Converters

Stage 1:

• Assume Q was on in previous stage, $$V_C = 0$$
• Constant current charge $$V_C$$
• $$V_{AB}$$ drops towards 0

Stage 2:

• D is ON, once $$V_{AB} &lt 0$$
• $$V_C$$ oscillates
• $$\Delta V = I_0 \sqrt{\frac{L}{C}}$$
• $$\omega_0 = \frac{1}{\sqrt{LC}}$$
• $$V_C$$ drop below 0, and turns $$D_2$$ ON

Stage 3:

• $$D_2$$ ON, shorts C
• $$V_L$$ increase linearly
• conduct as i reaches 0
• time: $$\tau_3 = \frac{-I’ L}{E}$$

Stage 4:

• Q takes over from D2
• stage 4 finishes when I reached $$I_0$$
• time: $$\tau_4 = \frac{I_0 L}{E}$$

Stage 5:

• Q conducts, D turns off
• Duration $$\tau_5 is controllable$$

ZVS Waveform

### Inverter

PWM

modulation index = $$V_{b} / V_{DC}$$ (peak voltage)

Current peak = $$\frac{V I }{2 E}$$

Current RMS =   $$\frac{V_{rms} I_{rms} }{E} \cdot \frac{1}{\sqrt{2}}$$

Power triangle:

$$P = \frac{E V \sin(\delta)}{X}$$