H63PED Power Electronic Design

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”: \(V_{line-line,RMS} = 415 V\) , \(V_{line-line,Peak} = V_{line-line,RMS} \cdot \sqrt{2} V\)

Bridge Rectifier Commutation

Commutation Chart
Commutation Chart

OverLap Commutation Voltage table


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

resonant converter
resonant converter

 

Stage 1:

ZVS Stage 1
ZVS Stage 1
  • Assume Q was on in previous stage, \(V_C = 0\)
  • Constant current charge \(V_C\)
  • \(V_{AB}\) drops towards 0

Stage 2:

ZVS Stage 2
ZVS 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:

ZVS Stage 3
ZVS 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:

ZVS stage 4
ZVS 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:

ZVS stage 5
ZVS stage 5
  • Q conducts, D turns off
  • Duration \(\tau_5 is controllable \)

ZVS Waveform

ZVS waveform
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:

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