Power Electronics Test 6

In a set up chopper,

\({V_0} = \frac{{{V_s}}}{{1 - D}}\)

Given that, V₀ = 660 V

V­_s = 220 V

\(\Rightarrow 660 = \frac{{220}}{{1 - D}}\)

\(\Rightarrow D = \frac{2}{3}\)

T_off = 100 μs

\(D = \frac{{{T_{ON}}}}{{{T_{ON}} + {T_{off}}}} = \frac{2}{3}\)

\(\Rightarrow {T_{ON}} = 200\;{\rm{\mu s}}\)

Concept:

For Single – quadrant (type A) Chopper

Average value V_o = ∝ Vs

R.M.S value \({V_{rms}} = \sqrt \propto {V_s}\)

where \(\propto = \frac{{Ton}}{T}\)

We known the form factor \(F.F = \frac{{{{\left( {{V_o}} \right)}_{rms}}}}{{{{\left( {{V_o}} \right)}_{avg}}}}\)

Calculation:

Form factor \(= \frac{{\sqrt \propto {V_s}}}{{ \propto {V_s}}} = \frac{1}{{\sqrt \propto }}\)

\(\propto = \frac{{1.5}}{{1.5 + 2}} = \frac{{1.5}}{{3.5}} = 0.429\)

Concept:

For step-down chopper

Average output V₀ = α V_S

Where α = duty cycle

\(\alpha = \frac{{{T_{On}}}}{{{T_{on}} + {T_{off}}}}\)

Time period \(T = {T_{on}} + {T_{off}} = \frac{1}{f}\)

Calculation:

Given

V₀ = 140 V

V_S = 200 V

T_on = 70 μsec

\(\alpha = \frac{{{V_0}}}{{{V_S}}} = \frac{{140}}{{200}} = 0.7\)

\(\frac{{{T_{on}}}}{{{T_{on}} + {T_{off}}}} = 0.7\)

0.3 T_on = 0.7 T_off

⇒ 0.3 × 70 = 0.7 T_off

T_off = 30 μ sec

Time period T = T_on + T_off = 70 + 30

= 100 μ sec

Frequency \(f = \frac{1}{T} = \frac{1}{{100}} \times {10^6}\)

= 10 kHz

Now frequency is increased by 25%

Hence f¹ = 1.25 × 10

= 12.5 kHz

New time period \({T^1} = \frac{1}{{{f^1}}} = \frac{1}{{12.5}} \times {10^3}\)

= 80 μ sec

Conduction Time is constant

T_on = 70 μ sec

New duty cycle

\(\alpha ' = \frac{{{T_{on}}}}{{{T^1}}} = \frac{{70}}{{80}}\)

New Average output voltage

V₀ = α‘V_S

\( = \frac{{70}}{{80}} \times 200\)

= 175 V

Concept:

In a buck converter,

The output voltage (V₀) = DV_in

\(\frac{{{\rm{\Delta }}{V_0}}}{{{V_0}}} = \frac{{\left( {1 - D} \right){\pi ^2}}}{2}{\left( {\frac{{{f_c}}}{{{f_s}}}} \right)^2}\)

Where D is the duty ratio

V_in is the input voltage

V₀ is the output voltage

ΔV₀ is the output ripple voltage

f_s is the supply frequency

f_c is the frequency of the converter

\({f_c} = \frac{1}{{2\pi \sqrt {LC} }}\)

L is the inductance of the converter

C is the capacitance of the converter

Calculation:

Given that, input voltage (V_in) = 30 V

Duty ratio (D) = 0.5

Output voltage (V₀) = DV_in = 30 × 0.5 = 15 V

Supply frequency (f_s) = 10 kHz

Inductance (L) = 1 mH

Capacitance (C) = 10 μF

The frequency of the converter,

\({f_c} = \frac{1}{{2\pi \sqrt {LC} }} = \frac{1}{{2\pi \sqrt {1 \times {{10}^{ - 3}} \times 10 \times {{10}^{ - 6}}} }} = 1.591\;kHz\)

\(\frac{{{\rm{\Delta }}{V_0}}}{{{V_0}}} = \frac{{\left( {1 - D} \right){\pi ^2}}}{2}{\left( {\frac{{{f_c}}}{{{f_s}}}} \right)^2}\)

\( \Rightarrow {\rm{\Delta }}{V_0} = 15 \times \frac{{\left( {1 - 0.5} \right){\pi ^2}}}{2}{\left( {\frac{{1.591}}{{10}}} \right)^2}\)

Concept:

The efficiency of a chopper is given by,

\(\eta = \frac{{{P_o}}}{{{P_{in}}}} \times 100\)

Where P_o is the output power delivered to load and it is given by,

\({P_o} = \frac{{V_{or}^2}}{R}\)

V_or is the rms value of output voltage

R is the load resistance

P_in is the input power and it is given by,

P_in = V_in I₀

V_in is the input voltage

I₀ is the average load current

In a step-down chopper,

The average load voltage, \({V_0} = D\left( {{V_{in}} - {V_d}} \right)\)

The average load current, \({I_o} = \frac{{{V_0}}}{R}\)

The rms value of load voltage, \({V_{rms}} = \sqrt D \left( {{V_{in}} - {V_d}} \right)\)

D is the duty cycle

V_d is the voltage drop across the switch in ON condition

Calculation:

Given that, input voltage (V_in) = 230 V

Voltage drop (V_d) = 2 V

Duty cycle (D) = 0.4

Load resistance (R) = 10 Ω

The average output voltage, V₀ = 0.4 (230 – 2) = 91.2 V

The average output current, I₀ = 9.12 A

The rms value of the output voltage, \({V_{rms}} = \sqrt {0.4} \left( {230 - 2} \right) = 144.2\;V\)

Output power, \({P_0} = \frac{{{{144.2}^2}}}{{10}} = 2079.36\;W\)

Input power = 230 × 9.12 = 2097.6 W

In a boost converter, for continuous conduction mode

\(K > \frac{{2fL}}{R}\)

K > (1 – D)²D

For the minimum value of K,

\(\frac{d}{{dt}}\left[ {{{\left( {1 - D} \right)}^2}D} \right] = 0\)

\( \Rightarrow D = \frac{1}{3}\)

The minimum value of K is \( = {\left( {1 - \frac{1}{3}} \right)^2} \times \frac{1}{3} = 0.148\)

Given that, f = 100 kHz

Inductance, L = 10 μH

\(K > \frac{{2fL}}{R}\)

\( \Rightarrow 0.148 > \frac{{2 \times 100 \times {{10}^3} \times 10 \times {{10}^{ - 6}}}}{R}\)

⇒ R > 13.51 Ω

Minimum load current \({I_0} = \frac{{{V_o}}}{R} = \frac{{60}}{{13.51}} = 4.44\;A\)

For V_s = 2.7 V:

The duty ratio, \(D = 1 - \frac{{{V_s}}}{{{V_0}}} = 1 - \frac{{2.7}}{8} = 0.663\)

The average inductor current is, \({I_L} = \frac{{{V_0}{I_0}}}{{{V_s}}} = \frac{{8\left( 1 \right)}}{{2.7}} = 2.96\;A\)

The variation in inductor current to meet the 40 percent specification is,

ΔI_L = 0.4 × 2.96 = 1.19 A

The inductance, \(L = \frac{{{V_s}D}}{{{\rm{\Delta }}{i_L}f}} = \frac{{2.7 \times 0.663}}{{1.19 \times 200 \times {{10}^3}}} = 7.5\;\mu H\)

For V_s = 4.2 V:

The duty ratio,

The average inductor current is, \(D = 1 - \frac{{{V_s}}}{{{V_0}}} = 1 - \frac{{4.2}}{8} = 0.475\)

The variation in inductor current to meet the 40 percent specification is,

ΔI_L = 0.4 × 1.90 = 0.762 A

The inductance, \(L = \frac{{{V_s}D}}{{{\rm{\Delta }}{i_L}f}} = \frac{{4.2 \times 0.475}}{{0.762 \times 200 \times {{10}^3}}} = 13.1\;\mu H\)

Concept:

In a class – A chopper,

Average load voltage V_o = D V_dc

Minimum value of load current \(={{I}_{o}}-\frac{\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}}{2}\)

Maximum value of load current \(={{I}_{o}}+\frac{\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}}{2}\)

Average load current \({{I}_{o}}=\frac{{{V}_{o}}}{R}\)

\(\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}=\frac{{{V}_{dc}}}{L}D~\left( 1-D \right)T\)

Calculation:

Given that, Source voltage (V_dc) = 100 V

R = 0.5 Ω

L = 1 mH

T_ON = 1 ms

T = 3 ms

\(D=\frac{{{T}_{On}}}{T}=\frac{1}{3}\)

\({{V}_{o}}=100\times \frac{1}{3}=33.33~V\)

\({{I}_{o}}=\frac{{{V}_{o}}}{R}=\frac{33.33}{0.5}=66.66~A\)

\(\text{ }\!\!\Delta\!\!\text{ }{{I}_{L}}=\frac{100}{1\times {{10}^{-3}}}\times \frac{1}{3}\times \left( 1-\frac{1}{3} \right)\times 3\times {{10}^{-3}}\)

= 66.66 A