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Project 1

Let the output ripple voltage be less than 1% => $\frac{δ V_{o}}{V_{o}} = 0.01$ $(deltaV_o)/V_o=0.01$ and switching frequency fs=100kHz1. BUCK CCM:Vi=40V to 50V , Vo=30V, Po=500 WD= Vo/Vi, $R = \frac{V_{o}^{2}}{P_{o}} = \frac{30^{2}}{500} = 1.8 o h m$ $R=V_o^2/P_o=30^2/500=1.8 ohm$ => range of D= $\frac{30}{40} \to \frac{30}{50} = 0.75 \to 0.6$ $30/40 to 30/50=0.75 to 0.6$ Caclulation of circuit parameters:Indutance: …

KYATHAM UMA MAHESHWAR REDDY
updated on 24 Feb 2022

Let the output ripple voltage be less than 1% => $\frac{δ V_{o}}{V_{o}} = 0.01$ $(deltaV_o)/V_o=0.01$ and switching frequency fs=100kHz

1. BUCK CCM:

Vi=40V to 50V , Vo=30V, Po=500 W

D= Vo/Vi, $R = \frac{V_{o}^{2}}{P_{o}} = \frac{30^{2}}{500} = 1.8 o h m$ $R=V_o^2/P_o=30^2/500=1.8 ohm$

=> range of D= $\frac{30}{40} \to \frac{30}{50} = 0.75 \to 0.6$ $30/40 to 30/50=0.75 to 0.6$

Caclulation of circuit parameters:

Indutance:
$L_{min} = \frac{(1 - D) * R}{2 * f_{s}}$ $L_min=((1-D)**R)/(2**f_s)$

subustition the duty ratio in above formula, we get

$L_{min} = 2.25 μ H, 3 μ H$ $L_min=2.25muH, 3muH$

For continuous conduction we need to take inductance value greater than the highest value of Lmin obtained so that inductor current is continuous for the varying input voltage. so lets take $L = 5 μ H$ $L=5muH$

Capacitance:
$C_{o} \geq \frac{1 - D}{8 * f_{s}^{2} * L * \frac{δ V_{o}}{V_{o}}}$ $C_o>=(1-D)/(8**f_s^2**L**(deltaV_o)/V_o$

substituting the values of D, we get $C_{o} \geq 62.5 μ F, 100 μ F$ $C_o>=62.5muF,100muF$

so lets take $C_{o} = 150 μ F$ $C_o=150muF$

Simulation:
Input is varied from 40V to 50V at t= 2sec and at the same instant duty ratio is also varied from 0.75 to 0.6 so as to maintain the required output voltage. Total simulation time is 3sec.

input variation:

Duty ratio variation:

Simulation results:

Output voltage:

output voltage is slightly less than 30V because of not considering the ON state drops of diode and voltage.

Zoomed plot:

The ripple voltage is 29.04-29.12=0.08V which is than 1 percent of output voltage => 0.01*30=0.3V , so the desgined capacitor satifies the assumed design consideration.

Inductor current:

Inductor current is well above 0A which implies that the inductor current is continous which make this converter as Buck converter in continuous conduction mode. The ripple is inductor current is increases when input voltage changes from 40V to 50V.

Ripple current in inductor, $δ I_{L} = \frac{(V_{i} - V_{o}) * D}{L * f_{s}}$ $deltaI_L=((V_i-V_o)**D)/(L**f_s)$
so at 2sec product of (Vi-Vo)*D also increases
=> 10*0.75 to 20*0.6 =>7.5 to 12V

since deltaIL is proportional to Vi-Vo * D , ripple also increases.

Power output:

The above block diagram depicts the formula V^2/R to calculate output power. Since voltage is slightly less, power is also less than 500W (desired).

(10). BUCK CCM:

Vi=400V to 500V, Vo=300V, Po=5000W

D= Vo/Vi, $R = \frac{V_{o}^{2}}{P_{o}} = \frac{300^{2}}{5000} = 18 o h m$ $R=V_o^2/P_o=300^2/5000=18 ohm$

=> range of D= $\frac{300}{400} \to \frac{300}{500} = 0.75 \to 0.6$ $300/400 to 300/500=0.75 to 0.6$

Caclulation of circuit parameters:

Indutance:
$L_{min} = \frac{(1 - D) * R}{2 * f_{s}}$ $L_min=((1-D)**R)/(2**f_s)$

subustition the duty ratio in above formula, we get

$L_{min} = 22.5 μ H, 36 μ H$ $L_min=22.5muH, 36muH$

Capacitance:
$C_{o} \geq \frac{1 - D}{8 * f_{s}^{2} * L * \frac{δ V_{o}}{V_{o}}}$ $C_o>=(1-D)/(8**f_s^2**L**(deltaV_o)/V_o$

substituting the values of D, we get $C_{o} \geq 26.59 μ F, 42.5 μ F$ $C_o>=26.59muF,42.5muF$

so lets take $C_{o} = 47 μ F$ $C_o=47muF$

Simulation results:
Input is varied from 400V to 500V at t= 2sec and at the same instant duty ratio is also varied from 0.75 to 0.6 so as to maintain the required output voltage. Total simulation time is 3sec.

Input variation:

Duty ratio variation:

Simulation results:

Output voltage:

transients when input is changed at 2 sec.

output voltage is slightly less than 30V because of not considering the ON state drops of diode and voltage.

zoomed in plot:

The ripple voltage is 298.7-299=1.3V which is than 1 percent of output voltage => 0.01*300=3V , so the desgined capacitor satifies the assumed design consideration.

Inductor current:

Inductor current is well above 0A which implies that the inductor current is continous which make this converter as Buck converter in continuous conduction mode. The ripple is inductor current is increases when input voltage changes from 400V to 500V.

Ripple current in inductor, $δ I_{L} = \frac{(V_{i} - V_{o}) * D}{L * f_{s}}$ $deltaI_L=((V_i-V_o)**D)/(L**f_s)$
so at 2sec prodcut of (Vi-Vo)*D also increases
=> 100*0.75 to 200*0.6 =>75 to 120V

since deltaIL is proportional to Vi-Vo * D , ripple also increases.

Power output:

The above block diagram depicts the formula V^2/R to calculate output power. Since voltage is slightly less, power is also less than 5000W (desired).

(2). Boost Converter: CCM

Vi= 20V to 40V, Vo=30V, Po=600W

=> R= $\frac{V_{o}^{2}}{P_{o}} = \frac{40^{2}}{600} = \frac{8}{3} o h m$ $V_o^2/P_o=40^2/600=8/3 ohm$

Circuit:

$V_{o} = \frac{V_{i}}{1 - D}$ $V_o=V_i/(1-D)$ , D= duty ratio

=> D= $D = \frac{V_{o} - V_{i}}{V_{o}}$ $D=(V_o-V_i)/V_o$

substituting the values of Vi is the above equation, we get a range of duty ratio

$\Rightarrow D = \frac{40 - 30}{40} \to \frac{40 - 20}{40}$ $=>D=(40-30)/40 to(40-20)/40$

D=0.25 to 0.5

Calculation of circuit parameters:

Inductance:
$L_{min} = \frac{D \cdot {(1 - D)}^{2} * R}{2 * f_{s}}$ $L_min=(D*(1-D)^2**R)/(2**f_s)$

substituting the values with the limit values of D, we get

$\Rightarrow L_{min} = 1.667 μ H, 1.875 μ H$ $=> L_min=1.667muH, 1.875muH$

Capacitance:
$C_{o} = \frac{D}{f_{s} * R * \frac{δ V_{o}}{V_{o}}}$ $C_o=D/(f_s**R**(deltaV_o)/V_o)$

substituting the values with the limit values of D, we get

$\Rightarrow C_{o} = 187 μ F, 93.75 μ F$ $=> C_o=187muF,93.75muF$

We need to take Co value greater than the above values obtained so that the output ripple is in limits
so let $C_{o} = 220 μ F$ $C_o=220muF$

Simulation:

Model:

Input is varied from 20V to 30V at t= 2sec and at the same instant duty ratio is also varied from 0.5 to 0.25 so as to maintain the required output voltage. Total simulation time is 3sec.

Input variation:

Duty ratio variation:

results:

Output voltage:

zoomed plots, ripple voltage before and after 0.5sec

ripple voltage voltage in both the cases (39.1-38.8=0.3V) & (39.1-38.9=0.2V) is less than the assumed ripple of 1% => 0.01*40=0.4V. So the design considerations are met with designed capacitor.

Inductor current:

Zoomed plots, showing the ripple cuurents both before and after 0.5sec

In both the cases inductor currents are well above 0A which implies that current is continuous for the varying inputs.

The ripple is inductor current and average current decreases when input voltage changes from 20V to 30V.

Ripple current in inductor, $δ I_{L} = \frac{(V_{i}) * D}{L * f_{s}}$ $deltaI_L=((V_i)**D)/(L**f_s)$
so at 2sec prodcut of (Vi-Vo)*D also increases
=> 20*0.5 to 30*0.25 =>10V to 7.5V

since deltaIL is proportional to Vi* D , ripple also increases.

Also average current, ILavg= $\frac{I_{o}}{1 - D}$ $I_o/(1-D)$ , Io (output current) is same in both cases but 1-D varies and is inversely proportional to (1-D)
at 0.5 sec D changes from 0.5 to 0.25