Solucionario Daniel Hart Electronica De Potencia Checked B1 ⚡ Must Try

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A buck converter has an input voltage of 50 V, output voltage of 25 V, switching frequency of 20 kHz, and load resistance of 10 Ω. Determine the duty ratio, inductor value for continuous current, and the output voltage ripple if C=100 μF.

Step 1 – Duty Ratio (D)
Formula: ( V_o = D \cdot V_in )
( D = 25/50 = 0.5 ) (Correct in all versions)

Step 2 – Inductor for Continuous Conduction Mode (CCM)
Official solution manual often uses:
( L_min = \frac(1-D)R2f )
( L_min = \frac(0.5)(10)2 \times 20,000 = \frac540,000 = 125 \mu H )

Checked B1 Correction:
The correct formula from Hart’s textbook (Eq. 3-14) is:
( L_min = \frac(1-D)R2f ) yes – but many unofficial solutions forget to use the load resistance in ohms. Here it’s fine. However, they often misprint units as mH. Checked version explicitly writes ( 125 \times 10^-6 H ) and adds a note: "For boundary between CCM and DCM, choose L > 1.25x Lmin → use 150 μH." solucionario daniel hart electronica de potencia checked b1

Step 3 – Output Voltage Ripple
Official: ( \Delta V_o = \fracV_o (1-D)8LCf^2 )
( \Delta V_o = \frac25(0.5)8(125e-6)(100e-6)(400e6) ) — Wait, this is where errors creep in.

Checked B1 correction:
Frequency squared: ( f^2 = (20,000)^2 = 4 \times 10^8 )
Denominator: ( 8 \times 125e-6 \times 100e-6 \times 4e8 )
First: ( 125e-6 \times 100e-6 = 1.25e-8 )
Then ( 1.25e-8 \times 4e8 = 5 )
Times 8 = 40
So ( \Delta V_o = 12.5 / 40 = 0.3125 V ) (or 1.25% ripple).

The unchecked manual mistakenly used ( 200 \times 10^-6 F ) and got 0.156 V. The checked B1 highlights this with a red box and corrects it.

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A three-phase bridge rectifier has an AC line-to-line voltage of 208 V rms at 60 Hz, feeding an RL load with R=20 Ω and L very large. Find average output voltage, average load current, and RMS current through a diode. If you can't find a solution manual, consider

Official (flawed) solution:
( V_o,avg = \frac3\sqrt3 V_LL,peak\pi ) — wrong! That’s for peak, not RMS.

Checked B1 correct approach:
Formula: ( V_o,avg = \frac3\sqrt2 V_LL,rms\pi ) wait — actually, correct formula from Hart (Eq. 2-28):
( V_o,avg = \frac3\sqrt3 \sqrt2 V_LL,rms\pi ) — Let's simplify:
( V_LL,peak = \sqrt2 \times 208 = 294.16 V )
Then ( V_o,avg = \frac3 \times 294.16 \times \sin(\pi/3)\pi ) — but better:
Known constant: ( V_o,avg = 1.35 \times V_LL,rms ) for three-phase bridge.
So ( V_o,avg = 1.35 \times 208 = 280.8 V ).

Then ( I_avg,load = 280.8/20 = 14.04 A ).
Diode RMS current: ( I_D,rms = I_load / \sqrt3 = 14.04 / 1.732 = 8.1 A ).

The checked B1 adds a note: "Some manual versions wrongly use ( I_load/2 ); the correct factor for a three-phase bridge is ( 1/\sqrt3 ) per diode." If you are a professor or TA :

Professors know the "solucionario" exists. However, using it correctly is key:

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