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Energy and power in a boost converter

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Paul E. Schoen

Jan 1, 1970
0
I posted this at the end of the gaps thread, but that turned into a
cussfest, so here's another shot at trying to understand the energy and
power transfer in a simple boost converter I have built.

Basically, I can predict the maximum current in the inductor, and hence the
energy stored, vs frequency. Using LTspice with a 61 ohm load, I found that
at 200 kHz and 70% duty cycle the maximum inductor current with 12 VDC at
10 uH is 4.4A (Energy = 97 uW-sec * 0.2 = 19.4 W), and I get 40 volts (26.2
W). At 100 kHz, I can get 48 volts (37.7W) with a maximum inductor current
of 8 A (32 W). The actual inductor current in the first case, which is
running in continuous mode, includes a DC component of 650 mA from the 12
volt source. Adding that gives a power contribution from the battery of 7.8
watts in the first case and 9.4 watts in the second.

The maximum output will be generated when the inductor starts charging
again after its energy has been discharged into the output capacitor, so
there will be no "dead time". With 12 volts, the inductor charges to 8.4 A
in 7 uSec, and it takes 3 uSec to charge the output capacitor, for 70% duty
cycle. The output is about 48 VDC into 61 ohms, or 38 watts. I calculate
the average input power to be about 70% of sqrt(8.4*8.4/2) * 12 V = 35.3
watts, plus the 780mA * 12V = 9.4W from the battery, or 44.7. I'm guessing
at this, but the simulator measured input watts to be 43, so I'm close.
This is 82% efficiency.

I'm running simulations in LTspice, and I think they are pretty much
correct, but I am still a little puzzled. In the continuous mode operation
at 200 kHz, I can see the DC component through the inductor as a 388 mA
minimum current. I get input power of 28.77 W and output of 25.77 W or
89.5% efficiency. In the discontinuous mode at 100 kHz, I get 38.7 watts
out, 43.1 watts in, and 89.8% efficiency. However, I have a hard time
grasping how it can output 38.7 watts when there is almost no inductor
current (It's actually negative) 10% of the time, and peak energy of 320
uW-Sec at 100 kHz or 32 watts. Maybe I'm simplifying the calculation too
much. The true power is probably the integral of the peak energy
(0.5*I^2*L) over the entire waveform, times frequency. OK, when I do that,
I get an average of about 98 uJ, but a peak of 316 uJ = 31.6 W.

Paul
 
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