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The circuit for the step-up boost converter operates by varying the amount
of time in which inductor
receives energy from the source.
In the basic block diagram the operation of the boost converter can be seen
that the output voltage
appearing across the load is sensed by the sense / error amplifier and an error
voltage is generated that
controls the switch.
Typically the boost converter switch is controlled by a pulse width modulator, the switch remaining on of
longer as more current is drawn by the load and the voltage tends to drop
and often there is a fixed
frequency oscillator to drive the switching.
Boost converter operation
The operation of the boost converter is relatively straightforward.
When the switch is in the ON position, the inductor output is connected to ground and the voltage Vin is
placed across it. The inductor current increases at a rate equal to Vin/L.
When the switch is
placed in the OFF position, the voltage across the inductor changes and
is equal to
Vout-Vin. Current that was flowing in the inductor decays at a rate equal to (Vout-Vin)/L.
Figure: 3.25 circuit diagram of Boost regulator during switch off condition
Referring to the boost
converter circuit diagram, the current waveforms for
the different areas of the
circuit can be seen as below.
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Figure: 3.26 Input and output waveforms of Boost regulator
It can be seen from the waveform diagrams that the input current to the boost converter is higher than the
output current. Assuming a perfectly efficient, i.e. lossless, boost converter, the power out must equal the
power in, i.e. Vin
⋅
Iin = Vout
⋅
Iout. From this it can be seen if the output voltage is higher than the input
voltage, then the input current must be higher than the output current.
In reality no boost converter will be lossless, but efficiency levels of around 85% and more
are achievable
in most supplies.
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