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In the diagram of the current waveforms for the buck converter / switching regulator, it can be seen that
the inductor current is the sum of the diode and input / switch current. Current either flows through the
switch or the diode.
It is also worth noting that the average input current is less than the average output current. This is to be
expected because the buck converter circuit is very efficient and the input
voltage is greater than the
output voltage. Assuming a perfect circuit, then power in would equal power out, i.e. Vin
⋅
In = Vout
⋅
Iout. While in a real circuit
there will be some losses, efficiency levels greater than 85% are to be
expected for a well-designed circuit.
It will also be seen that there is a smoothing capacitor placed on the output. This serves to ensure that the
voltage does not vary appreciable, especially during and switch transition times. It will also be required to
smooth any switching spikes that occur.
Boost regulator
One of the advantages of switch mode power supply technology is that it can be used to create a step up
or boost converter / regulator.
Boost converters or regulators are used in many instances from providing
small supplies where higher
voltages may be needed to much higher power requirements.
Often there are requirements for voltages higher than those provided by the
available power supply -
voltages for RF power amplifiers within mobile phones is just one example.
Step-up boost converter basics
The boost converter circuit has many similarities to the buck converter. However the circuit topology for
the boost converter is slightly different. The fundamental circuit for a boost converter or step up converter
consists
of an inductor, diode, capacitor, switch and error amplifier with switch control circuitry.
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