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Read description first.
Important:
-Wait a couple of seconds for things to settle down.
-The green waveform is the input.
-The orange waveform is the actual output.
-The blue waveform is the output waveform shifted into the positive side of the graph to demonstrate the point of having a boost + buck converter in one package, and that is the circuit is capable of spitting out voltages above or below the input.
-Don't worry about voltmeter connected to the battery. It is done to shift the waveform to the positive half of the graph.
-If you want to make adjustments to the output, then disable the voltmeter and probe the output voltage, output of the comparator and the 2 inputs of the comparator.
You don't have to read the rest of the description if you don't want to.
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This is a DC-DC Step-Up/Step-Down switching mode voltage converter. It's output is negative because this is the single switch inverted output topology.
The feedback resistors are configured to give an output voltage of 7.5V (-7.5 on the scope, because of the polarity of the output; this is one way of creating negative voltages). The circuit can output voltages above or below the input voltage, because this is a boost + buck converter in one package.
Because of the inverted output, the feedback system is rather complicated. Also, the fed back voltage is biased by 300mV because initially, the output voltage is zero and that won't get the MOSFET switching.
If you probe the output of the comparator, The duty cycle looks like it is increasing on the scope, but that signal is fed to a PMOS, and the PMOS only turns on when the signal is low, so the duty cycle is infact decreasing as the desired voltage is reached.
This circuit doesn't have current sensing.
The switching frequency is very low, just like in my previous buck and boost converter circuits. I've used a low frequency so that the sim-speed can be kept high. In the real world, switching frequencies of 100kHz and above are used.
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