buck boost

I do not understand exactly what you are asking. I see two possibilities.

1) You want to move both the inductor and the switch as is shown in the wikipedia article on the boost converter: In this case the circuit will act as described by wikipedia. 2) You simply want to move the switch to being in series with the inductor. In this case, there will be problems. While the switch is closed the inductor is storing energy in its magnetic field. (E = L * I**2 / 2) The inductor will attempt to maintain the current when the switch is opened. This will result in a voltage spike across both the inductor and the switch. Most likely the switch contacts will arc (with a mechanical switch) or the voltage spike will destroy a semiconductor switch.

With the switch between input and coil, the output is inverted. Positive input supply produces negative output voltage, since the grounded coil produces reverse voltage when the switch opens, to try to keep the current going the same way.

If you put the switch between coil and ground, and connect the output diode at that same end of the coil, the voltage across the coil still reverses when the switch opens, but this reversal adds voltage to the input supply during the switch off time, so the output is not inverted, but higher voltage than the input. For instance, when the input is positive and the switch is on, the switch grounded end of the coil is negative with respect to the input end. When the switch opens, the voltage across the coil reverses and the switched end becomes more positive then the input end, so the voltage across the coil adds to the input voltage like two batteries in series.

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Regards,

John Popelish

"voltage coil" does not refer to any node. Try rewording your question, so that we know what you are asking.

Best,

remove the word "voltage"

Still, the coil is not a node. A coil has two electrical nodes. I imagine that you are referring to the node that is not ground. You mean leaving the inductor where it is (you didn't say anything about moving it) and connecting the switch across it (in parallel with it)? You shouldn't do that. The switch would short circuit the input voltage source. That's never good. The input current would be huge if the voltage source was close to ideal.

Best,

uh, no. That would be silly.

Ha ha ha.

I'll post a scan, if I find time (and find the correct driver to install my scanner).

I suppose a picture is worth 1E3 words...

Michael