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I have trouble understanding something in the circuit below. It is inverting converter, seems quite easy, but the part I don't get is when the switch is OFF we have forward-biased diode, i.e. plus on anode - minus on cathode. Why then we have minus on the output pin that is THE EXACT SAME wire that is plus on the diode's anode? I don't get how can we state that anode is positive at the same time as other part of same wire is stated to be negative.

This is not a typo obviously, since it is popular circuit and everywhere it is like that.

enter image description here

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  • \$\begingroup\$ Out of curiosity, which book is this from? \$\endgroup\$
    – Nick Alexeev
    Sep 14, 2014 at 4:06
  • \$\begingroup\$ Nick, David A. Bell - Electronic Devices and Circuits \$\endgroup\$
    – ScienceSamovar
    Sep 14, 2014 at 4:13

2 Answers 2

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The charging and discharging of an inductor is the key to understanding this circuit and is the reason why it is marked (correctly) with the appropriate polarity.

enter image description here

Consider the two cases

SWITCH ON

Current flows into the inductor. The current will increase with time and the energy supplied to the inductor will be stored in its magnetic field. This will produce the polarity shown above.

SWITCH OFF

With no current to sustain the magnetic field it will collapse and the changing magnetic field will INDUCE a voltage in the opposite direction to the original current. If this was a relay coil or motor we would refer to it as 'back emf'.

The capacitor,load and diode.

From the load's point of view it effectively sees the voltage drop across the coil added to the battery voltage. The diode in the circuit prevents the load from 'seeing' the battery voltage directly when the inductor is being charged (switch closed) because at this point it is reverse biased. The capacitor is charged up when the switch is OPEN and will maintain the voltage across the load during the time the inductor is being charged (SWITCH CLOSED).

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  • \$\begingroup\$ I understand the concept of opposite voltage induced in the inductor, but I have troubles understanding why there is + sign on the anode of the diode, can you clarify it for me? I think I get something, but not entirely... :( \$\endgroup\$
    – ScienceSamovar
    Sep 14, 2014 at 9:11
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    \$\begingroup\$ @ScienceSamovar With the switch CLOSED the voltage across the inductor is in the same direction as the input voltage. This makes the cathode of the diode positive (with respect to the common connection or ground) but the output across the capacitor is negative (wrt ground), hence +l<-. When the switch is OPENED the voltage across the inductor reverses making the cathode of the diode more negative than the output across the capacitor (the node of the diode). This is difficult to label as such. BUT this means that the anode of the diode must be more positive than the cathode. Hence -<+ \$\endgroup\$
    – JIm Dearden
    Sep 14, 2014 at 13:38
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The + & - only show the polarity of voltage across a given component at that moment in time. They are not necessarily relative to ground or to the + & - across other components. Think of them in matched pairs as "which end of this component is at a higher voltage relative to the other end".

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  • \$\begingroup\$ Exactly. A hint is the brackets that enclose the +/- marks, indicating they are relative to each other. \$\endgroup\$
    – Spehro Pefhany
    Sep 14, 2014 at 4:20
  • \$\begingroup\$ eh... I kinda get it, it is like in this example with resistors: s018.radikal.ru/i511/1409/bb/9f490d57d189.png But! What I don't get that in that circuit we have something like this, and it doesn't make much sense: s020.radikal.ru/i716/1409/c1/52db4baafc07.png Or do I miss something? \$\endgroup\$
    – ScienceSamovar
    Sep 14, 2014 at 4:30
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    \$\begingroup\$ @ScienceSamovar the trick is in the direction of the current in the inductor. In (b) the inductor is charged with a current flowing from top to bottom. When you open the switch like in (c) the inductor tries to keep its current flowing and the only way it can do that is by inducing a negative voltage across it, causing the diode to be forward biased. \$\endgroup\$
    – jippie
    Sep 14, 2014 at 5:09
  • \$\begingroup\$ Hm... now it makes a little sense. One more question - do I understand correctly that on the anode of diode we have also a negative voltage, but it is not so negative as on cathode side(that's why we say it's positive)? \$\endgroup\$
    – ScienceSamovar
    Sep 14, 2014 at 5:15

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