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Shown below is the circuit diagram of a boost converter. What would if I interchange the position of the inductor and capacitor in this circuit? Considering they are both storage elements, there shouldn't be a change right?

enter image description here

With reference to this explanation, why does the inductor discharge when the MOSFET is OFF and charge when the MOSFET is on? Since the inductor is always connected to the positive supply, shouldn't it charge even when the MOSFET is off? (or) When specifically does the inductor discharge, and why? (The link mentioned something about back emf being developed but I couldn't quite understand that)

In general, what usually happens when an inductor and capacitor are interchanged?

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  • \$\begingroup\$ You can't swap. The goal is to develop a DC voltage across the load. Placing an inductor in parallel (when swapped with C1) shorts it out to 0V. \$\endgroup\$ – glen_geek Aug 3 '19 at 16:06
  • \$\begingroup\$ Operation of basic boost converters is thoroughly covered by many many articles and websites. Have you done any background study at all? \$\endgroup\$ – Andy aka Aug 3 '19 at 17:03
  • \$\begingroup\$ I did, I couldn't find an article that thoroughly explains the questions that I've asked here. \$\endgroup\$ – noorav Aug 3 '19 at 17:15
  • \$\begingroup\$ Why would you expect to find an article on a completely arbitrary question? \$\endgroup\$ – Chu Aug 3 '19 at 18:39
  • \$\begingroup\$ What would happen if I interchange the position of the resistors and LEDs in my TV? Considering they are both passive elements, there shouldn't be a change right? \$\endgroup\$ – user253751 Jul 6 '20 at 11:29
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Since the inductor is always connected to the positive supply, shouldn't it charge even when the MOSFET is off? (or) When specifically does the inductor discharge, and why?

Suppose for the moment that the converter is functioning correctly; then C1 has \$V_{\text{OUT}}\$ across it, which is greater than \$V_{\text{IN}}\$. Therefore, the voltage difference between the terminals of L1 is opposite of the direction of the current. This voltage opposes the flow of current. Therefore L1 is discharging (current is decreasing); it is giving up stored energy to force the current to continue in that direction.

What would if I interchange the position of the inductor and capacitor in this circuit? Considering they are both storage elements, there shouldn't be a change right?

They both have the property of storing energy but that doesn't mean they interact with the circuit in the same way. If you replace L1 with a capacitor, then the DC input cannot flow through it; it becomes charged to the input voltage and nothing happens after that.

In general, what usually happens when an inductor and capacitor are interchanged?

In general? You'll get a non-working circuit.

If you interchange inductors and capacitors specifically in an RLC network (a circuit composed only of resistors, inductors, and capacitors) you might see a reversal of its frequency response (for example, a low-pass filter becomes a high-pass filter and vice versa).

In fact, if we "squint" and delete some elements from this circuit, it looks like a low-pass filter:

schematic

simulate this circuit – Schematic created using CircuitLab

There's a reason for this: low-pass filtering is precisely what we want to get a stable DC voltage. If you use just this circuit, you'll remove ripple, but you won't get a higher voltage. The additional active components (MOSFET and control circuit) take advantage of the inductor's property of making current flow opposite voltage differences (temporarily) to create higher voltage.

If we swap L1 and C1 then we get a high-pass filter, which is a perfectly reasonable circuit but in this application does nothing, because we want a DC input and a DC output, but a high-pass filter blocks DC.

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  • \$\begingroup\$ "In general, what usually happens when an inductor and capacitor are interchanged?" Doesn't the low-pass / high-pass filter example illustrate the principal of duality? Can the components in the circuit in question be rearranged to construct a working dual? \$\endgroup\$ – Charles Cowie Aug 3 '19 at 15:35
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    \$\begingroup\$ @CharlesCowie Probably, but I don't know how to do that accurately. Maybe someone could write that answer. \$\endgroup\$ – Kevin Reid Aug 3 '19 at 15:40
  • \$\begingroup\$ I also strongly suspect it could be done, but don't want to spend hours confirming my suspicion and writing an answer. Thank you. \$\endgroup\$ – Charles Cowie Aug 3 '19 at 15:50
  • \$\begingroup\$ Oh okay. I'm sorry but I still fail to understand this part. MOSFET ON- inductor charges, MOSFET OFF- inductor discharges. Why won't it charge even when the MOSFET is off because it's always connected to a positive source. As far as I've learnt, the inductor discharges only when power is disconnected from it, right? \$\endgroup\$ – noorav Aug 3 '19 at 18:17
  • \$\begingroup\$ @noorav The input may be positive but the input voltage is less than the output voltage. Remember, the inductor “doesn't know anything” about the negative side of the circuit, only the parts of the circuit it is directly wired to. You must compare the voltages on the two ends of the inductor to learn whether it is charging or discharging. The MOSFET & the diode switch the right side of the inductor between being connected to Vin- (allowing the rightward inductor current to increase because the right side of the inductor is lower voltage) and to Vout+ (the opposite condition). \$\endgroup\$ – Kevin Reid Aug 3 '19 at 18:42

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