So I watched this video to understand how DC-DC boost converter works.

Here's circuit diagram:

enter image description here

And I don't get, when switch is off, and polarity in the inductor is such that positive is on the right, negative is on the left.

Then in order for diode to conduct, anode should be at higher potential than cathode. In other words, voltage across inductor needs to be higher than voltage across capacitor.

Now video said, there's gonna be a spike in voltage in the inductor after switch is off due to collapsing magnetic field to keep current constant, but how big is this spike? And wouldn't charge, and therefore voltage across capacitor be so big, that the spike in inductor will no longer be higher, so diode would not conduct at all?


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    \$\begingroup\$ When the switch is on, current flows through inductor and through the switch to ground. When the switch is turned off, the current continues to flow in the same direction through the inductor, but now it has nowhere to go except through the diode to the capacitor and load. The INTUITIVE way to understand this is that inductors do not allow current flowing through them to change suddenly. Once the switch turns off, the voltage at the switch will ramp up very rapidly until the diode is forward biased. \$\endgroup\$ – mkeith Feb 22 '18 at 3:32
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    \$\begingroup\$ The voltage spike is as big as it needs to be to maintain the amount of current. In the case of a boost converter this means it's just barely big enough to make the diode conduct. (In the case of switching a relay this means it's just barely big enough to destroy your transistor) \$\endgroup\$ – user253751 Feb 22 '18 at 3:41
  • \$\begingroup\$ @mkeith the change in voltage at inductor defined by the formula L di/dt, right? is current decreasing over time as switch is off? \$\endgroup\$ – Jack Feb 23 '18 at 20:09
  • \$\begingroup\$ Yes. V = L * di/dt. Since V is opposite in direction to current flow, current flow will be decreasing. \$\endgroup\$ – mkeith Feb 24 '18 at 1:06
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    \$\begingroup\$ Yes. The way I would say it is that the polarity of the voltage across the inductor reverses. The current in an inductor doesn't change instantly. The rule is V=L*di/dt. Or you could say di/dt = V/L. So the slope of the current depends on voltage and inductance. If the voltage reverses for too long, it can eventually reverse the current, but in the video, the professor said to assume continuous conduction (CCM), which means that the current never stops or reverses. \$\endgroup\$ – mkeith Apr 23 '18 at 0:15

The whole system starts when the FET is on, i.e. Vds = 0v in this situation. For a certain period called Duty Cycle the inductor is gonna charge up with the increasing current that flows through it, and like that it will store energy. As the FET is turned off, the current would normally decrease, but, as there´s a inductor it will try to force the current to same direction it was flowing due to Lenz´s Law. As a result of that, the voltage in the inductor will have changed its polarity in order to keep current flowing to the same direction and then the voltage will be greater than the input(9V) voltage. If the circuit is working under Steady State(SS) conditions then the current that was flowing through the inductor will have the same variation.

In discharge, as the voltage across the FET is now greater than input voltage, the diode will be successfully biased. The spike you talked about is going to be as big as your duty cycle and your load, because inductor tries to keep the current constant whatever resistance it sees ahead, then causing the voltage to go up. Regarding the duty cycle, in ideal models the appropriate formula is given by: Vo/Vi= 1/(1-D), which states that the greater your Duty Cycle, the greater your output voltage.

  • \$\begingroup\$ "Vo/Vi= 1/(1-D)" - No, duty cycle determines the ratio of output current to input current. Output voltage is then determined by the load. With no load the theoretical output voltage is infinite (for any duty cycle other than 0% and 100%). \$\endgroup\$ – Bruce Abbott Feb 22 '18 at 14:28
  • \$\begingroup\$ Duty cycle also takes part of controlling duty cycle, alongside load resistance, and I mentioned that on my answer. Also in TI papers which are famous for this application they state the duty cycle takes part of the output voltage from that formula, which is a ideal model formula. And it makes sense since it'll have stored more energy in the inductor the greater the duty cycle is. \$\endgroup\$ – Flávio Alegretti Feb 22 '18 at 19:27
  • \$\begingroup\$ @FlávioAlegretti wait, what? If inductor's voltage is higher, shouldn't current go towards lower potential, i.e. voltage source of 9V? Current goes from "+" to "-", doesn't it? \$\endgroup\$ – Jack Feb 24 '18 at 3:05
  • \$\begingroup\$ I know what you mean, but the point is that the inductor changes its polarity. When charging, the voltage in "left side of the inductor" is positive. When the fet is turned off, the inductor will face a decreasing current, and then the inductor polarity will change in order to keep the current flowing to the same direction. Voltage is gonna change its direction because V= L* - di/dt, and then the inductor voltage will be added to the input voltage, something like a "series battery connection". Then it will not discharge towards the battery, but the load. \$\endgroup\$ – Flávio Alegretti Feb 25 '18 at 3:30
  • \$\begingroup\$ @FlávioAlegretti just so it's clear, when FET is on, the left side of inductor is positive, and right side is negative, so the current flows from left to right. When FET is off, it's the same!! Otherwise current won't flow from left to right. So where does this notion that the left side of inductor is negative and right side is positive? Because in that case, the current will flow towards the input battery! (current direction is from "+" to "-" terminal). In other words, the inductor polarity doesn't change.... \$\endgroup\$ – Jack Apr 22 '18 at 3:45

An inductor works on the following fundamental principle: -

\$V = L\frac{di}{dt}\$

Where \$\frac{di}{dt}\$ is the rate of change of current in the inductor

If you apply a voltage across an inductor (by grounding one end for instance), current will climb at the rate V/L amps per second and, while doing so, energy will be stored in the magnetic field. When the inductor is disconnected from that ground connection, the stored energy pushes a current out of the inductor in the same direction.

That current tries to maintain its value but it can't and so current starts to fall. This means that the rate of change of current (\$\frac{di}{dt}\$) is negative.

This generates a negative voltage across the inductor terminals \$V = L\frac{-di}{dt}\$.

The input side of the inductor is "tied" to the incoming supply voltage hence, the switched side of the inductor (previously at ground) generates a voltage that is greater than the incoming voltage - this is the voltage reversal seen across the inductor i.e. this is a negative voltage compared to when the inductor was grounded by the transistor.

This "greater voltage" rapidly rises (in order to push current out) and when this rapidly rising voltage equals the voltage across the output capacitor (plus one diode drop) it finds a "load" to dump the current into. From this point on the output voltage acquires a level suitable to carry on pushing current into the capacitor until all the magnetic energy previously stored is depleted.

  • \$\begingroup\$ also, just outta curiosity, isn't dc-dc boost and transformers that step up voltages increase the rate of flow of electrons? So basically battery source will be depleted of its electrons faster? Like you don't get higher voltage for free, because higher voltage essentially means more electrons that are ready to go. No? In the case of dc boost, higher voltage corresponds to more electrons stored in the capacitor. \$\endgroup\$ – Jack Feb 23 '18 at 20:11
  • \$\begingroup\$ Power in (voltage x current) = power out (voltage x current) plus conversion losses so if you step up the volts, the input current to the converter also increases compared to the output current. \$\endgroup\$ – Andy aka Feb 23 '18 at 20:34
  • \$\begingroup\$ but isn't polarity across inductor such that on the positive side of diode there's gonna be negative voltage? And in order for diode to conduct it'd have to be positive and bigger than voltage across capacitor, which is on the cathode side of the diode? here's pic: i.imgur.com/bya5P92.jpg \$\endgroup\$ – Jack Feb 24 '18 at 2:41
  • \$\begingroup\$ Your picture shows the inductor voltage incorrectly. There is a bigger voltage on the right hand side compared with the left hand side when the transistor opens. Imagine a spring tied to a hook on a wall; you pull the spring end down to ground then you release it - where does the recoil take the loose end of the spring? \$\endgroup\$ – Andy aka Feb 24 '18 at 9:49
  • \$\begingroup\$ upwards? Also, the polarity on inductor is correct when switch is open (off), see the video youtube.com/watch?v=vmNpsofY4-U&t=5m \$\endgroup\$ – Jack Mar 2 '18 at 23:36

When the FET is ON, current is stored in the inductor as a magnetic field.

When the FET is OFF, this magnetic field collapses and induces current back into the inductor winding, in the same direction. It has to go somewhere, so the fast reacting diode forces it to the output side where a filter capacitor smooths it out into a DC voltage.

These circuits can be very efficient because the power source is able to use the diode as well, so this circuit boost the voltage higher than the input voltage.

You could say the diode adds the idle flow current to the inductors current, thus raising the voltage. That is the simple answer. The math wizards will complain about "but this goes this way", etc.

The value of the inductor has much to do with how much boost you can get (as well as FET 'ON' time, or duty cycle), but high inductive values (>1mH) are not as efficient (Q) due to the DC resistance of the inductor. (Q = L/R)

You can boost the voltage a thousand times if you like, but the current available will be reduced by the same amount, minus conversion losses.


I think you might be misunderstanding what exactly an inductor is.

The voltage across an inductor determines the rate of change of current through the inductor.

When the FET is on, the input voltage is applied across the inductor, which causes the current to increase until the FET is switched off.

When the FET is switched off, the current will flow through the diode into the capacitor. It will flow, because current through an inductor doesn't just stop instantaneously, and that current has to go somewhere.

The capacitor terminal will normally (or soon) be at a higher voltage than the source. This difference is applied across the inductor, causing the current to slow down. This is the "voltage spike", but the particular voltage is not chosen by the inductor. It's just the voltage that results from stuffing current into the capacitor. The voltage at the inductor terminal has to be the capacitor terminal voltage plus the diode voltage drop, and realistically a little more due to stray resistances.


Sinking current with 0V on FET when released creates a potential in the opposite direction at the same current unless clamped by the cap or battery and diode, when forward conducting. The current decays at a rate of L/ESR, where ESR is all the loop series resistances including the diode and cap.

Consider the FET and diode as a SPDT switch with a ramp up current and ramp down current that raises the Cap voltage.


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