I guess you have the wrong idea sir. The voltage of inductor doesn't change, but the 'current' itself should be change. And from the picture that TS posted, it's written that vL(t)=-Vout(t) , and the picture shown the vL(t), not the iL(t) polarity. In case of your answer is for iL(t), but in the picture TS has shown above, it's for showing vL(t)Hi,
The second diagram is incorrect because it shows the polarity of the inductor the same as it was before the field started to collapse.
Once the field starts to collapse, the voltage reverses and thus charges the capacitor positively.
So when the transistor turns off, the diode conducts and the voltage across the inductor changes from +L- to -L+, with the positive on the right hand side now. That's the way inductors behave in circuits like this.
I think you made a mistake here. In the second diagram, the polarity of inductor voltage is what we can choose, it can be arbitrary, no matter what it is in reality.Hi,
The second diagram is incorrect because it shows the polarity of the inductor the same as it was before the field started to collapse.
Once the field starts to collapse, the voltage reverses and thus charges the capacitor positively.
So when the transistor turns off, the diode conducts and the voltage across the inductor changes from +L- to -L+, with the positive on the right hand side now. That's the way inductors behave in circuits like this.
I am not sure where you are getting that information from, but once the transistor turns off the inductor must switch voltage polarity. If that did not happen, the diode would not be able to conduct. In other words, if you are right then show how the diode can conduct with a positive voltage at it's cathode with say 5v across the capacitor (as in a 10v to 5v buck converter).I guess you have the wrong idea sir. The voltage of inductor doesn't change, but the 'current' itself should be change. And from the picture that TS posted, it's written that vL(t)=-Vout(t) , and the picture shown the vL(t), not the iL(t) polarity. In case of your answer is for iL(t), but in the picture TS has shown above, it's for showing vL(t)
And i completely agree with that. The voltage DOES NOT change the polarity, but the current DOES (because the characteristic of inductor). It's a little bit confusing i guess at the start, but once i learned, it's more interesting
Are we talking about the same circuit?I think you made a mistake here. In the second diagram, the polarity of inductor voltage is what we can choose, it can be arbitrary, no matter what it is in reality.
If the result is positive, then the chosen polarity is correct, otherwise, it is the other way around.
Nahhh, that's the tricky part. I'm completely agree with you.Hello again BramLabs, anhnha,
I am not sure where you are getting that information from, but once the transistor turns off the inductor must switch voltage polarity. If that did not happen, the diode would not be able to conduct. In other words, if you are right then show how the diode can conduct with a positive voltage at it's cathode with say 5v across the capacitor (as in a 10v to 5v buck converter).
About the actual polarity of inductor, you are right. However, I just disagree with what you said here that the the second diagram is incorrect.The second diagram is incorrect because it shows the polarity of the inductor the same as it was before the field started to collapse.
Once the field starts to collapse, the voltage reverses and thus charges the capacitor positively.
Well, yes, during the transistor ON phase the inductor is more + to the left and - to the right, as a function of its current flow and impedance creating a voltage drop.It means that, we should dissipate 5V at the inductor.
So in the easy ways, it should be :
10V (input) - 5V (inductor) = 5V (load)
Then we get the load 5V as you mentioned before.
Wait a minute, are you talking about the OFF state or the ON state? In the transistor OFF state the 10V is no longer there. The only voltage source present is the induced voltage in the inductor - on the left and + on the right, which drives against the voltage on the capacitor to deposit a little more charge on it (by flowing current through the diode).And let's talk about the polarity.
Input 10V --> (+) at top, and (-) at bottom or ground.
And if the inductor it's in (-) at the left of the inductor and (+) at the right of the inductor as you have been said before (off state switch), because you said that the voltage polarity change, then if we applied KVL, it would be 10V + 5V, then we get = 15V
I'm not sure what you're trying to say here, but the formulas are reasonable, but only as they apply to the diagram, and the formulas are ONLY talking about the transistor ON phase. They show that during the ON phase, the inductor voltage is + on the left and - on the right.In this case, 10V (input) - 5V (output).
And we get vL = Vi - Vo
And how's the polarity ?
If we subtitute from the equation above, we get :
Vi - vL = Vo
Then we should have a different polarity so that they aren't adding each other. But they have to substract each other to make 10V(input) - 5V(inductor) = 5V (load)
"because the inductor BECOMES the voltage source" that's the part when you're wrong sir.I'm with MrAl on this. Voltage in a circuit is not arbitrary. It is clearly defined by physics.
Here is another way to look at it: the role of the inductor switches from a passive, driven component to an active one that supplies the voltage.
During the ON phase of the transistor, the voltage source drives a current through the inductor into the capacitor. The leading edge of the inductor is as positive as the supply voltage - transistor voltage drop, and because of the inductor's impedance, its trailing edge must be more negative.
During the OFF phase of the transistor, the collapsing field induces a current in the inductor. In order for this current to flow, the trailing edge of the inductor flips to positive and the leading edge flips to negative because the inductor BECOMES the voltage source. And as MrAl points out, the only way current will flow in the transistor OFF phase is to turn on the diode, which requires the trailing edge of the inductor to become more positive than the charge on the capacitor plus the voltage drop of the diode. This is the only way the diode will switch on.
Here is an AAC article about induced voltage that includes the diagram below, showing the voltage flip on the inductor. Note the arrows are showing electron flow instead of classic current flow, but the voltage polarity is the same.
I think my explanation wasn't good at all. Sure, we can't mix ON phase to the OFF phase, but we can't definitely only say when it's OFF phase or it's ON phase. We should analyze it when OFF state FIRST and then ON state. Not only OFF state or ON stateWell, yes, during the transistor ON phase the inductor is more + to the left and - to the right, as a function of its current flow and impedance creating a voltage drop.
Wait a minute, are you talking about the OFF state or the ON state? In the transistor OFF state the 10V is no longer there. The only voltage source present is the induced voltage in the inductor - on the left and + on the right, which drives against the voltage on the capacitor to deposit a little more charge on it (by flowing current through the diode).
I'm not sure what you're trying to say here, but the formulas are reasonable, but only as they apply to the diagram, and the formulas are ONLY talking about the transistor ON phase. They show that during the ON phase, the inductor voltage is + on the left and - on the right.
When the switch opens the 10V is no longer there and the collapsing field on the inductor turns it into a voltage supply with - on the left and + on the right.
I think your problem may be that you're mixing up the ON phase and the OFF phase.
What ??vL = 1/L integral vL dt
OK, the second diagram isn't "wrong" IF YOU TAKE IT IN CONTEXT. It doesn't show the polarity of the voltage on the inductor. It shows the polarity CONVENTION of their equations. If you read on from those diagrams, they explain that in their text. The +/- labeling in the diagrams is simply a convention to explain the polarities in their equations. And during the OFF phase, the drawing and equations below show that the polarity reverses and becomes - to the left and + to the right, the negative of what they show in the drawing above. -v(t) means that the voltage during the OFF phase is the reverse of the labeled convention.About the actual polarity of inductor, you are right. However, I just disagree with what you said here that the the second diagram is incorrect.
The polarity of inductor voltage is what we choose before solving the circuit. The actual polarity will be known after getting the value.
For example, with the circuit above, we choose the left end is plus and the right end of inductor is minus.
When the switch is off we have the picture on the right.
In this case, VL = -Vout < 0. So the actual polarity of inductor voltage is just reverse what we assume initally.
I would disagree with saying that the second diagram is wrong here.
View attachment 103711
http://ecee.colorado.edu/copec/book/slides/Ch2slides.pdf
If the inductor is a current source it is also a voltage source. No current will flow in the circuit unless the right end of the inductor is more + than the charged voltage on the capacitor."because the inductor BECOMES the voltage source" that's the part when you're wrong sir.
Inductor doesn't act as an voltage source, but it acts as an current source sir when charged
And capacitor is the one who becomes a voltage source when charged^^
Oh, i already fixed it sir. ahahahahWhat ??
The voltage across an ideal inductor is VL = L* ΔI/Δt
This means that to produce the voltage across an inductance, the applied current must change. If the current is kept constant, no voltage will be induced, no matter how large the current. Conversely, if it is found that the voltage across an inductance is zero this means that the current must be constant but not necessary zero.
So in BUCK convert the voltage across the inductor changes his polarity. And here you have the proof
I build this circuit on the breadboard
View attachment 103707
And for F = 10KHz I set the duty to 56% to obtain 5V output at 39Ω load.
I also measure voltage across the inductor and use Rs resistor to measure inductor current.
The blue trace shows the voltage across the inductor. As you can see the voltage across the coil changes his polarity.
View attachment 103708
When the switch is ON the voltage across the inductor is equal to VL ≈ +4.8V. But when the switch is OFF (D1 ON) the voltage is VL ≈ - 6V.
And this voltage must be negative because the inductor current ramp-down (as we can see on the read trace) and this is why the rare of change (ΔI/Δt) is negative so the voltage also must be negative.
Or we can view this way:
The goal of the inductor is to resist changes in current flow.
During the ON-time the upper end of the inductor gets set to a higher voltage than its lower end. And the current flow from left to right.
But when the switch opens (OFF), the input DC source gets disconnected from the inductor. But the inductor wants to keep the current to flow in the same direction as it was when the switch was closed (from left to right). And this can only happen when the inductor change his polarity (the inductor becomes a sort of “voltage source" now). And thanks to this the current will keep on flowing in the SAME direction during the switch OFF as it was flowing before. And this is why we get the apparent sign reversal for voltage.
View attachment 103712
However in fact voltage reversal does not always occur . That is because the primary requirement is only that the
inductor current somehow needs to keep flowing in the SAME direction.
I agree.What ??
The voltage across an ideal inductor is VL = L* ΔI/Δt
This means that to produce the voltage across an inductance, the applied current must change. If the current is kept constant, no voltage will be induced, no matter how large the current.
If the voltage across the inductor is zero, no current will flow, because the voltage at the right end of the inductor needs to be greater than the voltage on the capacitor and the voltage on the left end of the inductor needs to be more negative than the voltage drop of the diode in order for current to flow.Conversely, if it is found that the voltage across an inductance is zero this means that the current must be constant but not necessary zero.
This is all true, except for the fact that, as your simulation and diagrams say, there must be a polarity switch on the inductor, as it goes from a passive component in phase 1 to an active component in phase 2.So in BUCK convert the voltage across the inductor changes his polarity. And here you have the proof
I build this circuit on the breadboard
View attachment 103707
And for F = 10KHz I set the duty to 56% to obtain 5V output at 39Ω load.
I also measure voltage across the inductor and use Rs resistor to measure inductor current.
The blue trace shows the voltage across the inductor. As you can see the voltage across the coil changes his polarity.
View attachment 103708
When the switch is ON the voltage is across the inductor is equal to VL ≈ +4.8V. But when the switch is OFF (D1 ON) the voltage is VL ≈ - 6V.
And this voltage must be negative because the inductor current ramp-down (as we can see on the read trace) and this is why the rare of change (ΔI/Δt) is negative so the voltage also must be negative.
Or we can view this way:
The goal of the inductor is to resist changes in current flow.
During the ON-time the upper end of the inductor gets set to a higher voltage than its lower end. And the current flow from left to right.
But when the switch opens (OFF), the input DC source gets disconnected from the inductor. But the inductor wants to keep the current to flow in the same direction as it was when the switch was closed (from left to right). And this can only happen when the inductor change his polarity (the inductor becomes a sort of “voltage source" now). And thanks to this the current will keep on flowing in the SAME direction during the switch OFF as it was flowing before. And this is why we get the apparent sign reversal for voltage.
View attachment 103712
However in fact voltage reversal does not always occur . That is because the primary requirement is only that the
inductor current somehow needs to keep flowing in the SAME direction.
Woah, now it's looked more confusing. ahahahaha...OK, the second diagram isn't "wrong" IF YOU TAKE IT IN CONTEXT. It doesn't show the polarity of the voltage on the inductor. It shows the polarity CONVENTION of their equations. If you read on from those diagrams, they explain that in their text. The +/- labeling in the diagrams is simply a convention to explain the polarities in their equations. And during the OFF phase, the drawing and equations below show that the polarity reverses and becomes - to the left and + to the right, the negative of what they show in the drawing above. -v(t) means that the voltage during the OFF phase is the reverse of the labeled convention.
View attachment 103713
But this is not the simulation resultas your simulation and diagrams say
But here I was talking about general case. Notice that at steady state VL = 0V but IL can be larger than 0A.If the voltage across the inductor is zero, no current will flow