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Inductor question

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zesla

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Hi guys,

Sorry but I came across to a problem when I was studying about Boost converters.

The Problem is That I am not able to properly undrstand what will happen to the voltage of a an inductor when the voltage across it is oppened immediately.
considering the below circuit the circuit A is understandable to me and it shows that the inductor is charging up and storeing the energy in its magetic field,

But Why while we open the key (the circuit B) the voltage going through the diode to the capacitor is going to be much higher than the input voltage?
I know that the diode allows the stored current to go to the cap while the key is opened and of course does not allow it to come back at other conditions but my question is tha I want to know if the inductor itself generates a voltage which is higher than the input voltage? If so does this voltage reaches to the MAX at just one cycle?

I tried it practically to see what will happen. I used an 33mH inductor and a 1N4007 diode and a 220uF/500V electrolyte capacitor. I noticed each time I turn the key on and then off just a rather small incasement will happen to the cap voltage. the input voltage was 10V DC and I could cause the capacitor voltage to reach to the 20V DC by then time switching or more.
Before doing this test I thought that by just switching on and off for once the inductor will generate the MAX posibble voltage for the cap but After doing the test I am not able to see what does happen and of course why...?

Thanks for any help
 

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The inductor is like a rubber band. Hold the left end 10volts in the air, then take the right end and pull it to ground. Let go and the elastic band will shoot up until it hits something. (the capacitor) The energy stored in the rubber band will push the capacitor up a little every time.
In a switching power supply this will happen 100,000 times a second. It may take 100 times to pump the capacitor up.
 
it is due to the parallel charging to 10v, and then series(additive) discharging in the inductor, that is the theory anyway, I find it easier to understand using "capacitor voltage multipliers", (IKD the diff)
 
The inductor stores energy equal to ½LI². The capacitor stores energy equal to ½CV². When the switch is opened the stored inductive energy is transferred to the capacitor. To determine the amount of capacitor voltage increase for each operation of the switch you just set these two terms equal to each other and solve for ΔV (Δ = incremental change). This gives ΔV² = LI²/C or ΔV = √(LI²/C). (Edit: This neglects any loses such as from the diode forward voltage drop.)

Obviously the capacitor voltage increase for each switch cycle thus depends upon the inductor and capacitor values, and the inductor maximum current just prior to the switch opening.

Incidentally, there is no theoretical limit as to how high the voltage can go with no load. Each switch cycle will increase the voltage per the equation. That's why such a circuit typically has a feedback loop to adjust the duty-cycle (PWM) of the switch and regulate the output voltage.
 
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Hi guys,

Sorry but I came across to a problem when I was studying about Boost converters.

The Problem is That I am not able to properly undrstand what will happen to the voltage of a an inductor when the voltage across it is oppened immediately.
considering the below circuit the circuit A is understandable to me and it shows that the inductor is charging up and storeing the energy in its magetic field,

But Why while we open the key (the circuit B) the voltage going through the diode to the capacitor is going to be much higher than the input voltage?
I know that the diode allows the stored current to go to the cap while the key is opened and of course does not allow it to come back at other conditions but my question is tha I want to know if the inductor itself generates a voltage which is higher than the input voltage? If so does this voltage reaches to the MAX at just one cycle?

I tried it practically to see what will happen. I used an 33mH inductor and a 1N4007 diode and a 220uF/500V electrolyte capacitor. I noticed each time I turn the key on and then off just a rather small incasement will happen to the cap voltage. the input voltage was 10V DC and I could cause the capacitor voltage to reach to the 20V DC by then time switching or more.
Before doing this test I thought that by just switching on and off for once the inductor will generate the MAX posibble voltage for the cap but After doing the test I am not able to see what does happen and of course why...?

Thanks for any help

Hi there z,


The boost circuit will cause an output voltage quite high
without any load but with a load the output is limited, so
we'll assume there is always some load.

Lets say that we turn on the power to the circuit and at the
same time we have the switch S closed. That puts the battery
right across the inductor and the current in the inductor is:
di=v*dt/L
That simply means that the current in the inductor ramps up
to some value in some small amount of time dt.
Now later some time we open the switch, and that causes the inductor
voltage to rise up quite a bit and very fast until diode D conducts
Once diode D conducts capacitor C conducts and so the current
through the capacitor C is about the same as the inductor current.
That means the voltage across the capacitor starts to build up.
That means current starts to flow through the load R and so it starts
to take some of the current from the inductor, so the capacitor charges
up more slowly. All this time the inductor has been supplying current
to the cap and load, so it's stored energy decreases. If we left the
circuit that way too long, the inductor energy would be taken up by
the cap and load, then finally the load until the inductor looked like
a wire and only allowed the battery voltage to get to the load.
Instead of letting the inductor energy dissipate completely or just
as it starts to dissipate completely, if we turn the switch back on
it again causes current to flow the inductor and more storage of energy
in the inductor. The diode D is now reverse biased so the output load
current is supplied by the cap alone. But, next time we turn off the switch,
we have more energy to pump to the output, and the diode again conducts and
more energy gets to the output so the voltage goes up a little again,
In this manner, when the switch is on the load gets its current from the cap
and so the cap voltage decreases, and with the switch off the inductor current
again charges the cap and supplies the load. Thus the cap voltage is always
increasing a little and decreasing a little, with its average value equal to
the desired output voltage. The output cap current also goes up and down
somewhat so the cap has to be rated to be able to handle that change.

There are a few differences between this circuit and a buck circuit (which you
should probably study first if you havent already) and one of the main
differences is that output voltage decreases when the switch is on rather than
increases like the buck. This presents some interesting control loop problems.
Another important difference is that because we have to have the switch 'on for
a certain minimum time to achieive enough energy flow input to output, there is
a critical duty cycle point of operation. In a buck converter, the longer we
have the switch on the more energy we get to the output, but in this circuit
that only works up to a point, and once we reach that point the energy actually
*decreases* in the output, so the output voltage would actually *fall* instead of
get higher. This critical duty cycle depends on a number of factors involving the
losses of the circuit, but the main idea is to make sure that the duty cycle can
never go over that limit, because if it does, the phase of the feedback signal
reverses and the system goes to a very low output value! Crazy, but that's one
of the complexities that come up with the boost circuit with feedback.

So anyway, the main points to remember are:
1. When the switch is on, energy is stored in the inductor and current rises as di=v*dt/L.
2. When the switch is open, energy moves from the inductor to the output cap and load.

For #1 above, the voltage across the inductor is equal to the battery voltage.
For #2 above, the voltage across the inductor assumes whatever the diode, cap, and load
happen to have across them at the time.

So when the switch is on, the battery dictates the voltage across the inductor, but when
the switch is off, the external circuit (diode, cap, load) dictate the voltage across the
inductor.
 
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Ok thank you very much guys, Special thanks to Carl and MrAl for their almost voluminous inputs.
I am still trying to learn about inductors with focusing to my question here. I will put my findings here soon while I ended up the papers I am reading.
But a question which came across while reading one of the saids papers is as below:
Does inductors and coils have abut inductance (L) while are connected to a DC source at the steady state too? I know that the reactant of the inductors or capacitors is just for AC, but what about L??

What about a capacitor in DC? does it has any capacitance in DC too?

Thanks a bunch.
 
Flux = L* I
It works for DC.

thanks but two reasons which cause me to think that in DC there would not be any inductance in an inductor are:
An inductor is a short circuit in Dc so it is like a piece of small wire.
The other reason is the first paragraph of this:
Inductance - Wikipedia, the free encyclopedia
We know that self-inductance is due to CHANGES of the current going to the coil which causes changes in the magnetic field of the coil, then as the above link states if inductance is equal to the self-inductance then the inductance would not be a case of the matter in DC steady state.

makes sense?
 
thanks but two reasons which cause me to think that in DC there would not be any inductance in an inductor are:
An inductor is a short circuit in Dc so it is like a piece of small wire.
The other reason is the first paragraph of this:
Inductance - Wikipedia, the free encyclopedia

makes sense?
You are confusing inductance with inductive reactance.

An inductor always has inductance which is fairly constant over it's design range of current and frequency. The inductive reactance, on the other hand, varies with frequency and is zero at DC. But if you try to change the value of the DC current you will immediately notice the inductive reactance as it will resist any change in the current.
 
thanks but two reasons which cause me to think that in DC there would not be any inductance in an inductor are:
An inductor is a short circuit in Dc so it is like a piece of small wire.
The other reason is the first paragraph of this:
Inductance - Wikipedia, the free encyclopedia
We know that self-inductance is due to CHANGES of the current going to the coil which causes changes in the magnetic field of the coil, then as the above link states if inductance is equal to the self-inductance then the inductance would not be a case of the matter in DC steady state.

makes sense?


Hi again,

There are basically at least two ways to define 'inductance'.

The first way appears in that link you provided which may make L appear to depend on current i because of the di/dt factor. If we solve for L we get:
L=v*dt/di
and here we can see that letting di=0 wont be valid, so we have to know what dt/di is and take the limit to find the inductance L. The assumed way to know what di and dt are is to supply a *test* current or voltage, then measure the other, then compute the slope, then compute the self inductance L. If we apply a test voltage for example we get some reasonable value for dt and di, but if we let that voltage go to zero that doesnt mean that L goes to zero, because in a linear inductor (more about that later) the slope is considered constant and doesnt change with current.

The second way is much more involved (in that link scroll down the page farther to see other derivations), and depends on the physical construction of the inductor such as the coil radius, wire diameter, core material, etc., but does not depend on current or voltage. The self inductance is considered the same no matter how much or how little current is flowing in it. The test voltage in the first method allows us to measure the self inductance at some test points, that's all. This second method relies on known characteristics of the materials used to build the inductor, so we dont have to apply a test current.

There are secondary effects however, in the non linear inductor, where the inductance is not constant (di/dt changes with current). This kind of inductor usually has a core made out of a ferromagnetic material. That kind of material allows us to obtain a much larger inductance value with fewer turns of wire, but the drawback is that the core can only take a certain amount of excitation before it no longer acts like a magnet, so the inductance falls sharply (search: 'magnetic domains'). This point where this happens is usually called "saturation", or "core saturation".

Right now you are probably concerned mostly with the linear inductor so you dont have to worry about the non linear one yet. You may encounter inductors who's inductance changes with current though, so you know that these are not linear inductors but are non linear and probably have some kind of core material other than air.
 
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