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.