Fringe's some coils buck and some coils don't

Hey Marcel,

The edit button is available for 1Hr, then it gets turned off. I had to turn it off, because the Trolls, when they used to come here, go through and Delete their own posts all over the show, in an attempt to make my forum messy and inconsistent. Typical of those Looser Trolls!

The Linear trace is Lenz's Law in the Coils, its the same representation as the Neo Magnet falling down through the Copper Pipe.

As the Coils Interact with each other, many different configurations, give different results as you know, then the Scope will show different results as your experiment progresses.

Yes, your comment on the Buck Boost Converter, it is related! I have found this video very useful:

Here is the video Transcript:

0:00
from the University of British Columbia
0:02
and Martin Ordonez and this is power
0:10
[Music]
0:15
today we're going to talk about the back
0:18
post apology and we have an invited
0:21
speaker who's going to introduce a topic
0:24
hi my name is Ignacio Vanessa pregnant
0:26
and I'm a researcher at UVC today we're
00
going to be talking about the path post
01
converter which is one of the
03
fundamental topologies in power
04
electronics and can generate a negative
06
voltage starting from a DC input voltage
09
let's get started
0:41
the back post converter is a switch
0:44
topology that takes a DC input voltage V
0:47
in and transforms it into a DC output
0:51
voltage V out which has reverse polarity
0:55
when compared to the input voltage the
0:59
output voltage can be smaller equal or
1:02
larger than the input voltage but it's
1:05
polarity is always negative respect to
1:07
the input ground the circuit takes the
1:11
input voltage source and uses two
1:13
complementary switches as one and as two
1:16
two alternatively connect either the
1:19
input by closing s1 and opening s2 or
1:22
the output by opening s1 and closing as
1:27
two to the inductor l which by
10
interacting with the capacitor C
13
produces a DC output voltage be out with
17
reverse polarity which is applied to the
19
load in a synchronous buck boost
1:42
converter the switch s2 is implemented
1:46
using a diode which will automatically
1:48
turn on when the switch S one
1:51
implemented with a controllable switch
1:53
such as a MOSFET or IGBT is turned off
1:57
in the synchronous buck boost converter
2:00
the diode is replaced by another
2:02
controllable switch driven by a
2:05
complementary signal this modification
2:08
provides additional regulation features
2:10
when compared with the asynchronous two
2:13
in order to maintain this regulation
2:16
capabilities the asynchronous buck boost
2:19
converter can be designed to operate in
2:21
continuous conduction mode there is the
2:25
operating range is selected to
2:28
continuously guarantee a positive
20
current through the inductor which
23
ensures the diary saw is forward biased
26
unable to carry the current if this
29
condition is not met the equations that
2:42
describe the behavior of the converter
2:44
will change this switch topology under
2:49
continuous conduction mode has two
2:51
different states when the control signal
2:54
s is high the controllable switch s1 is
2:58
turned on and connects the input voltage
3:01
to the inductor creating a current flow
3:04
this is maintained during a period of
3:07
time called on time T on after which the
3:11
control signal s changes to a low state
3:14
which causes the switch s1 to turn off
3:18
and forces the diode to carry the
3:20
inductor current this is maintained
3:23
during a period of time called off time
3:26
T off the cycle of T on followed by T
30
off repeats continuously with the
33
circuit alternating between its on and
35
off States the sum of 1 on time and 1
3:40
off time is called switching period and
3:43
it's reciprocal is called the switching
3:46
frequency one of the fundamental
3:48
parameters of the converter the ratio
3:51
between the on time T on and the
3:54
switching period t is called duty cycle
3:57
D the off time can then be found as 1
4:01
minus D times the switching period t the
4:06
input to output voltage ratio found in a
4:08
buck boost converter during steady state
4:10
can be determined by examining the
4:13
voltage across the terminals of the
4:15
inductor in on state when the switch S 1
4:18
is closed the input voltage is directly
4:21
applied to the inductor and maintained
4:24
until the end of the on time
4:26
when the switches one is turned off
4:29
during the off period the diode connects
43
the inductor to the output voltage which
46
has a reverse polarity respect to the
48
input gram this voltage is maintained
4:41
until the end of the switching period
4:43
when the switching sequence is restarted
4:47
in steady state the voltage balance
4:50
principle requires the integral of the
4:52
inductor voltage to be zero over a
4:55
switching period this means that the
4:57
area below the voltage curve is equal to
5:00
zero expanding the integral for the
5:03
union of times and considering the
5:05
voltage is constant during these times
5:07
the integral becomes the product of the
5:11
on state voltage be in times the on time
5:15
minus vo times the off time representing
5:20
T on and T off as a function of the
5:23
switching period and duty cycle leaves
5:26
the equation only as a function of the
5:28
input and output voltage the switching
50
period and the duty cycle solving for
54
the duty cycle gives a input to output
57
voltage ratio since T can vary between 0
5:42
and 1 the voltage at the output can be
5:45
lower equal or higher than the input
5:47
voltage the current in inductor il can
5:52
be found as a function of the output
5:54
current IO which equals the diode
5:57
current ID minus the capacitor current
6:00
IC the first step to find the inductor
6:03
current is to consider its average value
6:05
in steady state the average current in
6:09
the capacitor is zero since there is no
6:12
net change in the voltage and the
6:14
average diode current equals the average
6:16
inductor current times 1 minus T since
6:20
the diode only carries the inductor
6:22
current during the off time therefore
6:25
the average output current equals the
6:28
average inductor current times 1 minus T
61
which means the average inductor current
64
equals the average output current
67
determined by the output voltage and
6:40
sister over 1 minus T the variation in
6:44
the ignitor current around the average
6:46
value is controlled by the inductor
6:49
voltage waveform we found previously and
6:51
in that differential equation which
6:54
relates the derivative of IL with the
6:57
voltage applied to the inductor during
7:00
the on time the input voltage is applied
7:02
to the inductor since this is a positive
7:06
constant value the current increases
7:08
linearly during the on time when the
7:11
switch is turned off the output voltage
7:14
which has a reverse polarity is applied
7:17
to the inductor since this is a constant
7:20
negative voltage the inductor current
7:22
will decrease linearly to the same
7:25
starting point to start the sequence
7:27
again if the inductor equation is
70
integrated during the on time since the
72
voltage is constant the amplitude of the
75
inductor current ripple is from to be
77
equal to the voltage to the on time
7:40
times the length of the on time scaled
7:43
by the inductor value plugging in the
7:46
value of the inductor voltage during
7:48
this time they end and the value of the
7:51
on time the do recycle' times the
7:54
switching time and then replacing the
7:57
switching time by 1 over the switching
7:59
frequency the amplitude of the inductor
8:02
current ripple is found as a function of
8:05
the converter parameters this equation
8:08
can be used to find the ripple in the
8:10
inductor current or it can be arranged
8:14
to selecting that for value for a given
8:16
desired current ripple the voltage in
8:21
the capacitor BC is equal to the output
8:24
voltage vo the mean value of the
8:28
capacitor voltage is equal to the main
80
output voltage and by using the input to
83
output relationship we can see that the
86
average capacitor voltage equals the
89
input voltage times T over 1 minus T
8:42
during the end time the current the
8:45
capacitor IC is the exact opposite to
8:48
the load current during the off time the
8:51
capacitor current
8:53
equals the inductor current minus the
8:55
load current the voltage in the
8:57
capacitor is related to the capacitor
9:00
current by the capacitor differential
9:02
equation the capacitance times the
9:05
derivative of the capacitor voltage with
9:07
respect to time is equal to the current
9:10
in the capacitor since the capacitor
9:12
current has a constant negative value
9:15
during the on time the capacitor voltage
9:18
decreases linearly during the off time
9:21
the current decreases linearly starting
9:25
from the inductor current tick minus the
9:27
load current then the capacitor voltage
91
is a concave parabola with a maximum
94
occurring when the capacitor current
96
crosses zero the derivative is zero
99
these cycles repeat every switching
9:42
period with an amplitude of Delta V out
9:44
which is given by the charge
9:47
accumulating in the capacitor scaled by
9:49
the capacitor value this charge Delta Q
9:53
can be approximated by the area below
9:56
the capacitor current during the on time
9:58
which equals the on time T on x the load
10:03
current io representing T on as function
10:07
of the switching period and the recycle
10:09
replacing the output current by the
10:12
output voltage divided by the load
10:14
resistance and finally replacing the
10:17
switching period by the switching
10:19
frequency and the output voltage by the
10:22
input voltage times the input to output
10:24
voltage ratio the expression that
10:27
approximates the output voltage ripple
10:29
is then found by rearranging terms this
103
expression shows the output voltage
105
ripple as a function of the converter
108
parameters and can be also used to size
10:41
the appropriate capacitor required to
10:43
maintain a desired level of ripple let's
10:47
analyze a design example to use all the
10:50
formulas we have just derived the
10:53
proposed back boost converter takes an
10:55
input voltage of 24 volts and converted
10:58
to an output voltage with reverse
11:00
polarity of to a bolt the switching
11:03
frequency is 100 kilohertz
11:06
and the minimum load resistance which
11:08
corresponds to the maximum loading
11:10
condition is 1.2 ohms the maximum ripple
11:14
allowed in the inductor is 20% of the
11:17
average inductor current under maximum
11:19
load and the maximum ripple in the
11:21
capacitor voltage is plus or minus 2% of
11:25
the average output voltage consider the
11:28
input voltage of 24 volt and the output
110
voltage of 12 volt the desired value for
114
the duty cycle can be obtained from the
116
input to output voltage ratio equation
118
which indicates the duty cycle must be
11:41
equal to the output voltage over the sum
11:44
of input and output voltages in this
11:47
case 12 volts over 12 volts plus 24
11:50
volts which equals 1 over 3 or 0.33 the
11:56
average inductor current at the maximum
11:58
load condition 1.21 is equal to the
12:02
output current determined by V out and R
12:05
out over 1 minus T which is 15 amp ere
12:09
the ripple in the adductor is limited to
12:12
20 percent of the maximum average
12:14
inductor current this is 20 percent of
12:18
15 amperes or 3 ampere from the equation
12:22
for the inductor current ripple solving
12:25
for the inductor value and plugging in
12:27
the known values such as the input
120
voltage 24 volts duty cycle 0.33 desire
125
inductor current ripple 3 ampere and
127
switching frequency 100 kilohertz gives
12:41
the required value for the inductor of
12:44
24 micro Henry the voltage ripple in the
12:49
capacitor is limited to plus or minus 2%
12:52
or 4% in total of the average output
12:57
voltage that is 0.40 8 volt using the
13:02
capacitor voltage ripple equation
13:04
solving for the capacitor value and
13:06
plugging in the known values such as
13:09
input voltage 24 volts duty cycle 0.33
13:14
desired output voltage ripple
13:17
0.48 volt switching frequency
13:20
hundred kilohertz and minimum load
13:22
resistor 1.2 ohms gives a required value
13:26
for the capacitor 69 micro farad since
132
the analysis we have done is valid only
134
for continuous conduction mode it is
136
important to determine the minimum load
139
necessary to achieve this condition when
13:42
the average inductor current decreases
13:44
due to a higher load resistance the
13:47
minimum value of the inductor current
13:49
will become closer and closer to zero
13:52
when the average current is to load the
13:56
inductor current will touch zero this is
13:59
a boundary of continuous conduction mode
14:01
and while all the equations previously
14:03
derived are still valid at this point
14:05
the analysis would lose validity if the
14:09
load current was further reduced the
14:12
boundary of discontinuous conduction
14:14
mode happens when the average inductor
14:17
current is equal to half of the current
14:19
ripple which in this case is 1 point 5
14:22
ampere using the relationship between
14:26
the output voltage and the current in
14:29
the load the load in the boundary of
141
discontinuous conduction mode is found
143
to be 8 ohms with this our design is
146
finalized as all the main parameters and
149
waveforms of the asynchronous buck boost
14:41
converter for this example happening
14:44
found and selected thank you for
14:46
watching this tutorial we hope it will
14:48
help you with your upcoming designs
14:52
today we talked about the back post
14:54
topology and we presented a design
14:56
example if you want to see more videos
14:59
on power electronics please check our
15:00
channel
15:04
[Music]

Note the basic waveforms:

This quote, I think may give a bit more insight:

7:11: When the switch is turned off the output voltage which has a reverse polarity is applied to the inductor since this is a constant negative voltage the inductor current will decrease linearly to the same starting point to start the sequence again.

The Inductor, the Actions and Interactions that occurs, thanks to Faradays Law, Lenz's Law, Amperes Law and others, gives us a Gold Mine, all we need do is dig and find the Nuggets.

I hope this helps some My Friend?

Best Wishes,

   Chris