Yes, it seems your device is producing
results.  Very good!


In order for the sulfation to be fully
restored allow the battery voltage to
increase to the 14.4 Volts - 15 Volts
range.

At some point while being restored,
the battery voltage will stabilize and
cease to increase at those higher voltages.

13.8 Volts is effective with batteries that
are in extremely good condition (no sulfation)
as it will prevent the battery becoming
sulfated by self-discharge.



---  Chuck Carpenter  wrote:
>
> I've built and have running a breadboard version of the 10kHz single coil
> desulfator design by Bill Roth.
>
> Initial testing with a 12V 12AHr VRLA battery looked very good.
>
> Currently testing is underway using a badly sulfated lawn tractory battery
> of about 34AHr capacity..


   The nominal voltage of the HF float charger is 13.8.

   When the desulfation process continues to that voltage
   level, I'd consider that the battery is probably
   desulfated.

  That would be verified with a discharge/charge test.
>
> More later...

I've built and have running a breadboard version of the 10kHz single coil
desulfator design by Bill Roth.

Initial testing with a 12V 12AHr VRLA battery looked very good.

Currently testing is underway using a badly sulfated lawn tractory battery
of about 34AHr capacity.  I got if from my friend at a local auto parts
supply store.  His store is a collection point for battery pick up. It is
clean, hydrated and shows no sign of physical damage. Apparently, just neglect.

The battery measured about 12.3V.  Didn't appear to have damaged cells. I
connected my B&D Smart Charger and attempted to charge it.  Ran it for
about 5 hrs at a 2A rate with no change.  Looked like a good candidate for
desulfting.

The Roth 10kHz single coil desulfator was connected to the battery in
combination with a Harbor Freight float charger.  Initial voltage was about
13.1 starting at 1800z 21Nov.

Using a Tektronics 7623 100MHz o'scope, I captured images of the voltage
and current pulses.  These are posted in the Photo section, ChuckC folder.

As of today, the voltage has slowly increased to 13.55 and continues to
rise.  The nominal voltage of the HF float charger is 13.8.  When the
desulfation process continues to that voltage level, I'd consider that the
battery is probably desulfated. That would be verified with a
discharge/charge test.

More later...




Chuck Carpenter, W5USJ, Point, -|- Rains Co., TX -|- EM22cv -|- 72/73
54 years -|- 19 as K2OFN and 35 as W5USJ -|- Most fun = QRP since 1984.
Website = http://www.w5usj.com

Yes, those are good suggestions.

Also investigate the degree of
radio frequency interference
generated in both cases.

There are several very good tutorials
available, as well as Applications Notes
from the various chip manufacturers,
which offer excellent discussion on the
design and implementation of switching
circuits and switching power supplies.

The "Desulphator" circuit is in reality
a 'boost' switching circuit which has
very wide application today in consumer
electronics devices.






---  William Roth  wrote:
>
> One might consider actually trying the circuit without the cap and inductor
> before insisting upon their necessity based upon "standard engineering
> practice".
>
> One might actually want to  look at real waveforms and measure actual
> efficiency as well.
>

It is getting to the point where the Couper is just one inductor and
transistor,
which is now comparable in simplicity with the resistive approach. I
doubt
the capacitive approach can compete in terms of simplicity.

At first I felt it was obvious that if one wanted an over-voltage one
had to
inject energy, so that one needed and inductor or capacitor.
(I am not certain the capacitive approach shows promise on the grounds
that if it did, then power supplies would be capacitive based rather
than
inductive.)

However, all this changed with the unexpected resistive approach where
it
became clear that in this particular application that one didn't need to
inject energy into the battery; this makes me think that (again in this
application alone) it may be an error to think in terms of a circuit
whose
purpose is to inject energy.
    It might be that the inject is not actually doing much, and that it
is
the battery that is doing the work.

The 'bypass capacitors' located at the
power connections to chips are needed
to accommodate power 'surge' demands of
the chips and to prevent the generation
of surge impulses being reflected back
into the circuit board wiring which
might adversely affect other chips on
the board.

The 'bulk capacitance' in the switching
circuit augments the power source in that
the capacitor stores significant energy
which will be instantly available, at very
low impedance, to the demands of the switching
circuit.  It enhances the efficiency of the
switching circuit and prevents any surge
transients being developed on the power line
input to the switching circuit.

The desired 'transients' at the output of the
switcher are more efficiently generated without
losses in the power input wiring from the power
source.

All switching devices utilize this technique
to minimize generation of 'transients' in the
wrong places (source) and to maximize the 'desired
transients' in the right places (load.)






---  Anwar Shiekh  wrote:
>
> >
> > By the way, the inclusion of the "bulk capacitance"
> > very close to the active switching components of the
> > desulphator circuit is "standard engineering practice."
> >
> Buffer capacitors (like on RAM cards)?
> These are there is suppress transients, while here transients are
> desirable.
>
> Maybe (probably) I have misunderstood.
>

Inductor 2 and Bulk Cap

Upon power up current flows through the 1000uh inductor and begins charging up
the bulk capacitor.  At the same time the capacitor on the control side begins
charging up through the 330 ohm current
limit resistor.  The 555 timer output doesn't reach 12 volts for quite some time
base upon the time constant of the cap and resistor. So during startup the FET
will be seeing some pulses below
and at its threshold voltage. This capacitor can be made smaller than Couper
specified.  I am using a 4.7uh electrolytic with no problems.  I also reduced
the current limit resistor to 200 ohms to allow Vcc to the 555 to come up
faster. 100 ohms might work even better.

During this start up period the peak current through the switching inductor can
reach 10 amps and for a while (several miliseconds) the switching is in
continuous mode  (the inductor current
never falls to zero) until the bulk cap fully charges, switching gond on for
while and the circuit stabilizes/settles down.

The bulk cap  has to provide 5-6 amps of pulse current.  I have heard that this
cap can get quite warm under certain circumstances.   It has to handle the
switching current and also creates some
loss (100-200mw) in the circuit (depending upon type and ratings).

While it may be  "standard engineering practice" to provide a bulk capacitor in
many other circuits, in the Couper circuit
it doesn't appear to be necessary. The battery itself is the power source and
should have plenty of current available unless it is in terrible condition.
Where a battery cannot provide adequate current,  a bulk capacitor may be
useful.  Transient suppression is not an issue.  Using a bulk cap in this
particular circuit because it is "standard engineering practice" may be like
adding salt to your steak before tasting it.  It may not need any salt.

If the cap is removed along with the 1000uh inductor, the cost of these parts is
saved and the circuit is more efficient by eliminating any losses associated
with the cap and the 1000uh inductor.  There appears to be no loss in
functionality or performance with these parts eliminated.

My testing was done with a 4.7uh inductor at 5-20 khz while keeping peak current
through the inductor at or below about 5 amps. I did not test with a 220u
inductor.

Bill


--- In randrdesulfatorforum@yahoogroups.com, Anwar Shiekh <shiekh@...> wrote:
>
> >
> > By the way, the inclusion of the "bulk capacitance"
> > very close to the active switching components of the
> > desulphator circuit is "standard engineering practice."
> >
> Buffer capacitors (like on RAM cards)?
> These are there is suppress transients, while here transients are
> desirable.
>
> Maybe (probably) I have misunderstood.
>

Any "circuit" which contains the
necessary "components" will "resonate"
when "stimulated" by a sharp pulse.

In most cases, this "damped oscillation"
is coincidental and has no bearing on
the "Process" being performed.  It is
simply a "parasitic" event that is most
often harmless.

Desulphation is accomplished by sufficient
"overvoltage" being impressed upon the
crystallized sites of Lead Sulphate in
the plate structures that "normal"
charging voltage isn't able to restore.
Pulses are extremely effective because of the
low duty cycle and very high peak currents.
During the desulphation process the battery
under treatment does not become dangerously
"hot" and the plates are not dangerously
"stressed."  In order to avoid damage to
the battery, the desulphation process should
proceed slowly.

Normal charging is able to restore the "newly
formed" amorphous lead sulphate to active
plate material while releasing sulphuric acid
back into the electrolyte solution.  This
process occurs at the "normal" charging voltage.

"Resonance" in battery desulphation by pulsing
is simply a parasitic response.  It is not the
"source" of the conversion of crystallized lead
sulphate into useful active plate material.

Ed's observations are correct regarding the
charge and discharge of the pulsing inductor.

By the way, the inclusion of the "bulk capacitance"
very close to the active switching components of the
desulphator circuit is "standard engineering practice."
This permits the "peak" charging current to be "maximized"
leading to greater efficiency of the switching circuit.
'I squared R' losses associated with circuit lead
lengths are minimized.

What may appear to be "saturation" current through the
inductor while "charging" is very much dependent upon the
source impedance, the resistance of wiring from source
to inductor, and finally the resistance of the switching
device while "turned on."  Therefore, in any given circuit,
it may be well below the rated characteristics of the
inductor specifications.



---  "Bill"  wrote:
>
> Andy,
>

>
> But before we get too wrapped up in this HV pulse,  let's also consider that
Alastair Couper suggested that the desulfation was a result of the battery
"resonating"  when a current pulse was applied and because of this resonance,
sulfur ions begin to bounce about and return those stuck to the plates back into
solution.
>
> It seems to me then, that the focus should be to maximize this resonance with
whatever type of circuit is being used.
>
> Bill
>

Bill wrote:
> Ed,
>
>
>
> --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:
>
>>Bill wrote:
>>
>>>Ed,  you are going strictly by Faraday's law that works on paper on an ideal
model. If you have ever done any real design work on this kind of stuff then you
know that. There are other variables involved here. Battery
inductance/capacitance/ resistance,  and other and parasitics ...
>>>
>>
>>Bill,
>>
>>I have asked what accounts for the discrepancy between the
>>formula and the observations.  I pointed out a number of
>>different possibilities.  How can you interpret that as going
>>"strictly by Faraday's law that works on paper on an ideal model"?
>>How can mentioning other possibilities mean going strictly
>>by Faraday's law?
>>
>>Faraday's law applies to real inductors in real circuits,
>>not just to inductors on paper on an ideal model.  It is
>>a disservice to suggest otherwise, or to suggest that it
>>does not apply to discharge.  Given that it is a law of
>>physics, and that the observations do not match what the
>>law indicates, the obvious question is "what accounts
>>for the difference?".
>>
>>Only you can determine that.  The rest of us, including you,
>>can speculate, but you are the one with the setup.  You
>>have the option of pursuing it and trying to figure it
>>out.
>
>
> Unless I have misunderstood,  you are claiming that the voltage "spike" seen
across the battery is the V in V= L* di/dt.  And where L is the inductor in  a
Couper Circuit.

Yes, it's the V produced by the inductor, minus any circuit losses.
Investigating the discrepancy between the observations and V=L*di/dt
would determine what else is going on.

>
> So we have an applied voltage of 12v an inductance of 4.7uH and
> a 6 microsecond pulse to the inductor.  In 6 us the current in the inductor
rises to 6 amps. When the pulse ends the electromagnetic flux field collapses
inducing a current that flows through the catch diode and then to the battery
which is now the "load".

Right.  But the 6us is irrelevant to the discharge time - 6us is the
charge time.

>
> The battery is already at 12 Volts.  But also has an unknown amount of series
inductance( inductive reactance), parallel capacitance ( capacitive reactance),
series resistance as well as Warburg Impedance due to ion currents. So the
battery is not only a load but an integral part of the circuit.
>
> Simply stated I doubt that simple application of Faraday's law where the only
L considered  is the 4.7uf inductor on the circuit board can accurately
calculate the amplitude of the voltage spike seen across the battery.  I suspect
that the battery may play an integral role in generation and amplitude if this
spike and that there may be a point of diminishing return on amplitude  due to
the nature of the battery.

Yes!!! That is the point of the question: what accounts for the
discrepancy between the observation and the formula.  Whatever
is going on in the rest of the circuit has to account for the
difference, unless the observation is incorrect.  That has been
the point all along.

>
> My observation that the voltage spike does not increase in amplitude with an
increase in current past about 5 amps does not IMO necessarly mean there is
something wrong with the circuit, sloppy testing, etc.  The circuit is well
constructed and the layout is consistent with accepted standards for SMPS
prototyping.  The test equipment is more than adequate and the designer/tester
is a bit more than a electronics hobbyist, tinkerer or armchair theorist.
>
> I haven't seen where anyone has tried inductors of this low of a value or have
tried peak currents above 6 amps up into in the 15 amp range. If they have the
data is not commonly available.  I certainly  cannot find where any half way
systematic testing was done and where the peak voltage readings were recorded
along with inductance values, peak current, pulse widths, etc.
>
>
>
>
>
>
>
>

Reinserting what you wrote, and snipped from this post:

  >>>To suggest a "problem" eg. (blown diode,etc) saturated inductor ,etc
  >>>because the voltage spike peaks at a certain peak current in the
  >>>inductor  based upon that alone is jumping to a premature conclusion
  >>>while ignoring other possibilities.
>
>>Bullshit. That is purposely misleading. Quoting my prior post:
>>"In any event, the word problem was used to refer to the
>>observation that increasing current through a non-saturated
>>inductor does not result in higher amplitude on discharge.
>>That is inconsistent with the way inductors work. Therefore
>>either the circuit is somehow limiting the amplitude, or
>>the observation is somehow incorrect or maybe a combination
>>of those things.  Something has to account for V = L di/dt
>>not holding true in the observation."
>
>
> I will not digress here to that kind of language here, but if we were face to
face you would get an ear full and them some.

Irrelevant to the question.

>
> You seem to be more interested in being "theoretically right" than to offer
anything helpful in a practical manner.

I asked a question.

I don't care about _me_ being "theoretically right", as you put it.
What I care about is determining why the observation does not match
V=L*di/dt

>
> I see no reason to continue any further along these lines.

Not if it is about personal antagonism. But if it's about figuring
out what is going on, it should be of interest to all.

Ed

>
>
>
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>
>

Ed,



--- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:
>
> Bill wrote:
> > Ed,  you are going strictly by Faraday's law that works on paper on an ideal
model. If you have ever done any real design work on this kind of stuff then you
know that. There are other variables involved here. Battery
inductance/capacitance/ resistance,  and other and parasitics ...
> >
>
> Bill,
>
> I have asked what accounts for the discrepancy between the
> formula and the observations.  I pointed out a number of
> different possibilities.  How can you interpret that as going
> "strictly by Faraday's law that works on paper on an ideal model"?
> How can mentioning other possibilities mean going strictly
> by Faraday's law?
>
> Faraday's law applies to real inductors in real circuits,
> not just to inductors on paper on an ideal model.  It is
> a disservice to suggest otherwise, or to suggest that it
> does not apply to discharge.  Given that it is a law of
> physics, and that the observations do not match what the
> law indicates, the obvious question is "what accounts
> for the difference?".
>
> Only you can determine that.  The rest of us, including you,
> can speculate, but you are the one with the setup.  You
> have the option of pursuing it and trying to figure it
> out.

Unless I have misunderstood,  you are claiming that the voltage "spike" seen
across the battery is the V in V= L* di/dt.  And where L is the inductor in  a
Couper Circuit.

So we have an applied voltage of 12v an inductance of 4.7uH and
a 6 microsecond pulse to the inductor.  In 6 us the current in the inductor
rises to 6 amps. When the pulse ends the electromagnetic flux field collapses
inducing a current that flows through the catch diode and then to the battery
which is now the "load".

The battery is already at 12 Volts.  But also has an unknown amount of series
inductance( inductive reactance), parallel capacitance ( capacitive reactance),
series resistance as well as Warburg Impedance due to ion currents. So the
battery is not only a load but an integral part of the circuit.

Simply stated I doubt that simple application of Faraday's law where the only L
considered  is the 4.7uf inductor on the circuit board can accurately calculate
the amplitude of the voltage spike seen across the battery.  I suspect that the
battery may play an integral role in generation and amplitude if this spike and
that there may be a point of diminishing return on amplitude  due to the nature
of the battery.

My observation that the voltage spike does not increase in amplitude with an
increase in current past about 5 amps does not IMO necessarly mean there is
something wrong with the circuit, sloppy testing, etc.  The circuit is well
constructed and the layout is consistent with accepted standards for SMPS
prototyping.  The test equipment is more than adequate and the designer/tester
is a bit more than a electronics hobbyist, tinkerer or armchair theorist.

I haven't seen where anyone has tried inductors of this low of a value or have
tried peak currents above 6 amps up into in the 15 amp range. If they have the
data is not commonly available.  I certainly  cannot find where any half way
systematic testing was done and where the peak voltage readings were recorded
along with inductance values, peak current, pulse widths, etc.











>
> Bullshit. That is purposely misleading. Quoting my prior post:
> "In any event, the word problem was used to refer to the
> observation that increasing current through a non-saturated
> inductor does not result in higher amplitude on discharge.
> That is inconsistent with the way inductors work. Therefore
> either the circuit is somehow limiting the amplitude, or
> the observation is somehow incorrect or maybe a combination
> of those things.  Something has to account for V = L di/dt
> not holding true in the observation."

I will not digress here to that kind of language here, but if we were face to
face you would get an ear full and them some.

You seem to be more interested in being "theoretically right" than to offer
anything helpful in a practical manner.

I see no reason to continue any further along these lines.

Or put another way; if you have discovered a violation
of the laws of Electrodynamics, this would be a
breakthrough in Physics, and so rather unlikely.

Bill wrote:
> Ed,  you are going strictly by Faraday's law that works on paper on an ideal
model. If you have ever done any real design work on this kind of stuff then you
know that. There are other variables involved here. Battery
inductance/capacitance/ resistance,  and other and parasitics ...
>

Bill,

I have asked what accounts for the discrepancy between the
formula and the observations.  I pointed out a number of
different possibilities.  How can you interpret that as going
"strictly by Faraday's law that works on paper on an ideal model"?
How can mentioning other possibilities mean going strictly
by Faraday's law?

Faraday's law applies to real inductors in real circuits,
not just to inductors on paper on an ideal model.  It is
a disservice to suggest otherwise, or to suggest that it
does not apply to discharge.  Given that it is a law of
physics, and that the observations do not match what the
law indicates, the obvious question is "what accounts
for the difference?".

Only you can determine that.  The rest of us, including you,
can speculate, but you are the one with the setup.  You
have the option of pursuing it and trying to figure it
out.



> To suggest a "problem" eg. (blown diode,etc) saturated inductor ,etc because
the voltage spike peaks at a certain peak current in the inductor  based upon
that alone is jumping to a premature conclusion while ignoring other
possibilities.

Bullshit. That is purposely misleading. Quoting my prior post:
"In any event, the word problem was used to refer to the
observation that increasing current through a non-saturated
inductor does not result in higher amplitude on discharge.
That is inconsistent with the way inductors work. Therefore
either the circuit is somehow limiting the amplitude, or
the observation is somehow incorrect or maybe a combination
of those things.  Something has to account for V = L di/dt
not holding true in the observation."

>
> So let's agree to disagree.

But we don't disagree on the overall issue, unless you are claiming
that Faraday's law does not apply.  We have both speculated on
other possibilities that account for the observation that increasing
current through the inductor does not increase the pulse amplitude.

  From an earlier post, you seem to believe that V=L*di/dt does not
apply to the discharge of the inductor.   If that is your belief,
then we do indeed disagree.

Ed

>
> Bill
>
>
>
>
> --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:
>
>>Bill wrote:
>>
>>>--- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@> wrote:
>>>
>>>
>>>
>>>
>>>>V = L di/dt is not presumption.
>>>
>>>
>>>You apparently missed my point which has little to do with the common
>>>and well known formula above.     Your V and my V are different so you
>>>are arguing a different point altogether.
>>>
>>>The formula above does not calculate the amplitude of the over voltage
>>>spike across a battery being pulsed.    Formula above calculates the
>>>applied voltage to an inductor given a known L and di/dt.  In that  case
>>>V=12.5v which is  the battery voltage.
>>
>>That's not correct Bill.
>>
>>It is the discharge of the inductor that produces the overvoltage
>>spike. The formula calculates the spike produced by the inductor
>>when the current is interrupted. That spike, minus any losses in
>>the discharge path, is what you see across the battery.
>>
>>The overvoltage you see occurs when the inductor discharges,
>>and V = L*di/dt most definitely applies to that, in spite of
>>what you think.
>>
>>Ed
>>
>>
>>
>>>Apples and  oranges.
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>
>
>
>
>

--- In randrdesulfatorforum@yahoogroups.com, Anwar Shiekh <shiekh@...>
wrote:
>
>
> On Nov 19, 2009, at 3:49 AM, Bill wrote:
>
> > Andy,
> >
> > So far my testing of 4 different circuits types suggests that the
> > HV pulse seen across the battery is not solely related to an
> > inductor used in the circuit.
> >
> > We get a similar hv pulse when using a capacitive discharge circuit,
> > and by initiating current flow and abruptly stopping it ("resistive"
> > approach). This suggests to me that the battery itself plays a large
> > role in this HV pulse . Interestingly, the HV pulse amplitude is
> > similar in each method (50 - 70 volts) given similar applied
currents.
> >
> Yes, this part suggests the battery may play a dominant role.
>
> > But before we get too wrapped up in this HV pulse, let's also
> > consider that Alastair Couper suggested that the desulfation was a
> > result of the battery "resonating" when a current pulse was applied
> > and because of this resonance, sulfur ions begin to bounce about and
> > return those stuck to the plates back into solution.
> >
> The battery resonating makes sense, but how this 'breaks up' the
> crystals is unclear to me. For the moment I
> feel that it is the high voltage that is doing the trick.
>
> > It seems to me then, that the focus should be to maximize this
> > resonance with whatever type of circuit is being used.
> >
>
> Then we have a lot of work to do, for the battery only rings for a
> very short time.
>

When I have some additional time I want to explore this resonance thing.
The battery resonance (per Couper) is in the 2-6  mhz range.  I am
thinking that this could be possibly be detected and exploited/enhanced
via a Phase Locked Loop IC.  2-6  mHz  is well within the range  of the
74H4046 PLL IC.

Andy,

So far my testing of 4 different circuits types  suggests that the
HV pulse seen across the battery is not solely related to an inductor used in
the circuit.

We get a similar hv pulse when using a capacitive discharge circuit, and by
initiating current flow and abruptly stopping it ("resistive" approach). This
suggests to me that the battery itself plays a large role in this HV pulse . 
Interestingly, the HV pulse amplitude is similar in each method (50 - 70 volts)
given similar applied currents.

But before we get too wrapped up in this HV pulse,  let's also consider that
Alastair Couper suggested that the desulfation was a result of the battery
"resonating"  when a current pulse was applied and because of this resonance,
sulfur ions begin to bounce about and return those stuck to the plates back into
solution.

It seems to me then, that the focus should be to maximize this resonance with
whatever type of circuit is being used.

Bill



--- In randrdesulfatorforum@yahoogroups.com, Anwar Shiekh <shiekh@...> wrote:
>
> OK, I'm not here to start an argument, and agree that in a switch
> opening it is indeed the fast current cut that induces a high voltage.
> That said; since the resistive approach also induces a high voltage,
> maybe the battery is the dominant factor in this particular application?
>
> We have all read Horowitz and Hill, so please no 101 jokes.
>
>
>
> On Nov 18, 2009, at 6:27 PM, ehsjr wrote:
>
> > Bill wrote:
> > > --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@>
> > wrote:
> > >
> > >
> > >
> > >>V = L di/dt is not presumption.
> > >
> > >
> > > You apparently missed my point which has little to do with the
> > common
> > > and well known formula above. Your V and my V are different so you
> > > are arguing a different point altogether.
> > >
> > > The formula above does not calculate the amplitude of the over
> > voltage
> > > spike across a battery being pulsed. Formula above calculates the
> > > applied voltage to an inductor given a known L and di/dt. In that
> > case
> > > V=12.5v which is the battery voltage.
> >
> > That's not correct Bill.
> >
> > It is the discharge of the inductor that produces the overvoltage
> > spike. The formula calculates the spike produced by the inductor
> > when the current is interrupted. That spike, minus any losses in
> > the discharge path, is what you see across the battery.
> >
> > The overvoltage you see occurs when the inductor discharges,
> > and V = L*di/dt most definitely applies to that, in spite of
> > what you think.
> >
> > Ed
> >
> > >
> > > Apples and oranges.
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> >
> >
> >
>

Ed,  you are going strictly by Faraday's law that works on paper on an ideal
model. If you have ever done any real design work on this kind of stuff then you
know that. There are other variables involved here. Battery
inductance/capacitance/ resistance,  and other and parasitics ...

To suggest a "problem" eg. (blown diode,etc) saturated inductor ,etc because the
voltage spike peaks at a certain peak current in the inductor  based upon that
alone is jumping to a premature conclusion while ignoring other possibilities.

So let's agree to disagree.

Bill




--- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:
>
> Bill wrote:
> > --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@> wrote:
> >
> >
> >
> >>V = L di/dt is not presumption.
> >
> >
> > You apparently missed my point which has little to do with the common
> > and well known formula above.     Your V and my V are different so you
> > are arguing a different point altogether.
> >
> > The formula above does not calculate the amplitude of the over voltage
> > spike across a battery being pulsed.    Formula above calculates the
> > applied voltage to an inductor given a known L and di/dt.  In that  case
> > V=12.5v which is  the battery voltage.
>
> That's not correct Bill.
>
> It is the discharge of the inductor that produces the overvoltage
> spike. The formula calculates the spike produced by the inductor
> when the current is interrupted. That spike, minus any losses in
> the discharge path, is what you see across the battery.
>
> The overvoltage you see occurs when the inductor discharges,
> and V = L*di/dt most definitely applies to that, in spite of
> what you think.
>
> Ed
>
>
> >
> > Apples and  oranges.
> >
> >
> >
> >
> >
> >
> >
>

-- Best regards,
Derek Calanchini
Owner
Creative Network Solutions
Phone: 916-852-2890
Fax: 916-852-2899

Bill wrote:
> --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:
>
>
>
>>V = L di/dt is not presumption.
>
>
> You apparently missed my point which has little to do with the common
> and well known formula above.     Your V and my V are different so you
> are arguing a different point altogether.
>
> The formula above does not calculate the amplitude of the over voltage
> spike across a battery being pulsed.    Formula above calculates the
> applied voltage to an inductor given a known L and di/dt.  In that  case
> V=12.5v which is  the battery voltage.

That's not correct Bill.

It is the discharge of the inductor that produces the overvoltage
spike. The formula calculates the spike produced by the inductor
when the current is interrupted. That spike, minus any losses in
the discharge path, is what you see across the battery.

The overvoltage you see occurs when the inductor discharges,
and V = L*di/dt most definitely applies to that, in spite of
what you think.

Ed


>
> Apples and  oranges.
>
>
>
>
>
>
>

I'm interested in testing desulfator circuits.
I work as an equipment mechanic so have access to various types of batteries
from small battery backup batteries, to riding mower batteries, to car batteries
to big tractor trailer batteries, as well as ones from our local scrap yard.

I'm familiar with electrical and some electronics but not an EE, thus I don't
have the resources to build several types of circuits to see what works.  If you
are in the opposite situation where you have built several circuits but don't
have big batteries to test, please contact me offlist.
I live about an hour south of Atlanta, Georgia, USA in Moreland, Georgia.
Thank you,
William

--- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:


>
> V = L di/dt is not presumption.

You apparently missed my point which has little to do with the common
and well known formula above.     Your V and my V are different so you
are arguing a different point altogether.

The formula above does not calculate the amplitude of the over voltage
spike across a battery being pulsed.    Formula above calculates the
applied voltage to an inductor given a known L and di/dt.  In that  case
V=12.5v which is  the battery voltage.

Apples and  oranges.

Bill wrote:
> Ed,
>
> If you have done all this testing, then where is the data ?

Over the years I've posted test results in the other desulfator
group. So have others.

>
> I am not willing to search a defunct newsgroup with a grumpy owner  for
piecemeal data that is scattered over numerous disorganized posts.  And where
... when you ask a legitimate question,  you either get ignored or you get
chided/berated for asking it because the answer might be in an obscure post
somewhere on the board.
>

Your choice.

> As far as the "problem" goes.  You are presuming that there will be an
infinitely linear response in the amplitude of the voltage "spike" across the
battery relative to the peak current through the inductor.
>

V = L di/dt is not presumption.

> Can you offer test data that shows this to be the case ?
> Maybe some graphs or charts of your test readings?
>
> A battery is not a typical load and may not respond according to the formulas
we generally use in SMPS design.
>
> Also consider (in case you missed it) the 220uf bulk capacitor and the 1000uf
inductor were removed from the circuit.
>

Bill, it is simple. The formula does not match the observations
you reported. You and I and anyone else can speculate as to what
might cause it, but only you can determine where the discrepancy
lies, and then only if you wish to.

Ed




>
> Bill
>
>
>
> --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@...> wrote:
>
>>Bill wrote:
>>
>>>
>>>
>>> > >
>>> > > Let me add to this for the sake of additional clarity.
>>> > >
>>> > > The over voltage pulse height varies with pulse width to the FET
>>>ONLY from 0 - 2.2us. This correlates to a peak current through the
>>>inductor from 0 - 4.5 amps.
>>> > >
>>> > > Increasing the pulse width past 2.2 us does not increase the height
>>>of the overvoltage pulse any further. However it does increase the
>>>average current used by the circuit.
>>> > >
>>> > > The inductor is is rated at about 15 amps peak current so there is
>>>no chance that it is saturating at 4.5 amps.
>>> > >
>>> > > Bill
>>> > >
>>> >
>>> > Hmmm...that pulse amplitude limitation is very interesting!
>>> > It brings up a bunch of questions. You say that increasing current
>>> > through the inductor beyond some point (4.5A) does not result in
>>> > any increase in discharge pulse amplitude. That implies inductor
>>> > saturation. However, you state it is not saturation.
>>> >
>>> > Ok, if it is not saturation, what limits the pulse amplitude?
>>> > Is the FET turn off slow? Does the FET driver drive the FET
>>> > on and just allow it to turn off, or does it drive it off too?
>>>
>>>The FET Driver is a TC4428.  It drives the FET off in less than 40ns
>>>
>>> > Is the peak current measured or computed or both? How is
>>> > peak voltage measured? Have you experimented to determine
>>> > where inductor saturation does occur? Is there any possibility
>>> > that the output circuit or measuring circuit is "eating"
>>> > the output pulse - some kind of series L or parallel C or
>>> > bad scope probe or??? Maybe a flaky diode?
>>>
>>> I do not necessarily see this as a problem or a malfunction but more likely
>>>how the battery is responding to the circuit.
>>>
>>>The inductor is a 2020.472NL.  Per the data sheet the saturation current
>>>is 28 amps
>>>@25c. We are not even approaching that,  so I am not prepared or willing
>>>to hit it that hard to find the saturation point.  The waveform shows a
>>>nice peak and fall off typical of correct operation.
>>>
>>>The peak current through the inductor is measured via a .1 ohm resistor
>>>in series
>>>with the source to ground (like a sense resistor in a typical SMPS.)
>>>
>>>The peak voltage I am referring to is the over voltage "spike" that is
>>>seen when read directly across the battery with a 100 mhz oscillisocpe
>>>with a high quality 10:1 probe.
>>>
>>>The diode is a FFFP10UP20STU from Farichild.  It is an "ultrafast"
>>>silicon recovery  type rated at 200V and 10 amps.  There is no evidence
>>>that it has failed.
>>>
>>> >
>>> > This is a really neat problem! I hope you are able to track
>>> > down what is going on. :-)
>>> >
>>>
>>>Like I said,  I am not sure that there actually is a problem.  Since no
>>>one has ever done similar testing on an original Couper circuit ( at
>>>least not that I am aware of),    there is no baseline data for reference.
>>>
>>> Bill
>>
>>
>>A lot of similar testing has been done over the years - read
>>the other desulfator group entries. I've done it myself,
>>numerous times.  In fact, I have a "test bed" desulfator
>>used specifically to test various inductors for performance,
>>including saturation, in the Couper original circuit. I'm
>>not sure how similar that is. But I don't know how "similar"
>>fits anyway.  I would think you need identical testing to make
>>valid comparisons to a benchmark.
>>
>>In any event, the word problem was used to refer to the
>>observation that increasing current through a non-saturated
>>inductor does not result in higher amplitude on discharge.
>>That is inconsistent with the way inductors work. Therefore
>>either the circuit is somehow limiting the amplitude, or
>>the observation is somehow incorrect or maybe a combination
>>of those things.  Something has to account for V = L di/dt
>>not holding true in the observation.
>>
>>Ed
>>
>>
>>>
>>>
>>>
>>>
>>>
>>
>
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>
>

I just cranked the pulse width up to 6us.  This gives a "measured"
peak current through the inductor of 12 amps.   The current waveform shows no
signs of saturation.  Current starts at zero,  ramps up to 12 amps in 6 us and
then falls back to zero in about 60ns.

This clearly shows that there is not a saturation issue, so saturation is off
the table as far as the perceived/theoretical problem goes.

The inductor is barely warm, running at precisely 5kHz. The average current is
176ma.  The diode is dissipating some heat but is right in line with what SPICE
shows.

There is a 45V pulse measured directly across the battery.

Bill


--- In randrdesulfatorforum@yahoogroups.com, "Bill" <cyber.roth@...> wrote:
>
> Ed,
>
> If you have done all this testing, then where is the data ?
>
> I am not willing to search a defunct newsgroup with a grumpy owner  for
piecemeal data that is scattered over numerous disorganized posts.  And where
... when you ask a legitimate question,  you either get ignored or you get
chided/berated for asking it because the answer might be in an obscure post
somewhere on the board.
>
> As far as the "problem" goes.  You are presuming that there will be an
infinitely linear response in the amplitude of the voltage "spike" across the
battery relative to the peak current through the inductor.
>
> Can you offer test data that shows this to be the case ?
> Maybe some graphs or charts of your test readings?
>
> A battery is not a typical load and may not respond according to the formulas
we generally use in SMPS design.
>
> Also consider (in case you missed it) the 220uf bulk capacitor and the 1000uf
inductor were removed from the circuit.
>
>
> Bill
>
>
>
> --- In randrdesulfatorforum@yahoogroups.com, ehsjr <ehsjr@> wrote:
> >
> > Bill wrote:
> > >
> > >
> > >
> > >  > >
> > >  > > Let me add to this for the sake of additional clarity.
> > >  > >
> > >  > > The over voltage pulse height varies with pulse width to the FET
> > > ONLY from 0 - 2.2us. This correlates to a peak current through the
> > > inductor from 0 - 4.5 amps.
> > >  > >
> > >  > > Increasing the pulse width past 2.2 us does not increase the height
> > > of the overvoltage pulse any further. However it does increase the
> > > average current used by the circuit.
> > >  > >
> > >  > > The inductor is is rated at about 15 amps peak current so there is
> > > no chance that it is saturating at 4.5 amps.
> > >  > >
> > >  > > Bill
> > >  > >
> > >  >
> > >  > Hmmm...that pulse amplitude limitation is very interesting!
> > >  > It brings up a bunch of questions. You say that increasing current
> > >  > through the inductor beyond some point (4.5A) does not result in
> > >  > any increase in discharge pulse amplitude. That implies inductor
> > >  > saturation. However, you state it is not saturation.
> > >  >
> > >  > Ok, if it is not saturation, what limits the pulse amplitude?
> > >  > Is the FET turn off slow? Does the FET driver drive the FET
> > >  > on and just allow it to turn off, or does it drive it off too?
> > >
> > > The FET Driver is a TC4428.  It drives the FET off in less than 40ns
> > >
> > >  > Is the peak current measured or computed or both? How is
> > >  > peak voltage measured? Have you experimented to determine
> > >  > where inductor saturation does occur? Is there any possibility
> > >  > that the output circuit or measuring circuit is "eating"
> > >  > the output pulse - some kind of series L or parallel C or
> > >  > bad scope probe or??? Maybe a flaky diode?
> > >
> > >  I do not necessarily see this as a problem or a malfunction but more
likely
> > > how the battery is responding to the circuit.
> > >
> > > The inductor is a 2020.472NL.  Per the data sheet the saturation current
> > > is 28 amps
> > > @25c. We are not even approaching that,  so I am not prepared or willing
> > > to hit it that hard to find the saturation point.  The waveform shows a
> > > nice peak and fall off typical of correct operation.
> > >
> > > The peak current through the inductor is measured via a .1 ohm resistor
> > > in series
> > > with the source to ground (like a sense resistor in a typical SMPS.)
> > >
> > > The peak voltage I am referring to is the over voltage "spike" that is
> > > seen when read directly across the battery with a 100 mhz oscillisocpe
> > > with a high quality 10:1 probe.
> > >
> > > The diode is a FFFP10UP20STU from Farichild.  It is an "ultrafast"
> > > silicon recovery  type rated at 200V and 10 amps.  There is no evidence
> > > that it has failed.
> > >
> > >  >
> > >  > This is a really neat problem! I hope you are able to track
> > >  > down what is going on. :-)
> > >  >
> > >
> > > Like I said,  I am not sure that there actually is a problem.  Since no
> > > one has ever done similar testing on an original Couper circuit ( at
> > > least not that I am aware of),    there is no baseline data for reference.
> > >
> > >  Bill
> >
> >
> > A lot of similar testing has been done over the years - read
> > the other desulfator group entries. I've done it myself,
> > numerous times.  In fact, I have a "test bed" desulfator
> > used specifically to test various inductors for performance,
> > including saturation, in the Couper original circuit. I'm
> > not sure how similar that is. But I don't know how "similar"
> > fits anyway.  I would think you need identical testing to make
> > valid comparisons to a benchmark.
> >
> > In any event, the word problem was used to refer to the
> > observation that increasing current through a non-saturated
> > inductor does not result in higher amplitude on discharge.
> > That is inconsistent with the way inductors work. Therefore
> > either the circuit is somehow limiting the amplitude, or
> > the observation is somehow incorrect or maybe a combination
> > of those things.  Something has to account for V = L di/dt
> > not holding true in the observation.
> >
> > Ed
> >
> > >
> > >
> > >
> > >
> > >
> >
>