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Time for a new design
This new design fills the gap between the low power kick back and the direct drive desulfators and has features that a lot of you will
like.
Here is a list of its features:
- It is as simple as the kick back circuit but capable of >30 amp pulses.
- No custom inductors!! All parts are off the shelf.
- Powered from the battery.
- Pulse width and rate fully variable.
- Very efficient (75 to 85% depending on load)
- No audible sound
- Very fast rise and fall times
Looks a lot like the old kick back circuit.
D1 - 12 volt zener diode. In this circuit it is mainly for protection of the driver and 555. I use a 1N4742A 1 Watt diode but a 1/2
watt will do also.
D2,3,4 - this is a 40 volt 3 amp (or greater) schottsky diode. Almost any switching type diode will work. Don't use the standard
power supply rectifier like a 1N5404. They are too slow. the short list of parts numbers is:
HER302 thur HER308
MBR340 thur MBR360
SR304 thur SR306
1N5822, 1N5824
50WQ10
50SQ060, 080
D2 - (listed twice) - You can push this circuit to over 100 amps peak with really good caps. If you do, you need to use the 50SQ060
or 50SQ080 for D2.
L1,2 - 20uh and 2.5 amp saturation current minimum. Here is where powdered Iron cores work very well. Pulse Engineering,
Renco, etc. make a line of "simple switcher" inductors that work well. Read the early posts in this thread for PE part numbers. The
Micrometals -8, -26 and -56 toroid cores that are about 3/4 inch in diameter work well here. Ferrites actually should not be used
here.
C6,7 - The capacitance is not important, the ESR is. The lower the ESR the more output current you will get. If you can achieve .01
ohm ESR for these caps you can get over 100 amps peak out of the circuit.
R3 - indicated as 10 ohms in the schematic. Value not critical. You can use 100 ohms if you want. This resistor is just part of the
protection of the 555 and mosfet driver.
Q1 - an IRF1404 works great. Lowest "ON" resistance wins. The part needs a Vds of at least 30 volts (40 and above is better)
Beware of counterfeit Mosfets from China.
Go forth and desulfate
Mark
I have corrected my stupid mistakes. I have made a png of the table of components. Can you check for stupid errors again please.
I have kept the redundant protection diod as I need the trace to go through it I could replace it with a 0R I guess.
I cannot be certain of the inductor without physical examination, as the suppliers website does not really indicate much.
[edit] I have changed the images to reflect the latest pcb version. Our local supplier does not have any current rated inductors
except one 0.1mH, which from what you [Mark] have written may be large enough. What is the part number of the tiny little
inductors in the first post. They look good. I fear that the high voltage nova caps wont cut it. I think I need to build a rig to
measure ESR, which doesnt look easy. Detecting bad caps is not so bad, but measuring ESR looks a bit tricky. If anyone could look
at http://www.emcesd.com/tt020100.htm and http://repairfaq.ece.drexel.edu/sam/captest.htm to help me find a method to
determine the ESR in ohms using a square wave genreator and a scope, it would be much appreciated. (I can see the hint of it ,but
dont quite get it yet...)
Cheers.
Jasper
Here's a new twist guys:
There seems to be efficiencies to be gained using a transformer run direct drive with a half wave voltage doubler front end.
http://www.creative-science.org.uk/multipliers.html
Because of the ac sinewave we dont see high inrush currents into the caps thus we can eliminate the current limit resistor.
Further the voltage doubler means we only need half the supply V from the transformer => lower transformer resistance losses. I
calc. that with a transformer that uses 14 gauge wire at 10" per turn needing 16 turns secondary wire (for 16VAC RMS) we should
see around .04 Ohm secondary resistance. Add that to perhaps .04 Ohm for a 15000uF Low ESR cap as the first cap in the doubler
circuit => .08ohm total. Thus for a given current draw of say 3amps we have a power loss of i*i*r = 3*3*.08= .72 Watt. Compare
to say a 3 ohm "limit" resistor we get 27 Watts lost! An improvement of 97%!!!.
With 16VRMS the voltage doubling works out to about 38VDC. (32*1.4 peak) - (6V Cap charging losses). This is as simulated in
Proteus ISIS with a 4000uF secondary cap.
For the second cap I used a 4000uf .01 ESR ....annd estimate about .03ohm for the MOSFETS and wiring losses and .01 for a 13V
desulphated batt. This still delivers up to (38 -13)/.05 = 500 Amps per pulse at a 60Hz frequency, 10us pulse.
If the 38V peak pulse bottoms out around 25V (based on pulsewidth), avg energy delivered is : 0.5 * capacitance * V *V = 0.5
*.004*(38-25) *(38-25) = 0.338 Joule per pulse. At 60Hz => .338*60 = 20 Watts into the .05ohm load! or 20/5 = 4Watts of pulse
energy into the batt!!
For a heavily sulphated batt of perhaps .1 ohm Int Res. We have a pulse of (38- 13)/.14 = 178A. And a larger pulsewidth =
200usec due to slower discharge. This simulates to a peak of 38V and a base of 29V (on a 13V batt) for a 9V change.
Energy = .5*.004*81= .162Joule per pulse = 9.72 Watts into the .14 ohm load => .1/.14 * 9.72 = 7 Watts into the batt!!!
I have completed assembly of your Voltage Doubler Desulfator Pulsor and have it connected to a healthy battery rated at 435 CCA.
I have also connected a voltage adjustable 2 amp battery charger with amp meter to this battery. The voltage is set at 13.5 volts.
Current draw with both pulsor and battery connected is 590 ma. The battery alone is 90ma. The control circuitry operating behind a
100 ohm register at 12.6 volts draws 93 ma. The output voltage measured on the + and - bus at the output terminal of the PCB is
20.98 volts. Voltage measured across battery terminals is 15.47 volts. The output voltages were measured thorough a peak
detector circuit consisting of a 1N4148 diode and a .05 ceramic cap. The frequency measured off pin 3 of the 555 timer is 947 hz. I
am sorry to say I do not own a scope. Some day maybe. I constructed this circuitry on a 4" x 6" PCB which provides a generous
amount of space for components and traces. The traces on the power side are 3/8" wide and two of them are 3/4" wide. I hope this
is adequate. What do you think? If not I can lay solder wick along them. These wide traces are about the limit for a 25 watt
soldering iron. The leads from the PCB to the battery consists of #12 stranded wire 6" long. Heat sinks purchased from Radio
Shack were placed on both the mosfet and diode D2, thankfully.
The following is the list of components I used:
Mosfet - IRFB3307
Diode D2 - Schottky Power Rectifier, MBR40250, 250 Volts, 40 Amp, DigiKey #MBR40250GOS-ND, I hope this was a good choice.
Diodes 3,4 - Schottky Rectifier, 5Amp, 50SQ080.
Inductors - Pulse Electronics PE-53120NL, DigiKey #553-1588-5-ND.
Capacitors - 4 @ 35V, 3300uf, low ESR .013, Nippon Chemi-con, KY series.
NE555N
Mosfet Driver IC - TC4426CPA, DigiKey listed 3 series of these, EPA, CPA, AEPA, I flipped a coin, came up C. You get the picture.
DigiKey #TC4426CPA-ND. If this was not the
best choice, it is easily replaced and fairly cheap. I later looked at the 2 data
sheets. AEPA may be the latest edition, I am not sure.
First let me say that I am not an electronic engineer. I have some basic skills, but thats about it. I am not new to kit building nor
new to battery desulfation. However, the last couple of months have been my first attempt at building my own battery desulfator.
This will be my third one. The first two appear to be successful, at least one battery thinks so. My test equipment consists of two
DMM's, A Fluke 27 and a Fluke 179. The 179 along with the normal menu will read frequency, temperature and capacitance. I built
this desulfator with the hope of obtaining more desulfation power. The first two were to slow slow slow.
My observations: The circuit appears to be operating, but less than optimal. The mosfet became very hot. Temp reached 150 deg
F. I had to blow on the heatsink to keep it cool to finish taking measurements. I hooked up an 80mm computer case fan to cool it.
The temp went down to 95 deg F. Very nice. I then checked the voltage drop across Diode D2, it was .039 volts. That concerns me,
because not much current flowing. Voltage on base of mosfet is .096 volts. Whats going on here? This circuit is drawing
approximately 400 ma, but not much output to battery. Problem could be in the timing circuit not properly turning on the mosfet. I
will replace both ic's in timing circuit. I hope the problem is not with the mosfet or diode itself. Mark, what do you think?
Mark, I want to thank you and others like yourself who contribute so much knowledge and help to others on this forum. I have read
a good portion of the posts on this forum and am amazed at the willingness to help.
Thank you,
Cyber Guy
Richard
The basic circuit will work very well with only minor changes at 24 Volts.
- You do not need the 7812 regulator. A 12V .5 watt zener diode works well at 24 volts with a 500-600 ohm .5 watt series
resistor.
- The IRF 540 is a bad choice. It will get very very hot due to its high on resistance. An IRFB4110 is a far better part. You can
start with the IRF540 and change later. Watch the temperature with the 540.
- D4 appears to be in backwards and has no reason to be there at all. If you must use a 7812 regulator change D4 to a resistor,
200 to 400 ohms .5 Watt.
The original simple circuit works very well at 24 volts with only component changes. Change the series resistor that feeds the zener
diode and change the Mosfet to a IRFB4110. And of course you will need C5 and C6 to be rated at 30 Volts or above.
Simple is better. Please do not hesitate if you have more questions.
Mark
I use the big blue cap (51,000uF), a power supply and low ohm resistors to simulate a battery. This is much more stable than using
a real battery,and I can change the "internal resistance" very easily.
These are all the same parts we are use to using. The 555 timer circuit is right out of a kick back design. The TC4426 replaces the
PNP transistor Mosfet driver. (You can replace the TC4426 with the PNP Transistor or an inverting totem pole driver if needed.)
Heres how it works.
When the Mosfet is off, the two caps are charged up to the battery voltage (13V) through the inductors. When the Mosfet turns on
the capacitors are now in series, generating a high current 24 volt potential that discharges into the battery. It is a pretty simple
design and generates a lot of current. The desulfating pulse voltage will always be double the battery voltage minus resistive losses
(ESR, on resistance,etc.) As the battery voltage rises so will the output of this circuit.
The following photos were measured under the following conditions:
- Pulse width = 7 microseconds
- Pulse rate = 1700 Hz
- Battery internal resistance = .25 ohms
- Battery voltage = 13 volts
This shows 22 volts peak (9 volts above the battery voltage)
This shows 34 Amps peak current pulse (measured with a pearson current transformer)
The equasion to calculate the peak current (with no losses) is as follows:
Vin X Iin / (Vpeak - Vbat) X DF = peak current
Vin = Input voltage to the circuit (13 volts)
Iin = Average input current to the circuit (.371 amps)
Vpeak = Peak voltage seen across the battery (22 volts)
Vbat = battery voltage (13 Volts)
DF = Duty factor (.000007 X 1700 = .0119)
This means that the calculated peak current with no losses should be 46 amps.
Another way to calculate peak current is:
(Vpeak - Vbat) / battery resistance = peak current.
Recap of peak current measurenents and calculations:
Calculated no loss - 46 amps
Measured pearson CT - 34 amps
Calculated from voltage and current - 36 amps
So far I am thrilled with the performance and the simplicity of this circuit.
More details in the next posts
The following posts defines the parameters for the parts used in the Voltage Doubler desulfator.
This formula yelds 36 amps.
I tried to do away with inductors all together but I could not get the performance with out them. But the good news is that they are
not critical.
The inductors do two things in this circuit:
1. Provide a DC path to charge the capacitors. This is such a long time that there is no inductance involved, just the coil resistance,
which is very low.
2. When the Mosfet turns on, the inductors "get out of the way" of the current flow.
I experimented using resistors instead of inductors. 1 ohm 2 watt resistors gave the best results. This came at the cost of lower
output, less efficiency, and slower rise time.
Again the two inductors are not critical. I suggest a minimum inductance of 100 uH. I used as low as 10 uH but had a little lower
output than when I use more inductance. The difference in output between a 220 uH and a 1000uH inductor is about 1 volt. All of
the pictures in the previous post used 1000uH inductors. The biggest reason that I used 1000uH inductor is that I had a bunch of
them.
The inductors should not be these ultra miniture types that will fit on a pencil eraser. They will saturate too quickly and will soak
power out of the circuit. The 1000uH inductors I used were a Pulse Electronics PE-53614. The core is about 1/2 inch in diameter
and 1/4 inch high.
It does not matter if they are Ferrite or Powdered Iron.
Radio Shack sells a 150 uH choke that would work just fine. I suggest that you can get a better deal on Ebay.
Again, the most importance purpose that these inductors have is to be high impedance (get out of the way) when the Mosfet fires.
They store no energy. In fact D2 and D3 are there to absorb any inductive kick back they produce. If you would like to have a short
negative output pulse after the desulfating pulse, just remove D2 and D3
Resistance is not your friend. It makes energy disappear (into heat). The ESR of C6 and C7 will make a huge difference in the
output of this circuit. I started with 2200uF 25V caps. When the Mosfets fired the voltage across them instantly dropped like a rock.
A clear effect of ESR. Next was 6800uF 25V caps. Better, less drop due to ESR. Finally, 6800uF 63 volt caps. Less than a volt drop
due to ESR.
Q1 - Just use a good Mosfet with very low on resistance. The IRF1404 is very common and cheap.
You will see more improvement with selecting the lowest ESR caps (there are two of them, twice the ESR.)
Terry
What sets the lower limits of the amount of capacitance is how much voltage droop you can stand during the pulse. The formula
for this is dV/dT = I/C. (d= delta or "change in"). This tells you how quickly the voltage will drop for a given current
and capacitance. For instance, a 10 uF cap under a 30 amp load will drop 3 volts per microsecond. Way too much droop for this
application. A 100 uF cap would drop .3 volts/uS and 1000uF cap would drop .03 V/uS. (I like the .03V/uS number) This is the
electrical nature of a capacitor regardly of the type. Now add to this the effect of ESR. Look at the scope photos in the previous post
of the voltage wavefrom of the 100 uF and 2200 uF caps. The wave shapes are pretty much the same but there is 2 volts difference
in the amplitude. This is the effect of ESR. The 100 uF cap has a higher ESR than the 2200uF cap and losses more voltage.
Why did I use a 6800uF 63 volt cap when 1000 uf is all the capacitance I should really need? Because the bigger cap has a much
lower ESR.
I got in a few more inductors to try in the circuit. So far I have tried three off the shelf parts and a few hand would inductors.
I have tested:
- Pulse engineering, PE-53830, 77uH
- Pulse Engineering, PE-53614, 1070uH
- Renco, RL-1960, 680uH
The difference in the outputs were in the tenths of a volt. They all worked well.
The hand wound inductors that did not work well were the high permiability cores. They saturated and drew power from the circuit.
I suggest inductors that have a saturation current of at least .5 amps. This is based on my circuit that draws about .4 amps with a
1700 Hz, 7uS, 30 amp output pulse.
Hi Mark,
Its a pleasure for me to include LTSpice of your circuit
Here is the circuit I used for simulation and some comments about components
Capacitors C2 C3 ESR is 0.02 and L1 and L2 ohmic resistance is 10 Ohm
Mosfet driver is a conventional totem pole because TC4426 it is not available for me
Mosfet IRF7468 is a 40V/0.016 Ohm and it is the model I have at hand
Battery model is a simplified one that I frequently use for desulfator circuits but it is adapted at 13V/0.25 Ohm according to your
data
and here are results: V(bat+) is pulse voltage at +Battery and I(V1) is pulse current
Peak current is 34A and its waveform and peak I believe is very well adapted to your Pearson meter
Peak battery voltage presents an overshoot and it is not exactly as your scope one
Inductance values is according to your description... but if you need futher details please let me know
-Francesc
More waveforms
Mark, let me start off by saying that I think it is great that you
are experimenting with improvements. Keep it up.
A few points:
1. Try your run caps again with much shorter and/or thicker
connections. There is a generous amount of inductance
in those tiny wires with fast rise-times. Ground the case.
2. Two or more smaller caps (i.e., 3500 uf + 3500 uf) are
almost always better (and easier to obtain) than one
large (6800 uf) cap. Use the shortest leads possible.
Old PC supplies are good sources for low-ESR caps,
and small ceramic bypass caps in parallel with the
electrolytics will lower ESR even more.
3. Use a larger, counter-wound or thick film (10W or more)
resistor in the output leg. A low-value shunt is even
better. I use 20 mil, 3/8" wide copper strips, but any
copper plane surface is usable. Stack 'em for more
current, or use thicker/wider strips. Calibrate with
an accurate ammeter. The arrangement shown in
your photos constitutes a low-value choke in series
with the pulser output.
4. Use _Large_ leads on the output (AWG #6 or bigger).
The shorter, the better. Composition does not seem
to matter much, but fine strands are easier to handle.
(Think "welding leads").
5. Use only a lead-acid battery for a load. Preferably,
one the approximate size as your target application.
Use the _same_ battery for your comparisons, and
keep the test times short. (It is sort of aggravating
that the battery under test becomes de-sulfated and
skews the readings).
If your target is NiCad, your environment changes.
Pulsers have value here, but in a different manner.
Simulated battery loads are just that - simulated.
I made gobs of different versions, based on the
models found around the various vendor and
white paper approximations in an attempt to
create a consistent test load for pulsers.
None worked as good as a genuine LA battery.
6. I found it easier to make a test jig:
http://hmin.tripod.com/al...dysm/pages/jigproto.html
Note the paralleled low-ESR caps and fabricated shunt.
All components are replaceable except the caps including the timer/driver board tucked under the pots.
7. See the timer and driver circuits posted at:
http://hmin.tripod.com/al...dysm/pages/desulf-1.html
I found it easier with the TL494-type timer chips - they
have an adjustable dead time, and one of the driver
transistors can be used in F/F mode to give an advance
scope trigger to move the waveform start to the center
of the screen. They also have a wider timing range and
more consistent and adjustable pulse width than 555-types.
Note that these are all earlier versions; I never got around
to posting the later attempts and finished versions, which
included using the on-chip comparators for much the same
functions as Don's PIC-based versions. Maybe sometime.
8. Note that the total energy in a voltage-doubler circuit is
going to be approximately the same as the conventional.
Whether the higher-voltage/lower-current is more useful
in this application is not something I am prepared to
address without extensive testing. If you can find some
roughly equivalent batteries and attach one of each
type pulser, monitor the results, and post them, it would
be wonderful. Good luck on finding equivalents.
9. It would also be helpful if, after you have changed the
resistors out for a shunt, if you would post the displays
for the shunt and the coil-type pickup (on the same screen).
It very well could be that the inductive pickup will be a
valuable tool, since it does not require that the output
lead be interrupted. I have a vague recollection of just
such an attachment that I tried in the past (in which I was
not terribly impressed with the results), but may have
missed something or had a biased notion of the method.
By all means, keep it up. I absolutely am not trying to
discount your efforts or discourage you from further
development, but merely trying to pass on experiences
and my limitless knowledge on the subjects.
*** Add the appropriate number of smileys here ***
Sorry for the length of this post - I don't seem to be
able to do short essays.
*** Add the appropriate number of smileys here, too ***
Cheers again.
Ok. I built the circuit and put it on a 7ah battery just because it is small enough to fit on my cluttered workbench. I put the scope
across a .05ohm resistor in the neg lead and I get a current waveform similar to what you posted but the question is how do you
figure out how much of that current to attribute to the actual desulphation process since the current there is also partially (mostly)
the current used to charge the capacitors and losses as well. Maybe I'm missing something in the waveform and which part to
measure. Me thinks maybe the current to charge is in one direction and the desulphation is in the other direction. Maybe I need to
measure only part of the waveform??
I finally put the scope across the battery itself and I can see that the voltage is pulsing up to about 20 volts. I suppose it's visible
there becasue the low capacity battery has some resistance?
I followed your formula which takes into account duty cycle and average current etc
Vin X Iin / (Vpeak - Vbat) X DF = peak current
(12v bat, 134ma avg, 20v peak, 1khzrep rate, 7us pw)
I think that gives me 27 amps?
But again, how much is actually going into the battery to desulphate?
Then it all went south. It was working fine on the little battery, all components running cool, so I decided to put it on a 33Ah
battery. Poof !! both diodes and the FET toasted immediately.
The FET was a 55A 75V 19mohm unit, the diodes fast recovery 3A (MR850). Not sure why the diodes went. I would have thought
the failure would have been the FET and the chokes.
Hmm now that I look at the circuit again - maybe I hooked it up backwards - (I give that about a .01% chance) That would explain
the diodes (both forward biased across the reversed battery but not sure about the FET) any ideas?
Fixed it up and put it back on the 7ah battery for more evaluation. I have a 6A poly fuse in the circuit now. I suppose it's ressitance
is messing everything up but I feel bettery with it in there for now.
I was using 19,ooo uf 40v caps becasue I have them.
How much horsepower is a common question. The answer is I don't know. It's more of a show engine. You run it for a few minutes
just to show it off and then it sits and looks pretty at model engineering shows. I don't like to rev it too high either (horsepoer is a
function of RPM) nor would I want to load it excessivly since I have about 3yrs work in it. The last thing I want to do is scatter it. It
sounds just like the real thing and runs pretty well. That's all I'm really interested in. It's about 4cuin displacement.
Sage
You say that's the peak current provided to the circuit (cap charge and dischare plus losses)
Ok, so help me out here. I'm not getting something becasue I don't see how we are calculating how much power we are
contributing toward actually desulfating the battery.
First off I'm assuming we are measuring average curent by running the circuit with a current meter in series and using that
measurement for average current.
And we are measuring the peak voltage with the scope across the battery (or where?). (I built the peak detector and it shows about
22v about what the scope across the battery shows)
For arguments sake let me assume some really silly numbers here to force the equation to an extreme. This usually helps me figure
out what's going on.
Lets say we have some really crappy capacitors that make a very low peak. Lets say the peak is only 12.1v for a 12v battery.
Lets say the circuit still draws an average current of 130ma - maybe becasue the capacitors are more like resistors or the timing is
such that the inductors actually draw current when the FET turns on - whatever. This is the value I measured so lets use it.
The duty factor comes from 7us pulse and 1Khz PRF. That is going to be fixed at .007 regardless of what else is going on.
So the formula is:
Vin x Iin /(Vpeak-Vbat) x DF
(are the brackets correct? see what follows)
12X.13/(12.1-12) x .007
1.56/(.1) x .007
1.56 / .0007
= 2228 amps !!!
Do I have to divide before multiply by DF
That would make it:
(1.56/.1) x .007
That gives 0.109 amps. Maybe that's better because it's almost the same as the average?
In any case, I guess I'm trying to figure out how much current is being pushed through the battery during the 1us pulse.
Maybe I should be using a current waveform. Problem is I think a current transformer shows both charge and discharge currents in
one big waveform making it impossible to separate and measure the current conributed by capacitor discharge into the battery
only.
What am I missing here?
Sage Sage Sage
You had it right the first time, 2228 amps is correct. If you keep the wattage the same but lower the voltage the current has got
to go up. The best way to understand this circuit is: wattage in equals wattage out. If I put in 1.56 watts and the output has a .007
duty factor then I must have 1.56w/.007= 223 watts (peak) out. If the output voltage is only .1 volts then the peak current will be
223W/.1V= 2230 amps peak. (Of course the Mosfets and the caps may not like this much peak current.)
Sage
Lets step through this.
1. Lets begin with the caps full charged. No current is flowing in the circuit.
2. The Mosfet turns on for 7 microsecs. The caps are now effectively in series and current begins to flow. What determines the
amount of current is the total series resistance of all the devices in the chain and the voltage difference between the caps and the
battery. To make it simple, all the caps, Mosfet,wiring, and battery have a total of .48 ohms and the voltage differential is 12 volts.
The current flowing is 12V/.48ohm = 25 amps.
3. Lets determine how much the caps discharge over the 7 microsec. pulse with the standard formula V/Sec = Amps/C. Volts/ 7e6sec. = 25A/6800e-6 Farads which equals 0.0257 volts. After the 7 microsec. pulse is over the caps return to their origional
voltage minus 0.0257 volts. Only 0.0257 volts is all the caps were discharged during the "ON" time. You will notice if you look at
the voltage across the caps with a scope that the voltage will drop several volts during the pulse but will instantly return back to the
original voltage after the pulse. This is due to the caps ESR, but not from being discharged.
4. One RC time constant (.01 ohm X 6800e-6) is about 100 microsecs. This is the time required to recharge the caps 63%. In 1
millisec. the caps are recharged 100%. The amount of energy needed to recharge the caps is equal to the amount drawn out of
them during the 7 microsec. pulse.
Sage, view the caps as energy storage devices. They turn very high peak current into continuous average current. If the duty
factor is .007 then the average power will be .007 times the peak power. Power in must equal power out (minus some loss due to
resistance).
Mark
Thanks for putting actual numbers to my ramblings. I understand all of that now.
To recap - I think you have demonstrated, and seem to agree, that the capacitors discharge somewhat in (7us) and charge very
quickly (100us) and they are fully charged in far less than 1ms (1khz). And that most of the time they are just sitting there doing
nothing fully charged drawing no power.
.
What's still in question (for me anyway) is the formula:
(Vin * Iavg) / ((Vpeak - Vbat) x (PRF x Pw))
If I put previously presented and agreed to numbers to it: (my results)
(12 * 0.13) / ((20 - 12) x (1000 x .000007))
= 28 amps
I think we agree that the caps are sitting drawing no current for most of the 1000hz time period of 1ms and therefore contribute
nothing to the average current during that time.
So it makes sense that they would be sitting even longer not drawing any current if we changed the freq to only 100hz (10ms) and
keep the 7us pulse the same.
Yet, if we plug 100hz into the formula:
(12*0.13) / ((20-12) x (100 x .000007))
= 280 amps
The calculated current goes up tenfold from 28 to 280 amps.
You must agree that the current should not change when nothing has been changed that will affect it i.e we've just added more
time sitting around not drawing any current.
I can see if we changed the PW from 7us to say 10us it would make a difference. As you have demonstrated the caps would
discharge more and require more charge time. That would contribute to a higher current calculation.
In short, I think the formula should not include the frequency component at all. Somehow just the PW.
Please remember that my goal here is to be figuring out how much current is going toward desulfating. Apparently that current is
flowing only when the caps are discharging back into the battery and that happens only during the 7us pulse and it makes no
difference what happens any other time (to me).
This formula for that particular current doesn't seem to work (IMHO) for the reasons stated above.
So, where have I gone wrong this time ?
Sage
Mark:
I performd one experiment I mentioned.
(original circuit parameters 1Khz rep rate, 7us PW and 130ma average current)
I set the frequency to 125Hz. The average current actually went down from 130ma to 90ma
I set the frequency to 2.5Khz and the current only went up a bit to 190ma.
So ,changing the frequency 10 fold in either direct makes little difference to the average current as I suspected because the
charging / discharging is all happening aound the 7us pulse not in the time inbetween pulses.
Sorry I'm not convinced (yet)
I don't think frequency should be part of the formula.
Still looking for a way to determine the desulfating current.
So what am I missing now?
Sage
Sage
Something is not right. I have done the same experiment when I built the prototype and again to prove out the equasion. This
equasion/relationship is not new. It is a intergal part of switch mode power supply design. I have absolute confidence in the
correctness of that equasion.
Another thing is that I am drawing a lot more current than you are. Depending on the resistance of the load, my 12 volt unit
draws between .25 and .4 amps. I suspect we my not be talking apples to apples here.
Even though I have a few "Honey dos" today I'll sneak down to the shop. Data later.
Mark
Sage
I set up my 12volt unit and increased the resistance of my load to match your average current.
1054 Hz,
7 microsec.
load resistance .3 ohms
average current .135 amps
I increased the rate to 1961 Hz and the average current went up to .245 Amps. (This tracks the equasion very well.)
Sage, when you changed the rate you went lower. Guess what, the zener regulator, depending on the supply voltage, will draw 30
to 60 mills. This distorted your measurement since the desulfator section was drawing less than the zener. I don't know what was
going on when you went to 2.5KHz. I do know the equasion is correct.
(This is much better than Honey Dos.)
Mark
Well now you've thrown another wrench into the works
- the zener.
I never installed the zener because the 555 is capable of pretty high voltage - more than the 20v peaks I'm seeing. It doesn't
change it's timing one bit regardless of the voltage you supply to it (one nice thing about a 555) and I wanted maximum drive to
the FETS not limited to 12v when the FET is seeing something higher. The biggest reason I didn't use one is that I didn't have one
handy.
I do have the 10ohm resistor to isolate the 555 a bit and I have a 0.1 and a 1000uf cap across the 555 to filter out any trash. It
runs reliably and never changes or adds any funny stuff to the drive waveform. The zener might be necessary under other
conditions to limit the voltage but I don't see the need so far in my circuit on 12v.
Zeners are VERY crappy regulators anyway. They draw more current depending on voltage (see my remarks below).
Lets leave that can of worms for a minute unless you can see big problems with it.
I did the other test I mentioned and put a 0.01 resistor in the positive leg of C7 (the cap that goes to the bat +ve). I'm sure having
it there upsets things a bit but I'm out to prove this freq vs current thing. It should measure all of the capicitor current in both
directions.
I ran the circuit and I get a nice pulse waveform up and down. The positive portion which is the square pulse 7us wide is
about 1.5volts. I guess that equates to 150A. Don't forget my caps are 19000uf and adding the resistor is probably messing things
up.
The odd thing is on the trailing, negative going narrow ringing spike. It goes down about 7 volts !!.
I don't know what that's all about but I think you explained it somewhere along the way. Must be a resistance thing because it can't
be 700amps.
In any case and most important (to me) I also changed the freq up and down 10 fold just like the previuos test and it makes
absolutly no difference to the size or shape of the cap current waveform (other than the frequency). The scope was triggered on the
front edge of the pulse and it did not change a bit. So unless you can explain otherwise I don't think repitition frequency has
anytihng to do with the current anywhere.
Concerning the zerer in the circuit - off the top, because you have a zener as a regulator the current suplying it is varying because
they are shunt regulators. Why?
The average VOLTAGE across the battery goes up with the higher frequency since there are more high voltage pulses spaced closer
together.
With the zener in the crcuit more voltage means more current as it tries to shunt the higher average voltage. In my case the
current does not change because the 555 just follows the average voltage and draws no more or less currrent. Same in the lower
freq aspect. Less pulses, lower average voltage less zener current.
This makes sense according to your finding - freq up current up. This is substantiated by another observation made as follows:
I recently built the peak detector circuit (diode and cap) and through a couple of transistors use it to drive an LED to give an
indication that there is boosted voltage (above vbat) present. This way I know if someting has gone wrong without connecting the
scope. From playing around with that circuit I know freq changes the voltage output of the peak detector so this is probably what's
going on with your zener reguator.
That's my explaination and I'm stickng to it
.
Try removing your zener and see what happens.
If you feeel the need for a regulator and your FET driver will accept lower than supply voltage on the logic inputs you might be
better off with a 5v or 9v three terminal regulator to power the 555. The current to the 555 wouldn't vary then. I'm using the old
totem pole driver becasue I didn't have the fancy chip (yet).
BTW. As someone else mentioed I think you should be using a real battery rather than your simulated one. Maybe it will make a
difference.
Now what
Sage
Sage
1. The zener in the 12 volt design is mainly to limit peaks and surges. The max voltage for 555 is 15 V amd the 4426 is 18V. If
eather of these chips fail it could take mosfet and a few other parts with it. Run without it if you want.
2. Read the thread about "Measurement Errors" using series resistors in these circuits. I made the same mistake. You aren't
putting out 150 amps, you are putting out a respectable 25Amps.
3. Of course you will not see a change in the peak current when you change the rate. The only way to effect peak current is to
change the supply voltage or the load resistance. These are things that I can do very easily with my battery simulator.
4. I have real batteries that I test desulfators with. I just don't think it wise to keep them in the same room as the very expensive
test equipment I use to develop desulfators (and other things) with.
Sage, I think you should build up a battery simulator.
Mark
I'll take my chances with the 15v rating on the 555 for now.
You made no mention of the possibility of the varying zener current corrupting your average current measurement (other than you
didn't want to remove it - fair enough). I did point out that the average VOLTAGE on the battery is changing with frequency (reread my post if you need to), therefore the average current measured in the circuit with a zener is suspect. Just like you thought it
might be fooling me when I adjusted the freq down (see your message).
I'm aware of measurement errors with resistors in the circuit. I pointed it out numerous places when I thought my readings were
suspect including the 150amps. The point was to see if the current waveform changes regardless of the impact of the series
resistance.
I also mentioned more than once that I measure the average current by putting an ampmeter on a ten amp range (for lowest
resistance) in series with the battey connection to the circuit. No comment was made that it was incorrect or that you had another
approach.
A light might have gone on a bit here when you said the PEAK crrent would not change.
I'm beginning to see how the average current might change. Just like I say the average voltage increase in the presence of more
high voltage pulses per second. I guess more current pulses would work the same way. Not sure what my thinking was before.
Thanks.
My point all along is that I've not been able to substantiate the formula with measurements. But who cares.
I have enough real batteries around to play with. They're all VRLA (sealed Gel cell) types and have seen no problems with them in
the presence of all my machine shop stuff. Liquid cells might be a different story although I have closed bucket of concentrated
H2SO4 around that I use for anodizing. Still no problems for years. Now, hydrochloric acid is a different story. I've saw rust
appearing on tools with only one use of it to etch circuit boards. I use it outside now. Nasty stuff it is.
If you want to post your battery simulator I might think about building it. I don't have any high power low resistance variable
resistors around if that's what it takes. I have a few 200w fixed values though.
I know I tend to write too much (for the purpose of clarity) and that a lot of people don't like to read everything I write. Not sure if
you fall into that catagory but I think we've tossed this subject around enough for now.
I'll continue on. I have a 33Ah battery on the circuit now. Let's see if it gets any better. It's at 12.2v now. Started at 12.5
yesterday. I guess I'll let it go to about 11.5 or so and then re-charge it to see if the "after charge" voltage rises.
Is that the process?
Sage
I thought it would be nice to have an LED to indicate operation of the desulfator rather than having to haul out the scope.. The
circuit below connects across the battery. It's an extension of the peak detector circuit presented earlier. As long as the desulfator
is producing boosted battery voltage pulses (about 18v or more) the peak detector will charge up and turn on the LED. It's nice to
tell at a glance if everything is normal.
R15 has been found to be optional. It's good circuit design to have it there but it seems to work ok without it. The LED current limit
resistor R6 should be chosen as high as possible but still have the LED brightness suit you.
Works well for me. Enjoy.
Sage
Well, this battery is worse than it was before (see below).
For general interest here's the trick I use to load test batteries.
Instead of having huge power resistors to load batteries I use a cheap 12vdc to 110vac inverter.
If you load the inverter with a 60watt light bulb. The current is about 6amps on the DC side.
If you use a 100watt bulb the DC current is about 10amps.
If you use a 60watt compact flourescent bulb which consumes only 23watts, the DC load is 1.7amps - a pretty close fit to the 20
hour rate for a 33ah battery (1.65amps) which is what I have.
This way you don't have any smoking hot resistors to play around with and you probably have a lot of different load values in your
cupboard.
(15, 25 , 40, 60, 100, 200 watt light bulbs)
BTW this is not a constant current load (either is a resistor) since the DC current actually goes UP as the battery discharges. This is
because the inverter attempts to regulate 110vac on the output. Constant power out means constant power in so as DC voltage
goes down (battery dies) current must go up. But it's a pretty simple way to get a high power DC load with some variability
(change bulb wattage).
There is one more handy built in feature. Most of these inverters will sound an alarm and shut off at 10.5 volts saving your battery
and allowing you to leave it within ear-shot and keep track of the full discharge time waiting for the alarm to sound.
In any case this battery I've been desulfating under the 6amp load originally would settle at about 12.6 volts fairly steady for
several minutes.
I didn't run it dead because I knew it wouldn't last.
After desulfation and a re-charge it drops like a rock to 2 volts. The inverter doesn't even begin to run on the battery.
I didn't expect this but it appears the desulfator has damaged this battery - well, it was crap before . I guess it's really crap now. I
wasn't expecting the desulfator to "make a silk purse from a sows ear" but I didn't think it would get worse.
I'm wondering if there is now a bad connection inside the battery. It's actually dangerous to load test batteries that might have a
bad connection inside since the connection could cause a spark inside and the battery could explode. I think I'll abandon this
battery.
I have lots of other batteries. I'll try again on another one maybe in better condition.
Sage