Tesla's Charging Circuit and it's Application to Pulse Motors

[TinselKoala]

Umm

E = (Iw2)/2. So where I = 0.00098655969 and
where w = 230.2666 then E = 26.155 @ 2200 rpm

Where w = 157 then E = 12.1588 @ 1500 rpm

And where w =  104.6666 then E = 5.4039 @ 1000 rpm

Need more data now.


...

Heh... not to be too harsh or anything.. but _significant digits_ only, please! You cannot expect me to believe that you have measured your current or the values that go into calculating it to the degree of precision that a number like 0.00098655969 implies. You are here saying that the actual value of the current is 0.00098655969 amps, and NOT 0.00098655968 amps or  0.00098655970 amps.... or any other value. So I say to you that your current value is wrong! I could not say this if you had cited your current value as 1 milliamp, though. You would then be right! because the 0.001 value represents the degree of accuracy of your measurement. If you had given it as 0.001000000, you would once again be wrong.

The answer of any computation cannot legitimately include more digits of precision than the _least_ precise value that goes into the calculation. Those extra digits are just fantasy; the odds of them being exactly correct are smaller and smaller the more of them there are.

[Farmhand]

Heh... not to be too harsh or anything.. but _significant digits_ only, please! You cannot expect me to believe that you have measured your current or the values that go into calculating it to the degree of precision that a number like 0.00098655969 implies. You are here saying that the actual value of the current is 0.00098655969 amps, and NOT 0.00098655968 amps or  0.00098655970 amps.... or any other value. So I say to you that your current value is wrong! I could not say this if you had cited your current value as 1 milliamp, though. You would then be right! because the 0.001 value represents the degree of accuracy of your measurement. If you had given it as 0.001000000, you would once again be wrong.

The answer of any computation cannot legitimately include more digits of precision than the _least_ precise value that goes into the calculation. Those extra digits are just fantasy; the odds of them being exactly correct are smaller and smaller the more of them there are.

Umm What current ? That's not a current calculation. That is the energy in the rotor. I did the calculations and those are the figures I got as the result. I don't understand what else I could do except leave off some numbers, which would surely give a less accurate result, that's all it can do. Can you explain to me how I could do the calculations without getting decimal points ?

The hall sensor I realize it's an open collector device, and it will be supplied 5 volts to the supply pin of course, (as the drawing shows) I couldn't expect it to work without any input. The CMOS chip has a high impedance so I don't see how much current will flow to require the pull up resistor.

I don't get the replies, but that's OK because I can work on alone. No problem.

Don't worry about being Harsh, as long as it is said when I am misunderstood I don't mind at all.

I mean to say you provided me with the formula, can you show me then how I should have calculated it without all the decimal points ?

I don't understand what it is that I didn't show that was so confusing ? The formula is there. Are the calculations correct or not ? Please say so either way.

Umm

E = (Iw2)/2. So where I = 0.00098655969 and
where w = 230.2666 then E = 26.155 @ 2200 rpm

Where w = 157 then E = 12.1588 @ 1500 rpm

And where w =  104.6666 then E = 5.4039 @ 1000 rpm

Need more data now.

You also spoke of a reed which I don't use, so I can only assume you are not taking much notice of what I post.

If you don't want to help then fair enough, just say so and I'll stop wasting time asking questions here and go elsewhere. I won't be offended, you are busy with debunking, I know.

MileHigh has bowed out and said he no longer wants to help, which is fair enough. He's not here to help us only to debunk, that is clear now.

I'll just use a solderless board and spend time working things out for myself.

Cheers

.

[TinselKoala]

Umm What current ? That's not a current calculation. That is the energy in the rotor. I did the calculations and those are the figures I got as the result. I don't understand what else I could do except leave off some numbers, which would surely give a less accurate result, that's all it can do. Can you explain to me how I could do the calculations without getting decimal points ?

Sorry, I thought I was looking at a current calculation. The "I" confused me. But the point still stands. You calculated that MoI figure somehow. From measurements and weights, and somewhere in there you made a measurement that is only precise, to say 3 digits. Like 0.756 grams or something, or 0.625 millimeters. This means that any calculation you do with that measurement can't be more precise than that. And I used the rounding to 0.001 in your example as the extreme case. You probably can round to three sig digs. Please look up "significant digits" or figures in the Wiki.
http://en.wikipedia.org/wiki/Significant_figures

The hall sensor I realize it's an open collector device, and it will be supplied 5 volts to the supply pin of course, (as the drawing shows) I couldn't expect it to work without any input. The CMOS chip has a high impedance so I don't see how much current will flow to require the pull up resistor.

I think you are misunderstanding how open-collector devices work. The power supply pin to the Hall chip does not produce a "high" signal at the open collector output, it only supplies the chip's sophisticated electronics inside. The presence or absence of a triggering event closes or opens the collector-emitter channel of the sensor's output transistor. That is all. Just like the block diagram shows. The CMOS chip that you are driving will not consume much current at its high impedance inputs, and so you can raise the value of the pullup to limit current flow thru the Hall open-collector output transistor to the minimum value consistent with triggering. When the sensor is not sensing, the collector-emitter channel will be open, and so the voltage at the junction of the collector and the pullup resistor will be at the positive rail voltage (logic HI). When the sensor triggers, the channel closes and so the voltage at the collector-resistor junction falls to near the negative rail voltage (zero, logic LO). It is a bipolar transistor in there, wired as the block diagram shows! You have to provide a voltage for the transistor to switch, there is no internal connection to the collector. Open collector. Just as if it were a photodiode or phototransistor with no base lead at all.
The emitter pin is of course shared by the bipolar transistor output and the chip's power supply ground.

I don't get the replies, but that's OK because I can work on alone. No problem.

Don't worry about being Harsh, as long as it is said when I am misunderstood I don't mind at all.

I mean to say you provided me with the formula, can you show me then how I should have calculated it without all the decimal points ?

I don't understand what it is that I didn't show that was so confusing ? The formula is there. Are the calculations correct or not ? Please say so either way.

The calculations are "correct" but wrong, and hopefully you will understand after you read up on "significant digits". Just because your calculator has all those digits doesn't mean they are always meaningful.

You also spoke of a reed which I don't use, so I can only assume you are not taking much notice of what I post.

I spoke of Allegro 2-wire Hall sensors, which function like reed switches in that they only need two wires, instead of three, to produce a signal. Your schematic would be usable directly with one of these sensors, probably. I can only assume that you are looking deliberately for ways to misunderstand what I'm telling you. I thought you had enough electronic knowledge to know right away how a bipolar transistor is used in a circuit, and how to read Allegro's data sheet for your sensor. Sorry, my mistake.

If you don't want to help then fair enough, just say so and I'll stop wasting time asking questions here and go elsewhere. I won't be offended, you are busy with debunking, I know.

How do you possibly get "you don't want to help" out of my trying to explain to you how to use your sensor, in direct response to your question?
That is going to take some explaining for me to understand.

MileHigh has bowed out and said he no longer wants to help, which is fair enough. He's not here to help us only to debunk, that is clear now.

I'll just use a solderless board and spend time working things out for myself.

Cheers

.

Have fun.

[MileHigh]

I honestly don't see any need for the drama.  The issue of significant figures is something I didn't mention to not break the flow.  I am going to guess the accuracy of the measurements is about one in 50 or thereabouts.  So that's one and 4/5 significant figures.  This issue of significant figures should always be a "background process" running in any experimenter's head.

Okay so you add the pull-up resistor.  In your CMOS Cookbook they tell you in the very beginning that CMOS inputs can't float.  The input is just a capacitor plate.  There is a perv analogy here.  Without the pull-up resistor you are floating there is a good chance the output will go crazy with the slightest 'twitch' of the input.

As an ex hardware guy with my eyes closed I would use a 10K pullup resistor.  I would still read the Hall sensor datasheet through, because the last time I read one was probably in 1982.  So 10K with confidence and due diligence.  ha ha  Then some due diligence with your scope.

As a generic comment, if you use a project board and have long wire leads flying in the air then you have to check for ringing at the destination.  You have to put your scope ground on the receiving chip's ground so you see what the signal looks like from its perspective.  Let's assume your setup is powered by a battery so no ground loops.  If there is a lot of ringing that could flip the input gate of the receiving chip.  The usual solution is the magical 50-ohm series resistor at the source.  Sometimes tiny capacitors help if you have a pesky signal.  In this particular setup there is just one chip so you likely won't have to worry about ringing when connecting the CMOS gates to each other.

For me here is what the nitty-gritty would be all about:  First you establish what your pull-up resistor is going to be.  Then I would be very curious to see how the Hall sensor responds to various magnetic field stimuli so I would have to play with that for a while.  So moving magnets around and looking at the collector waveform to see how it switches on the scope.  The key questions being how fast is the signal gradient and what is the sensitivity vs. distance and angle like for various magnets.  I would need to do that to develop an innate feel for how this particular model of Hall sensor responds.  I would be tempted to change the pull-up resistor value and see if that can pull out some extra sensitivity.  Doing that might not necessarily be desirable because I am guessing you might see heightened noise.  So it would just be exploratory.

The truth is that the slew rate is not too important because you are driving a CMOS input and the CMOS amp output will switch very fast.

Then after all this crap you settle back and end up using a boring old 10K pull-up resistor (no point in wasting milliamperes) and you know that in your setup the switching at the CMOS NAND gate output is clean and sharp like a knife like it should be and everybody is happy.  lol

The little 100 uF cap should be right by pin 14 of he 4011 chip to supply the teeny spike of switching current when called upon to do so.

Once perfected, seal the thing up in a big thick black gooey and oily mass of potting compound from hell and start a crowd-funding project!   

P.S.:  Not A and Not B = Not (A or B)    Muhahaha...

[Farmhand]

I'm just playing with youse, I know how the open collector output works, I've been using SG 3524 and TL494 chips for some time now a couple of years. I was just demonstrating how frustrating some people can be.

I keep being misunderstood. It peeves me.

I asked you MileHigh several times for my calculation results to be checked, no response on that.

Never mind. I'm Busy.

..