[witsend]

Sophicentaur, just quickly, I have not read through your post.  I do mean resistor.  It was an error.  And I have been so careful not to take offense.  I have NOT been OFFENDED in any way at all. If you're referring to the fact that JerryGG38 did not ask me directly - it is simply because I need to be involved in any discussion. 

Where do you say that I've reacted badly.  I am trying my very best to answer questions.

[lyner]

jg38

However the zener diode in the mosfet will pass this voltage to the shunt resistor. therefore the shunt resistor could see a spike of 48 volts minus the zener voltage. Let us say that the zener voltage is specified at 10 volts (each mosfet has different zeners)

The zener acts presents zero resistance for voltages in excess of its Vz. I think that you won't find a huge voltage across the 'shunt' resistor, the voltage will be IR, where I is, at most,the original current flowing and R is 0.25ohms. How could it be more than that, when 'held down' by such a low resistance? But the action of D1 should catch the voltage spike long before this happens - it only needs to be 0.7V above the potential of the + battery terminal for it to conduct all the current round through the Rl again.

[lyner]

witsend
When I say you have reacted badly, I mean badly for the discussion - you just say you don't understand the Science or the Maths, when it suits you but you expect everyone else to trust your judgment because it is obvious to you. Or, you bat on about conventional Science not being willing to listen and just being reactionary for the sake of it.
It really isn't the way to conduct a Scientific dialogue.

[jerrygg38]

Ok JerryGG38 - then we're on the same page.  But that voltage spike can and in fact does exceed 34 volts.  It depends on the duty cycle.  Now Vern is absolutely spot on.  The amount of energy that is returned by the counter electromotive force may very well have been stored on the resistor in the first instance.  And, also correct, is that it never seems to exceed the amount of energy that was first delivered during the ON period of the switching cycle.  But here's the thing.  It always returns some very small fraction less.  Not much difference.  If the duty cycle is on for 10% or 90% - however much energy is first stored is then returned - less that fraction in that spike.  I'm sure you're right.  It's probably because of the diode in the MOSFET or even the diode in parallel to the resistor - or, indeed, both.

So.  If it was stored - or - if the energy that delivered the counter electromotive force was courtesy extra energy from the battery, then it would be evident how?  We measure the voltage across the resistor - during the 'ON' period of the duty cycle to follow Ohm's Law.  In other words the amount of voltage divided by the Ohm's value of the load resistor, over the time of the duty cycle, conforms to whatever would have been determined according the same measurement applied to a simple load placed in series with a battery without the complication of a switch.

So.  If the energy was stored at some extra cost from the battery, where do we find this extra energy?  Is it something that's there, but hidden?

I think the inductance has complicated the problem. You could have used a non-inductirve resistor for your experiment. The inductance merely adds complications but has little meaning for this circuit.

  What was the value of discharge resistor for the battery. RL is ten ohms and the duty cycle was 3.7 percent as shown on page 5.

  The corresponding draw down resistor should have been

  R = (100/3.7) (10 ohms) = 270 ohms

A 270 ohm resistor across the battery should draw down the battery in the same time as your switching circuit. The inductance does not change the draw down very much.

  What was the value of your draw down resistor?

[jerrygg38]

quote author=sophiecentaur link=topic=23243.msg257259#msg257259 date=1244654562]
jg38

However the zener diode in the mosfet will pass this voltage to the shunt resistor. therefore the shunt resistor could see a spike of 48 volts minus the zener voltage. Let us say that the zener voltage is specified at 10 volts (each mosfet has different zeners)

The zener acts presents zero resistance for voltages in excess of its Vz. I think that you won't find a huge voltage across the 'shunt' resistor, the voltage will be IR, where I is, at most,the original current flowing and R is 0.25ohms. How could it be more than that, when 'held down' by such a low resistance? But the action of D1 should catch the voltage spike long before this happens - it only needs to be 0.7V above the potential of the + battery terminal for it to conduct all the current round through the Rl again.
[/quote]

You are correct in general. I am just doing a worst case possibilitiy in which the shunt resistance and the wiritn path back to the battery has enought inductance to allow for such a large spike. The actual circuit will respond based upon many parameters. Most of the spiking will occur at the junction of D1 and Q1 and not at the shunt resistore.

  However the whole discussion is meaningless. They will get the same results even if the resistor RL has no inductance whatsoever. I am afraid that the inductance spiking merely complicated a more basic error.
Right now I believe that the 3.7 duty cycle required a draw down resistor of 270 ohms to replace the switched 10 ohms. I just asked Witsend for the value of their draw down resistor. I think they used the wrong resistor.

[witsend]

If you cannot rely on the samples being correct then you can't do valid calculations with them. Sophiecentaur

I think that Spescom dealt with this problem in the paper.  They got Fluke to send a guarantee that the instrument was capable of sampling within the frequency range that we were testing. 

[Vern]

Vern you answered this too quickly. Energy is conserved. Voltage times current times time.

I realized I misspoke while at lunch.

[lyner]

If you cannot rely on the samples being correct then you can't do valid calculations with them. Sophiecentaur

I think that Spescom dealt with this problem in the paper.  They got Fluke to send a guarantee that the instrument was capable of sampling within the frequency range that we were testing. 

Fluke answered the question you asked them. It is not their job to tell you that it was not the right question to ask.
My question concerns the highest frequency in your waveforms. That requires an answer of the numerical kind. The relevant frequency is not the fundamental oscillation frequency of your circuit. It will be much higher. What is it?

[jerrygg38]

These posts are coming too fast to cope with AND make the dinner! Yeah - I know - get your priorities right man!


Tell


jg38 (And Vern, who just wrote the same thing whilst I was cogitating)
I have just had second thoughts about the operation of the diode D1 when the Mosfet switches off. The only path for current to flow is, in fact, in a loop through the diode and the resistor RL. All the magnetic energy will be dissipated in the resistance. The Mosfet is off so the battery is no longer in circuit. That's correct, isn't it? How can charge return to the battery - apart from through some parasitic component? The Drain Source capacitance is a few thousand pF, according to the data sheet.

Ah ha. A few thousand pf. Yes. When you have a high accuracy analog to digital converter a few thousand pf hurts.

As far as the scope is concerned, a 20 MHz scope is too powerful for this simple circuit. We are only dealing with 2.4 KHz. If you go 100 times all your need is a 0.25 MHz scope for good results. It won't show up the small spikes through the mosfet but who cares?

  A 60 year old scope with 1 MHz would be far too good for this circuit.
It looks like an old fashioned simpson voltmeter would work just as well.

The whole discussion is reduced to absurdity I am afraid. You are correct that the only path for the resistive discharge is through the diode and itself.
  It is hard to understand how anyone who can operate the fancy equipment for the test could come up with such incorrect answers. I am beginning to laugh at the meaningless ness of this discussion.

  I return to Union Square Park in 1956 in NYC to the man with the talking coconut. The coconut said that he had a simple switching circuit
that could power the world. The people did not believe the man with the talking coconut but every night he returned and stated that he had a simple switching circuit that could power the world. It was funny then and it is still funny today.

  Sorry Witsend. I cannot stop laughing!!! Sorry to offend but I cannot stop laughing.

[witsend]

We look to use inductive resistors to ensure that there is counter electromotive force.  However, to the best of my knowledge, the only non-inductive resistors used were for the shunt.

What was the value of discharge resistor for the battery. RL is ten ohms and the duty cycle was 3.7 percent as shown on page 5.
The corresponding draw down resistor should have been
R = (100/3.7) (10 ohms) = 270 ohms
A 270 ohm resistor across the battery should draw down the battery in the same time as your switching circuit. The inductance does not change the draw down very much.jerryGG38

I do not know what a corresponding draw down resistor is.  But if you mean a control then we did not use a control for the experiment as we were advised that battery draw down rates were meaningless.  However, we did these type of tests for BP.  And also, the rate at which our battery discharged in the experiment in that paper was indeed consistent with that amperage draw down.

What was the value of your draw down resistor?

What resistor?

[witsend]

 Sorry Witsend. I cannot stop laughing!!! Sorry to offend but I cannot stop laughing.

Not at all.  Glad to know your at least amused.

[jerrygg38]

Vern you answered this too quickly. Energy is conserved. Voltage times current times time.

I realized I misspoke while at lunch.

I am beginning to get punchdrunk from this discussion. Right now I cannot stop laughing

[lyner]

jg38

I think the inductance has complicated the problem. You could have used a non-inductirve resistor for your experiment. The inductance merely adds complications but has little meaning for this circuit.

I see where you are coming from but wasn't the whole point to show that an inductance has the magical property of regenerating energy? Of course it would have made sense to buy (for a couple of quid) a high quality non-inductive resistor as a control. But there a lot of other things that could have been done in order to isolate the flaws and to account for the anomaly (Occam's Razor). I don't think the exercise was aimed in that direction, though.

AND you didn't read the info about the free-running frequency, which was about 150kHz!!
Really, I've got time to do the dinner PLUS read all the facts. I multitask so well I could be a woman!

It strikes me that the people who helped you with this venture, witsend, may not have been as commited as you were. It is much easier to agree with someone who is fired with enthusiasm  than to dig deep into the theory and spot the flaw.

[witsend]

Sophiecentaur and JerryGG38 - make up your minds.  Nothing is fast enough for Sophiecentaur and everything is too fast for JerryGG38.  The assurance we had was that the accuracy of the flukemeter was adequate for the frequency of the experiment.

May I please have an answer from someone regarding the question - if energy was first stored in the load resistor, then where do we find that extra energy.  The voltage measured across the load resistor conforms to Ohm's Law.

[jerrygg38]

We look to use inductive resistors to ensure that there is counter electromotive force.  However, to the best of my knowledge, the only non-inductive resistors used were for the shunt.

What was the value of discharge resistor for the battery. RL is ten ohms and the duty cycle was 3.7 percent as shown on page 5.
The corresponding draw down resistor should have been
R = (100/3.7) (10 ohms) = 270 ohms
A 270 ohm resistor across the battery should draw down the battery in the same time as your switching circuit. The inductance does not change the draw down very much.jerryGG38

I do not know what a corresponding draw down resistor is.  But if you mean a control then we did not use a control for the experiment as we were advised that battery draw down rates were meaningless.  However, we did these type of tests for BP.  And also, the rate at which our battery discharged in the experiment in that paper was indeed consistent with that amperage draw down.

What was the value of your draw down resistor?

What resistor?

Something is wrong. You have a circuit which turns on and off at a 3.7 percent duty cycle. Since the load resistor was 10 ohms, the circuit basically looks like a 270 ohm resistor to the battery.
  If you connect the circuit to one battery and at the same time connect a 270 ohm resistor to another sister battery, both batteries should draw down about the same rate. Then you can compare how much better the switching circuit did compared to its equivalent 270 ohm resister. The switching circuit should discharge the battery faster than the 270 ohm resistor.,
  That is a very simple test. Ultimately the inductance means nothing to the switching circuit. The circuit is ultimately a 270 ohm resistor.

[jerrygg38]

jg38

I think the inductance has complicated the problem. You could have used a non-inductirve resistor for your experiment. The inductance merely adds complications but has little meaning for this circuit.

I see where you are coming from but wasn't the whole point to show that an inductance has the magical property of regenerating energy? Of course it would have made sense to buy (for a couple of quid) a high quality non-inductive resistor as a control. But there a lot of other things that could have been done in order to isolate the flaws and to account for the anomaly (Occam's Razor). I don't think the exercise was aimed in that direction, though.

AND you didn't read the info about the free-running frequency, which was about 150kHz!!
Really, I've got time to do the dinner PLUS read all the facts. I multitask so well I could be a woman!

It strikes me that the people who helped you with this venture, witsend, may not have been as commited as you were. It is much easier to agree with someone who is fired with enthusiasm  than to dig deep into the theory and spot the flaw.

But the funny thing is that the circuit works the same if we had an inductor or just a resistor. The measurements would be the same. So let them repeat the same experiment using only a resistor as a test. If the results are the same,then the problem is the experimental measurements. Which it is.

[witsend]

Something is wrong. You have a circuit which turns on and off at a 3.7 percent duty cycle. Since the load resistor was 10 ohms, the circuit basically looks like a 270 ohm resistor to the battery.
  If you connect the circuit to one battery and at the same time connect a 270 ohm resistor to another sister battery, both batteries should draw down about the same rate. Then you can compare how much better the switching circuit did compared to its equivalent 270 ohm resister. The switching circuit should discharge the battery faster than the 270 ohm resistor.,
  That is a very simple test. Ultimately the inductance means nothing to the switching circuit. The circuit is ultimately a 270 ohm resistor.jerryGG38

Yes indeed.  You've got it in one.  We did those tests for BP.  Over and over and over.  It took us more than 5 weeks - working late into the night.  Indescribably boring.  

BUT when we wanted to include them for the article in QUANTUM we were specifically advised by ALL AND SUNDRY that to compare battery draw down rates, under any circumstances whatsover, would be of no value because battery vagaries were such that any reference to this as PROOF of anythig at all would be meaningless.

[witsend]

May I please have an answer from someone regarding the question - if energy was first stored in the load resistor, then where do we find that extra energy.  The voltage measured across the load resistor conforms to Ohm's Law.

this is copied from further down.  Please could someone answer this question.

[jerrygg38]

May I please have an answer from someone regarding the question - if energy was first stored in the load resistor, then where do we find that extra energy.  The voltage measured across the load resistor conforms to Ohm's Law.

this is copied from further down.  Please could someone answer this question.

Unfortunately we are discussing experimental errors while we are not there to observe the experiment. All we are doing it looking for the errors. You are going to have to have your friend and coauthor to study what we have said and to look for the errors himself.

  I pointed out that you should have used a 270 ohm resistor. You said you did. However first you did not know what resistor to use. Then you said that others denied that such a technique was satisfactory. We are going around in circles.
   You are unwilling to admit that an error was made. I know you spent a lot of time on this project but an error was made. You have to rely upon others to admit an error was made but they will not admit it.
   In anay event I am tired of this topic. How about posting your ideas on current flow. Or should I start the topic?

[witsend]

I see where you are coming from but wasn't the whole point to show that an inductance has the magical property of regenerating energy? Sophiecentaur

The whole point was to use the counter electromotive force from an inductive resistor.  I really did not think that anyone needed to be persuaded about the properties of counter electromotive force.  That has never been an issue until Jerry discounted its contribution to the experiment.