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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)
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?
Vern you answered this too quickly. Energy is conserved. Voltage times current times time.
If you cannot rely on the samples being correct then you can't do valid calculations with them. SophiecentaurI 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.
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.
Quote from: jerrygg38Vern you answered this too quickly. Energy is conserved. Voltage times current times time.I realized I misspoke while at lunch.
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.
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 beenR = (100/3.7) (10 ohms) = 270 ohmsA 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.jerryGG38I 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?
jg38QuoteI 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.
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.