LTJT - poynt99 Tests #2

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GibbsHelmholtz	Re: LTJT - poynt99 Tests #2 « Reply #100 on: 2011-02-28, 02:33:55 »
Group: Guest	Quote from: poynt99 on 2011-02-28, 02:19:17 Hopefully, Gibbs is in agreement as well. I would hope that the efficiency will increase as the value of CSR decreases. .99 what the heck, leave me out of this. You and prof do your thing. I have no problem. lol I will chime in and out like a magic man.

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #101 on: 2011-02-28, 02:58:44 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	OK, fine...be like that Gibbs. Professor, et al: It is very constructive to think of a battery as "a reservoir" that loses energy, and every other component in the circuit as "sinks" that gain energy. In this respect, think of the energy given up or lost from the battery as negative energy, and energy gained or dissipated by the circuit components as positive energy. The CSR placed in series with the battery is a component of the circuit and can be thought of as a load actually. Therefore it GAINS energy. Meanwhile, our battery loses energy (most of the time). Normally, we would have no issue conceiving of the power lost from the battery; it is simply the battery voltage times the battery current. If we used a current probe to measure the battery current, this would all be quite straight forward. However, we have resorted to using current sense resistors (CSR's), at least for the moment. The CSR placed in series with the battery (it matters not if the CSR is in the negative or positive leg of the battery) allows us to gauge the battery current, but in doing so it presents itself as a small load that eats up a bit of power. In the method I proposed to use for obtaining both the battery power and CSR1 power, the first measurement involves acquiring a power measurement that captures the voltage across both the battery AND CSR. I called this power Pitotal. This of course will not provide the true battery power figure we are looking for, so we have to subtract the CSR power from Pitotal in order to obtain the battery power figure. We know that the battery power is negative and it is much larger in amplitude than the CSR power, therefore we can safely assume that Pitotal will also be a negative value because the battery power figure dominates. We can therefore write: Pbattery = (Pitotal) - (Pcsr1) establishing the correct power polarities as discussed, we have: -(Pbattery) = -(Pitotal) - (+Pcsr1) if Pitotal measured to be 30mW and Pcsr1 measured to be 3mW, then we have the following: -Pbattery = -30mW - 3mW -Pbattery = -33mW .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #102 on: 2011-02-28, 03:22:17 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	We can look at it another way. Let's assume we have a battery and 3 circuit components (C for component, not capacitor); C1, C2, and C3. The power of each is Pbat, PC1, PC2, and PC3. There are no other losses. Let's assume that PC1=3mW, PC2=20mW, and PC3=10mW. We know that Pbat must therefore =33mW. So we can write the power balance equation as follows: Pbat = PC1 + PC2 + PC3 33mW = 3mW + 20mW + 10mW If we decide to capture the voltage across the battery AND one of the circuit components in the same measurement because they are in series, say PC1 for eg., then we have to rearrange the power balance equation; Pbat - PC1 = PC2 + PC3 and substituting in the values: 33mW - 3mW = 20mW + 10mW 30mW = 30mW The true battery power though is 33mW. So you see that although it may seem strange that we actually added the CSR1 power to the battery in the above example, mathematically, it is sound. This example works backwards from where we started in the actual test. Hope that helps. .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

PhysicsProf	Re: LTJT - poynt99 Tests #2 « Reply #103 on: 2011-02-28, 03:45:24 »
Group: Professor Hero Member Posts: 3017	I must be really slow or something tonight. You write: Quote I would hope that the efficiency will increase as the value of CSR decreases. .99 Now for me, I would hope that the calculated efficiency will STAY THE SAME as the value of the measuring resistor CSR decreases. That's the whole point of this exercise of subtracting out the effect of the CSR, in order to determine the COP / efficiency with the measuring resistor removed. Can you agree with that? Next, You write, Quote Let's assume that PC1=3mW, PC2=20mW, and PC3=10mW. We know that Pbat must therefore =33mW. So we can write the power balance equation as follows: Pbat = PC1 + PC2 + PC3 33mW = 3mW + 20mW + 10mW Suppose we want to determine the input Power, Pin, in the case where there is no CSR1. That would be handy, wouldn't it? save us some trouble. But with this method we need a measuring resistor to get the current. So next best thing, we reduce the ohms in stages, from say 1 ohm (giving 3mW dissipation) to 1/2 ohm (giving 1.5 mW lost) to 1/10 ohm (giving just 0.3 mW dissipated, essentially negligible). Having done that, down to just 0.3 mW in CSR1, what is the input Power Pin? And, for the sake of discussion, let's say the output Power stays constant somehow, at 20 mW. So, what do you get for the COP = Pout/Pin for the two cases? Is your n the same as COP? One of us is missing something, and I hope your answers will clarify -- starting with my first question -- Now for me, I would hope that the calculated efficiency will STAY THE SAME as the value of the measuring resistor CSR decreases. That's the whole point of this exercise of subtracting out the effect of the CSR, in order to determine the COP / efficiency with the measuring resistor removed. Can you agree with that?

GibbsHelmholtz	Re: LTJT - poynt99 Tests #2 « Reply #104 on: 2011-02-28, 03:48:24 »
Group: Guest	Quote from: GibbsHelmholtz on 2011-02-28, 00:45:15 Yes, the power from CSR1 comes from the battery to acquires the output. However, if we reduced or cut that power out, we can save some energy and have the same output. The professor is right. On the other hand, I think we have a problem with the method. CSR1 is subtracted from the whole cycle means it's being subtracted both the inductor charging and inductor discharge cycle. On the inductor charging cycle, the battery provide power to charge the coil and CSR1. On the discharge cycle, we have LED, CSR1, and CSR2 but... also there are two power source: inductor discharge and battery EMF. Therefore, we must add 1/4 CSR1 power to the output, 1/2 LED power to input, 1/2 CSR2 to input. How did I come up with this? This is just how I roll. lol Oops, I made a mistake. $:-\$ Sorry Poynt, this is why. I don't like to give people false hope. I did my analysis based on 2 separate cycles (inductor charging and discharging), but the input power is already took account for both cycles. Therefore, the only thing left is add that 1/4 CRS1 to the ouput. Wow... I have to think about this.

humbugger	Re: LTJT - poynt99 Tests #2 « Reply #105 on: 2011-02-28, 04:03:31 »
Group: Guest	Poynt et al, As you include more and more miniscule dissipation components in your measurements and calculations, it is only natural that your measurements will converge on 100% efficiency. If you include the resistance of the inductor winds, the core loss, the transistor dissipation and all, you will certainly end up with 100% efficiency every time. As you approach perfect unity COP, your measurement error will begin to dominate over the ever-decreasing unaccounted-for loss percentage. Remember, the scopes are 8-bit and, after doing scope-math, I'd be surprised if you had better than 2 or 3% accuracy, considering you are multiplying two numbers each with an unknown quantization error and offset error and probe cal errors, too. For instance, to be absolutely proper, if you are going to include the shunt losses, you should equally consider the battery impedance. Just because the loss is internal to the battery itself, it still makes the measured "input power" appear to be less than it actually is (from the ideal voltage source within the battery) if you ignore the internal battery loss. Using worn-down penlight cells, you might be surprised at how high the internal resistance is. I think we can all safely say, at this point, that the claims of Lawrence citing COPs (or FLEETs) of hundreds of percents are completely disproven. As we finagle over the last few percentage points above or below 100%, even the best scope measurements are going to be inadequate. With so much effort being put in by all parties involved, this casual onlooker is beginning to wonder what the point is. Other than an exercise in finding the limits of accurate measurement using scopes and in accounting for and identifying every tiny loss factor, that is. Plus, of course, its kind of fun, iand a good exercise in measurement approach and communication. Soon you'll be breaking out the 0.01% tolerance Kelvin-sensing shunts and the NIST-traceable calibrated ten-digit accurate DVMs Humbugger P.S. [edit] Please don't take my comments as being demeaning or dismissive. I think the group is making very useful solid progress in eliminating confusion and errors generally found in these kinds of exercises. It's a good thing, as Martha Stewart might say.

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #106 on: 2011-02-28, 04:15:40 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: PhysicsProf on 2011-02-28, 03:45:24 I must be really slow or something tonight. Not at all, you are asking good questions. Quote Now for me, I would hope that the calculated efficiency will STAY THE SAME as the value of the measuring resistor CSR decreases. That's the whole point of this exercise of subtracting out the effect of the CSR, in order to determine the COP / efficiency with the measuring resistor removed. Can you agree with that? We first need to establish what YOU mean by efficiency. As far as I know, we have only alluded to what the metric for n is in our tests, and I have always assumed it is based on the LED power in relation to the INPUT power. If this is what you mean also, then we are on the same page, and we can continue with an answer to your question. In the last or second last post, I mentioned that the CSR1 resistor that is in series with the battery is actually a small load on the battery, i.e. it dissipates real power. The amount of power it dissipates is related to the voltage across it and it's value of resistance. For a certain, fairly wide range, the larger it's resistance, the more power it will eat up. As it eats up more power, there is less power available for the intended load, i.e. the LED, therefore the efficiency n must decrease. Does that makes sense and answer your question as to why the efficiency will go up as the value of CSR1 goes down? Quote So next best thing, we reduce the ohms in stages, from say 1 ohm (giving 3mW dissipation) to 1/2 ohm (giving 1.5 mW lost) to 1/10 ohm (giving just 0.3 mW dissipated, essentially negligible). Having done that, down to just 0.3 mW in CSR1, what is the input Power Pin? And, for the sake of discussion, let's say the output Power stays constant somehow, at 20 mW. The output power to the LED can not remain constant if other components in the circuit are changed to consume more power, as I described above. Remember what happens to the collector LED when you reconnect the LED load that is in the secondary circuit, which we have both disconnected in favour of the general JT connection? I hope things are becoming clearer. .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #107 on: 2011-02-28, 04:24:07 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Hum, At this point, we are interested only in determining the battery power and the LED power. The measurement I made today indicates an n of 99.5%. That figure does not include the CSR2 nor any of the other circuit components such as the transistor etc. I think our goal is to factor those out so we only have P_LED/P_BAT, and nothing more. Any ideas why my n measurement came out so high? Thanks, .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

humbugger	Re: LTJT - poynt99 Tests #2 « Reply #108 on: 2011-02-28, 04:28:31 »
Group: Guest	Quote from: poynt99 on 2011-02-28, 04:24:07 Hum, At this point, we are interested only in determining the battery power and the LED power. The measurement I made today indicates an n of 99.5%. That figure does not include the CSR2 nor any of the other circuit components such as the transistor etc. I think our goal is to factor those out so we only have P_LED/P_BAT, and nothing more. Any ideas why my n measurement came out so high? Thanks, .99 Zipons?

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #109 on: 2011-02-28, 04:40:05 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	I'll try the measurement again tomorrow if I have time, and this time I'll post the wave forms. In the mean time, do you see any flaws in my method for getting the power measurements? .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

PhysicsProf	Re: LTJT - poynt99 Tests #2 « Reply #110 on: 2011-02-28, 05:32:13 »
Group: Professor Hero Member Posts: 3017	Good comments -- we're getting closer I think. Will digest and respond tomorrow. Meanwhile -- Quote from: poynt99 on 2011-02-28, 04:40:05 I'll try the measurement again tomorrow if I have time, and this time I'll post the wave forms. In the mean time, do you see any flaws in my method for getting the power measurements? .99 Great -- now, could I convince you to replace the 1 ohm CSR1 with a 1/2 ohm, so we can see empirically what happens to the calculated efficiency? BTW, I would consider the 1 ohm CSR2 in your schematic to be part of the output load, and the interest to me would be COP = Pout/Pin, where Pout is Pled + Pcsr2, and Pin is still under discussion. PS -- I still seek a way to calculate the COP that does not depend on our choice of resistance for CSR1, which after all is only there so that one can evaluate the input current. « Last Edit: 2011-02-28, 06:22:13 by PhysicsProf »

MileHigh	Re: LTJT - poynt99 Tests #2 « Reply #111 on: 2011-02-28, 05:42:41 »
Group: Guest	Poynt: I have only skimmed through this but I can make one recommendation. Since you briefly fluttered into over unity territory and are getting numbers like 99.9%, I think it's time to measure the values of your shunt resistors with your best multimeter if you haven't done so already. You have Rosemary in a tizzy! MileHigh

muDped	Re: LTJT - poynt99 Tests #2 « Reply #112 on: 2011-02-28, 08:01:08 »
Group: Tinkerer Hero Member Posts: 3055	Quote from: PhysicsProf on 2011-02-28, 05:32:13 ... BTW, I would consider the 1 ohm CSR2 in your schematic to be part of the output load, and the interest to me would be COP = Pout/Pin, where Pout is Pled + Pcsr2, and Pin is still under discussion. ... Since during the output pulse time interval CSR1 and CSR2 are series connected within that discharge loop, would not the total output power be: Pout = Pled + Pcsr2 + Pcsr1 or, since CSR1 = CSR2, Pout = Pled + 2(Pcsr2) ? --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #113 on: 2011-02-28, 13:21:56 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	I would tend to agree with both of you, however, we can keep adding components on and on. At some point we have to draw the line. CSR1 is a necessary evil at the moment, but soon I hope to show that it can be substantially reduced in size (while still obtaining an accurate measurement) so as to minimize these losses. .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

PhysicsProf	Re: LTJT - poynt99 Tests #2 « Reply #114 on: 2011-02-28, 17:10:55 »
Group: Professor Hero Member Posts: 3017	Quote from: poynt99 on 2011-02-28, 13:21:56 I would tend to agree with both of you, however, we can keep adding components on and on. At some point we have to draw the line. CSR1 is a necessary evil at the moment, but soon I hope to show that it can be substantially reduced in size (while still obtaining an accurate measurement) so as to minimize these losses. .99 Ah, here we are getting somewhere -- reduce CSR1 as I have repeatedly requested so that it's effect on the system is finally insignificant. That is the solution I've proposed and I think it is a good one. After all, the measuring resistor should be a minor perturbation on the DUT so that we do not need to quibble over its effects on the measurement. Humbugger wrote: Quote As you include more and more miniscule dissipation components in your measurements and calculations, it is only natural that your measurements will converge on 100% efficiency. If you include the resistance of the inductor winds, the core loss, the transistor dissipation and all, you will certainly end up with 100% efficiency every time. I detect a certain bias here that is disconcerting. We are trying to determine EXPERIMENTALLY whether or not a simple system can demonstrate overunity, more power out than in. It is not a foregone conclusion that as we include "miniscule dissipation components in your measurements and calculations, it is only natural that your measurements will converge on 100% efficiency." That is what we are in process of determining, experimentally. Not by pre-determined conclusion without the need for experiments ("Ipse dixit" authoritarian style). There is something else here that needs to be considered also -- we may vary the circuit components such as the toroid windings and get "better" results. That is, if there is a real effect here, we may expect the COP to move further above 1. I don't know whether .99 is willing to keep trying things; but I am. Meanwhile, I have a card in hand I need to disclose... I took my little JT circuit which showed n over unity last week, to the University to test on the Tektronix 3032. My colleague allowed me to use his scope; unfortunately, I did not have access to his computer to allow me to upload records from the 3032 to the computer. But I did take a photo of the set-up (attached). The red waveform represents the POWER. I took measurements much as did .99 later; however, as I said before, the output from the emitter I put into the circuit at the point he labels V2 rather than going directly to ground. I will test his exact set-up later hopefully this week as I travel back to the University. You're not the only one taking measurements, .99... Here are the results in brief from last Friday, 25 Feb 2011. The input leg of my JT circuit showed mean Pin = 67.1 mW. This includes the power dissipated in the 1 ohm CSR1 which I estimated at 6%, or 4 mW. The output leg of the JT, including the LED and a 1-ohm measuring resistor, showed mean Power Pout = 74.1 mW. So the uncorrected n = 74.1/67.1 = 1.10. If I subtract or add 4 mW from CSR1 to the Pin, still n >1. But it is uncomfortably close to one. So I keep trying this and that, to see what the effect is on the COP. I have added a capacitor to the circuit which appears this weekend to increase the COP; results on that later as I can get back to the 3032. IF the power in the output leg can be verified as greater than Pin, how is this possible? I still believe in conservation of mass-energy. So to me, this result would imply an input of energy from an unidentified source, rather than a violation of the laws of Physics. And that would be interesting! ------------------------ LTJT - poynt99 Tests #2 JT3032Fri25Feb2011.png (1363.72 kB, 1280x800 - viewed 795 times.)

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #115 on: 2011-02-28, 17:52:18 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Indeed professor, it is strange to be hovering even around the 100% mark as in my second (corrected) calculation. Hopefully, with the wave forms posted, and a little more time examining the computation method step by step, we can discover why the efficiency is coming out much higher than expected. .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

PhysicsProf	Re: LTJT - poynt99 Tests #2 « Reply #116 on: 2011-02-28, 19:34:55 »
Group: Professor Hero Member Posts: 3017	Agreed. And I look forward to your next set of measurements with a lower-ohms CSR1 which should tell us much also.

humbugger	Re: LTJT - poynt99 Tests #2 « Reply #117 on: 2011-02-28, 21:05:47 »
Group: Guest	Quote Humbugger wrote: "As you include more and more miniscule dissipation components in your measurements and calculations, it is only natural that your measurements will converge on 100% efficiency. If you include the resistance of the inductor winds, the core loss, the transistor dissipation and all, you will certainly end up with 100% efficiency every time." I detect a certain bias here that is disconcerting. We are trying to determine EXPERIMENTALLY whether or not a simple system can demonstrate overunity, more power out than in. It is not a foregone conclusion that as we include "miniscule dissipation components in your measurements and calculations, it is only natural that your measurements will converge on 100% efficiency." That is what we are in process of determining, experimentally. Not by pre-determined conclusion without the need for experiments ("Ipse dixit" authoritarian style). Yes, Professor, your "bias detector" is well calibrated. Please don't be too disconcerted, however. There are very good reasons for my "bias", as I hope you are aware. I make no pretense of having even a shred of doubt that the laws of energy conservation are entirely applicable here. I'm not engaged in the experiments except as an onlooker and occasional commenter. I think that anyone educated in physics or electronics or mechanics probably, in their heart of hearts has a bias, as well. In general, to be biased toward a belief that the accepted "rules" of energy conversion and transfer apply to this circuit is quite healthy and normal from a science perspective. After all, the whole OU movement is looking for "exceptions" to the rules and, as far as I know, none have been demonstrated or applied in any practical device to date. If we weren't biased, why would the phrase "Exceptional claims require exceptional proof" be so prevalent? Do you not agree that claims of overunity performance are "exceptional"? A healthy bias toward the enormous history and solid well-documented body of "conventional scientific knowledge" is the reason science today demands accurate investigation and well-designed experiment to vet exceptional claims, after all. Without that bias, the rigor of investigation and the practical advancements in technology that have resulted would definitely suffer, I believe. Humbugger P.S. I'm sure this response will win me no popularity contests here, but I'm not too worried about that. My overall comment remains: I do not trust the accuracy of oscilloscope-based measurements using math on irregular waveforms to be much better than about 5% overall, and that's when immaculate bench practices are used. When all the error sources are added up, I'd much sooner trust a good multi-digit calibrated DVM looking at DC equivalent input and output currents and voltages derived by passive low-pass filters and precision shunts. Especially in the milliwat regions.

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #118 on: 2011-02-28, 22:40:05 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Hum, I would agree. I also agree with the weariness of using 8-bit conversion on relatively low-level signals. The DC equivalent method is what I am shooting for next. However, I am still surprised at the efficiency numbers both the professor and I are getting with this particular arrangement, and with different scopes. Flirting with the 100% area is quite unusual, isn't it? Hopefully, additional careful testing will reveal what is going on. .99 --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

humbugger	Re: LTJT - poynt99 Tests #2 « Reply #119 on: 2011-03-01, 00:33:46 »
Group: Guest	Yes, it does seem kind of strange getting so close to 100% n, but it wouldn't really surprise me much if the efficiency was as high as 95% at the low powers involved. I^2R losses are bound to be very minimal, the transistor base loss the same and I bet the Vsat of the transistor is pretty low under these conditions, too. It might be interesting to take a close look at Vsat, which I'm guessing is the largest loss factor in the present circuit at the current levels involved. I mean just to get a ballpark as the the known losses there as a reference point. If you have 5 or 10% lost power in the transistor, you'll know to be surprised (spell that SUSPICIOUS) if you get n of 90-95% or more right off the bat. You guys are doing a great job of whittling it down to the nitus gritonius by factoring in the shunts, etc. It is definitely a worthwhile exercise and I can see no glaring flaws anywhere. If the scope-based tests show n>95% or especially if they consistently show >100%, it's time to break out the RC filters, precision Kelvin shunts and a high-accuracy DVM. Humbugger

Magluvin	Re: LTJT - poynt99 Tests #2 « Reply #120 on: 2011-03-01, 01:18:03 »
Group: Guest	Something to think about All the parts in the circuit consume some power. Considering the input power vs output power is not a complete story. If you can add up all the power used in the circuit by each component, those can be all added to the output figure, which is a good thing. ;] Even a reed switch has a certain on resistance value that when it closes, voltage is present across that reed resistance, current flows, power dissipated. The real figures may be better than realized. ;] Mags

GibbsHelmholtz	Re: LTJT - poynt99 Tests #2 « Reply #121 on: 2011-03-01, 02:41:15 »
Group: Guest	I've found something... ... but neither Kirchhoff or Faraday can handle this.

GibbsHelmholtz	Re: LTJT - poynt99 Tests #2 « Reply #122 on: 2011-03-01, 04:19:09 »
Group: Guest	Considered the JT pix I attached. The charging and discharging cycle are as shown. We know that the current direction for both cycles are the same. All little minor details are omitted to see the bigger picture. Now, let's start at point A and apply the "modify KVL" ( E + dB/dt coil = 0 ). Let's go counter clockwise for charging. -dB/dt coil + Vbat = 0 Now for the discharging, let's go clockwise. -Vbat +....woohoo, it's now positive dB/dt coil = 0 Phew...that was close call. Let's try Faraday (E = dB/dt coil ). Let's go counter clockwise for charging. Vbat = -dB/dt Now for the discharging, let's go clockwise. -Vbat = dB/dt ===> Vbat = -dB/dt !! ------------------------ LTJT - poynt99 Tests #2 Kirchhoff vs. Faraday.jpg (45.35 kB, 1092x661 - viewed 670 times.)

PhysicsProf	Re: LTJT - poynt99 Tests #2 « Reply #123 on: 2011-03-01, 05:37:20 »
Group: Professor Hero Member Posts: 3017	Good -- my digital power supply finally arrived today. I'm using a separate digital voltmeter to measure the output voltage more accurately; I'm wondering how reliable the built-in ammeter is for measuring the input current; it reads down to 1 mA. Perhaps as a starting point, it may give some information. More learning experience! Ive been thinking more about the connection of the emitter direct to ground in .99's circuit, a concern I raised before. It seems to me that this represents output power which is not accounted for, just thrown away. I think we can do better. .99 at one point mentioned adding a 1 ohm resistor in series with the emitter -- and I think that is an important step to take. (Can you do it?) Call it CSR3. Would not Pcsr3 ADD to the output power and increase n?

poynt99	Re: LTJT - poynt99 Tests #2 « Reply #124 on: 2011-03-01, 13:04:50 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: PhysicsProf on 2011-03-01, 05:37:20 Good -- my digital power supply finally arrived today. I'm using a separate digital voltmeter to measure the output voltage more accurately; I'm wondering how reliable the built-in ammeter is for measuring the input current; it reads down to 1 mA. Perhaps as a starting point, it may give some information. More learning experience! It wouldn't hurt to double-check both the voltage and current with a known good DMM. Quote I've been thinking more about the connection of the emitter direct to ground in .99's circuit, a concern I raised before. It seems to me that this represents output power which is not accounted for, just thrown away. I think we can do better. .99 at one point mentioned adding a 1 ohm resistor in series with the emitter -- and I think that is an important step to take. (Can you do it?) Call it CSR3. Would not Pcsr3 ADD to the output power and increase n? We need all the components' "return" paths to go through the CSR1, otherwise we will not obtain a true INPUT power figure. Perhaps redrawing the circuit slightly will shed some light on this. We don't need ground symbols in this circuit, so I have removed them. They only cause confusion it seems. You may now notice that CSR2 is in series with CSR1, but don't be concerned about that. Please also note that I have corrected the calculation for Pvbat in the diagram notes; i.e. Pitotal and Pcsr1 must be added together to obtain Pvbat. Indeed I can use a 1 Ohm in the emitter to see what power our transistor is dissipating. I'll try this when I get to testing the circuit again, today or tomorrow. .99 ------------------------ LTJT - poynt99 Tests #2 LTJT2_schema02.gif (80.59 kB, 1803x1472 - viewed 637 times.) --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

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