I built another JT, this time air-core.
The inductors are only about 3uH each. I used #28 AWG, and a rubber (packing) facet washer as my former.
Fo is about 1.1MHz! >:-)
There is no secondary. The LED is powered like the standard JT off the collector, and goes to GND through a 1 Ohm for current measurement. The emitter has it's own 1 Ohm as well.
It's only about 25% efficient, unless my two power measurements aren't telling the whole story. Note also the strangeness with the negative current and positive voltage; p(t) is zero rather than negative. ???
I used a bit of averaging (4) to clean up the traces a little.
I'll post a schematic later. I suspect this answers the core/saturation question guys. ;)
.99
Poynt,
I still have a nagging feeling about those negative current measurements and the scope channel. Even if the LEDs when reverse-biased are leaky like you say.
For example, lets take a look at output_view.png. At about 1.7 main divisions from the left of the capture, the current transitions through zero and starts to change direction. However, the corresponding voltage is still positive, about +2.0 volts, and is sloping downwards. Why should the current change direction at this point in time? It doesn't make sense to me. You see the same thing on most of your captures.
In this particular case you show negative current but zero power. What that suggests to me is that the current is zero, and the data for the channel internal to the scope is correct, and the mathematics are correct, but for some reason when that data for that particular channel is given to the module that has to massage it and generate the display, that module is adding a negative offset to the displayed data.
Of course, when the LED is not forward biased enough, the current should be nearly zero, not negative.
If you have any doubts yourself, you might consider simply swapping the channels and doing the same capture. Will the math be the same? Will the displayed channels be the same or will you see an offset migrate to the channel that is now displaying the voltage? I will restate the simple test I stated before, just reverse bias an LED in series with the resistor (like the LTJT secondary setup) with your DC power supply and measure the leakage with your best multimeter and see if it is in agreement with the DSO.
MileHigh
Yes, MH I agree. There is something going on here, and at this frequency I am not too surprised.
Maybe that LED is not leaking that much. Could this be a probe problem? The skew looks ok, but there is a DC offset in the current reading.
The scope is also ignoring the fact that the current is negative, because the resulting power trace stays at zero, indicating that the current should be zero.
My simulation also corresponds quite well with this unit, except for that negative current. I am going to try my good 500MHz probes and see if there is a difference. I'll post the results if they look any different.
.99
OK, found the problem causing the offset in CH2 in the OUTPUT scope.
Have a look at the scope shot... C.C
.99 >:-)
Ha! Ha! Ha! It happens to the best (of us) also!
Hey,
I've been posting all kinds of shots with this setting....you guys missed it too! O0
I've downloaded the latest firmware version for the TDS3012B and installed it on both scopes, now I am going to try and calibrate the high frequency adjustment on the probes. One probe is showing zero current where there is zero, and the other is showing a slight positive current. Hopefully, the HF cal will fix this.
Then we'll be set to do some of these tests over again. >:( LOL.
.99
Here is a schematic of this P9901 JT.
It doesn't get more minimalist than this I suspect. Once the two CSR resistors are removed, it's down to 4 components; Q1, Rb, LED, air-core transformer.
.99
PS. I will be re-measuring all these units with my other good probes on the OUTPUT scope.
The scopes have been updated to latest firmware, OUTPUT scope CH2 coupling changed to "DC", and bad probe eliminated (two new probes used in their place). The CH2 current in the OUTPUT scope now looks correct.
Here are the test results once again after the "cleanup":
input_mean_2nd.png indicates an average INPUT power of 41.84mW.
output_mean_2nd.png indicates an average OUTPUT power of 22.3mW.
n=53.3%
.99
Quote from: poynt99 on 2011.02.13, 15:13:22
Here is a schematic of this P9901 JT.
It doesn't get more minimalist than this I suspect. Once the two CSR resistors are removed, it's down to 4 components; Q1, Rb, LED, air-core transformer.
.99
PS. I will be re-measuring all these units with my other good probes on the OUTPUT scope.
Would not the "load" LED and series resistor be better
isolated from the "source" if placed directly across the
primary winding?
Quote from: Dumped on 2011.02.13, 21:17:52
Would not the "load" LED and series resistor be better
isolated from the "source" if placed directly across the
primary winding?
When driven with a 1.5 volt battery or less, there is no need to worry as the forward drop is at least 2.0 volts for a standard red led and higher for other types.
You are correct for higher source voltages as this can flow into the led during the off state of the transistor. I like to keep the load current circulating around the inductor of use, as that is most efficient, and prevents transients from rocking the supply. You will see on my bench most of the circuits I use are drawn and tested that way.
CSR1 should really be moved to the negative leg of the battery, and the emitter grounded. This way you can watch current flowing out of the battery, and during the inductor charge and discharge cycles, including any current that might flow back into the battery (as some may claim).
The resistor CSR1 in the emitter is the worst possible place as it is a potential degenerative element, and does not let you see the whole picture. Also it should be a fraction of an ohm (0.1) to optimize efficiency.
Quote from: Dumped on 2011.02.13, 21:17:52
Would not the "load" LED and series resistor be better
isolated from the "source" if placed directly across the
primary winding?
The LED load can be "discharged" through to ground or Vbat, it works both ways. However, there is more power delivered to the LED when it is connected to ground, which is why all the schematics show it this way.
Do you know why more power is delivered to the LED when it goes to ground?
.99
Quote from: poynt99 on 2011.02.13, 21:54:16
The LED load can be "discharged" through to ground or Vbat, it works both ways. However, there is more power delivered to the LED when it is connected to ground, which is why all the schematics show it this way.
Do you know why more power is delivered to the LED when it goes to ground?
.99
Are you referring to "more power" as in "greater efficiency" or more power through current leakage from the source supply through the inductor?
Quote from: ION on 2011.02.13, 22:08:47
Are you referring to "more power" as in "greater efficiency" or more power through current leakage from the source supply through the inductor?
What I mean by "more power" is not only more power delivered to the LED, along with more power used from the supply, but higher efficiency as well.
.99
Quote from: poynt99 on 2011.02.13, 22:29:25
What I mean by "more power" is not only more power delivered to the LED, along with more power used from the supply, but higher efficiency as well.
.99
I'd like to hear of this. Are you stating a truth for all JT's of this topology?
Quote from: ION on 2011.02.13, 22:40:26
I'd like to hear of this. Are you stating a truth for all JT's of this topology?
It seems to hold true for both air-core and core-type versions of this configuration. However, in the simulations at least, there is a bigger difference with the air-core version. But in both cases, more power is dissipated in the LED when it is grounded vs. connected to the battery.
In sims with the core, the Pin did not go up at all, but the efficiency went up 12% when the LED was grounded vs. connected to Vbat. The more current we can get through the LED, the brighter it will be and the more power it will dissipate. When the LED is pulsed by IK into the Vbat terminal, there is a lower potential difference seen by the inductor.
Try it yourself ION with your BO unit. Which way produces higher intensity in the LED? Is there any difference in Pin?
.99
Quote from: poynt99 on 2011.02.14, 01:31:01
It seems to hold true for both air-core and core-type versions of this configuration. However, in the simulations at least, there is a bigger difference with the air-core version. But in both cases, more power is dissipated in the LED when it is grounded vs. connected to the battery.
In sims with the core, the Pin did not go up at all, but the efficiency went up 12% when the LED was grounded vs. connected to Vbat. The more current we can get through the LED, the brighter it will be and the more power it will dissipate. When the LED is pulsed by IK into the Vbat terminal, there is a lower potential difference seen by the inductor.
Try it yourself ION with your BO unit. Which way produces higher intensity in the LED? Is there any difference in Pin?
.99
Are you calculating input power over the full cycle or just during the on time of the transistor? With the LED to ground, there is current pumped out of the battery during the inductor discharge cycle because the current loop includes the battery, but unfortunately is in the wrong direction, and does not have a regenerative effect, just the opposite. Because of this, if you are not calculating power over the full cycle, the circuit will appear to be drawing the same current from the supply yet delivering more power to the LED.
I have simulated this both ways and tested it on the bench. To my observations, the current drain from the battery never reverses, is always in the same direction i.e. out of the battery even during the inductor discharge cycle, although it is about half the inductor charge drain. This must be included in the power calculation.
With the LED connected across the driven winding, it absorbs the energy without forcing the battery to also supply current as current no longer circulates through the battery during the inductor discharge cycle.
I must state again I prefer a 0.1 ohm resistor in the negative lead of the battery as opposed to the 1.0 ohm emitter resistor, as it more accurately allows assessment of battery load effects during both inductor charge and discharge cycles. Also 0.1 Ohm is pretty close to the battery impedance for the sims.
I agree, that a smaller CSR placed in series is the best way to go for current sensing.
Regarding this LED placement issue, it is less complicated than what you seem to be looking at ION.
What I'm doing when simulating this, is starting with the basic JT connection just as you showed in your excellent writeup about Testing the Efficiency of Blocking Oscillators (http://www.overunityresearch.com/index.php?topic=602.msg9541#msg9541).
(http://www.overunityresearch.com/index.php?action=dlattach;topic=602.0;attach=3202)
You show D1 going to Vbat. You will need to replace this diode with a LED that has a VF of at least 2V. Shunting a normal diode to ground will clamp the collector too hard, and it won't oscillate. At least that has been my findings.
While on your bench, try the LED with both connections (i.e. gnd and Vbat); which produces the highest intensity? How does Pin compare in each case?
What I am doing in the simulation is plotting the average power from the battery AND the average power in the LED, for each case. Power transfer efficiency to the LED is greater in the case when it is tied to ground rather than Vbat. Whether the battery power goes up or not seems to depend on the build and whether a core is used. However, that aside, n always seems to increase with the ground connection.
Does this make sense? If not, I will make a diagram and try to explain it.
Btw, I don't believe I have mentioned battery current reversal in this discussion. That is a different issue I believe, but since you mention it, current won't return to the battery in either case of LED connection.
.99
Looking back, I would not have used the emitter resistor, but I wanted the circuit to be somewhat close to the device under discussion.
I needed to use a string of 6 1N5817 Schottky diodes to get above the battery voltage of 1.5 Volts or the circuit would not oscillate.
We agree, there is no current pumped back into the battery with either method. What I'm saying is you will observe a triangle wave current drain using the grounded led technique when measured at the source battery.
The current goes from 140 to 70 mA during the discharge cycle, when it should go from 140 to close to 0 mA when the transistor switches off at the end of the charge cycle. During the "off" period the current ramps from 140mA to 70 mA which is the series current from the power supply through the led not accounted for if only evaluating the charge cycle.
Quote from: ION on 2011.02.14, 14:10:42
The current goes from 140 to 70 mA during the discharge cycle, when it should go from 140 to close to 0 mA when the transistor switches off at the end of the charge cycle. During the "off" period the current ramps from 140mA to 70 mA which is the series current from the power supply through the led not accounted for if only evaluating the charge cycle.
I am not evaluating any particular part of the cycles actually.
I trust you have been following along somewhat with the oscilloscope measurements on the LTJT etc? For these tests, the average power both from the battery and to the LED is measured using MEAN[v(t) * i(t) = p(t)]. I am doing precisely the same measurement in the simulation over several (minimum 10) cycles to obtain an accurate measurement. Analysing any particular part of the cycle becomes irrelevant this way. What you get is what you get, an
average, and it is accurate. Does this make sense?
So, making the average power measurements for BOTH the battery power AND the LED power,
the LED always exhibits more power when it is tied to ground, vs. being tied to Vbat. As well, the efficiency is always higher this way. At least that is my finding.
I encourage you to try the experiment on your bench. I will be surprised if you see no improvement in efficiency with the ground connection.
Ok Thanks I'll double check it on the bench.
Referring to P9901 schematic in reply #6 and my statements in the prior posts;
I double checked the waveforms and my analysis and will now state that the method of sensing current in the emitter resistor CS1 misses a full 1/3 of the power being delivered from the battery during the off cycle of the transistor or discharge cycle of the inductor.
To get the full picture of battery power to the load, you need an independent shunt through which all current during charge and discharge cycles can be seen.
Then, rather than the clean cutoff off of current as seen in the emitter sensing method of CS1, you will see a triangle wave of current across that external shunt and the extra power drawn from the battery during the discharge cycle that has been missing from the calculation.
Continue to look only at current through CS1 as a indication of input power, and you will be missing part of the input power picture.
This is true for all cases where the LED is brought back to the battery for ground and bypasses the current sense resistor CS1. It does not apply where LED current is strictly circulated around the driven inductor as in post #16 above.
Attached is the corrected method of accounting for all power used from the battery source.
Quote from: ION on 2011.02.14, 23:05:11
...
To get the full picture of battery power to the load, you need an independent shunt through which all current during charge and discharge cycles can be seen.
...
That's my analysis and I'm sticking to it...
Well done!
Thank you for straightening out my error ION.
I will retest the P9901 shortly. That should make my 53% efficiency drop somewhat. :'(
.99
NOTE: All units other than the P9901 were tested as per single CSR method.
Here is some simulation data requested by ION regarding my claim about the differences in efficiency with the LED terminated at ground (GND) vs. when terminated at the + terminal of the battery (Vbat). I claim that the efficiency will always be higher when the LED is terminated at GND.
I tested two circuits, one with a core, and one with an air-core. The average power (mW) for each component is listed, and the core loss is determined by the difference between the battery power and the components power. Efficiency n is determined by LED power / Battery power. Rb is the 1k Base resistor, and Rs is the primary DCR.
With Core
GND Connection
BAT:68.7, LED:27.9, Q:12.6, Rb:0.85, Rs:0.30. Components Total: 41.65. Core Loss: 68.7 - 41.65=27.05. n=27.9/68.7=40.6%
Vbat Connection
BAT:69.3, LED:20.8, Q:15.3, Rb:1.04, Rs:0.36. Components Total: 37.5. Core Loss: 69.3 - 37.5=31.8. n=20.8/69.3=30.0%
Air Core
GND Connection
BAT:80.4, LED:49.6, Q:29.8, Rb:0.59, Rs:0.39. Components Total: 80.4. n=49.6/80.4=61.6%
Vbat Connection
BAT:68.6, LED:32.99, Q:34.3, Rb:0.833, Rs:0.40. Components Total: 68.6. n=32.99/68.6=48.1%
The two circuit diagrams illustrate the two connections. I hope it is fairly clear from the data which one yields a higher efficiency.
.99
Thank you for the testing POYNT. Your numbers tell the story. Looks like your efficiency went up rather than down???
There appear to be two questions:
first, why did the efficiency go up when current drain over the full cycle is considered.(external shunt)
The second question is: why is the ground connection more efficient than the Vbat connection ?
This is completely counter-intuitive considering in the grounded connection we see current drawn from the battery during the charge and the discharge cycle and there is no mechanism for recycling of power back to the battery.
While in the Vbat connection, the current flow during discharge is isolated from the battery, hence there is no extra drain during the discharge part of the cycle.
I admit I am now at a loss to explain this.
Anyone care to offer an explanation?
Here I present measurements again for the P9901 with the circuit corrected with both the emitter and LED currents going through a single 1 Ohm CSR for power measurement.
Scratch your head on this one guys. ??? hehehehe...
input_mean_corrected.PNG indicates an average INPUT power of
55.37mW.
output_mean_corrected.PNG indicates an average OUTPUT power of
45.52mW.
n = 82.2% O0
.99
NOTE: Test must be re-done. Voltage incorrectly taken across LED and CSR1.
Quote from: ION on 2011.02.15, 04:00:03
Thank you for the testing POYNT. Your numbers tell the story. Looks like your efficiency went up rather than down???
The real question now is: why is the ground connection more efficient than the Vbat connection ?
This is completely counter-intuitive considering in the grounded connection we see current drawn from the battery during the charge and the discharge cycle and there is no mechanism for recycling of power back to the battery.
While in the Vbat connection, the current flow during discharge is isolated from the battery, hence there is no extra drain during the discharge part of the cycle.
I admit I am now at a loss to explain this.
Anyone care to offer an explanation?
These were the tests regarding the GND vs. Vbat issue.
I already knew the efficiency would be better with the ground connection. I even stated that in my claim.
The post above however (corrected CSR issue), is the one where we thought the efficiency would go down....but it went up. ;)
.99
OK, I think I know where I went wrong on this last test (reply #25). I will have to do it again.
I'm fairly sure that the voltage measurement for the LED was taken across both the LED and the CSR, whereas it should only be across the LED. Kinda makes it difficult to do then. :(
I thought 80% was too good to be true. Damn, have to quit testing when I'm tired. :-[
.99
When you are recording the output power you are registering output power when the transistor is on and power is not actually flowing through the LED.
If I may be so bold as to suggest yet another setup (all relative to P9901_Schema01.gif): I assume that your probes have to have a common ground. So the transistor emitter node should be the probe ground. Now you can measure the collector-emitter voltage as the LED voltage. Battery positive to the emitter as your power supply voltage. Then emitter to real ground across the CSR for your current. The current will read "backwards" but who cares?
So you end up factoring out the power dissipated in the CSR like this. That makes sense when you think of it. Since you have a recording of the current waveform anyways, you have the option to factor it in if you want. If the DSO can't do the i^2R math on the current waveform alone, I suppose you could export it into Excel if you had to.
Think about that one, I think it makes sense. It's not perfect but it's an improvement.
*** LED Output power measurements ***
When the transistor is ON, you are still recording a very small amount of LED output power even though the LED is off. You are not recording any CSR power.
When the transistor is OFF, you are recording the LED output power. You are not recording any CSR power.
*** Input power measurements ***
When the transistor of ON, you are recording the input power. You are not recording any CSR power.
When the transistor is OFF, you are recording the input power. You are not recording any CSR power.
MileHigh
How do we get an accurate scope measurement of the battery voltage when the CSR is in the bottom leg of the battery?
One way I can think of is to use the scope to subtract (or add) it from the voltage across the CSR.
.99
Quote from: ION on 2011.02.15, 04:00:03
...
I admit I am now at a loss to explain this.
Anyone care to offer an explanation?
Transistor Losses.
Creative "tweaking" may minimize those.
Quote from: ION on 2011.02.15, 04:00:03
Thank you for the testing POYNT. Your numbers tell the story. Looks like your efficiency went up rather than down???
The second question is: why is the ground connection more efficient than the Vbat connection ?
This is completely counter-intuitive considering in the grounded connection we see current drawn from the battery during the charge and the discharge cycle and there is no mechanism for recycling of power back to the battery.
While in the Vbat connection, the current flow during discharge is isolated from the battery, hence there is no extra drain during the discharge part of the cycle.
I admit I am now at a loss to explain this.
Anyone care to offer an explanation?
I think that when the transistor is off and the coil is discharging through the LED to ground, a discharge circuit is created via the supply battery back to the coil, so the battery voltage is added (series aiding) to the discharge which results in more power dissipated in the load.
Hoppy
Quote from: Hoppy on 2011.02.15, 09:35:31
I think that when the transistor is off and the coil is discharging through the LED to ground, a discharge circuit is created via the supply battery back to the coil, so the battery voltage is added (series aiding) to the discharge which results in more power dissipated in the load.
Hoppy
I think you're on the right track here Hoppy.
Yesterday in a chat with ION, I was trying to remember something I was going to point out, then I forgot. What I was going to mention was a "trick" I could use to not only make the efficiency worse, but it serves to help prove the point that it is about potential difference.
Imagine adding another battery or variable power supply to the circuit and you discharged through the LED into that rather than GND or Vbat. All are good AC grounds, but the DC level seems to make the difference. If I set my second power supply to say 3V, the power into the LED will be even less than it was when tied to Vbat.
It is somewhat analogous to attempting to discharge one capacitor into another, when both are already at the same voltage. It just doesn't happen.
The primary inductor becomes a current source when it is discharging its energy, so it doesn't really care where it is terminated, but in order to have more current for longer (and less voltage), it needs to see not only a low impedance, but a potential far away from that at the other end, which is tied to the Vbat.
So without our variable voltage supply, the place that allows the greatest potential difference is of course GND.
So what would happen if we set our variable voltage supply to say -3V, and discharged through to that?
.99
Quote from: poynt99 on 2011.02.15, 05:10:29
How do we get an accurate scope measurement of the battery voltage when the CSR is in the bottom leg of the battery?
One way I can think of is to use the scope to subtract (or add) it from the voltage across the CSR.
.99
No takers?
OK, here is proof that my solution would work, provided we can get the scope to do
this and the multiplication simultaneously:
.99
Quote from: poynt99 on 2011.02.15, 13:30:20
So what would happen if we set our variable voltage supply to say -3V, and discharged through to that?
.99
I assume you mean the cathode of the LED to the neg of the 3V supply with pos to ground? In this case we have 3V plus the supply battery voltage across the LED and inductor. In effect three supplies in series across the load. In this case the LED will be permanently lit assuming the LED has a forward drop less than the combined voltage of the two supplies.
Hoppy
Quote from: Hoppy on 2011.02.15, 14:25:10
I assume you mean the cathode of the LED to the neg of the 3V supply with pos to ground? In this case we have 3V plus the supply battery voltage across the LED and inductor. In effect three supplies in series across the load. In this case the LED will be permanently lit assuming the LED has a forward drop less than the combined voltage of the two supplies.
Hoppy
Yes.
It would appear that as long as the potential is close to Vbat, the power to the LED (and n) will be diminished compared to when the LED is connected to GND.
.99
Darren , here are my numbers taken with a true RMS FLUKE DVM measuring across 1 Ohm shunt resistors per my earlier posted "corrected" schematic.
This shows more current flowing into the LED in the Vgnd position at the expense of greater current drain from the battery.
It also shows greater efficiency with the Vbat connection.
What am I missing?
edit: I'm going to rerun this test using simple RC filters on all measured parameters
Quote from: ION on 2011.02.15, 15:01:29
Darren , here are my numbers taken with a true RMS FLUKE DVM measuring across 1 Ohm shunt resistors per my earlier posted "corrected" schematic.
This shows more current flowing into the LED in the Vgnd position at the expense of greater current drain from the battery.
It also shows greater efficiency with the Vbat connection.
What am I missing?
ION,
I think I may have an answer to that question, but I will need a little time to verify my hypothesis.
.99
ION,
How did you measure the voltage across the LED?
.99
Darren
Here is the latest excel chart of bench measurements. I think I understand why it is more efficient to return Vled to ground. My theory is that this method uses the switcher as a bias supply for the LED, and allows the battery to light the LED directly, therefore in a way bypassing the need for the switcher to deliver all the current to the Vled. In other words, if you had a 2.0 volt battery and used a 1.5 volt led the switcher would not be necessary and the efficiency would be very high.
The Vbat connection forces the switcher to deliver all of the current to the LED, and will show a lower efficiency than the Vgnd method that allows a leak path for current(through the inductor) from the battery directly to the LED, thus raising apparent efficiency.
I used 1k and 2.2 uF R-C filters for the measurements and further isolated the forward pulse from the Vled with a Schottky diode outside the filter loop. This diode loss is not accounted for in the spreadsheet.
If this work is just creating more confusion, I'll back out of it.
Quote from: Hoppy on 2011.02.15, 09:35:31
I think that when the transistor is off and the coil is discharging through the LED to ground, a discharge circuit is created via the supply battery back to the coil, so the battery voltage is added (series aiding) to the discharge which results in more power dissipated in the load.
Hoppy
Correct.
Thereby "masking" transistor switching losses.
Enhancing transistor switching efficiency will likewise
result in more power dissipated in the load.
Transistor power losses must be evaluated and
minimized by "tweaking" base drive pulse shape
and magnitude. Each transistor is different and
can be effectively "tuned."
Quote from: ION on 2011.02.15, 15:01:29
Darren , here are my numbers taken with a true RMS FLUKE DVM measuring across 1 Ohm shunt resistors per my earlier posted "corrected" schematic.
This shows more current flowing into the LED in the Vgnd position at the expense of greater current drain from the battery.
It also shows greater efficiency with the Vbat connection.
What am I missing?
edit: I'm going to rerun this test using simple RC filters on all measured parameters
Very good question.
Some configurations utilize a suitably sized capacitor
across the (Rseries + LED) load in order to enhance
(photon production) efficiency.
The LED itself is quite interesting. Some will produce
photons (albeit dimly) at a forward potential well below
what is considered the "normal." Testing a batch will
reveal some surprising irregularities.
Quote from: Dumped on 2011.02.15, 17:13:20
Correct.
Thereby "masking" transistor switching losses.
Enhancing transistor switching efficiency will likewise
result in more power dissipated in the load.
Transistor power losses must be evaluated and
minimized by "tweaking" base drive pulse shape
and magnitude. Each transistor is different and
can be effectively "tuned."
It sounds like you are making things too complicated. You just want the transistor to switch on and off as fast as possible. If you calculate how much base current is required to switch the load on completely and then add an extra 10% "insurance" current then you are sure that the transistor will be 100% switched on.
I am uncomfortable with the suggestion that transistors are individually different and have to be "tuned." It plays into the whole notion of "secret sauce" unlocking something "special." The fact is that if you are going to hold up a microscope to each individual transistor then they are different. However, the differences are so minuscule that normally they can be ignored.
MileHigh
Quote from: ION on 2011.02.15, 17:01:08
Darren
Here is the latest excel chart of bench measurements. I think I understand why it is more efficient to return Vled to ground. My theory is that this method uses the switcher as a bias supply for the LED, and allows the battery to light the LED directly, therefore in a way bypassing the need for the switcher to deliver all the current to the Vled. In other words, if you had a 2.0 volt battery and used a 1.5 volt led the switcher would not be necessary and the efficiency would be very high.
The Vbat connection forces the switcher to deliver all of the current to the LED, and will show a lower efficiency than the Vgnd method that allows a leak path for current(through the inductor) from the battery directly to the LED, thus raising apparent efficiency.
I used 1k and 2.2 uF R-C filters for the measurements and further isolated the forward pulse from the Vled with a Schottky diode outside the filter loop. This diode loss is not accounted for in the spreadsheet.
If this work is just creating more confusion, I'll back out of it.
Thanks ION for these tests.
It could be a few factors that is contributing to the increase in efficiency, including Fo and duty cycle. The important issue to be realized and resolved though, is that the GND connection always seems to yield a better efficiency over all, so we should use that connection when building this configuration of JT.
Are we in Agreement?
.99
ION,
Do you have any alternative suggestion to this (http://www.overunityresearch.com/index.php?topic=729.msg10837#msg10837) for obtaining Vbat on an oscilloscope using normal probes?
Thanks,
.99
Quote from: MileHigh on 2011.02.15, 17:30:56
...
I am uncomfortable with the suggestion that transistors are individually different and have to be "tuned." It plays into the whole notion of "secret sauce" unlocking something "special." The fact is that if you are going to hold up a microscope to each individual transistor then they are different. However, the differences are so minuscule that normally they can be ignored.
MileHigh
Sadly, that is indeed a commonly held belief.
In order to accommodate conveniently the variations
in transistor quality within any given "type" certain
"trade-offs" must be made in circuit design which
invariably result in diminished performance uniformity.
Consumer Electronics.
But, those who have learned how to squeeze the very
best from each individual transistor are well aware of
the tricks and techniques which unlock superior performance.
Naturally, this does require "re-tuning" when transistors are
replaced but it is a small price to pay.
Linear Amplifier applications are much less critical. Switching
applications are more challenging.
Quote from: poynt99 on 2011.02.15, 17:58:09
ION,
Do you have any alternative suggestion to this (http://www.overunityresearch.com/index.php?topic=729.msg10837#msg10837) for obtaining Vbat on an oscilloscope using normal probes?
Thanks,
.99
Darren that is an excellent method and will work well in the sims. For the DSO it will also work well provided Vbat- is your scope ground. I don't foresee any problem.
Personally, in the sims and on the bench with a scope, I prefer making Vbat- the absolute ground of the circuit for measuring input power and make the negative end of the shunt resistor on the LED side the absolute ground for output measurements. But that's just my preference.
If you want to make I/O power measurements simultaneously using the same scope, it is a bit more of a problem. As MH said you can make the emitter absolute ground and measure back to Vbat- and to Vled - measuring across each of the shunts. The battery current measurement will be inverted in polarity, but this is of no consequence and can be flipped in the math calc.
Quote from: ION on 2011.02.15, 18:42:59
Darren that is an excellent method and will work well in the sims. For the DSO it will also work well provided Vbat- is your scope ground. I don't foresee any problem.
Personally, in the sims and on the bench with a scope, I prefer making Vbat- the absolute ground of the circuit for measuring input power and make the negative end of the shunt resistor on the LED side the absolute ground for output measurements. But that's just my preference.
If you want to make I/O power measurements simultaneously using the same scope, it is a bit more of a problem.
Indeed, thanks.
I am hoping the solution I've proposed will work with two simultaneous scopes, and all grounds connected to the circuit ground, i.e. to the bottom side of the CSR as shown in my diagram. It's only a question if the scope will perform the subtraction and multiplication without issue.
.99
Darren, try this in your sims or on the bench:
Leave Vbat at 1.5 volts. Set the forward drop of the LED to 1.4 volts. Now disconnect the 1K drive resistor and with the circuit not oscillating, what is the efficiency?
Quote from: ION on 2011.02.15, 19:07:35
Darren, try this in your sims or on the bench:
Leave Vbat at 1.5 volts. Set the forward drop of the LED to 1.4 volts. Now disconnect the 1K drive resistor and with the circuit not oscillating, what is the efficiency?
What we're left with is simply a battery driving a LED, and the power is quite low. Even so, the efficiency should be fairly high, not having as many lossy components.
However this is no longer a JT. ;)
.99
Quote from: poynt99 on 2011.02.15, 17:52:36
Thanks ION for these tests.
It could be a few factors that is contributing to the increase in efficiency, including Fo and duty cycle. The important issue to be realized and resolved though, is that the GND connection always seems to yield a better efficiency over all, so we should use that connection when building this configuration of JT.
Are we in Agreement?
.99
Given that we start with a given level of energy to be discharged by the inductor, there are just two paths for the discharging current. If we take the grounded LED path back to the inductor via the supply battery, the level of power dissipated in the very low battery internal resistance will be negligible. If we take the flyback route to the inductor, again the power dissipation is negligible in the circuit wiring, so in theory I cannot see how one path can be so much different to the other. If anything, the return to Vbat should have the edge in efficiency IMO being the shortest path. However, if the battery exhibits high resistance due to sulfation, then this could change the situation. All this ignores the energy taken from the battery whilst the forward voltage of the LED is exceeded during inductor discharge in LED grounded config. This energy dissipates in the load thereby depleting battery capacity and increasing the overall load power dissipation in comparison with the return to Vbat config. I see it as a trade-off between lost battery capacity and reduced power dissipation in the LED load. Take your pick.
Hoppy
Quote from: poynt99 on 2011.02.15, 19:40:58
What we're left with is simply a battery driving a LED, and the power is quite low. Even so, the efficiency should be fairly high, not having as many lossy components.
However this is no longer a JT. ;)
.99
Actually I would guess the efficiency would be close to 100% and that was the point.
I was trying to make a case that part of the high efficiency is that the battery pre-biases the LED to the threshold of turn on with the Vgnd configuration and this is the unseen extra efficiency, that I believe will diminish when increasing the string of LED's hence the Vdrop.
I am only trying to identify a mechanism for the increased perceived efficiency and am theorizing that part of this higher efficiency is "borrowed" from "normal" battery/LED current.
I'll continue to research this privately, and yield that for a single LED with a Vdrop close to the battery voltage there is a measured increase in efficiency using the Vgnd method, so that is the best method for the JT's under discussion.
Quote from: ION on 2011.02.15, 22:07:56
Actually I would guess the efficiency would be close to 100% and that was the point.
I was trying to make a case that part of the high efficiency is that the battery pre-biases the LED to the threshold of turn on with the Vgnd configuration and this is the unseen extra efficiency, that I believe will diminish when increasing the string of LED's hence the Vdrop.
This is easy enough to test in the simulations. I'll let you know my findings.
.99
ION,
Here are the results of the tests with a core-type unit (3C90 material). I ran 3 tests; 1LED, 2LEDs, 3LEDs in series. Efficiencies computed in each case with LED(s) terminated at GND and at Vbat. All LED powers added together in cases where there is more than one.
1xLED
GND Connection:
Pbat: 61.3, Pled: 25, n=40.8%
Vbat Connection:
Pbat: 63.7, Pled: 18.08, n=23.38%
2xLED
GND Connection:
Pbat: 75, Pled: 12.5+12.5 n=33%
Vbat Connection:
Pbat: 74, Pled: 10+10, n=27%
3xLED
GND Connection:
Pbat: 79.8, Pled: 8.16+8.16+8.16, n=30.6%
Vbat Connection:
Pbat: 78.1, Pled: 7.07+7.07+7.07, n=27.2%
I suspect this trend will continue while adding LEDs in series, until finally, the two efficiencies will equal out at 0%.
In all the cases above, a GND connection shows a better efficiency. Unless this simulation is not telling the truth, I'd be inclined to always connect my LEDs to GND, if I was indeed a Joule Thief enthusiast that is. ;)
.99
This confirms some of my suspicions, the Vgnd shows a marked decrease in efficiency about 7% per added LED initially.
There really shouldn't be a decrease in efficiency if indeed the effect of a discharging inductor current source was supplying all the current. This is obviously not the case and the Vgnd configuration yields evidence that a partial voltage source (the battery) is part of the inductor discharge cycle.
Notice for the Vbat connection where the true current source effect of a discharging inductor is the only force at play, the efficiency starts rising about 3.5% per added LED initially as Ohmic losses no longer predominate.
Nevertheless the Vgnd has the added advantage of a jump start in efficiency by pre biasing the LED with the battery voltage source.
At some point the efficiency curves will equalize, at it looks like it will be at the 4 LED scenario.
Thank you for running these tests as they have provided some evidence for what I believe is the mechanism of the higher efficiency of the Vgnd configuration.
As they say, one picture is worth 1000 words, and three points can show a trend.
Of course this chart is meaningless unless you have been following the thread.
Just for fun, I've tested this "efficiency issue" out to 25 Volts or so. That is the equivalent of about 8 LEDs in series. I checked 3 points of clamping; 2V, 12V, and 25V.
For this test, I created a "self-controlled switch", i.e. like a diode or LED, made from a Switch part in SPICE. The Switch parameters are as follows:
RON = 40 Ohms
ROFF = 1M Ohm
VON = 2V, 12V, and 25V
VOFF= VON-2V
This switch part produces realistic results compared to the LED part I was previously using, only it is far easier to use than the LED.
Here are the results:
LED: 2VON, 0VOFF
Vbat
BAT:57.4, LED:16.07, n=28%
Gnd
BAT:62.24, LED:34.77, n=55.86%
LED: 12VON, 10VOFF
Vbat
BAT:81.0, LED:22.40, n=27.65%
Gnd
BAT:82.83, LED:24.72, n=29.84%
LED: 25VON, 23VOFF
Vbat
BAT:83.3, LED:22.10, n=26.53%
Gnd
BAT:84.3, LED:23.12, n=27.43%
It would appear that the results support my earlier remark that the efficiencies don't equalize until both are at 0%. These tests, if valid, indicate that for every case (any number of series LEDs) the n will be higher with the GND connection. Wave forms can be posted on request.
Simulation note: When simulating with high threshold voltages (i.e. > 8V) , the "minimum step size" must be set to 3ns or less, in order for the simulation to run properly, otherwise it can not sample the high transients present.
.99
ION,
You can implement this self-activating switch in PSpice by selecting the "S" part in the ANALOG library. Connect it as shown, and adjust your switch points as desired. A 2V difference worked well for me.
The two terminals on the left are the "sense", and the two on the right are the "switch".
Double-click the device and set the "RON" value to 20 Ohms, and leave "ROFF" at the 1e6 (1M) value.
Go to the "Edit Simulations Settings" and set the "Maximum Step Size" to 3ns. The runs will take longer, but you should have no convergence problems then.
.99
I wrote up a blog on my experiences with air core coils and JT efficiency.
http://watsonseblog.blogspot.com/2011/02/2011-feb-9-air-core-coil-and-jt.html
One question I have about the schematic. It says 2N2222 for the transistor. Does this have a plastic case or metal case? If it's metal then it can be a 2N2222, but if it's plastic, it should say PN2222 or MPS2222 or similar on the case. Without the true part number, it's impossible to replicate the circuit.
Thanks.
Thenks POYNT, tried the switch and it works fine. I'll have to ask a few questions later regarding the switching point vs what I see in the sims.
acmefixer:
Although I have not studied the data sheets that closely, the plastic case should be the same electrically, but temperature ratings and MTBF's may differ. Some data sheets show Ic of 1 amp others 800 mA and thermal resistance junction to case may differ.
Regarding JT utility and efficiency, there is no one size fits all for JT's. Best to define the goal or application, then proceed to make make a best fit. Highest frequency is not always the best and may waste unnecessary power in the switch and in radiation. Depends on the degree of miniaturization, weight, allowable EMI, input voltage, load characteristics and other factors.
Quote from: acmefixer on 2011.02.17, 16:59:55
...
One question I have about the schematic. It says 2N2222 for the transistor. Does this have a plastic case or metal case? If it's metal then it can be a 2N2222, but if it's plastic, it should say PN2222 or MPS2222 or similar on the case. Without the true part number, it's impossible to replicate the circuit.
Thanks.
There are a number of transistor types which work well
for this project. Any of the "2222" packages may be
used - just be aware that some of them may perform better
than others because of slightly different characteristics.
The new family of Low Vce(sat) or BISS transistors work
very well too. They have higher current rating capability
with equally high Beta within a the smaller SOT package.
One that I like is the NSS60601 (6 Amperes/60 Volts)
in the SOT223. Other similar types are available from
other manufacturers at Mouser or DigiKey etc...
The transistor type used is not nearly as important as
how well the circuit is "tweaked" or "tuned" for optimum
efficiency.
Quote from: ION on 2011.02.17, 18:09:43
Thenks POYNT, tried the switch and it works fine. I'll have to ask a few questions later regarding the switching point vs what I see in the sims.
Certainly ION. If you have any questions, either post them here or send me a PM. ;)
.99
Here is a corrected update as to the proper CSR setup for measuring the INPUT and OUTPUT power for the general JT configuration. This is now in agreement with ION's diagram.
I will try to redo the measurements for the air-core P9901 JT today. I will try to include the CSR1 and CSR2 dissipation as well.
.99
I've done a test using the above circuit, and I've realized a much better and simpler way to obtain the actual battery power and actual LED power, than what I proposed here (http://www.overunityresearch.com/index.php?topic=729.msg10831#msg10831). The power dissipation in the CSR resistors is also easily obtained. O0
QuoteNotes:
1) P11= probe, scope 1, CH1. P12= probe, scope 1, CH2. P21= probe, scope 2, CH1. P22= probe, scope 2, CH2.
2) INPUT power is obtained as follows:
2.1) Use the scope MATH to produce MEAN[(V1*V2)]. This is the total power of Vbat and CSR1 together. We will call it Pitotal.
2.2) Use the scope MATH to produce MEAN[(V2*V2)]. This is the power of CSR1 alone. We will call this Pcsr1.
2.3) Pvbat is computed by: Pitotal - Pcsr1.
3) OUTPUT power is obtained in much the same manner:
3.1) Use the scope MATH to produce MEAN[(V3*V4)]. This is the total power of LED and CSR2 together. We will call it Pototal.
3.2) Use the scope MATH to produce MEAN[(V4*V4)]. This is the power of CSR2 alone. We will call this Pcsr2.
3.3) Pled is computed by: Pototal - Pcsr2.
4) The efficiency n of the battery power delivered to the LED alone is then: n = 100(Pled / Pvbat).
5) If we were to insert a 1 Ohm CSR resistor in series with the transistor emitter, we could measure and calculate the transistor power dissipation in a similar fashion. We already have the voltage across the transistor, V3, so the other required voltage measurement would be that of a CSR3 resistor labeled "V5". Note, V5 is not the same as V2.
My results for the P9901 circuit are as follows:
Pintotal = 41.2mW
Pcsr1 = 3.39mW
Pvbat = 37.81mW
Pototal = 24.4mW
Pcsr2 = 0.69mW
Pled = 23.71mW
n(vbat to led) = 23.71/37.81 =
62.71%.99
Thanks, .99, for these test results and the clear explanation of how you obtained them.
The "other shoe" is of course the JT with the ferrite core, rather than the air core.
I'm looking forward to your measurement results using the ferrite core -- same method otherwise.
It was with the ferrite core in an almost identical set-up that I found a surprising (approx) n = 1.3, as I reported last Friday,
http://www.overunityresearch.com/index.php?topic=717.msg11327#msg11327
so I'm particularly looking forward to your further results with the ferrite core this time. I may have done something wrong, but would like to compare with your results. Of course, we are using different toroidal-windings.
Coming up soon professor.....I'm heading over to do them right now.
.99
Great, looking forward to the results .99
A comment about your schematic -- I have the wire from the emitter of the 2n2222 going back to the point you have labeled V2, rather than going to ground. As it stands, I don't see how you account for this "loss" from the circuit, with emitter output going direct to ground.
Other than that, and the fact that I used a ferrite core, our circuits look the same.
The results are posted professor.
See this post (http://www.overunityresearch.com/index.php?topic=717.msg11405#msg11405).
Regarding your emitter connection, that needs to be connected as shown on my diagram, where it is going to ground, not directly to the negative battery terminal. All currents going in or out of the battery must go through the supply current-sensing resistor, otherwise your INPUT power measurement will not be correct.
.99
OK, I've discovered that Pitotal must be added to Pcsr1, not subtracted. Let's try again:
My corrected test results for the P9901 circuit are as follows:
Pintotal = 41.2mW
Pcsr1 = 3.39mW
Pvbat = 44.59mW
Pototal = 24.4mW
Pcsr2 = 0.69mW
Pled = 23.71mW
n(vbat to led) = 23.71/44.59 = 53.2%
.99
Quote from: poynt99 on 2011.02.27, 22:51:41
OK, I've discovered that Pitotal must be added to Pcsr1, not subtracted. ...
.99
No, I think you were right the first time. You measure the total input, then SUBTRACT the power dissipated in the measuring resistor. The power dissipated in the measuring-resistor is not available for the rest of the circuit.
What would the power be in the input circuit if you reduced CSR1 to 1/2 ohm? then to 1/4 ohm, etc.
Perhaps the most important aspect of this exercise is getting the measurements right, subtracting when one should subtract, getting the ground connections right, etc. I'm not sure you/we are there yet.