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Author Topic: Joule Thief - P9901  (Read 43487 times)

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It's not as complicated as it may seem...
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.
« Last Edit: 2011-02-15, 04:46:38 by poynt99 »
   

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It's not as complicated as it may seem...
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
   

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It's not as complicated as it may seem...
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
   
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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
   

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It's not as complicated as it may seem...
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
   

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...

I admit I am now at a loss to explain this.

Anyone care to offer an explanation?



Transistor Losses.

Creative "tweaking" may minimize those.


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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
   

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It's not as complicated as it may seem...

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
   

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It's not as complicated as it may seem...
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
   
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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
   

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It's not as complicated as it may seem...
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
   
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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


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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It's not as complicated as it may seem...
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
   

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It's not as complicated as it may seem...
ION,

How did you measure the voltage across the LED?

.99
   
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It's turtles all the way down
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.


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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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."


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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.


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For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.
   
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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
   

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It's not as complicated as it may seem...
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
« Last Edit: 2011-02-15, 19:00:53 by poynt99 »
   

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It's not as complicated as it may seem...
ION,

Do you have any alternative suggestion to this for obtaining Vbat on an oscilloscope using normal probes?

Thanks,
.99
   

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...

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.


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It's turtles all the way down
ION,

Do you have any alternative suggestion to this 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.


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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It's not as complicated as it may seem...
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
   
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It's turtles all the way down
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?


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

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It's not as complicated as it may seem...
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
   
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