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Author Topic: LTJT - poynt99 Tests #2  (Read 89556 times)
Group: Professor
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Now let's look at the case where the secondary output circuit is cut open circuit, and the collector LED re-connected as the only load. Note that the LED goes through the 1 Ohm to ground.

colout_input_mean.PNG indicates an average INPUT power of 41.53mW.

colout_output_mean.PNG indicates an average OUTPUT power of 37.1mW.

n=89.3%

As you can see, it is far more efficient to load the JT from the collector than from a secondary winding with 110 Ohms series resistance.

Also of note, comparison of this unit with my P9901 air-core unit, you can see that a ferro-magnetic core is not used to make the device function (via core saturation), but rather to make it much more efficient, i.e. 90% vs. 50%. Q. How does it do that? A. By increasing the inductor Q.

.99

   This is interesting.  When you take measurements, please include a measurement of the power dissipated in the input-current-measuring-1-ohm-resistor [CSR1 -- did I get your definition correct this time?] for the above circuit.

That is, please re-calculate:

colout_input_mean.PNG indicates an average INPUT power of 41.53mW.

when you have subtracted the power dissipated in this current-measuring resistor; it is not essential to the operation of the circuit, but is useful for measuring the current.

And then re-calculate n, if you would. 

Looking forward to your measurements.
   
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  I would add that I'm personally learning a lot from a study of this "simple" JT circuit, using my own oscilloscope now.  The V*I math function on it is particularly useful. 

And I'm learning a lot from the civil discussion we're having, and for that I thank you (particularly .99).
   
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OK, so I did with my own circuit what I'm asking .99 to do.

I disconnected the "LT" part of the LTJT circuit, so just looking at the JT circuit.  See attached photo of the set-up.
Connections across the "input" CSR we've been discussing (CSR1) and across the battery.  The green waveform on the DSO is the total power input (Vbatt * I-csr1). 

Next:

  Red LED from the collector then across a 1-ohm measuring resistor (call it CSRb).  Result is shown in the second photo.

It's a bit rough to calculate the energy into the circuit and the energy out to the LED-CSRb  -- my DSO does not give the MEAN for the math function.  I calculate energy as explained previously, taking the area under the green Power waveform, for one cycle.  I hope .99 will do the measurement the other way this weekend...  I may try to go to the University and use the Tek 3032 also...

Long story short, I gotta run!  but I find that the NET input power to the circuit is a little less than the output into the LED-CSRb ... 
I probably made some mistake.  n is about 1.3 weird... 
really gotta run


   
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 :o my mechanical model predicted a theoratical 1.5
   

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It's not as complicated as it may seem...
I'll be re-testing this one as well, now that I've nailed down a good procedure.

Stay tuned...today.

.99
   

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It's not as complicated as it may seem...
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. The power dissipation in the CSR resistors is also easily obtained.  O0

Quote
Notes:

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.

The test results for my LTJT with a core and no secondary circuit (normal JT) are as follows:

Pintotal = 29.05mW
Pcsr1 = 1.72mW
Pvbat = 27.33mW

Pototal = 31.3mW
Pcsr2 = 0.66mW
Pled = 30.64mW

n(vbat to led) = 30.64/27.33 = 112%

So either I have a measurement anomaly, or.....  ;D

.99
   

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It's not as complicated as it may seem...
Does anyone see any flaw in the measurement method?

It seems correct to me.

.99
   
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I see a possible flaw in the calculation for the power dissipation in the LED. The forward resistance is not constant as the LED is being pulsed. I think the the duty cycle needs to be factored into the calculation.

Hoppy
   

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

Are you aware of the method being used for these tests? We are using the scope to multiply instantaneous voltage and current to get instantaneous power. Then we use a "measurement" function in the scope to provide the MEAN of that instantaneous MATH power trace.

All phase skewing and duty cycle allowances are taken care of by this method. There is no need to "do" anything, other than make the final calculation for efficiency.

Make sense?

.99
   
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Yes, I understand the method being used but I don't understand the MEAN[(V3*V4)} expression as defining the LED power dissipation.

Hoppy
   

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How are you accounting for the fact that CSR1 current flow
is both "input" (transistor on) current and also "output"
(transistor off) current?

Is there any way to fully isolate the two?



---------------------------
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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It's not as complicated as it may seem...
Yes, I understand the method being used but I don't understand the MEAN[(V3*V4)} expression as defining the LED power dissipation.

Hoppy

Hoppy, it isn't.

Check that line again:

Quote
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.

.99
   

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It's not as complicated as it may seem...
OK, I have found at least one problem. The INPUT power Pitotal must be ADDED to Pcsr1, not subtracted. This is because the power in the two are in opposite directions, and hence sign. So, let's try this again:

The corrected test results for my LTJT with a core and no secondary circuit (normal JT) are as follows:

Pitotal = 29.05mW
Pcsr1 = 1.72mW
Pvbat = 30.77mW

Pototal = 31.3mW
Pcsr2 = 0.66mW
Pled = 30.64mW

n(vbat to led) = 30.64/30.77 = 99.5%

This still seems rather high. Surely there is more than 0.5% of the power being dissipated in the transistor and base resistor...

.99
   
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Yes thanks, I have mis-read this line.

Hoppy
   

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It's not as complicated as it may seem...
How are you accounting for the fact that CSR1 current flow
is both "input" (transistor on) current and also "output"
(transistor off) current?

Is there any way to fully isolate the two?

We are interested in the net power delivered by the battery, and the net power dissipated in the LED. Having a current sense resistor in series with each along with the voltage across each (including it's CSR) allows us to determine the power in each.

Does that make sense?

.99
   

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We are interested in the net power delivered by the battery, and the net power dissipated in the LED. Having a current sense resistor in series with each along with the voltage across each (including it's CSR) allows us to determine the power in each.

Does that make sense?

.99

Yes, it makes sense if CSR1 is included in the "input"
inductor charge current path,

then

included with CSR2 for the "output" inductor discharge
path where the "source cell" and the "Inductor"
work series aiding to furnish "load" current which
puts the two resistors in series for that time.


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

Group: Administrator
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Posts: 3172
It's not as complicated as it may seem...
Yes, it makes sense if CSR1 is included in the "input"
inductor charge current path,

then

included with CSR2 for the "output" inductor discharge
path where the "source cell" and the "Inductor"
work series aiding to furnish "load" current which
puts the two resistors in series for that time.


So, what is your analysis...does it work correctly as I have it set up?

.99
   
Group: Professor
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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. The power dissipation in the CSR resistors is also easily obtained.  O0

The test results for my LTJT with a core and no secondary circuit (normal JT) are as follows:

Pintotal = 29.05mW
Pcsr1 = 1.72mW
Pvbat = 27.33mW

Pototal = 31.3mW
Pcsr2 = 0.66mW
Pled = 30.64mW

n(vbat to led) = 30.64/27.33 = 112%

So either I have a measurement anomaly, or.....  ;D

.99

Interesting, and in the same direction as my n=1.3 measurement on Friday....  Must be thoroughly checked!

Quote
Quote from: poynt99 on 2011-02-27, 23:51:41
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.  But I'm encouraged by the results (not ecstatic yet).
   

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

I am quite certain that all currents must go through the CSR which is in series with the battery in order to capture all that is either leaving or entering the battery. That is why your transistor emitter must be tied to the circuit ground as shown.

The addition or subtraction of the power in the CSR is not so easy to see. The voltage drop across that resistor is actually reversed from what you might think. It caught me too, and Humbugger thankfully corrected me on that. So actually we are subtracting the two powers, but it so happens that one of them is a "negative" power so to speak.

You can think of the battery CSR as an internal resistance in the battery. In such a case, this resistance is part of the total power dissipation in the battery. It is the same case here.

Perhaps someone else can explain it better, I am quite properly burnt out at the moment.  :D

If you are still in much doubt professor, I can show you with a simulation where it is quite clear.

.99
   
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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
   

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It's not as complicated as it may seem...
OK guys, let's start with the basics.

In order to determine the average power dissipation in a circuit component, we must measure the voltage across, and current through that component. If we have a running instantaneous product of those two measurements, which is p(t), we can have the oscilloscope compute the MEAN of p(t) to indicate the average power.

Are we in agreement on that?

.99
   
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OK guys, let's start with the basics.

In order to determine the average power dissipation in a circuit component, we must measure the voltage across, and current through that component. If we have a running instantaneous product of those two measurements, which is p(t), we can have the oscilloscope compute the MEAN of p(t) to indicate the average power.

Are we in agreement on that?

.99

That's one method, valid yes, but not the only method.  Can we also agree that the resistor will HEAT UP and we can measure the rate of heating to determine the power dissipated in the resistor?  Further, is this heat from the resistor CSR1 going to add to the light output of the LED?  (Certainly not.)

Also, .99, please PREDICT what will happen to n when you replace CSR1 with a 1/2 ohm resistor... use your simulation if you wish.  THEN make the measurement with a 1/2 ohm CSR1.
   

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It's not as complicated as it may seem...
I should have been more specific, sorry.  ;)

Using the oscilloscope method (so far your questions have only pertained to oscilloscope measurements), are we in agreement with the above?

.99
   
Group: Professor
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Posts: 2884
OK guys, let's start with the basics.

In order to determine the average power dissipation in a circuit component, we must measure the voltage across, and current through that component. If we have a running instantaneous product of those two measurements, which is p(t), we can have the oscilloscope compute the MEAN of p(t) to indicate the average power.

Are we in agreement on that?

.99

As I said, I am in agreement with this method, with the other questions I raised unanswered as yet.
   

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It's not as complicated as it may seem...
Hopefully, Gibbs is in agreement as well.

I would hope that the efficiency will increase as the value of CSR decreases.

.99
   
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