author=picowatt link=topic=8341.msg478241#msg478241 date=1458786487]
Tinman,

A better way would have been to keep all scope grounds on the battery minus and use both scope channels and probes to measure both ends of the 10R base resistor.  With both scope channel vertical gains set identically (ca 20mv/div) invert one channel and then add them together (or use a math function to subtract one channel from the other).  This allows you to make a differential measurement across the 10R with minimal capacitive loading (particularly if you use 10X probes).


PW

Ok,the results of the test above.
first up,there is no way that we can use 20mV/div--just way to low,so i have used 5v/div
Second--why invert one channel?,as if we keep both channels non inverted,then we just subtract the lowest value from the highest value to get our voltage drop across the 10ohm resistor--but anyway,done as you asked.

Circuit below with scope placement.
I have a wheel that has a selection of set resistance values which i have used for these test.
The first scope shot is with a 1Kohm resistance value(+ the 10 of course).
The second scope shot is with a base resistance value of 820ohms-just the next value down.
The third scope shot is with a base resistance of 550ohms-next value down.

Here is an idea.
How about you put together a JT yourself,and draw your own timing diagrams,and carry out your own testing.
Wouldnt that be a hoot.

It's not going to happen and you are making the claim and three people have already told you that you are wrong.  So give it a go and try to prove that you are correct.

It's not going to happen and you are making the claim and three people have already told you that you are wrong.  So give it a go and try to prove that you are correct.

Name those three people MH that have said i am wrong with what i claimed in regards to the reduction of base resistance increasing the light output--name them.
Name those that will argue against the scope and light meter--i dare ya.

Brad

See-your doing it again--changing things to suit your needs.
You know very well that the idea behind adjusting the base resistance as the voltage in the battery drop's,was to maintain the highest amount of light output as possible.
I have proven beyond doubt that this happens,and your theory about V/R limit's-and all other associated garbage was quickly dismissed.

Now you have started on these wakadoo claims about my JT running to fast(frequency to high),and mine is operating differently to Mag's JT--!!!wow!!!
Then there is this giant killer spike,which in actual fact only has a voltage value of 100mV--yes,100mV-->so massive ,not to mention that the time period for this !!!massive!!! spike is that small it's not funny. And then have you dismiss the emitter/base junction capacitance as being the body that could indeed produce this very brief spike of current,is more than comical.

You are doing nothing more than trying your best to backpeddle from further mistakes on your behalf--you do this all the time--try to change the direction of the subject at hand,to turn attention away from your mistakes.

It's not working for you MH.

Decreasing the base resistance as the battery voltage drop's,dose indeed increase the brightness of the LED--proven.
Decreasing the base resistance dose increase the current flowing through L2--proven
Increasing the current through L2 dose increase the magnetic field strength--proven by the fact that the LED gets brighter.

Everything i stated has been proven.
Sorry MH,but you lost another one.

Brad

More wilful ignorance on display.

Look at this comment:

And then have you dismiss the emitter/base junction capacitance as being the body that could indeed produce this very brief spike of current

You just plucked your famous junction capacitance out of thin air to explain that nasty reverse current spike?  So you aren't going to try to THINK and investigate it and try to figure out what is going on?  You don't even have a rational line of reasoning.  You know that there is such a thing as junction capacitance and you saw the nasty reverse current spike and you just made a "magical" connection to "explain" the reverse current spike?

You "assigned" "junction capacitance" to the problem and it is now "explained" and you just walk away from it?   The junction capacitance is only a few PICOFARADS so it can't possibly explain what you are observing.  Where is your common sense when it comes to picofarads and how much current you can get to flow through a few bloody PICOFARADS?

You have to do better than that.

And you are avoiding the issue of the brightness adjustment instead of just being honest about it.   The V/R limit is absolutely real so it makes no sense at all for you to dismiss it.

author=picowatt link=topic=8341.msg478241#msg478241 date=1458786487]
Tinman,



Ok,the results of the test above.
first up,there is no way that we can use 20mV/div--just way to low,so i have used 5v/div
Second--why invert one channel?,as if we keep both channels non inverted,then we just subtract the lowest value from the highest value to get our voltage drop across the 10ohm resistor--but anyway,done as you asked.

Circuit below with scope placement.
I have a wheel that has a selection of set resistance values which i have used for these test.
The first scope shot is with a 1Kohm resistance value(+ the 10 of course).
The second scope shot is with a base resistance value of 820ohms-just the next value down.
The third scope shot is with a base resistance of 550ohms-next value down.

Tinman,

What you want to end up displaying is the voltage difference across the 10R resistor. 

Connecting the two probes as you have done is the first correct step in doing so.  With both channels' VPD settings (and probe cals) identical (try 2 VPD for the 10R), you should see identical waveforms on CH1 and CH2.

Next, you will want to invert one channel as you have also correctly done.

The next step, which you omitted, is to then ADD the two channels together.  You will now have just one trace displaying the voltage difference between the two ends of the 10R resistor.

This method effectively gives your scope a single channel differential input and is valid for both analog and digital scopes.  Inverting one channel and then adding them together is the same as subtracting the inverted channel from the other channel.

Using your digital scope, you should be able to use one of its math functions to directly perform the subtraction of one channel from the other.  However, being unfamiliar with your scope, I do not know if a difference trace will be displayed, which I would want to see (I am less interested in just "numbers in the boxes").

You may have to increase the vertical sensitivity (VPD), but as you do so you must keep both channel VPD settings identical at all times.  As you change the VPD settings, switch between the normal and ADD display to ensure that you are not clipping the channel inputs with too much VPD gain.  If necessary, increase the value of the 10R to 100R to increase the voltage difference across it.

PW   

Tinman,

What you want to end up displaying is the voltage difference across the 10R resistor. 

Connecting the two probes as you have done is the first correct step in doing so.  With both channels' VPD settings (and probe cals) identical (try 2 VPD for the 10R), you should see identical waveforms on CH1 and CH2.

Next, you will want to invert one channel as you have also correctly done.



This method effectively gives your scope a single channel differential input and is valid for both analog and digital scopes.  Inverting one channel and then adding them together is the same as subtracting the inverted channel from the other channel.



You may have to increase the vertical sensitivity (VPD), but as you do so you must keep both channel VPD settings identical at all times.  As you change the VPD settings, switch between the normal and ADD display to ensure that you are not clipping the channel inputs with too much VPD gain.  If necessary, increase the value of the 10R to 100R to increase the voltage difference across it.

PW

The next step, which you omitted, is to then ADD the two channels together.  You will now have just one trace displaying the voltage difference between the two ends of the 10R resistor.

Nope-this scope cannot do that unless the math function is used,and then the math trace is in with the other two traces. CH1 + CH2 is only available in the math function.

Using your digital scope, you should be able to use one of its math functions to directly perform the subtraction of one channel from the other.  However, being unfamiliar with your scope, I do not know if a difference trace will be displayed, which I would want to see (I am less interested in just "numbers in the boxes").

I can bring up the math trace-as in scope shots below. But the scope will not display the math trace value. This has been an issue than many have tried to solve for the past 3 year's,and to date,no one has been able to figure out how to get the math value displayed on the scope.


Brad

If you can display the math trace, forget about inverting one of the channels and just display a math trace of CH1-CH2...

Try to understand what it is you are trying to make your scope display, the voltage difference between the ends of the 10R.  If you cannot get better vertical display resolution of the math channel, increase the 10R to 100R if necessary to get more difference signal.



ADDED:  I don't recall that harsh step in the CH2 waveform being present in your previous attempt.



PW

Why do these traces look so much different than your previous traces made across the 10R?  Is this the same test condition as in your earlier attempt?

No-this is a different JT i was testing at the time of your request,so rather than swap back to the other JT,i simply did the same thing on this JT to give you examples of what i can and cannot do with the scope as quickly as i could.

Also, can the CH1 and CH2 displays be turned off leaving only the math trace?

No they cannot-the math trace switches off as soon as you switch of one channel.

Brad