Strange voltage measurements - can anyone explain?

[PhysicsProf]

Consider the "reverse-JT" circuit that we have developed -- schematic attached.

The resistor to the base is 2kohms, L1 and L2 are approx 80 uH (I wound a new toroid like the old one, will measure L's precisely soon).  MPS2222 transistor. Cap is 150 pF.
Vbatt is 2.48 volts, from 2 rechargeable AA's.
NOTE: I have replaced the 1 ohm CSR2 with 50.6 Kohms, and the red LED still lights! (though dimly)

I have been puzzling over voltage measurements I've taken on the "output leg" of this circuit -- let's start with measurement of V3, from P21G-to-V3.

V3, from P21G-to-V3 = 0.00 mV  (auto-ranging DMM) 

Now, the voltage drops across the 50.6 Kohm resistor and LED.   To be clear, I attached the DMM such that both voltages will read as POSITIVE voltages, and this means the black probe is connected to the point P22T and the red probe is connected to P21G .

Then the voltage across the 50.6 Kohm Resistor is +5.1 volts.

-- that is correct, I have not made a mistake.  I cannot explain why the voltage is POSITIVE and so large relative to Vbatt and request efforts to explain.

Next puzzle, with the black probe on point P22T and the red probe of the DMM on point V3,

the voltage across the LED is +0.6V and it is dimly (but very visibly) lit up, glowing red.

I checked the direction of the LED -- it is correct, as shown in the attached schematic.  I applied a voltage of 2.48V with negative on point P22T (where the black probe of the DMM was connected) and positive on point P21T (where the red DMM probe was connected) -- and the LED did NOT light.  Yet it lights up in this circuit.  EXPLAIN THAT, IF YOU CAN.  I would really appreciate a valid explanation.

(The LED lit up when the battery connections were reversed.)


Next, I replaced the 50.6 Kohm resistor with a 468ohm resistor, and these are the voltages -- with the SAME probe connections as described above:


Then the voltage across the 468 ohm Resistor is +1.03 volts.

The voltage across the LED is +0.90 volts and it is brightly lit up, glowing red.

The voltage from P21G (black probe now) to P21T, that is, V3, is:  V3= 0.03mV (nearly zero, as before).

I'm puzzling over this and request your advice/explanations, all.

[Room3327]

Hi Prof,
I'll take a stab at this if you don't mind.
With a .6 volt battery, when the transistor is on almost all of the battery voltage is dropped across the transistor and there is very little if any across the inductor this is your measurement of V3.  That puts almost no voltage across the LED and CSR, leaving you a measurement of almost nothing. when the transistor turns off the inductor now discharges through the LED and CSR. The BEMF now tries to keep the current flowing in the same direction and discharges into the LED and CSR. The larger CSR is the lower the current and higher the voltage will be giving you the reading of 5.1 volts across the CSR and .6 volts across the Led which work on current not voltage. At .6 v across the Led there will be little current flowing through it and it will be dim.  Making the CSR 468 ohms now increased the current through the Led and CSR, more current through the LED causes a larger voltage drop across it and a brighter LED, giving you your voltages of 1.03 and .9.
Of course I could be wrong.
Room

[PhysicsProf]

Hi Prof,
I'll take a stab at this if you don't mind.
With a .6 volt battery, when the transistor is on almost all of the battery voltage is dropped across the transistor and there is very little if any across the inductor this is your measurement of V3.  That puts almost no voltage across the LED and CSR, leaving you a measurement of almost nothing. when the transistor turns off the inductor now discharges through the LED and CSR. The BEMF now tries to keep the current flowing in the same direction and discharges into the LED and CSR. The larger CSR is the lower the current and higher the voltage will be giving you the reading of 5.1 volts across the CSR and .6 volts across the Led which work on current not voltage. At .6 v across the Led there will be little current flowing through it and it will be dim.  Making the CSR 468 ohms now increased the current through the Led and CSR, more current through the LED causes a larger voltage drop across it and a brighter LED, giving you your voltages of 1.03 and .9.
Of course I could be wrong.
Room

As I said, the batteries' voltage is 2.48 Volts ( not 0.6 volts).  Further, 0.6 volts should not be enough to light up a red LED...

Your explanation of the increased voltage drop across the LED from 0.6 to 0.9 V as the CSR is reduced from 50.6 Kilo-ohms to 468 ohms is reasonable QUALITATIVELY, and I realized this before, but a QUANTITATIVE  explanation is lacking -- and that is what is important!

Also, you missed the point about the SIGN or DIRECTION of the voltage on the LED and the CSR, which are reversed from each other -- THAT reversal I would really like to see explained!  

Correspondingly, the NET voltage across the LED and CSR is zero -- which suggests to me that the currents in the two elements are OUT OF PHASE.  Right?  But how can that be -- these two are connected in SERIES and the LED presumably allows current in only ONE direction.

How do you get 0.6V in one direction from the LED-CSR junction, across the LED, while the CSR voltage drop is 5.1volts in the OPPOSITE direction -- with the further constraint that the TOTAL voltage across the LED-CSR together is ZERO?     Inquiring minds want to know...    I am certain that no simulation will give this OBSERVED result.

[Room3327]

Sorry Prof I missed your 2.48 volts, I was looking at the schematic.

[Allcanadian]

@PhysicsProf
The first thing we should remember is that voltage is relative and based solely on differential charge density so the first thing I do is designate the inductor polarity. L1 has a polarity the same as the source when charging and it is marked as such(+/-) but on discharge the voltage polarity reverses so I designate this reversal with this term (LD -/+)--Inductive discharge. Now once the inductance discharges the forward current with a reversed polarity will see the LED as a low resistance as it has crossed the bias theshhold of the LED and then it will see the resistor. The higest charge density with a negative sign may now fall between the LED and the resistor giving the appearance that the voltage has reversed in one of the elements. We may see (+)LED(-)resistor(+) because the charge density peaks between the two and what we measure as voltage is a relative measure of the difference in charge density across each element. You see this is why I do not rely solely on the standard equations and use electron flow notation exclusively, I also use fundamental physics which does not recognize Current nor Voltage as "something" but simply a measure of something. As such Voltage and Current mean very little to me and I would rather understand what the various forms of energy are actually doing in reality rather than what the measure of them is telling me.
You have to keep your eye on the ball and not the shadow of it because it's shadow is only a distorted reflection of the true reality of the situation.
Regards
AC

[PhysicsProf]

 I'm trying to understand your comment, AC, the gist of it seems to be:

We may see (+)LED(-)resistor(+) because the charge density peaks between the two and what we measure as voltage is a relative measure of the difference in charge density across each element.

Yes,  (+)LED(-)resistor(+)  is a decent summary of the observations.   But why does the LED light up under this condition?  Using a battery, the LED will only light up with the opposite condition, namely: (-)LED(+) (with minus at the P21T point as shown in the schematic).


And when you say "the charge density peaks between the two" -- just how does this occur?  an accumulation of electrons at this point??

Also, I like your approach of energy considerations -- I often use this.  So let me ask, from your model -- say I want to gather energy by putting a capacitor across the 50Kohm resistor.  Would this work, and to which point should I attach the negative terminal of the electrolytic cap?  What voltage will it reach?  (PS -- I've actually done this experiment, but wonder what your model will say.)

[BEP]

PhysicsProf,

If your LED is of the OLED variety it may glow simply because of the RF received with the LED leads. If this is true then voltage, current and current direction are almost immaterial. Also, you would be able to use another LED, of the same type, remove the leads, and hold the crippled LED near the signal source. If close enough, it will still illuminate.

All some OLEDs require is to be immersed in a strong electric field within a changing magnetic field.

http://www.zurich.ibm.com/st/spintronics/oleds.html

[BEP]

Before I forget....

If your LED is working as I suggest, there may be a noticeable color shift from normal color expected.

[PhysicsProf]

  OK - finally got my DSO probing things.  Very interesting. 

The voltage drop across the LED+CSR  has a mean value of close to Zero volts -- BUT, the oscillations are wild, swinging from +6 V to -6 V (approx) so that the Vpp is 12.7 V!  f= 1.4 MHz.

Surprising that averages out so close to zero.

Across the LED, Vpp is 3.44 volts, with the LED lighting up when the voltage is in the correct direction (no doubt).  Mean about 0.3 V per the scope.

Across the 50.6Kohm, Vpp is just about 5 V, while the mean V is about 3.2V (not quite in agreement with the DVM).

Most of the mysteries solved (IMO)...  by the strong AC component in the output leg of the circuit.

Sorry, busy here -- I should have pulled out the scope sooner.

[GibbsHelmholtz]

 
Across the 50.6Kohm, Vpp is just about 5 V, while the mean V is about 3.2V (not quite in agreement with the DVM).

Assuming that the diode is going a decent job and Vpp is when the LED lit up brightest.

5V/50Kohm = 10 microamp ?  makes me wonder what actually lights the LED, voltage or current.

[MileHigh]

Assuming that the diode is going a decent job and Vpp is when the LED lit up brightest.

5V/50Kohm = 10 microamp ?  makes me wonder what actually lights the LED, voltage or current.

This statement is a misnomer that's worth commenting on.

The LED is always lit by a combination of voltage and current.

If you have a voltage source, the voltage induces the flow of current.

If you have a current source, the current flow induces a voltage.

So if you are looking at a LED from a bird's-eye-view, you don't know if the power source lighting the LED is a voltage source or a current source.  It doesn't really matter, because either way you will be able to measure a voltage across, and a current through, the LED.

Also, voltage and current sources can mimic each other, you can look up "thevenin and norton equivalent circuits."

[ion]

http://en.wikipedia.org/wiki/Buck-boost_converter

Please study this stuff, it will clear up the confusion.

[BEP]

I take it no one here has any experience with LED phosphor conversion or phosphorescence of many LED types under high electric fields?

It seems to be clear the good Prof just needed to use the appropriate tool to measure the energy used by his LED but I'm still a bit surprised nobody asked what I was talking about.

This effect is seen more often now that improvements on the so-called white LEDs are more mature. It can be seen on many colors if the device is of recent manufacture.

This is why I haven't shown much interest in the JT and Dr. Stiffler's related work.

[poynt99]

Thanks WW.

I did mention the possibility of exploring the benefits of this effect some time ago, and no one then responded either.

Is there perhaps a benefit in terms of light to power ratio?

.99

[GibbsHelmholtz]

So if you are looking at a LED from a bird's-eye-view, you don't know if the power source lighting the LED is a voltage source or a current source.  It doesn't really matter, because either way you will be able to measure a voltage across, and a current through, the LED.

So you're saying a bird's-eye-view can not tell if the LED is 3V and 20mA or the LED is 3V and 10 microA? Seems like we can engineer it to 3V with 0 amp.  Light is energy you know.   

[PhysicsProf]

  What I've done today is to replace the LED with a diode -- still work great per the oscilloscope.

-->  Has anyone looked into the "Gabriel device"?  as discussed at OU by Feynman et al.   

Say, where is Feynman???  I miss his posts.

[BEP]

Thanks WW.

I did mention the possibility of exploring the benefits of this effect some time ago, and no one then responded either.

Is there perhaps a benefit in terms of light to power ratio?

.99

So far, my best understanding is there can be less energy used but not zero energy. One of the newer film type displays uses the same idea of a CRT only multiple CRTs. The energy usage is more like a leaking electrolytic capacitor in the tens of micro amp range. The phosphor usage enhances the amount of visible light or shifts the visible light.

[MileHigh]

So you're saying a bird's-eye-view can not tell if the LED is 3V and 20mA or the LED is 3V and 10 microA? Seems like we can engineer it to 3V with 0 amp.  Light is energy you know.   

No I was stating that you can measure voltage and current and record those two numbers.  That's voltage and current, not one or the other, if you are measuring power.  You can't produce power with voltage but zero amps or with amperage but zero volts.

Yes light is a manifestation of power, and in that sense it's the product of intensity times the amount of light flux.  Note again it's the multiplication of two variables.

So voltage times current gets transformed into light intensity times light flux, plus heat.

[GibbsHelmholtz]

So voltage times current gets transformed into light intensity times light flux, plus heat.

Okay, so let me try again to understand your thinking.  Let me write your statement with an equation.

I*V = V^2/R or I^2R = Light power + thermo power

I have an example below in the drawing.  The voltage is measured across the bulb with resistance R, DC steady state, so no need for complicated scope. The power measure then is IV or V^2/R or I^2R.  If we can measure all the light going out of the bulb and all the heat going out of the bulb, this should add up to the equation. 

Do we have an agreement here? Afterall, this is what energy conservation is about.

[exnihiloest]

...
The power measure then is IV or V^2/R or I^2R.  If we can measure all the light going out of the bulb and all the heat going out of the bulb, this should add up to the equation. 
Do we have an agreement here? Afterall, this is what energy conservation is about.

I also agree. In fact heat is infrared radiation not different by nature from visible light. They are electromagnetic waves whose only the ranges of frequencies are different.

In any case almost all the electrical energy used in a house ends in heat: washing machine, TV, dish washer, and even light (which is absorbed by the walls as heat, except for the part escaping by the windows). We could say that energy is used twice, for example if you watch TV in winter, you heat the room with the total consumed power in the same time. This is only because devices like TV have a near zero efficiency for producing the useful work: generating pictures whose the weak energy should be only their light, light which is in any case also converted in heat when it encounters matter. In fact the total and only power is still IV.

[GibbsHelmholtz]

exnihiloest 

Atleast you stick up for what you believed.  If it's wrong, so be it, right?  You have alot of potential.

[MileHigh]

Yes we have an agreement.  Following on Exnihiloest's point about the fact that heat is still EM radiation, you can define a frequency were you say above that frequency it's light and below that frequency it's heat.

[GibbsHelmholtz]

Good

I guess it's fair to share what I believed.  I believe that heat and radiation are two separate thing.  If one considers heat is momentum, then light is kinetic energy or the rate of momentum change.  Of course this concludes the equation as

light + heat = V^2/R + I^2R

We'll keep this for the record.  Only an experiment can concludes our views.  Thank you all. 

[MileHigh]

Radiated heat and light are the same thing.  What is considered "light" is really just a question of the wavelength sensitivity curve for the human retina.  Different species have different wavelength sensitivity curves associated with their sense of vision.

light + heat = V^2/R + I^2R

I am assuming a typo and you meant to say "light + heat = V^2/R OR I^2R.

[GibbsHelmholtz]

Radiated heat and light are the same thing.  What is considered "light" is really just a question of the wavelength sensitivity curve for the human retina.  Different species have different wavelength sensitivity curves associated with their sense of vision.

I am assuming a typo and you meant to say "light + heat = V^2/R OR I^2R.

radiated heat is infrared, heat is heat.  That is what I think.  Thank you for your kindness MH, it's not a typo.