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.
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
Quote from: Room3327 on 2011.04.28, 17:06:42
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.
Sorry Prof I missed your 2.48 volts, I was looking at the schematic.
@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
I'm trying to understand your comment, AC, the gist of it seems to be:
QuoteWe 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.)
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
Before I forget....
If your LED is working as I suggest, there may be a noticeable color shift from normal color expected.
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.
Quote from: PhysicsProf on 2011.04.28, 23:06:35
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.
Quote from: GibbsHelmholtz on 2011.04.29, 00:36:20
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."
http://en.wikipedia.org/wiki/Buck-boost_converter
Please study this stuff, it will clear up the confusion.
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.
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
Quote from: MileHigh on 2011.04.29, 03:02:07
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.
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.
Quote from: poynt99 on 2011.04.29, 23:12:14
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.
Quote from: GibbsHelmholtz on 2011.04.30, 00:51:40
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.
Quote from: MileHigh on 2011.04.30, 04:08:03
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 powerI 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. >:-)
Quote from: GibbsHelmholtz on 2011.05.01, 00:36:51
...
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.
exnihiloest O0
Atleast you stick up for what you believed. If it's wrong, so be it, right? You have alot of potential.
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.
(http://www.sciencebuddies.org/mentoring/project_ideas/Phys_img039.jpg)
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.
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.
Quotelight + heat = V^2/R + I^2R
I am assuming a typo and you meant to say "light + heat = V^2/R OR I^2R.
Quote from: MileHigh on 2011.05.01, 18:36:03
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.
Okay so if it is not a typo then it is wrong. Your theories about the two different energy forms translating into adding V^2/R and I^2R together to account for the two energy forms needs another look on your part.
V^2/R and I^2R are a way of expressing power. The root in both cases is P = VI and I = V/R. It's nothing more than an exercise in algebra to show the equivalency of V^2/R and I^2R. There is nothing separate and distinct about either term. It goes right back to me saying that there is no power in a voltage-only or a current-only setup.
MileHigh
(corrected my typo)
Quote from: GibbsHelmholtz on 2011.05.01, 16:21:57
We'll keep this for the record. Only an experiment can concludes our views. Thank you all.
;D
@GibbsHelmholtz
QuoteI 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
Hmm, I think I like this and of course I would agree heat and radiation are two separate things.
@All
An object at 20 Degrees C in an environment at the same temperature does not radiate anything but still has heat thus it should be obvious to even a child that heat and radiation are not the same thing. As well when we consider heat we should look to nature for answers, for instance supposedly the Sun is hot and this heat from the Sun is transferred to our little planet--- so why is all the space between the Earth and Sun so damn cold?(−270 °C). This is because people confuse things with measures of things, heat is not something it is a measure of something and this measure is relative. The Sun cannot conduct heat when there is nothing to conduct through such as the near perfect vacuum of space therefore the sun transfers energy in the form of radiation which has no "heat" as heat is solely a property of matter not empty space which is why space is cold.
As well we could apply this same thought to Voltage and Current and personally I find it absolutely mindboggling how some people refer to voltage and current as if they were something tangible. Voltage is the difference in charge density between two regions, Voltage is a measure of the difference and a measure of something is not tangible ---- it is a measure. Then we have Current and people tell me there is current flowing in their wires :D, well no there is no current flowing as an electrical current is a measure of the motion of the charges in, on and around the wires and it is the charges that are moving not a current. Current is a measure of the motion of charges and measures are not tangible things, the charges are tangible and can be measured in a relative sense. It is no wonder that nobody can seem to understand anything when most textbooks would have us to believe in things that have no substance to them.
As well I would question the level of understanding of anyone who tells us "things are the same" or "things are not different" when it should be obvious that if they have different names and different properties then they are different. Generalizing or marginalizing everything is not a very smart proposition in my opinion because I have found the devil is in the details, it is having an intimate understanding of things to such an extent that we understand the limitations in our understanding of things and then we learn new things.
Regards
AC
Quote from: GibbsHelmholtz on 2011.05.01, 13:05:09
...
Atleast you stick up for what you believed. If it's wrong, so be it, right? You have alot of potential.
I would like to be wrong. This would open fantastic gates. Unfortunately not one experiment until now can dismiss what I wrote. I deal with the physical world as it is and not as some dream it. Welcome in the real life :)
Quote from: GibbsHelmholtz on 2011.05.01, 16:21:57
...
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
...
This equation is outside of any science.
Heat is kinetic energy of particles. Acceleration of electrons radiates EM waves and inversely. The two phenomena are linked at the matter/free space interface and they exchange energy. In absence of current in a conductor, the mean kinetic energy of particles is in equilibrium with the ambient EM field: such a body radiates and absorbs the same quantity of EM energy, maintaining a certain amount of kinetic energy of particles inside. At room temperature, the EM field from all bodies have frequencies in the infrared region.
A current is a flow of electrons that collide the atomic lattice of the conductor, increasing the kinetic energy of particles therefore increasing the heat. This extra kinetic energy is equal only to
R*I2 and provides extra EM waves radiated by the conductor. Now the conductor radiates more than it absorbs from the environment.
The stronger the current, the stronger the heat, the higher the frequencies of the radiated EM waves. If the heat is strong enough, some EM waves enter the visible spectrum (light) while most of them stay in the infrared region (see Milehigh's post #21).
Heat is both kinetic energy and light (visible or not visible), depending on what you are looking at, matter or free space. It is an exchange of the same energy which must not be accounted twice.
AC and Gibbs:
Not sure if the following comment is about I^2R vs. V^2/R but let's have a look:
QuoteAs well I would question the level of understanding of anyone who tells us "things are the same" or "things are not different" when it should be obvious that if they have different names and different properties then they are different. Generalizing or marginalizing everything is not a very smart proposition in my opinion because I have found the devil is in the details, it is having an intimate understanding of things to such an extent that we understand the limitations in our understanding of things and then we learn new things.
P = VI, and I = V/R
Therefore:
P = V(V/R) = V^2/R.
P = VI and V = IR
Therefore:
P = (IR)I = I^2R.
So that means if you know the electrical resistance of a resistor, there are two ways to determine the power dissipated in the resistor. You can measure the voltage across the resistor to determine the power, or just as easily, you can measure the current through the resistor to determine the power.
This has nothing to do with "kinetic energy" vs. whatever. So any notion that you add V^2/R with I^2R is wrong.
Recognizing the similarities between things is just as important as recognizing the differences between things. A keen eye can distinguish between things that are relevant in a circuit vs. things that can be ignored in a circuit.
@exnihiloest
QuoteHeat is kinetic energy of particles. Acceleration of electrons radiates EM waves and inversely. The two phenomena are linked at the matter/free space interface and they exchange energy. In absence of current in a conductor, the mean kinetic energy of particles is in equilibrium with the ambient EM field: such a body radiates and absorbs the same quantity of EM energy, maintaining a certain amount of kinetic energy of particles inside. At room temperature, the EM field from all bodies have frequencies in the infrared region.
My God Man what you are suggesting smacks of witchcraft :D, particles forever moving about absorbing and radiating energy from who knows where, it must be the devils work. I'm sure Milehigh and friends will straighten all this out because they seem to know for certain that all this perpetual motion stuff is nonsense and it is a well known fact that all energy systems must be enclosed in a perfectly isolated box.
Regards
AC
Quote from: allcanadian on 2011.05.03, 04:03:11
@exnihiloest My God Man what you are suggesting smacks of witchcraft :D, particles forever moving about absorbing and radiating energy from who knows where, it must be the devils work. I'm sure Milehigh and friends will straighten all this out because they seem to know for certain that all this perpetual motion stuff is nonsense and it is a well known fact that all energy systems must be enclosed in a perfectly isolated box.
This is all just a manifestation of solar power. No sun and eventually everything goes to absolute zero and no more energy exchanges.
Quote from: MileHigh on 2011.05.03, 02:27:19
AC and Gibbs:
Not sure if the following comment is about I^2R vs. V^2/R but let's have a look:
P = VI, and I = V/R
Therefore:
P = V(V/R) = V^2/R.
P = VI and V = IR
Therefore:
P = (IR)I = I^2R.
So that means if you know the electrical resistance of a resistor, there are two ways to determine the power dissipated in the resistor. You can measure the voltage across the resistor to determine the power, or just as easily, you can measure the current through the resistor to determine the power.
This has nothing to do with "kinetic energy" vs. whatever. So any notion that you add V^2/R with I^2R is wrong.
Recognizing the similarities between things is just as important as recognizing the differences between things. A keen eye can distinguish between things that are relevant in a circuit vs. things that can be ignored in a circuit.
You're talking strictly about Ohm's law. How can you explain an LED with 3V 20mA and the same LED with 3V 10 microA .
Quote from: GibbsHelmholtz on 2011.05.03, 04:25:00
You're talking strictly about Ohm's law. How can you explain an LED with 3V 20mA and the same LED with 3V 10 microA .
Can you elaborate on that? I am not sure what your point is and I need some context.
If one assumes that you are talking about DC values, then there is a problem. You can't have two different currents for the same applied voltage.
Quote from: allcanadian on 2011.05.03, 04:03:11
@exnihiloest My God Man what you are suggesting smacks of witchcraft :D, particles forever moving about absorbing and radiating energy from who knows where
...
Only you, do not know where! ;D
And humanity also invented the wheel, didn't you know?
If you learned physics and made you aware of the current human knowledge?
Quote from: MileHigh on 2011.05.03, 04:37:24
Can you elaborate on that? I am not sure what your point is and I need some context.
If one assumes that you are talking about DC values, then there is a problem. You can't have two different currents for the same applied voltage.
Let's say I agree that you are 99.9% right. Now what? Does it means we don't have to do the experiment? The scientific method is not form a hypothesis, if all agree with you, skip all steps and jump to conclusion. If you saying what you based on is the truth from the books and experiments, by all means, bring the experiment that match my hypothesis to my attention for evaluation. Otherwise, we are entering the area where only talk, no action.
@Milehigh
QuoteThis is all just a manifestation of solar power. No sun and eventually everything goes to absolute zero and no more energy exchanges.
I would agree, now if we apply the term "solar" to not only our sun but every sun which includes stars then we could simply use a general term many in the past such as Tesla and Moray have used such as "Cosmic Energy". Cosmic meaning "Of or relating to the universe or cosmos, esp. as distinct from the earth" which implies that Tesla and Moray believed that their devices utilized "Solar" and/or "Cosmic" energy which relates not only to our sun but every star.
Now if we believe this solar or cosmic energy is not perpetual then we would have to assume that at some point in the future we could look up in the night sky and see no sun and no stars, that is the universe as we know it would have to be completely devoid of all stars. Personally I find this hard to believe and the fact that science has found that even at absolute zero all matter still has a very small energy exchange with "something" would suggest that the only way this form of energy would cease is if the universe as we know it was empty. Personally I find this thought would require a leap of faith I could never justify as I have been given no real proof nor reason to believe that the universe as we know it will just end.
Regards
AC
Quote from: GibbsHelmholtz on 2011.05.03, 12:27:28
Let's say I agree that you are 99.9% right. Now what? Does it means we don't have to do the experiment? The scientific method is not form a hypothesis, if all agree with you, skip all steps and jump to conclusion. If you saying what you based on is the truth from the books and experiments, by all means, bring the experiment that match my hypothesis to my attention for evaluation. Otherwise, we are entering the area where only talk, no action.
I am not sure what you are trying to say. You described using the same LED with three volts across it and two separate cases with different currents through the LED. I said that was normally impossible and and asked you to explain more. I have to repeat the request.
Quote from: MileHigh on 2011.05.03, 21:17:41
I am not sure what you are trying to say. You described using the same LED with three volts across it and two separate cases with different currents through the LED. I said that was normally impossible and and asked you to explain more. I have to repeat the request.
I already told you what I think. You said it's wrong so ok. I just think it's more economical to peform the experiment than to type it out. I know it's the opposite for you but I'm different... I'm sorry.
Quote from: GibbsHelmholtz on 2011.05.04, 01:35:11
I already told you what I think. You said it's wrong so ok. I just think it's more economical to peform the experiment than to type it out. I know it's the opposite for you but I'm different... I'm sorry.
I'm pretty sure I did the experiment about 30 years ago! lol
Try studying this link:
http://www.utc.edu/Faculty/Tatiana-Allen/ivcurve.html (http://www.utc.edu/Faculty/Tatiana-Allen/ivcurve.html)
Quote from: MileHigh on 2011.05.04, 01:45:55
I'm pretty sure I did the experiment about 30 years ago! lol
Well, that's a big help. lol
Edit: Can you describe what you did briefly? How you measured it, set up? I know 30 years is a long time but I think you have photographic memory or atleast your CPU is pretty strong for 30 years later. :) but no worry, anything will help.
Quote from: GibbsHelmholtz on 2011.04.29, 00:36:20
...
5V/50Kohm = 10 microamp ?
May I suggest you to invest in something like that:
http://sjsu.bncollege.com/wcsstore/ExtendedSitesCatalogAssetStore/714_71401_99_5256_NI/images/FULLIMAGE_85346.jpg
:)
Quote from: exnihiloest on 2011.05.04, 06:42:39
May I suggest you to invest in something like that:
http://sjsu.bncollege.com/wcsstore/ExtendedSitesCatalogAssetStore/714_71401_99_5256_NI/images/FULLIMAGE_85346.jpg
:)
OMG!!! :o Gibbs made a mistake! This is awsome. Nice catch. :)