How is Thermal voltage derived?

The Thermal voltage equation is Vt=kT/q
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Does anyone know how to derive the Thermal voltage equation?
Thanks,
Paul
Reply to
Paul
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| The Thermal voltage equation is Vt=kT/q | |
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| | Does anyone know how to derive the Thermal voltage equation? | | Thanks, | Paul
Read the first line: In semiconductors, the relationship between the flow of electrical current and the electrostatic potential across a p-n junction depends on a characteristic voltage called the thermal voltage.
Simplistically, Ohm's law will tell you that there must be a characteristic resisitance since you have a voltage difference (electrostatic potential) and a current through the junction, and hence it is not a perfect switch but a resistance and any resistor will radiate energy as heat. Being a semiconductor, this "resistance" is variable but it is quite real and your processor, having a lot of them, needs a fan to cool it. It is a "characteristic" thermal output because of the switching taking place, the voltage across the junction being greater when less current is passed as with any normal switch. In the perfect switch the resistance is zero and the voltage is across it zero when closed, the current having an undefined value controlled elsewhere, and when open the current is zero and the voltage undefined. Semiconductors are not perfect switches. Hope that helps.
Reply to
Androcles
I've read dozens of technical books on diode and semiconductor science, but have yet to find any details. Obviously the thermal voltage, which is ~ 25mV at 300K, is related to thermal fluctuations. Thermal voltage may be similar to kTC noise, if not directly related. kTC noise is Vn=sqrt(kT/C). Resistance term is irrelevant in kTC noise since the bandwidth is 1/(4RC), and therefore R cancels out in the Johnson noise voltage equation. As mentioned, the thermal voltage equation is Vt=kT/q, or in terms of kTC noise we would convert q to capacitance, which is ~ 5.6aF. If thermal voltage is kTC noise then why would the capacitance remain constant (~ 5.6aF @300K) for a given temperature regardless of the diode dimensions and size?
Regards, Paul
Reply to
Paul
A little more detail on the subject:
The diode IS term (saturation current) is can be derived by Vt/Rzb, where Vt is thermal voltage (~25mV @300K) and Rzb is the diodes zero bias resistance. IS term is the alternating current rms at thermal equilibrium (zero bias). So if the diodes zero bias resistance is 1Mohm, then IS = (kT/q)/1Mohm. Similarly, one could derive the thermal voltage as Rzb*IS, but I am searching for further detail.
If possible I would appreciate some forum recommendations that might have some active physicists and/or EE's able to answer such questions, who may even speak quantumese.
Regards, Paul
Reply to
Paul
| > | The Thermal voltage equation is Vt=kT/q | > | | > |
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| > | | > | Does anyone know how to derive the Thermal voltage equation? | > | | > | Thanks, | > | Paul | > | > Read the first line: | > In semiconductors, the relationship between the flow of electrical current | > and the electrostatic potential across a p-n junction depends on a | > characteristic voltage called the thermal voltage. | > | > Simplistically, Ohm's law will tell you that there must be a characteristic | > resisitance since | > you have a voltage difference (electrostatic potential) and a current | > through the junction, | > and hence it is not a perfect switch but a resistance and any resistor will | > radiate energy as heat. | > Being a semiconductor, this "resistance" is variable but it is quite real | > and your | > processor, having a lot of them, needs a fan to cool it. It is a | > "characteristic" | > thermal output because of the switching taking place, the voltage across the | > junction being greater when less current is passed as with any normal | > switch. | > In the perfect switch the resistance is zero and the voltage is across it | > zero when | > closed, the current having an undefined value controlled elsewhere, and when | > open the current is zero and the voltage undefined. Semiconductors are not | > perfect switches. | > Hope that helps. | | | I've read dozens of technical books on diode and semiconductor | science, but have yet to find any details. Obviously the thermal | voltage, which is ~ 25mV at 300K, is related to thermal fluctuations. | Thermal voltage may be similar to kTC noise, if not directly related. | kTC noise is Vn=sqrt(kT/C). Resistance term is irrelevant in kTC noise | since the bandwidth is 1/(4RC), and therefore R cancels out in the | Johnson noise voltage equation. As mentioned, the thermal voltage | equation is Vt=kT/q, or in terms of kTC noise we would convert q to | capacitance, which is ~ 5.6aF. If thermal voltage is kTC noise then | why would the capacitance remain constant (~ 5.6aF @300K) for a given | temperature regardless of the diode dimensions and size? | | Regards, | Paul
Capacitance is a function of plate surface area and separation. Those remain constant. You seem to be struggling with DC superimposed on AC. The Kirchhoff model here is a capacitor in parallel with a resistor.
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Reply to
Androcles
I understand basic concepts such as Kirchhoff model of plates, capacitance being a function of area and separation, and actually there are other parameters and hence it is not always a constant; e.g., diode capacitance. Also I know how to superimpose DC over AC. Perhaps my question was a bit too technical here.
Regards, Paul
Reply to
Paul
| > | | > | > | The Thermal voltage equation is Vt=kT/q | > | > | | > | > | > |
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| > | > | | > | > | Does anyone know how to derive the Thermal voltage equation? | > | > | | > | > | Thanks, | > | > | Paul | > | > | > | > Read the first line: | > | > In semiconductors, the relationship between the flow of electrical | > current | > | > and the electrostatic potential across a p-n junction depends on a | > | > characteristic voltage called the thermal voltage. | > | > | > | > Simplistically, Ohm's law will tell you that there must be a | > characteristic | > | > resisitance since | > | > you have a voltage difference (electrostatic potential) and a current | > | > through the junction, | > | > and hence it is not a perfect switch but a resistance and any resistor | > will | > | > radiate energy as heat. | > | > Being a semiconductor, this "resistance" is variable but it is quite | > real | > | > and your | > | > processor, having a lot of them, needs a fan to cool it. It is a | > | > "characteristic" | > | > thermal output because of the switching taking place, the voltage across | > the | > | > junction being greater when less current is passed as with any normal | > | > switch. | > | > In the perfect switch the resistance is zero and the voltage is across | > it | > | > zero when | > | > closed, the current having an undefined value controlled elsewhere, and | > when | > | > open the current is zero and the voltage undefined. Semiconductors are | > not | > | > perfect switches. | > | > Hope that helps. | > | | > | | > | I've read dozens of technical books on diode and semiconductor | > | science, but have yet to find any details. Obviously the thermal | > | voltage, which is ~ 25mV at 300K, is related to thermal fluctuations. | > | Thermal voltage may be similar to kTC noise, if not directly related. | > | kTC noise is Vn=sqrt(kT/C). Resistance term is irrelevant in kTC noise | > | since the bandwidth is 1/(4RC), and therefore R cancels out in the | > | Johnson noise voltage equation. As mentioned, the thermal voltage | > | equation is Vt=kT/q, or in terms of kTC noise we would convert q to | > | capacitance, which is ~ 5.6aF. If thermal voltage is kTC noise then | > | why would the capacitance remain constant (~ 5.6aF @300K) for a given | > | temperature regardless of the diode dimensions and size? | > | | > | Regards, | > | Paul | > | > Capacitance is a function of plate surface area and separation. Those | > remain constant. You seem to be struggling with DC superimposed | > on AC. The Kirchhoff model here is a capacitor in parallel with a resistor. | >
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| | | I understand basic concepts such as Kirchhoff model of plates, | capacitance being a function of area and separation, and actually | there are other parameters and hence it is not always a constant; | e.g., diode capacitance. Also I know how to superimpose DC over AC. | Perhaps my question was a bit too technical here.
Perhaps your understanding of basic principles is a bit less technical here.
| | Regards, | Paul
Reply to
Androcles
Sorry, I'm here to discuss science. Understandably some people are here so time will fly by at their jobs. I'm not here to pick fights. If you want attempt to derive the Thermal Voltage equation then great.
Anyhow, one consideration is the limitation of thermal noise bandwidth, which begins to fall significantly near 1THz.
Regards, Paul
Reply to
Paul
| > | | > | > | | > | > | > | The Thermal voltage equation is Vt=kT/q | > | > | > | | > | > | > | > | > | > |
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| > | > | > | | > | > | > | Does anyone know how to derive the Thermal voltage equation? | > | > | > | | > | > | > | Thanks, | > | > | > | Paul | > | > | > | > | > | > Read the first line: | > | > | > In semiconductors, the relationship between the flow of electrical | > | > current | > | > | > and the electrostatic potential across a p-n junction depends on a | > | > | > characteristic voltage called the thermal voltage. | > | > | > | > | > | > Simplistically, Ohm's law will tell you that there must be a | > | > characteristic | > | > | > resisitance since | > | > | > you have a voltage difference (electrostatic potential) and a | > current | > | > | > through the junction, | > | > | > and hence it is not a perfect switch but a resistance and any | > resistor | > | > will | > | > | > radiate energy as heat. | > | > | > Being a semiconductor, this "resistance" is variable but it is quite | > | > real | > | > | > and your | > | > | > processor, having a lot of them, needs a fan to cool it. It is a | > | > | > "characteristic" | > | > | > thermal output because of the switching taking place, the voltage | > across | > | > the | > | > | > junction being greater when less current is passed as with any | > normal | > | > | > switch. | > | > | > In the perfect switch the resistance is zero and the voltage is | > across | > | > it | > | > | > zero when | > | > | > closed, the current having an undefined value controlled elsewhere, | > and | > | > when | > | > | > open the current is zero and the voltage undefined. Semiconductors | > are | > | > not | > | > | > perfect switches. | > | > | > Hope that helps. | > | > | | > | > | | > | > | I've read dozens of technical books on diode and semiconductor | > | > | science, but have yet to find any details. Obviously the thermal | > | > | voltage, which is ~ 25mV at 300K, is related to thermal fluctuations. | > | > | Thermal voltage may be similar to kTC noise, if not directly related. | > | > | kTC noise is Vn=sqrt(kT/C). Resistance term is irrelevant in kTC noise | > | > | since the bandwidth is 1/(4RC), and therefore R cancels out in the | > | > | Johnson noise voltage equation. As mentioned, the thermal voltage | > | > | equation is Vt=kT/q, or in terms of kTC noise we would convert q to | > | > | capacitance, which is ~ 5.6aF. If thermal voltage is kTC noise then | > | > | why would the capacitance remain constant (~ 5.6aF @300K) for a given | > | > | temperature regardless of the diode dimensions and size? | > | > | | > | > | Regards, | > | > | Paul | > | > | > | > Capacitance is a function of plate surface area and separation. Those | > | > remain constant. You seem to be struggling with DC superimposed | > | > on AC. The Kirchhoff model here is a capacitor in parallel with a | > resistor. | > | >
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| > | | > | | > | I understand basic concepts such as Kirchhoff model of plates, | > | capacitance being a function of area and separation, and actually | > | there are other parameters and hence it is not always a constant; | > | e.g., diode capacitance. Also I know how to superimpose DC over AC. | > | Perhaps my question was a bit too technical here. | > | > Perhaps your understanding of basic principles is a bit less technical here. | | | Sorry, I'm here to discuss science.
Science is the observation, investigation and explanaton of natural phenomena; there is no way a semiconductor can be considered natural. Perhaps your misunderstanding of science is what the rest of the world calls technology.
| Understandably some people are | here so time will fly by at their jobs.
Perhaps some people are retired.
| I'm not here to pick fights.
Then don't start one when others are trying to help you understand the question you asked.
| If you want attempt to derive the Thermal Voltage equation then great.
I have no great desire to compute direct current * resistance or the heat radiated from a reactive capacitor.
| Anyhow, one consideration is the limitation of thermal noise | bandwidth, which begins to fall significantly near 1THz.
I'd be interested to see a semiconductor operating at that frequency.
Reply to
Androcles
I have no idea what you're referring to.
Vt (thermal voltage) is not derived from DC.
SD1T7 made by Virginia Diodes Inc. It's a GaAs THz diode. Would you like another part #?
Reply to
Paul
Nor apparently that you seem to be expecting rational information from a raving lunatic.
Reply to
jimp
Yes, this sure isn't the forum although there are people here who know the answer. I'd suggest a trip to sci.electronic.design. Any decent book on solid state physics and solid state device physics will do the job. Note Andro is a moron. Ohm's law does NOT apply to PN junctions. Although statistical distributions of "carriers" are a sort of half-baked theory in regard to these devices, it does produce a reasonable working "understanding" of what is going on, if not a complete and true theory.
PS. alt.electrical.engineering is more a forum to ask about the fine points of pole pig transformers than solid state device physics.
Reply to
Benj
Thanks for the info, and I'll check out sci.electronic.design.
Regards, Paul
Reply to
Paul
I'm not sure of a precise detailed derivation. But you might need to ask yourself how much work can be done on a single electron by a medium at an absolute temperature T.
Reply to
Igor
Hmm, good question. ... I'll think about it.
Regards, Paul
Reply to
Paul
| > | > > The Thermal voltage equation is Vt=kT/q | > | >
| > > Does anyone know how to derive the Thermal voltage equation? | > | > > Thanks, | > > Paul | > | > I'm not sure of a precise detailed derivation. But you might need to | > ask yourself how much work can be done on a single electron by a | > medium at an absolute temperature T. | | | Hmm, good question. ... I'll think about it.
Any such work will result in kicking said electron into a higher energy state from which it will fall to a lower energy state and emit a photon, or be ionized fronm the parent molecule. From Planck's E = h\nu, the the amount of work that can be done on a single electron is thus precisely known. How that helps with absolute temperature you should ask Baron Victor von Frankenstein's personal assistant, Igor, since it was his suggestion, although he isn't sure of anything.
Reply to
Androcles
In article , Paul wrote:
It is actually quite simple in terms of statistical mechanics. Temperature is defined as the average energy in a degree of freedom. That is
E=kT
where k, Boltzmann's constant, relates average energy to temperature temperature. Dividing by q expresses this average energy in electron-volts.
Calling this a derivation makes it sound much more complicated than it really is.
Bill
Reply to
Salmon Egg
That looks like the answer. I appreciate the help! According to WikiPedia,
+++ Boltzmann's constant k is a bridge between macroscopic and microscopic physics. +++
And later on it says,
+++ Given a thermodynamic system at an absolute temperature T, the thermal energy carried by each microscopic "degree of freedom" in the system is on the order of magnitude of kT/2 (i.e., about 2.07*10-21 J, or 0.013 eV at room temperature). +++
So then kT/q (Vt) express the electromotive force (voltage) per degree of free for each electron? I still don't understand why Vt is constant (for a given temperature) for all diodes. For example, if the diodes junction width is twice as long then shouldn't there be sqrt(2) times more noise voltage? That is, two noise voltage sources in series increases by the sqrt(2).
Regards, Paul
----------- Special thanks to Androcles, Benj, Igor, Jim Pennino, Roy, and Salmon Egg
Reply to
Paul

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