current operating vs voltage operating electronic devices

[kavan]

Hello Forum,
I have heard of current operating devices, like LEDs, and voltage operated devices...

What is the difference?

It seems to me that all electronic devices, once provided a certain voltage, draw the current they need....so what is this distinction between current operating and voltage operating?

thanks
kavan

 

[dougy83]

Hello Forum,
I have heard of current operating devices, like LEDs, and voltage operated devices...
What is the difference?

LEDs and other PN devices pass a current that is exponentially related to the applied voltage, as shown in the Shockley equation (). This means that for a very small change in voltage, there can be a very large change in current. The associated increase in power is nearly proportional to this increase in current.

The "change" mentioned above is important due to the fact that no two devices are the same, and will therefore have different operating points (which effectively changes the voltage in the equation).

For an LED, the large change in current results in a range of different brightness levels for different LEDs run from the same voltage; the large change in power may lead to its early demise due to overheating.

It seems to me that all electronic devices, once provided a certain voltage, draw the current they need....so what is this distinction between current operating and voltage operating?

Circuits are generally designed to accept a voltage and draw as much current as required, as you have stated. This is done by compensating for the aforementioned exponential behaviour.

 

[MikeMl]

Think about it in terms of dv/di, or the local slope of the V-I curve. There are really three types of devices; things that have a big change in current for a small change of voltage (LEDs, Diodes, transistors, regulated power supply) , things where the current is proportional to voltage (Ohmic Resistors), things where current remains constant while the voltage changes (current sink, current regulator).

 

[Ratchit]

kavan,

Hello Forum,
I have heard of current operating devices, like LEDs, and voltage operated devices...

What is the difference?

It seems to me that all electronic devices, once provided a certain voltage, draw the current they need....so what is this distinction between current operating and voltage operating?

You mean voltage or current operated devices. It means just what the phrase says, in that the voltage or current controls the device operation. For instance, a junction diode like a LED is a voltage controlled device. The voltage is controlling the current according to Schockley's equation. Specifically, the forward voltage is reducing the back-voltage caused by the diffusion of charge carriers across the junction barrier. This causes current to exist and the LED to shine. Now you can solve Schockley's equation to get voltage with respect to current, but that is not the way it physically works. It is the voltage that is making the diode work. The current is being controlled by the voltage and not the other way around.

A BJT is a voltage controlled current source (VCCS). The Vbe is controlling both the collector current (Ic) and the base current Ib in an exponential manner. Some folks think because the Ic is tracking the Ib, the BJT is a current controlled current source (CCCS). But it is the Vbe that is controlling both physically, due to Vbe reducing/increasing the back-voltage between the emitter and base junction, thereby modulating both Ic and Ib.

The device impedance has nothing to do with whether a device is a VCCS or a CCCS. Resistance is sometimes inserted in a circuit with the device in order to limit the current, so the circuit itself becomes a current or voltage source, but not the device. Again, it is the control that determines what the device is.

I only know of two current controlled devices, a gas discharge tube, and a magnetic amplifier.

Ratch

 

[Claude Abraham]

kavan,



You mean voltage or current operated devices. It means just what the phrase says, in that the voltage or current controls the device operation. For instance, a junction diode like a LED is a voltage controlled device. The voltage is controlling the current according to Schockley's equation. Specifically, the forward voltage is reducing the back-voltage caused by the diffusion of charge carriers across the junction barrier. This causes current to exist and the LED to shine. Now you can solve Schockley's equation to get voltage with respect to current, but that is not the way it physically works. It is the voltage that is making the diode work. The current is being controlled by the voltage and not the other way around.

A BJT is a voltage controlled current source (VCCS). The Vbe is controlling both the collector current (Ic) and the base current Ib in an exponential manner. Some folks think because the Ic is tracking the Ib, the BJT is a current controlled current source (CCCS). But it is the Vbe that is controlling both physically, due to Vbe reducing/increasing the back-voltage between the emitter and base junction, thereby modulating both Ic and Ib.

The device impedance has nothing to do with whether a device is a VCCS or a CCCS. Resistance is sometimes inserted in a circuit with the device in order to limit the current, so the circuit itself becomes a current or voltage source, but not the device. Again, it is the control that determines what the device is.

I only know of two current controlled devices, a gas discharge tube, and a magnetic amplifier.

Ratch

Countless threads on this issue have been generated so I wish to not start another one. Interested parties can research this matter on their own. Every OEM who produces LEDs & bjt's calls them "current controlled". Every OEM cannot be wrong. The Shockley diode equation is often introduced to the beginner in I vs. V exponential form as follows:

a) Id = Is*exp((Vd/Vt)-1), where Id = diode current, Vd = diode voltage, Vt = thermal voltage = nkT/q.

But said equation can also be expressed as follows:

b) Vd = Vt*ln((Id/Is)+1).

The b) form expresses voltage as a function of current, whereas a) expresses current as a function of voltage. In the forward direction, we do not force a constant voltage across LED terminals then allow the voltage to "control the current". Thermal runaway would occur incurring destruction of the LED. Instead we force a constant current through the LED and allow the voltage to be controlled by the current as well as temperature. LED devices are always current controlled. One cannpt control LED current with LED voltage.

Ratchit will argue this point to death, & I will not engage in a debate with anyone on this. Look up LED makers, bjt makers, & take their word for it. I & V are both needed for operation, but with LED & bjt devices, we must control current & V is incidental, controlled by current as well as temperature.

A gas discharge tube, & a magnetic amplifier are both indeed current controlled devices, but they are not the only such devices. An SCR gate to cathode junction is current controlled as are all p-n junctions in the forward mode per Shockley equation in form b). The form a) Shockley equation cannot be used in forward bias mode since Is increases grossly with temperature resulting in destruction. In the LED, Vd is controlled by Id & T.

I will elaborate if needed, but I will not engage in a war as to Id vs. Vd, which is in control, which is incidental. That issue is so well decided, it's akin to beating a dead horse. Cheers.

 

[MikeMl]

I applaud Claude's response. Ratch's points are unimportant when it comes to APPLYING the devices in question. Take that from an engineer who has been building electronic **** for 50+ years.

 

[ericgibbs]

I applaud Claude's response. Ratch's points are unimportant when it comes to APPLYING the devices in question. Take that from an engineer who has been building electronic **** for 50+ years.

I agree with Claude & MikeMI, I also have 50+ years experience in APPLIED electronics.

E.

 

[Ratchit]

Claude Abraham,

Every OEM who produces LEDs & bjt's calls them "current controlled". Every OEM cannot be wrong.

OEMs are in the business of selling their product, not necessarily educating their customers on the way their product really works. A concensus of opinion is not necessarily the correct answer. The correct answer comes with a good explanation of why it works the way it does. The "OEM" way of using base current to design BJT circuits still boils down to putting the correct Vbe on the trasistor.

But said equation can also be expressed as follows:

b) Vd = Vt*ln((Id/Is)+1).

The b) form expresses voltage as a function of current

Yes, I covered that point in my previous post, so you are repeating what I said.

In the forward direction, we do not force a constant voltage across LED terminals then allow the voltage to "control the current". Thermal runaway would occur incurring destruction of the LED. Instead we force a constant current through the LED and allow the voltage to be controlled by the current as well as temperature. LED devices are always current controlled. One cannpt control LED current with LED voltage.

I never said that LED circuits were designed by applying a constant voltage across them. I said that LED devices were controlled by the voltage across them. The physics of a junction diode prove that, as I mentioned in my previous post.

I & V are both needed for operation, but with LED & bjt devices, we must control current & V is incidental, controlled by current as well as temperature.

Yes, both current and voltage come into play, that is obvious. But that is irrelevant to the point I am making, which is that the voltage is controlling the current. Again, the physics of the diode show that to be true.

In the LED, Vd is controlled by Id & T.

Ansolutely untrue. Vd and Id have a one to one relationship to each other, but you cannot show by semiconductor physics how current affects voltage. It is the other way around.

MikeMl,

I applaud Claude's response. Ratch's points are unimportant when it comes to APPLYING the devices in question. Take that from an engineer who has been building electronic **** for 50+ years.

Then please show me the semiconductor physics of how current controls voltage in a diode. And don't confuse semiconductor physics with design methods.

Eric Gibbs,

I agree with Claude & MikeMI, I also have 50+ years experience in APPLIED electronics.

Ditto with respect my answer to MikeMl.

Ratch

 

[MikeMl]

...MikeMl,... don't confuse semiconductor physics with design methods.
...

My point, EXACTLY. This forum is primarily for helping hobbyists with their DESIGN issues (how do use a LED? how do I bias this transistor?). All that your pedantic minutia about semiconductor physics does is confuse them further...

OUT!

 

[Claude Abraham]

Ratchit, your "physics of junction devices prove that V controls I" is nothing more than form a) of the Shockley equation. Form b) & form a) are 1 & the same equation with variable order switched. A given current results in a corresponding voltage & vice-versa. When you say "V controls I", it is pure dogma. You state that V comes first then I is dependent on V but your "proof" is merely form a) of Shockley's equation, herein termed as "SE". I've explained numerous times that the forward voltage on the diode is a result of carriers crossing the p-n junction, becoming minority carriers (holes in n region, electrons in p region), and a finite time is required for recombination to occur, called "lifetime". During this time the accumulation of these minority charges results in a local E field. The E field tends to oppose new carriers approaching the depletion region where the charges are accumulated.

This is essentially a "potential barrier" since energy is needed to overcome said barrier. Equilibrium is attained. As charges recombine new charges enter depletion zone so that Id & Vd attain steady state values, i.e. equilibrium. Now if the external source (generator, microphone, whatever) increases its signal into the network containing the diode, current Id increases. Current is charge per time, so that more carriers per second enter the junction & then the depletion zone. The potential barrier, Vd, is as it was before the current increase. More charges entering the depletion zone, DZ herein, result in a larger accumulation of charges waiting to recombine. Charge density is increased, but lifetime is about the same so that more carriers have accumulated resulting in a larger potential barrier, & a larger Vd.

When forward biased, the current value Id, along with temp, & the geometry of the junction determine Vd. A change in Id precedes the change in Vd. The junction is modeled as a diffusion capacitance Cd. Remember Eli the ice man, that in a cap, I precedes E. The explanation I just gave is affirmed in any physics text. The forward current Id, cannot possibly be controlled by the barrier potential Vd. It is a barrier. All Vd is is the amount of work per unit charge needed to overcome said barrier. The source driving the network (mic, generator, etc.) provides all energy needed to move charges, i.e. the driving source controls the current. Along the way, a quantity of the total energy is lost at the junction overcoming the barrier. This voltage Vd is the loss incurred at the junction, it is not that which DRIVES Id the current.

If diode/LED/bjt/SCR/triac makers are only interested in selling their devices, what do they gain by propagating false info? These same OEM state that their JFET & MOSFET devices are voltage controlled. Of course, FET parts are voltage controlled because we must, due to device inherent physics, drive gate-source voltage Vgs directly, & gate current Ig is incidental. Ig is as important as Vgs, it just isn't the directly driven quantity, Vgs is that.

I just showed you the semicon phy as to how a diode works, how current changes precede voltage changes. This nullifies any contention that V controls I. You asked me to prove how I controls V in a diode, but please take note of this. My position is that V does NOT control I. That does not mean that I necessarily controls V. The external source, generator, mic, whatever, is what controls I. A change in generated signal results in a change in I which is a change in carriers per second of time. As a result Vd changes in accordance with form b) of SE.

Id & Vd are both controlled by the external source. But Id is impacted before Vd has a chance to react. Ultimately it is the external source that motivates the change, not Id. Think of a row of dominoes where the one at one end is pushed. If there are 5 dominoes, pushing on no. 1 results in no. 5 eventually falling. If no. 5 is Vd, & no. 4 is Id, then we know this. The finger pushing on no. 1 is ultimately what controls all 5 dominoes. Id, no. 4, changes & influences Vd, no. 5. Since domino no. 4 changes prior to no. 5, then no. 5 cannot be what controls no. 4. Thus Vd can never be what controls Id.

Is my explanation clear? BR.

 

[Ratchit]

MikeML,

My point, EXACTLY. This forum is primarily for helping hobbyists with their DESIGN issues (how do use a LED? how do I bias this transistor?). All that your pedantic minutia about semiconductor physics does is confuse them further...

OUT!

Well then, you should hop on the OP for asking a question not related to DESIGN issues. As well as other posts in other threads that do not pertain to DESIGN issues. All I did was answer and explain to the OP what he asked about. Others disagreed with my answer, which is OK. But if this form is only for DESIGN issues, then some more focused editing of the submitted posts is in order. Is it you who decides what this forum is about?

Ratch

 

[MrAl]

Hi,

We sort of have two threads with the same basic concept.

See this post for a complete description of voltage vs current operated devices:
https://www.electro-tech-online.com/threads/voltage-or-current-operated-devices.115713/#post1109589

It's in the other thread that is nearly the same as this one.

 

[Ratchit]

Claude Abraham,

Ratchit, your "physics of junction devices prove that V controls I" is nothing more than form a) of the Shockley equation. Form b) & form a) are 1 & the same equation with variable order switched. A given current results in a corresponding voltage & vice-versa. When you say "V controls I", it is pure dogma.

Not so, I gave a qualitative explanation that did not rely on Schockley's equation. That explanation showed that V controls I, and it stands on its own.

You state that V comes first then I is dependent on V but your "proof" is merely form a) of Shockley's equation, herein termed as "SE".

No, I did not. I said that V controls I. I never gave it a first or second standing, whatever that is. And I did not use Schockley's equation as a proof.

I've explained numerous times that the forward voltage on the diode is a result of carriers crossing the p-n junction, becoming minority carriers (holes in n region, electrons in p region), and a finite time is required for recombination to occur, called "lifetime". During this time the accumulation of these minority charges results in a local E field. The E field tends to oppose new carriers approaching the depletion region where the charges are accumulated.

That is a poor explanation of what most semiconductor book say happens. The electrons from the N-type and the holes from the P-type move into the other's territories. This movement is caused by diffusion. The electrons complete the bonding of a thin layer of P-type material, and the holes complete the bonding of a thin layer of N-type material. This causes negative ions to form in the P-type material and positive ions to form in the N-type material. This accumulation of positive and negative on either side of the junction causes an incrreasing back-voltage that reduces further current across the junction until it reaches zero current. The lifetime of the minority carriers is irrelevant for a qualitative explanation.

This is essentially a "potential barrier" since energy is needed to overcome said barrier. Equilibrium is attained. As charges recombine new charges enter depletion zone so that Id & Vd attain steady state values, i.e. equilibrium.

A forward voltage is need to overcome the back-voltage or "potential barrier". That is why voltage controls the current in a diode.

Now if the external source (generator, microphone, whatever) increases its signal into the network containing the diode, current Id increases. Current is charge per time, so that more carriers per second enter the junction & then the depletion zone. The potential barrier, Vd, is as it was before the current increase. More charges entering the depletion zone, DZ herein, result in a larger accumulation of charges waiting to recombine. Charge density is increased, but lifetime is about the same so that more carriers have accumulated resulting in a larger potential barrier, & a larger Vd.

The greater the forward voltage, the more current and the higher the back-voltage. Eventually equalibrium is reached. I still don't know what the lifetime of recombination has to do with a qualitative explanation.

When forward biased, the current value Id, along with temp, & the geometry of the junction determine Vd. A change in Id precedes the change in Vd.

The diode current is dependent on the forward voltage. The forward voltage is needed to reduce the barrier voltage so that diffusion can occur. No amount of explaining can prove otherwise.

The forward current Id, cannot possibly be controlled by the barrier potential Vd. It is a barrier. All Vd is is the amount of work per unit charge needed to overcome said barrier.

The forward current is dependent on the applied forward voltage and the opposing barrier voltage.

The source driving the network (mic, generator, etc.) provides all energy needed to move charges, i.e. the driving source controls the current. Along the way, a quantity of the total energy is lost at the junction overcoming the barrier. This voltage Vd is the loss incurred at the junction, it is not that which DRIVES Id the current.

You say "network", but I thought we were talking about a single diode. For a diode, the applied voltage has to supply energy to move current through the diode resistance, but it is the voltage that controls the current.

If diode/LED/bjt/SCR/triac makers are only interested in selling their devices, what do they gain by propagating false info? These same OEM state that their JFET & MOSFET devices are voltage controlled. Of course, FET parts are voltage controlled because we must, due to device inherent physics, drive gate-source voltage Vgs directly, & gate current Ig is incidental. Ig is as important as Vgs, it just isn't the directly driven quantity, Vgs is that.

It is not exactly false. Their devices do act functionally as current controlled devices when inserted into the appropriate circuit. But taken as a single device, they are voltage controlled. The OEM are in the business of selling design info, not theory.

I just showed you the semicon phy as to how a diode works, how current changes precede voltage changes. This nullifies any contention that V controls I. You asked me to prove how I controls V in a diode, but please take note of this. My position is that V does NOT control I. That does not mean that I necessarily controls V. The external source, generator, mic, whatever, is what controls I. A change in generated signal results in a change in I which is a change in carriers per second of time. As a result Vd changes in accordance with form b) of SE.

You gave a partly true statement of a diode works. You did not prove how current changes the voltage, and how the fact that the forward voltage opposes the back-voltage does not mean the device if voltage controlled.

Id & Vd are both controlled by the external source. But Id is impacted before Vd has a chance to react. Ultimately it is the external source that motivates the change, not Id. Think of a row of dominoes where the one at one end is pushed. If there are 5 dominoes, pushing on no. 1 results in no. 5 eventually falling. If no. 5 is Vd, & no. 4 is Id, then we know this. The finger pushing on no. 1 is ultimately what controls all 5 dominoes. Id, no. 4, changes & influences Vd, no. 5. Since domino no. 4 changes prior to no. 5, then no. 5 cannot be what controls no. 4. Thus Vd can never be what controls Id.

Id is controlled by the forward voltage in a exponential way. For all practical purposes, this happens instantly.

Is my explanation clear? BR.

I read the words, but they don't make sense. You are ignoring the fact that it is the foward voltage that reduces the barrier voltage and allows the diode current to continue. The physical relationship determines what is the control, and in a BJT or LED, which depend on diffusion, the voltage is in control because it regulates the diffusion.


Ratch

https://www.kevinaylward.co.uk/ee/voltagecontrolledbipolar/voltagecontrolledbipolar.html

 

[Claude Abraham]

Claude Abraham,



Not so, I gave a qualitative explanation that did not rely on Schockley's equation. That explanation showed that V controls I, and it stands on its own.



No, I did not. I said that V controls I. I never gave it a first or second standing, whatever that is. And I did not use Schockley's equation as a proof.



That is a poor explanation of what most semiconductor book say happens. The electrons from the N-type and the holes from the P-type move into the other's territories. This movement is caused by diffusion. The electrons complete the bonding of a thin layer of P-type material, and the holes complete the bonding of a thin layer of N-type material. This causes negative ions to form in the P-type material and positive ions to form in the N-type material. This accumulation of positive and negative on either side of the junction causes an incrreasing back-voltage that reduces further current across the junction until it reaches zero current. The lifetime of the minority carriers is irrelevant for a qualitative explanation.



A forward voltage is need to overcome the back-voltage or "potential barrier". That is why voltage controls the current in a diode.



The greater the forward voltage, the more current and the higher the back-voltage. Eventually equalibrium is reached. I still don't know what the lifetime of recombination has to do with a qualitative explanation.



The diode current is dependent on the forward voltage. The forward voltage is needed to reduce the barrier voltage so that diffusion can occur. No amount of explaining can prove otherwise.



The forward current is dependent on the applied forward voltage and the opposing barrier voltage.



You say "network", but I thought we were talking about a single diode. For a diode, the applied voltage has to supply energy to move current through the diode resistance, but it is the voltage that controls the current.



It is not exactly false. Their devices do act functionally as current controlled devices when inserted into the appropriate circuit. But taken as a single device, they are voltage controlled. The OEM are in the business of selling design info, not theory.



You gave a partly true statement of a diode works. You did not prove how current changes the voltage, and how the fact that the forward voltage opposes the back-voltage does not mean the device if voltage controlled.



Id is controlled by the forward voltage in a exponential way. For all practical purposes, this happens instantly.



I read the words, but they don't make sense. You are ignoring the fact that it is the foward voltage that reduces the barrier voltage and allows the diode current to continue. The physical relationship determines what is the control, and in a BJT or LED, which depend on diffusion, the voltage is in control because it regulates the diffusion.


Ratch

https://www.kevinaylward.co.uk/ee/voltagecontrolledbipolar/voltagecontrolledbipolar.html

I think the problem is that you will not differentiate source voltage from Vd. They are not the same. You start with a constant voltage source of 0 volts connected across the junction then increase said voltage until Vd is exceeded & current starts. But that is a self-fulfilling prophecy. There will be no current until the 0.65V barrier potential is exceeded, which I could have told you. We do not do that in the real world. If we start w/ a constant current source, CCS, with a switch across it in the closed position & a diode across the switch, forward biased direction, observe the following.

The CCS outputs current which takes a path through the CCS & closed switch. The diode is shunted by a perfect shorted switch, hence Vd=0 & Id=0. The switch is opened. Current continues through the diode. Vd=0 initially, but the current in the wires is the value of the CCS. We do not need a voltage to keep I going since the carriers are already energetic, being in the conduction band. Inductance will keep the carriers moving until energy is removed from the system via losses.

When the carriers arrive at the junction they cross and a depletion zone is formed. The accumulation of charge occurs & a local E field is incurred. The integral over the distance is the barrier potential. Vd formed as a result of I, existing energy, & charges crossing the junction. This barrier tends to oppose incoming charge flow. When the switch was closed, no energy loss occurred so the CCS outputted 0 volts to maintain the current. Now with switch open, the barrier voltage requires that the CCS outputs around 0.65 volts to maintain that same value of current, which it does.

The value of I is dictated by the CCS, & the value of V that the CCS will output is dictated by the barrier potential.

If the source is constant voltage, CVS, with a resistor R, a switch closed, shunted by a diode, we get the following. I will be V/R, Vd=0, Id=0. Switch opens resulting in charges transiting through p-n junction. When equilibrium is reached, I = (V-Vd)/R. If V is the CVS value, Vd is forward diode drop or barrier potential, as long as V >> Vd, then Vd exerts minimal influence over I. So Vd does have influence in controlling I if V is only marginally greater than Vd.

You seem to expend a lot of energy pointing this out. But we do not drive LEDs, bjt's, etc. with a voltage source of 0.71 volts for a 0.65V forward drop. If that were the case, then Vd would have substantial influence over I. You present a CVS through a resistor driving a p-n junction, then argue as CVS value just crosses the barrier potential threshold, I is created. But that scenario is direct voltage drive, which we never use. Again, for emphasis, if I connect a pure CVS (zero internal resistance) across a diode junction, then indeed Id is controlled by Vd for sure.

But if I connect a diode across a pure CCS, then Vd is controlled by Id for sure. We can force either modus operandi, but one of them produces great results while the other is catastrophic. When a diode is driven by a CVS much larger in value then Vd, with an adequate value of series resistor, Vd is controlled chielfy by Id, which is determined by CVS value & R value. Refer to my LED example above. With 1.80V Vd, 12V source, 1.0 kohm R, Vd is controlled by Id, but Id is controlled by Vsource (12V), & R (1.0 kohm), & slightly by Vd. Thus Id controls Vd, Vsrc and R control Id w/ slight influence by Vd.

Id & Vd are interactive, a change in 1 results in changing the other, then Vsrc & R are affected and another change happens. A drop in Vd due to ambient temp results in larger voltage across R, which increases I, then Vd goes up slightly due to increase in Id, then current drops slightly, etc. Equilibrium is then reached.

Ratch your explanation of semicon phy assumes a CVS driving the junction w/ I determined per SE form a). This can only happen if CVS is right across diode.

What does lifetime have to do with it? How can charges accumulate if lifetime is zero? The instant charges cross junction they instantly would recombine resulting in no accumulation. The barrier is a result of uncombined excess minority charge carriers. Before the lifetime expires they are not combined resulting in their E field forming a barrier to incoming charges. A charge crossing the junction "feels" the E field force of a charge which crossed moments ago yet has not yet recombined. A charge which crossed long ago and time has elapsed beyond that lifetime will not be felt by the newly arriving charge. Lifetime is all important. When a new charge arrives it feels the E force only from charges whose lifetime is not yet expired. Those charges whose lifetime has passed are already recombined and neutralized, hence they exert no barrier force to incoming new charges.

Ratch, you really need to take courses in semicon phy at the graduate level to know this. You keep introducing info counter to peer reviewed uni texts. Asking what lifetime has to do with anything is an admission that your background is too limtited to argue with the establishment.

You link to Kevin Aylward as a scholarly source?! He will not respond to me. He publishes on his web site w/o peer review & will not even print my criticism of his poorly constructed essays. I've written to him only to be ignored. He uses your argument. I is controlled by V because SE form a) says so. He also infers that since F = qE, that E fields cause charges to move, but he does not acknowledge that as an E field imparts energy to charge carriers, it gives up its own energy, needing replenished.

I will debate him anytime any place.

 

[Ratchit]

Claude Abraham,

I think the problem is that you will not differentiate source voltage from Vd. They are not the same. You start with a constant voltage source of 0 volts connected across the junction then increase said voltage until Vd is exceeded & current starts. But that is a self-fulfilling prophecy. There will be no current until the 0.65V barrier potential is exceeded, which I could have told you. We do not do that in the real world. If we start w/ a constant current source, CCS, with a switch across it in the closed position & a diode across the switch, forward biased direction, observe the following.

Yes, there will be a current at small forward voltages, albeit small. Only when the forward voltage approaches 0.7 volts does the current become significant. This is revealed in Schockley's equation. Real world? You are starting to talk about design proceedures, as opposed to diode theory.

The CCS outputs current which takes a path through the CCS & closed switch. The diode is shunted by a perfect shorted switch, hence Vd=0 & Id=0. The switch is opened. Current continues through the diode. Vd=0 initially, but the current in the wires is the value of the CCS. We do not need a voltage to keep I going since the carriers are already energetic, being in the conduction band. Inductance will keep the carriers moving until energy is removed from the system via losses.

Again the voltage across the diode will sustain the diode currect according to Schockley's equation. Inductance has nothing to do with anything.

When the carriers arrive at the junction they cross and a depletion zone is formed. The accumulation of charge occurs & a local E field is incurred. The integral over the distance is the barrier potential. Vd formed as a result of I, existing energy, & charges crossing the junction. This barrier tends to oppose incoming charge flow. When the switch was closed, no energy loss occurred so the CCS outputted 0 volts to maintain the current. Now with switch open, the barrier voltage requires that the CCS outputs around 0.65 volts to maintain that same value of current, which it does.

The carriers are at the junction since the diode was manufactured. So is the depletion zone. So is the accumulation of the ionic charge and E-field. We have been over the rest of this before. The applied voltage reduces the barrier voltage and allows more diode current to exist.

The value of I is dictated by the CCS, & the value of V that the CCS will output is dictated by the barrier potential.

The value of I is determined by the value of the voltage that the CCS puts across the diode according to Schockley's equation. There will a one to one correspondence between the voltage and current, with the voltage determining the current according to the physics of the diode.

If the source is constant voltage, CVS, with a resistor R, a switch closed, shunted by a diode, we get the following. I will be V/R, Vd=0, Id=0. Switch opens resulting in charges transiting through p-n junction. When equilibrium is reached, I = (V-Vd)/R. If V is the CVS value, Vd is forward diode drop or barrier potential, as long as V >> Vd, then Vd exerts minimal influence over I. So Vd does have influence in controlling I if V is only marginally greater than Vd.

No matter what the V is in the external circuit, the voltage across the diode determines the diode current.

You seem to expend a lot of energy pointing this out. But we do not drive LEDs, bjt's, etc. with a voltage source of 0.71 volts for a 0.65V forward drop. If that were the case, then Vd would have substantial influence over I. You present a CVS through a resistor driving a p-n junction, then argue as CVS value just crosses the barrier potential threshold, I is created. But that scenario is direct voltage drive, which we never use. Again, for emphasis, if I connect a pure CVS (zero internal resistance) across a diode junction, then indeed Id is controlled by Vd for sure.

No matter if you drive the diode circuit with a voltage or current source, Schockley's equation will hold even if the diode is embedded within the circuit. Of course we don't try to regulate the current by directly controlling the Vd, but Vd and Id will automatically follow Schockley's equation no matter how many resistors you insert in the circuit containing the diode.

But if I connect a diode across a pure CCS, then Vd is controlled by Id for sure. We can force either modus operandi, but one of them produces great results while the other is catastrophic. When a diode is driven by a CVS much larger in value then Vd, with an adequate value of series resistor, Vd is controlled chielfy by Id, which is determined by CVS value & R value. Refer to my LED example above. With 1.80V Vd, 12V source, 1.0 kohm R, Vd is controlled by Id, but Id is controlled by Vsource (12V), & R (1.0 kohm), & slightly by Vd. Thus Id controls Vd, Vsrc and R control Id w/ slight influence by Vd.

No, Vd will have a specific value determined by S's equation, but it will be Vd that controls Id. Within the diode itself, S's equation holds.

Id & Vd are interactive, a change in 1 results in changing the other, then Vsrc & R are affected and another change happens. A drop in Vd due to ambient temp results in larger voltage across R, which increases I, then Vd goes up slightly due to increase in Id, then current drops slightly, etc. Equilibrium is then reached.

Yes, temperature is a part of S's equation.

Ratch your explanation of semicon phy assumes a CVS driving the junction w/ I determined per SE form a). This can only happen if CVS is right across diode.

Not so. I am avering that the current is related to the voltage across the junction according to S's equation, and the voltage is not necessarily constant. There will always be a voltage across the diode even if the external voltage is applied to the circuit containing the diode and not directly to the diode itself.

What does lifetime have to do with it? How can charges accumulate if lifetime is zero? The instant charges cross junction they instantly would recombine resulting in no accumulation. The barrier is a result of uncombined excess minority charge carriers. Before the lifetime expires they are not combined resulting in their E field forming a barrier to incoming charges. A charge crossing the junction "feels" the E field force of a charge which crossed moments ago yet has not yet recombined. A charge which crossed long ago and time has elapsed beyond that lifetime will not be felt by the newly arriving charge. Lifetime is all important. When a new charge arrives it feels the E force only from charges whose lifetime is not yet expired. Those charges whose lifetime has passed are already recombined and neutralized, hence they exert no barrier force to incoming new charges.

You have everything wrong in the above paragraph. When a electron diffuses from the N-type into the P-type, it bonds with an atom with only 3 covalent bonds to make an atom with 4 covalent bonds. So that atom has the 4 covalent bonds it wants, but now the extra electron makes it into a negative ion. And the extra electron from the 5 electron atom in the N-type is gone, so now that atom is a positive ion. So the time it takes to make the transfer is unimportant. It is the transfer itself, not how fast it is done that determines the ion count and thereby the barrier voltage. As I said before, this ion generation happens as soon as the P-type and N-type are bonded together during manufacturing.

Ratch, you really need to take courses in semicon phy at the graduate level to know this. You keep introducing info counter to peer reviewed uni texts. Asking what lifetime has to do with anything is an admission that your background is too limtited to argue with the establishment.

I disagree. I think you should review what really happens when two different semiconductor types are placed in intimate contact with each other.

You link to Kevin Aylward as a scholarly source?! He will not respond to me. He publishes on his web site w/o peer review & will not even print my criticism of his poorly constructed essays. I've written to him only to be ignored. He uses your argument. I is controlled by V because SE form a) says so. He also infers that since F = qE, that E fields cause charges to move, but he does not acknowledge that as an E field i,parts energy to charge carriers, it gives up its own energy, needing replenished.

No, I do not consider him a scholarly source per se. It is a shame he does not defend his assertions.

I will debate him anytime any place.

Very good.

Ratch

 

[kavan]

Hello everyone,

Here the OP. I thank you for the various inputs which I think are all useful in different ways. As a beginner, I am still not comfortable with design and need some physics explanations too.
I will read through your answers carefully.

From a quick reading, it seems that voltage is what causes current, in general. A current operated device seems to be one where the flow of current in a specific part of the device enables the device to function the way it is supposed to function, fulfilling the purpose it was designed for. For instance, if a transistor is current operated, that current will make the transistor acts as a switch and block or pass current in some other area...Same reasoning goes for voltage controlled devices.
A LED is defined as a current operated device in the sense that the light brightness is dependent on the current in the LED, in a more correlated way than the voltage. That is the functional definition of why the LED is called current operated. But that current is surely generated by tweaking a voltage, even if the relation current-voltage is not linear...

Thanks,
Kavan

 

[MikeMl]

...
A LED is defined as a current operated device in the sense that the light brightness is dependent on the current in the LED, in a more correlated way than the voltage. That is the functional definition of why the LED is called current operated. But that current is surely generated by tweaking a voltage, even if the relation current-voltage is not linear...

No Kavan, the voltage across an LED is determined by the current flowing through the LED, and the LED's temperature. If you try to design a LED driving circuit that adjusts LED current by "tweaking the voltage" applied to the LED, you have a flawed design which is likely to go into thermal runaway and destroy the LED.

Think of it this way, when designing a LED circuit, the independent variable is the current, and the dependent variable is the resulting voltage across the LED.

 

[Ratchit]

kavan,

From a quick reading, it seems that voltage is what causes current, in general. A current operated device seems to be one where the flow of current in a specific part of the device enables the device to function the way it is supposed to function, fulfilling the purpose it was designed for. For instance, if a transistor is current operated, that current will make the transistor acts as a switch and block or pass current in some other area...Same reasoning goes for voltage controlled devices.
A LED is defined as a current operated device in the sense that the light brightness is dependent on the current in the LED, in a more correlated way than the voltage. That is the functional definition of why the LED is called current operated. But that current is surely generated by tweaking a voltage, even if the relation current-voltage is not linear...

I think you have the basic idea of operation. Remember, LEDs and BJTs are creatures of diffusion, i.e., they work by a difference in charge carrier type and concentration between P-type and N-type material. The holes from the P-type material and the electrons from the N-type material combine at the junction and annihilate each other. This process leaves behind ions that produce a barrier voltage that suppresses further combination. An external forward voltage lowers the barrier voltage so more holes and electrons can combine to produce a forward current. That is why junction devices are voltage controlled, even if driven by a current source. They must have an external voltage to lower the internal barrier voltage so that the diffusion process can continue. Their diffusion process is why the current is an exponential relationship to the voltage. In physics, diffusion equations are exponential. In FETs and tubes, current is controlled mainly by a voltage producing an electrostatic field, therefor the current relationship is linear to the voltage. So both junction and electrostatic devices are voltage controlled, but each in a different way.

Naturally, you would put a resistance in the LED circuit to limit the current, or use current limiting methods to design a LED circuit. But any external resistance or method you use will not change the principle upon which the device operates.

Ratch

 

[Claude Abraham]

Ratchit you keep re-iterating that voltage is needed to sustain LED current, but that point is not in contention. "Lowering barrier voltage" is a crutch you can't let go of. If an LED requires 1.80V at 10 MA forward current, then we know that a 10 mA current source w/ only 1.0 V of compliance will not work. The fact that the junction barrier potential must be overcome is not under discussion. Without the 1.80V, the 10 mA does not happen.

Your whole case is built on the fact that w/o overcoming barrier voltage, no current can be sustained. We already know that. But the current is controlled by the source powering the LED, & to a very small extent, the difference between source voltage Vg & LED forward drop Vd. A voltage source plus resistor approximates current drive, but Vd still has an influence. But once the Vd barrier is overcome, Vd is not in control of Id, the input source is. Id is basically (Vg-Vd)/R, so that if Vg >> Vd, then Vd has only a minor influence on Id.

The mere fact that voltage is needed to overcome the junction barrier does not make a device "voltage controlled". Using that logic every electrical device would be voltage controlled & no such thing as current controlled would exist. A resistor has barriers formed by charge carriers colliding with the lattice structure forming barriers & releasing photons in the 5 um IR wavelength (heat). Even mag amps which have current conduction has internal barriers due to winding resistance. To generate the magnetic fields for the mag amp, current must exist. But resistive wires in the windings result in barriers due to lattice resistive collisions. So to sustain current, a potential barrier must be overcome. That would make a mag amp voltage controlled using your narrow self-styled definition of said term.

Junction devices need voltage. W/o voltage no steady current could exist. The fact that the current which controls the LED brightness could not exist w/o voltage is accepted w/o argument. But the value of Vd under forward bias is literally determined by Id. Just as you cannot have Id w/o Vd, it is equally true in reverse. We already explained that Vd is determined by internal device geometry & doping, as well as Id & temp.

A FET is universally described as VC. But in order to charge up the gate-source a CURRENT is needed. This gate current Ig precedes Vgs. So a voltage controlled FET, requires that Ig charges the gate-source giving rise to Vgs which activates the FET change of state. The fact that Vgs cannot change w/o Ig does not nake FETs current controlled. Ig is indispensable but not directly driven. We supply Ig from a CVS letting Ig be incidental, Vgs directly controlled.

All electrical devices need I & V both. Which variable is the control variable is determined by which variable is directly forced on the device & which is incidental. But in all cases, the indirect incidental variable is indispensable. FETs cannot work w/o current, LED cannot work w/o voltage. But a FET driven from a CCS is destroyed. Gate-source is high resistance capacitor. A continuous current source charges the gate indefinitely until punch through occurs. But a CVS w/ low gate resistance provides high current at first, then current tapers off as Vgs approaches source value. Again Vgs cannot be attained w/o gate current Ig. That does not determine how FETs are classified.

All devices cannot work w/o one of the variables missing. Even the indirect uncontrolled variable is indispensable.

Kavan - voltage is NOT the CAUSE of current. Until you discard that heresy you will be stuck forever. The idea that voltage causes current is beyond a doubt the most common of all electrical heresies, so much so that I call it "heresy no. 1" of simply "H1".

 

[kavan]

Hello Claude and Ratchit,

I show my ignorance once again here. As far as H1 goes: well, I know that not every equation in physics implies a cause-effect relation between its variables. In some cases we can tell which one is the cause and the effect but most often we cannot. The equation only establishes a relationship. See stress and strain. There is a similar debate on which one causes the other....

Now, I thought I was pretty solid in thinking that a voltage causes current though. Let me explain why.
current is the flow of charges (either positive or negative, electrons, ions, etc....).
Voltage (potential difference) is more the expression of charge separation and the ability to cause current (if the material allows ,i.e. it is not an insulator).

consider the parallel plate capacitor: there is a voltage between the plates. a charge positioned between the plate will move due to that voltage and represent a current....

I don't like heresies so please help me deconstruct this one if it is one.

thanks
kavan