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.Hello Forum,
I have heard of current operating devices, like LEDs, and voltage operated devices...
What is the difference?
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.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?
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?
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
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.
Every OEM who produces LEDs & bjt's calls them "current controlled". Every OEM cannot be wrong.
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
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 & 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.
In the LED, Vd is controlled by Id & T.
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.
...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!
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 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.
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 i,parts energy to charge carriers, it gives up its own energy, needing replenished.
I will debate him anytime any place.
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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...
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...
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