Voltage or current operated devices?

[Ratchit]

BrownOut,

Voltage doesn't travel. Learn your basic physics.

It is a descriptive colloquial phrase. Learn your basic English.

I need to know nothing of your state of mind, only what's real and what's imagined.

That's the point. You cannot know what is imagined until you know my state of mind.

Poor example. I'm interested in devices, not chemistry. If you want to show an example, use a device. You're point isn't relevant, since I never made that claim. I claim that if a device cannot function without a current, then the current is necessary. I never calimed anything about existance. More song and dance.

Your interest is irrelevant for my illustrative example. You made the claim several times when you said that because Ib is present, Ib is essential. Not necessarily true in general, and not true in this specific case.

t is both small and finite. Small was your first criteria, now you change to finite. No device will ever have infinite or zero input reistance, so by your flawed logic, there can be no voltage or current devices. So we're back to that again. Nothing more than same old rationalization.

Why can't something be both, small and finite. No one said that any device was perfect. There can be both voltage and current activated devices, just not perfect ones.

You don't seem to know what you call it.

I called it a transconductance amplifier in both cases, didn't I? Did you become confused when I put an adjective in front of the name?

Ratch

 

[BrownOut]

It is a descriptive colloquial phrase. Learn your basic English.

I know english, and I know when I read something that's false. Voltage doesn't travel, and so something else must be responsible ( hint: it's current )

That's the point. You cannot know what is imagined until you know my state of mind.

I know what is imagined wether I know you or not.

Your interest is irrelevant for my illustrative example. You made the claim several times when you said that because Ib is present, Ib is essential. Not necessarily true in general, and not true in this specific case.

I never made that claim. That's just the way you distort what I've said repeatedly. I take it you can't find an example that uses devices, since this is a discussion about devices. Your poor example isn't going to convince anyone.

Why can't something be both, small and finite. No one said that any device was perfect. There can be both voltage and current activated devices, just not perfect ones.

That's my point exactly. The resistance is small and finite, thus the device is current controlled.

I called it a transconductance amplifier in both cases, didn't I? Did you become confused when I put an adjective in front of the name?

No, you didn't. I'm not confused, but you seem to be.

 

[Ratchit]

BrownOut,

I know english, and I know when I read something that's false. Voltage
------>E
doesn't travel, and so something else must be responsible ( hint: it's current )

No, not current, it is the electric field that supports and defines the voltage, and that propagates (travels) at the speed of light.

I know what is imagined wether I know you or not.

Thinking it may by so is not the same as knowing for sure.

I never made that claim. That's just the way you distort what I've said repeatedly.

Not explicitly, but implictly for sure.

That's my point exactly. The resistance is small and finite, thus the device is current controlled.

The resistance, while small and finite, is still too large be a current amplifier. It is not quite large enough to be a good transconductance amplifier, either. That is why I said "semi".

No, you didn't. I'm not confused, but you seem to be.

Are you saying I did not call it a transconductance amplfier in both cases? When both your quotes showed that I did?

Ratch

 

[BrownOut]

No, not current, it is the electric field that supports and defines the voltage, and that propagates (travels) at the speed of light.

No. The electric field is a product of charge, and nothing else. Charge travels, not the electric field, voltage or anything else. No charge=no E field= no voltage. Learn physics before trying to discussing it.

Thinking it may by so is not the same as knowing for sure.

I know what is real and what is imagined.

Not explicitly, but implictly for sure.

Not explicitly, implicitly or any other way.

The resistance, while small and finite, is still too large be a current amplifier. It is not quite large enough to be a good transconductance amplifier, either. That is why I said "semi".

The resistance is sufficiently small for a current amplifier.

Are you saying I did not call it a transconductance amplfier in both cases? When both your quotes showed that I did?

My quotes showed that you did not.

 

[Ratchit]

Brownout,

No. The electric field is a product of charge, and nothing else. Charge travels, not the electric field, voltage or anything else. No charge=no E field= no voltage. Learn physics before trying to discussing it.

An electric field can move charges by exerting a force on them. In this case, the electric field caused by Vbe. Charge travels very slowly compared with an electric field. Charge velocity is called the drift velocity, and its speed compares with cold molasses. It would take "forever" to put a voltage across a component connected to the source by a conductor if the voltage relied on charge movement. But voltage relies on the electric field. Learn physics.

Ratch

 

[BrownOut]

Brownout,



An electric field can move charges by exerting a force on them. In this case, the electric field caused by Vbe. Charge travels very slowly compared with an electric field. Charge velocity is called the drift velocity, and its speed compares with cold molasses. It would take "forever" to put a voltage across a component connected to the source by a conductor if the voltage relied on charge movement. But voltage relies on the electric field. Learn physics.

Ratch

Vbe doesn't come out nowhere. It can only be a product of charge, because only charge can create a potential. So to say Vbe moves charge is the same is saying Vbe comes from nowhere, which is impossible. Charge must enter the region before any Vbe can be devleoped. That's real physics, not the pseudo physics you're presenting. Just remember, no charge=no voltage. That principle will never fail.

 

[Ratchit]

BrownOut,

Vbe doesn't come out nowhere. It can only be a product of charge, because only charge can create a potential. So to say Vbe moves charge is the same is saying Vbe comes from nowhere, which is impossible. Charge must enter the region before any Vbe can be devleoped. That's real physics, not the pseudo physics you're presenting. Just remember, no charge=no voltage. That principle will never fail.

Vbe does come from somewhere. It is an external voltage applied to the emitter-base terminals. Without that voltage, no charge will flow. Remember the formula, Ic = Is*exp(Vbe/.026), where Is is the always present saturation current?

Ratch

 

[BrownOut]

External voltage cannot be "applied" to the E-B terminals. That's the way they teach in tech school. Engineers know that only charge can change a voltage. A voltage cannot change another voltage, thus no external voltage can change the junction voltage. Only a charge can change the junction voltage.

 

[Ratchit]

BrownOut,

External voltage cannot be "applied" to the E-B terminals. That's the way they teach in tech school. Engineers know that only charge can change a voltage. A voltage cannot change another voltage, thus no external voltage can change the junction voltage. Only a charge can change the junction voltage.

Of course an external voltage can be applied to the transistor B-E terminals. Do you realize what you are you saying? As I said before, an electric field can change the charge distribution, and that can change the junction voltage.

Ratch

 

[BrownOut]

An external voltage cannot be applied to a B-E junction. I know exactly what I'm saying. A voltage cannot change another voltage. Charge must be transfered.

 

[Ratchit]

BrownOut,

An external voltage cannot be applied to a B-E junction. I know exactly what I'm saying. A voltage cannot change another voltage. Charge must be transfered.

OK, now you got it right. Yes, one cannot reach into a junction and apply the voltage directly. But a electric field from Vbe can influence the junction charges and thereby the voltage. It is all tied together with the formula Ic = Is*exp(Vbe/.026) .

Ratch

 

[BrownOut]

The electric field that is Vbe arises from neutrialized charge. I doesn't influence the junction, the junction influences it.

No external E field can influence the junction E field, because, for one reason, the external source is connected by highly conductive wires. And so, and E field from the external source dissappears in the conductors ( or nearly so ). Since any E field must be continuous to exist in a region, there is no way it can magically reappear at the junction.

 

[MrAl]

MrAl,



Do you mean post #19 of this thread? Yes, I did. You did not specify whether you are trying to determine its functional or causal classification. Did you read my comments about a true current source not having little or no voltage across its input terminals? It doesn't matter much, however. By adding external circuitry, we can make a voltage source into a current source or vise versa. All I am saying is that at the quantum level, it is voltage that determines the charge flow.

Ratch

Knowledge of any kind is obtained by dividing subjects down more and more until we get at the root of the subject. It doesnt really help to divide it down and then suddenly throw it all back together again which only causes confusion.

We learn things about a device by learning more and more subdivisions of how it works. If we start to classify it with a more broad viewpoint we actually might end up saying less about it then we did before.

If we wanted to do that we would end up saying something like, "The transistor is an object", which only says that it is one of many objects in the universe and doesnt really boil down to much, does it? We have to start to say what kind of object it is and maybe how it is made and maybe how it works.

Case in point: the operation of the transistor; how it works. If we want to say that it is field controlled then so what, what does that buy us? Everything under the sun is field controlled so what good does that do? That doesnt tell us anything more about it in general than any other device, including a resistor. We can propose a theory of how the transistor works inside by mentioning how the electron shifts bands after the application of an electric field, but that's just quantum mechanics, and that happens in every device. We want some definitions that are going to help us to use the transistor in real life, and most of the people coming to this site will want that same thing. It might be ok to talk about it that way, but it should be held in proper context.

Question: Is the thermoelectric effect caused by heat or by movement of charge?
Answer: Heat initiates the movement of the charges but without the movement of charges we wouldnt have the thermoelectric effect.

If you want to say that a current source can not operate without an applied voltage to its input, that's ok, but then i can say that if you want to apply a voltage then you first have to separate charges.

What we say about the field causing the movement of charges will most likely be superseded by String Theory's vibrating strings anyway, so maybe i can say that it's not the field really it's the vibration of tiny strings. Do we then say that the transistor is "string controlled"?

Take two small metal balls and hold them together tightly and apply a voltage across the two. The two balls in contact create a short circuit which has zero capacitance. Move them even a tiny fraction of a millimeter apart and now we have capacitance to overcome if we wish to energize the two balls, and to do that we have to move charge where there was no charge previously, and that means once we apply the electric field it will initially short out and become zero volts until that capacitance is satisfied. Thus, any device that has two terminals that do not make a short circuit will have input capacitance. It's true that there is inductance too in series with the voltage source so the voltage source itself does not go to zero, but the two balls potential, even though we just applied an electric field, will go to zero before any current flows and as the voltage increases so does the current. So if we want to say that there is a theory that inside the transistor fields are making everything happen, we are taking a very limited viewpoint where we set things up the way we want to before we start analyzing anything. To really understand we have to let the transistor be as natural as possible, and when we do limit things we are going to allow in the discussion we have to make it clear that that is what we are doing, and also that it may not be the best view to take anyway.

Recap:
Does the transistor internally depend on fields to operate? Yes, but that's not the whole picture. We have to consider the movement of charges too, and also taking the field view doesnt help when we have to deal with a transistor in the real world. Perhaps trying to predict new kinds of devices the quantum view is great, but we arent trying to do that here.

 

[Ratchit]

BrownOut,

The electric field that is Vbe arises from neutrialized charge. I doesn't influence the junction, the junction influences it.

No external E field can influence the junction E field, because, for one reason, the external source is connected by highly conductive wires. And so, and E field from the external source dissappears in the conductors ( or nearly so ). Since any E field must be continuous to exist in a region, there is no way it can magically reappear at the junction.

How can any field arise from neutralized charge? If there is not a net charge, there is no field. How does a electric field appear across a capacitor connected to a voltage source with highly conductive wires? Why does the formula Ic = Is*exp(Vbe/0.26) work?

Ratch

 

[BrownOut]

How can any field arise from neutralized charge? If there is not a net charge, there is no field.

You already know the answer to that. A E field exists in an unbiased junction because of uncovered charge. No voltage can be measured at the terminals because the E field exists only in a state of equilibrium. The vbe field opposes that initial E field. If some of the charge is covered, the original field is reduced by vbe, and that's the voltage that exists at the device's pins.

Why does the formula Ic = Is*exp(Vbe/0.26) work

The formula relates the voltage at the terminals to collector current. It works because of the charge distribution in the base region, and the diffusion current that results.

 

[Ratchit]

BrownOut,

You already know the answer to that. A E field exists in an unbiased junction because of uncovered charge. No voltage can be measured at the terminals because the E field exists only in a state of equilibrium. The vbe field opposes that initial E field. If some of the charge is covered, the original field is reduced by vbe, and that's the voltage that exists at the device's pins.

OK, just to be correct, the charge at the junction is separated, not neutralized. So isn't that what I said earlier, when I said that the barrier voltage caused by the uncovered charges, the voltage from the current existing in the bulk NP slabs, and Vbe were in equilibrium with each other, and act as if they were independent voltage sources in series with each other? Even though the junction voltage cannot be measured at the E-B terminals, it still influences the charge flow.

The formula relates the voltage at the terminals to collector current. It works because of the charge distribution in the base region, and the diffusion current that results.

I have no trouble with that explanation, knowing that the voltage at the E-B terminals caused the base region charge redistribution because of the electric field it induced.

Ratch

 

[MrAl]

Hello again,

Ratch, i thnk you missed post #53.

 

[Ratchit]

MrAl,

Ratch, i thnk you missed post #53.

I read it, but I thought is was more of a philosophical discourse than something to settle an argument. Yes, fields are with us everywhere, and in the transistor internals, both field and charge movement abound. What else can I say?

Ratch

 

[BrownOut]

OK, just to be correct, the charge at the junction is separated, not neutralized.

yes, it's neutralized. It is also seperated.

So isn't that what I said earlier, when I said that the barrier voltage caused by the uncovered charges, the voltage from the current existing in the bulk NP slabs, and Vbe were in equilibrium with each other, and act as if they were independent voltage sources in series with each other?

With nonzero vbe, the junction voltage is no longer in equilibrium. It is lowered by charge flowing into the region.

I have no trouble with that explanation, knowing that the voltage at the E-B terminals caused the base region charge redistribution because of the electric field it induced.

There is no redistribution until external charge flows into the base region.

 

[Ratchit]

BrownOut,

With nonzero vbe, the junction voltage is no longer in equilibrium. It is lowered by charge flowing into the region.

Or the lowered voltage allows more charge to flow because there is less voltage to oppose the diffusion voltage.

There is no redistribution until external charge flows into the base region.

That kind of defines redistribution, doesn't it?

Ratch