Voltage doesn't travel. Learn your basic physics.
I need to know nothing of your state of mind, only what's real and what's imagined.
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.
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.
You don't seem to know what you call it.
It is a descriptive colloquial phrase. Learn your basic English.
That's the point. You cannot know what is imagined until you know my state of mind.
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.
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.
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?
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 )
I know what is imagined wether I know you or not.
I never made that claim. That's just the way you distort what I've said repeatedly.
That's my point exactly. The resistance is small and finite, thus the device is current controlled.
No, you didn't. I'm not confused, but you seem to be.
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 real and what is imagined.Thinking it may by so is not the same as knowing for sure.
Not explicitly, but implictly for sure.
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".
Are you saying I did not call it a transconductance amplfier in both cases? When both your quotes showed that I did?
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.
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.
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.
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.
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
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.
Why does the formula Ic = Is*exp(Vbe/0.26) work
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.
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.
Ratch, i thnk you missed post #53.
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?
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.
With nonzero vbe, the junction voltage is no longer in equilibrium. It is lowered by charge flowing into the region.
There is no redistribution until external charge flows into the base region.
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