Quote Claude: "Yes, gate current precedes gate-source voltage."
Claude, it took a long time until - I hope so - I was able to understand your way of thinking (and your definition of "control")..
To make it clear to me: I think, you would NOT agree that a dc motor is controlled by a dc voltage, correct?
Instead, you would say "current-controlled or field-controlled, or..."
That means: I am beginning to understand what you mean while saying "current precedes voltage" - however, for my opinion a rather "unconventional" way of decribing the working principles of transistors (BJT as well FET).
I am neither a scientist nor a physicist - I feel and think as an engineer.
That means: To me is the BJT a three-terminal device with a defined output signal (current Ic) and the possibility to "steer" (to "control") this output signal using the third terminal (base). For this purpose, I apply an electrical quantity to this node (base voltage against emitter potential).
In short: I am interested only in the external signal which is able to cause a change of the BJT`s output quantity.
And that`s what we are doing when we measure and record all the BJT parameters and characteristic curves (we cannot measure the diffusion voltage, for example).
More than that, for my opinion, this view is also in accordance with the general definition of the term "control":
The controlling quantity is applied to the external controlling input node (without the necessity to know in detail what happens inside the black box).
Coming back to my above example: That`s the reason we speak about a voltage-controlled dc motor. (I am not a specialist in motors, but I think its true).
Final question: Could you agree to the following?
The value of the collector current of a BJT is determined by an externally applied voltage between base and emitter terminals.
That`s what we can observe.
No! Externally applied voltage between base and emitter terminals will destroy the bjt. A bjt must be current driven. An externally applied current into the emitter is best. An externally applied current to the base results in beta dependency.
If I connected a 0.65 volt CVS across b-e junction, the ensuing emitter current is : Ies*exp((Vbe/Vt)-1). But Ies increases w/ temp non-linearly, while Vt increases as well in proportion to absolute temp. Thus Ies dominates. The 0.65 volts times the emitter current is power, which results in a temp increase. Ies goes up, Ie goes up, so power keeps going up, meaning Ies increases, Ie increases, power increases, temp increases, Ies increases, Ie increases ---------. We have a condition known as **thermal runaway**.
Instead, let us apply an external **current source** to the emitter terminal and force Ie to a fixed value, letting Vbe settle to whatever value Ies, and temp dictate. Ie is the value of the CCS, so Vbe=Vt*exp((Ie/Ies)+1). With Vbe*Ie we have power and temp increases. But notice here that Vt is in the numerator, and Ies is in the denominator. The slight rise in Vt is overwhelmed by the large rise in Ies, which forces Vbe downward. Instead of thermal runaway, the bjt settles at a safe value of Ie and Vbe. Of course it goes without saying that the current source value must be within the bjt limits, a 5 amp source into a tiny small signal device would destroy it.
Anyway, that is why bjt devices are always current driven, never voltage driven.
If the external source is a voltage source, it cannot be connected across the b-e junction. But inserting a resistor in between works fine, and is standard procedure. With a resistor, the thermal runaway condition cannot happen since the resistor limits the current.
If the supply is 12 volts, with a 10 kohm resistor, the current can never exceed 1.2 mA. As the bjt heats up, Ies increases, but the resistor voltage drop will eventually approach the value of the supply and equilibrium is reached. If Vbe is 0.70 V, then resistor drops 11.30 V, so Ie = 1.13 mA. At high temp a p-n junction can conduct large current with small voltage drop, so if Vbe drops to 0.60 V, Ie = 1.14 mA. If Vbe = 0.80 V, Ie = 1.12 mA. Resistors in series stabilize the operating point and prevent thermal runaway.
SO a bjt stage, can be driven or controlled by either a voltage source or current source. With a current source, one can drive the b-e junction directly. But with a voltage source, a resistor is needed. By inserting a resistor in series with the voltage source, we are fixing the emitter current to a specific value, and relying upon alpha for collector current, alpha being very predictable. If we insert the resistor on the base side and fix Ib, then Ic is equal to beta*Ib, which is a beta dependent circuit, not desirable in most applications.
A LIn topology 3-stage op amp or power amp has a differential front stage, common emitter 2nd stage, and emitter follower or complimentary feedback pair output buffer 3rd stage. The diff amp stage outputs a current at one collector of the diff pair. This current feeds the b-e junction of 2nd stage bjt. The 2nd stage output current = Ic2 = beta2*Ib2. In an op amp, global open loop gain is beta dependent. The Aol value varies w/ 2nd stage beta value, it's a fact. Do I make sense? BR.
Claude
