Here's something I do know! When you make a quote, as in all programming languages alike this, the syntax is {quote}<text>{/quote} (replace the curly brackets with normal brackets). When you end a command (or wrap or whatever), you always put a / in front of the command before the [.
Think it would be OK to start a new thread named "help with the basics" in the everything else forum where everybody can tell me what I don't know?
Hi keantoken, yes start a new thread, this one got a bit out of hand...
Have you already understood the 4.3ma ?
Mike
Have you already understood the 4.3ma ?
Mike
MikeB said:
Thread is Jacked up by the phenomena which is known as Threadjacking, but do you know who is really responsible for all this????😀 😉
mikeks said:
What, too high gain and supernatural feedback?
It is entirely possible to have too few words in a post 🙂
Okay, I looked at that post with 4.3mA and such by mikeB, and as always, I started seeing relations between the numbers....
If Rc1 and Rc2 don't modify the currents (because of the ideal transistors?), then the current through Re without the transistors would be 6mA. Then I saw that the input voltages after the voltage drop of Vbe would be -50mV (.7V-.65V), making current on each side -.05mA, combined in Re would be -.1mA+Ic1+Ic2. And since 6.0mA-4.3mA=1.7mA, then there has to be something taking 1.6mA from the emitters. Do these calculations have anything to do with voltage drop?
😕 😕 😕
If Rc1 and Rc2 don't modify the currents (because of the ideal transistors?), then the current through Re without the transistors would be 6mA. Then I saw that the input voltages after the voltage drop of Vbe would be -50mV (.7V-.65V), making current on each side -.05mA, combined in Re would be -.1mA+Ic1+Ic2. And since 6.0mA-4.3mA=1.7mA, then there has to be something taking 1.6mA from the emitters. Do these calculations have anything to do with voltage drop?
😕 😕 😕
I will start a new thread in the evreything else forum named "help with the basics".
No threadjacking!
No threadjacking!
keantoken said:
I am sorry, but your calculations reveal that you have no real notion of even the simplest electrical circuit, let alone one with a transistor - which makes most of our collective attempts at guiding you rather futile. As the saying goes, you can lead the horse to water but you can't make it drink.
Without the transistors, NO current would flow as there would not be a complete and closed circuit in sight, where it could do so. Current cannot just appear and dissapear from a node unless there is a way to flow in or out of a node, i.e. a CONNECTION.
Further, because we are now at a 'basics of the basics' stage here, I will possibly be just wasting time trying to explain basic transistor behaviour.
Assumptions: NPN, no base current, and Vbe=0.7V.
The first assumption is there so we don't have to complicate things by using negatives. The second and third are there because they make calculations simpler - one removes the relationship between Ib and Ic / Ie (Normally this is Ie=Ib+Ic and Ic=Beta*Ib). Therefore, it means Ic=Ie, the other is necessary because a transistor requires a Vbe over a certain threshold, in order to even work as an amplifier, i.e. for the above relationship between Ib, Ic, Ie to exist (even for the ideal case of Ib=0!). Also, Vc must be at least as high as Vb or more positive, for the transistor to amplify. As long as this is satisfied, the magnitude of Vc does not influence Ic, Ib and therefore Ie, for the ideal case. You will find that it also does not do so in any manner signifficant enough to affect basic calculations such as these, for real world transistors - i.e. real world transistors are sufficiently 'perfect' for this simplification to be valid.
The net result of all this is that the transistor will keep it's Ve, Vbe (0.7V) below Vb, as long as the emitter has a path for current that goes to a node with a potential >=0.7V less than Vb. This path will normally be a resistor to a power supply. Take note that 'ground' is also a power supply, with a potential of 0. How does the emitter keep itself at a potential 0.7V less than the base? By alowing current to flow from the collector into the emitter, and create a voltage drop in the emitter resistor, such that the emitter is 0.7V below base potential. This of course assumes that Vc > Ve in the first place, i.e. there is a path from the collector to a node with a potential higher than Ve - and also, see above, Vb. This very action is the root of the transistors ability to amplify.
Applied to the simple LTP example above, what is the emitter current?
The emitter voltage, assuming the transistor operates in it's active region as explained above, is always base potential less 0.7V.
Since base potential was given at -1V, this is -1.7V. Re is the path to a potential less than that, speciffically -6V. Therefore, the emitter CAN be 0.7V less than base so our initial assumption that Ve = Vb-0.7 is correct.
The difference of potentials, i.e. voltage across Re is therefore 4.3V. Because the resistor is 1k, and Ohms law states I=U/R, the current is 4.3mA - and this is the emitter current for both transistors. It has to flow from them, there is no other path!
Because both transistors are the same, the current here splits equally into the two collectors.
Finally, we have to check the other requirement for a transistor to operate as an amplifier, which is Uc>=Ub. Because we know, assuming infinite gain == no Ib, that Ic=Ie, and Ie=4.3mA/2, we can calculate the voltage drop on each Rc. This, again, using Ohms law, is Ic * Rc - again, this is the only way the current can flow, no other path is there! This comes out as 2.15V.
Because the voltage across Rc is dropped 'from' a potential of 6V (power supply), it subtracts from it giving us the voltage on the collectors. This would be 6-2.15 = 3.85V. Since we know Vb=-1V (given) this obviously means Vc is higher, therefore all the requirements for transistor operation are satisfied.
You may be surprised to learn that this extremely simplified calculation will give you results amazingly close to a real world implementation of the circuit - which is why such calculations are used in the first place!
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