# Basic electronics theory (response to Layman's issues)

Discussion in 'Architecture & Engineering' started by billvon, Nov 18, 2014.

1. ### billvonValued Senior Member

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Yes. That is capacitive, rather than inductive, coupling.
The formula for voltage induced in a coil due to the magnetic field from an electromagnet is V=It*M*w, where It is the electromagnet coil current, M is the mutual inductance between the two coils and w (omega) is the frequency.

Here's a challenge - calculate the voltage induced when the frequency is zero.

3. ### billvonValued Senior Member

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No, it can't. If the first transistor is on, then the collector is pulled down very close to ground (within about .4 volts for most bipolars.) The base of the second transistor must exceed .6 volts to turn on. Thus when the first transistor is on, the second one is off. (This by the way is how very early RTL inverters worked.)
Faraday's Law talks about voltages induced in conductors. Thus it is common to induce voltages in open circuits, and the voltage created is described via Faraday's Law.

5. ### LaymanTotally Internally ReflectedValued Senior Member

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Well, it would have been pulled closer to ground if it was tied directly to the base instead, like the amplifier that inverted the signal. It would be activated by a positive signal, the power source is positive and ground is more negative. Then the emitter would be more negative than the base or have positive bias across the base and the emitter, hence creating a positive voltage at the base when the first transistor is on. Just that subtle difference in the circuitry between it and the amplifier circuit would change it's operation. It is just the values of the components can differ for them to be exactly the same with a different power source.

7. ### TrippyALEA IACTA ESTStaff Member

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Re-read what I said, this time take your time and make sure you understand it.

Again, you haven't understood what I've said.

Correct.

The power regulatur of the hall sensor is irrelevant to anything outside the hall sensor.

Correct, now re-read what I actually said and take some time to understand it. The output draws its power directly from the power rail, not the regulated power supply.

Unless you're actually being that pedantic...

Oh well, that's something then, I guess.

It's a switching circuit, not an amplifier circuit - we can tell this because we could swap the transistors for relays and the motor would still work.

...
I don't know what to say to this... I'm actually lost for words. Have you done any of your own research in this discussion? Or are you making stuff up as you go along?

Why would 'it' (presumably here you mean an amplifier circuit) be neccessary? You realize that an amplifier circuit can only as high as the power rail, right?

:Roll:

Then it's both possible and probable that I have done electronics more recently than you (it's been a hobby for a good chunk of my life and I've studied an engineering degree specialized in electronics and electrical engineering).

Oi vey. And with that the conversation slides backwards, and we get right back to the point I made about the inductors in the circuit.

The motor will not work with all four inductors active at the same time, they need to be triggered as alternating opposing pairs, otherwise the rotor experiences equal attractive and repulsive forces.

Correct - moving charges create a magnetic field, moving magnetic fields cause charges to move.

No. A coil has the same effect as using a large number of paralell wires, the magnetic fields add. We wrap the wires in a coil because the way the magnetic fields add results in a dipolar magnetic field (rather than the circular on eassociated with a current carrying conductor.

Completely wrong. An electric motor requires a rotating magnetic field in order to operate.

Hey, I actually agree with you. Wow.

8. ### TrippyALEA IACTA ESTStaff Member

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But then you loose the feedback loop that means that when the first transistor is off, the second one is switched on, instead, they're both being switched by the same signal and are in the same state.

9. ### Aqueous Idflat Earth skepticValued Senior Member

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billvon gave you a tutorial on basic electronics. Trippy gave you examples of various applications of transistors supplementing what bill on said. Why not just thank them for taking the time to explain technical stuff you aren't very familiar with?

It looks like you haven't understood that the transistor amplifier in standard configuration uses a DC bias at the base, and a DC blocking capacitor to allow the bias voltage to add to the input signal, which you keep insisting must be "AC". Further, there are typically two circuits, input and output, in which the relationship is Vout = A Vin, where A is the gain of the stage in question.

The other configuration of bipolar transistors you haven't understood are the classes of circuits which have no such DC bias. In this case they simply saturate. They may or may not exhibit gain, but that's not a reason to call them amplifiers, as in the term "class A amplifier", or any other language which implies gain with low distortion.

billvon didn't point this out immediately but he covered it by about post #30 or so.

But given the superior scholarship of both billvon & Trippy, you be better off either accepting their explanations, or else asking for clarification as needed.

12v was a convention that came into use with the IBM PC and its clones. There are several reasons for using 12v. One of them is to add isolation between the transients in the fan and the digital devices in the PC.

One of your faults is a failure to recognize superior advice from folks better trained than you, and too much reliance on your own beliefs instead.
Here, your unfamiliarity with linear and nonlinear applications of bipolar transistors was your weakness.

Your posts reflect a lack of training in electronics engineering, a program billvon outlined for you in your opening vollies with him. Therein lies the rub.

That is not what he said.

It's impossible, and if it were possible, no such circuit would be required.
That is only partly true. The wire need not be straight.

No, not without external energy. How many times have you ignored this advice?

So far you alone are confused. Energy must be spent, from a power supply of some kind. Again: a permanent magnet is not a power supply.

Faraday's law is one of several rules dictated by Nature. They can't be violated. The rest of what you said makes no sense.

That makes no sense.

The term is "loop". No current flows without it. The more general term Is "network".
You still haven't understood practical voltage dividers.

Last edited: Nov 22, 2014
10. ### LaymanTotally Internally ReflectedValued Senior Member

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When
Transistors are very sensitive to have the correct amount of small voltages to create the bias in them. If the first transistor was off, it would have a very large voltage across the base and emitter. I think that would be too high of a voltage difference to activate it.

11. ### leopoldValued Senior Member

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coils do not make the field rotate.
they concentrate the field.
they also offer a varying resistance to AC called inductive reactance which increases as the frequency is increased.
a difference of potential no way implies a current.
a charged capacitor for example.

12. ### LaymanTotally Internally ReflectedValued Senior Member

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I specialized in interpreting circuit diagrams to know what type of signals would be present in them in order to troubleshoot them as a technician. Often times engineers have problems doing this, and I have heard of engineers not getting jobs as technicians because of this. Then you assume everything they said was correct, and everything I said was wrong. I think we have both learned some things here from each other, and it seems like it has been a good opportunity to refresh on the subject.

13. ### LaymanTotally Internally ReflectedValued Senior Member

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A capacitor works in a lot of the same way as the symbol dictates. A capacitor is an open circuit (gap in the line). Current cannot flow between this gap in a DC circuit because of this. Although, one plate in the capacitor can create a magnetic field that will affect the other plate. Then AC can pass through the capacitor as though that line is connected. The other plate can "feel" the changes in the voltage of the other side and alter the other side to match. Then in no way does a charged capacitor prove that a voltage difference doesn't allow for current. Current isn't allowed in a DC circuit with a capacitor, because there is a gap in the line preventing it. Then capacitors do not make use of rotating magnetic fields...

On the other hand, current could flow through a capictor with a constant voltage if the voltage is high enough to create a spark between the two plates, like lightning for instance.

14. ### TrippyALEA IACTA ESTStaff Member

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Key word there is often times. I've designed circuits and artwork from scratch, I was also the only person in a class of thirty to pick up a flaw in a circuit template designed by the instructor.

15. ### billvonValued Senior Member

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Depends on the drive signal. In this case since the output of the Hall sensor is almost certainly an open-collector low side driver, the two drive signals will be effectively identical.

Uh . . . yes. Small differences in circuits do change how it works. Which is why I think it's pretty dumb to say "but if you think this circuit looks like this one, your obviously suffering from some sort of brain damage!" If you are really evaluating a circuit on "how it looks" or how much it looks like some other circuit you will be wrong almost every time. You have to be able to analyze the circuit from first principles.

16. ### LaymanTotally Internally ReflectedValued Senior Member

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I would have a very difficult time making a circuit from scratch, but it would be a lot easier for me to understand the operation of circuit that is already made. Most of the time, calculations are not necessary to diagnose an electronic circuit. You have to try to be able to understand electron flow, but you guys really don't make it easy to describe it.

17. ### Aqueous Idflat Earth skepticValued Senior Member

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As I said, you lack the training to correctly follow the advice you're getting.
It sounds like you have problems with stereotyping. In any case, your preparation in the theory of electronic networks is inadequate.

Why would an engineer seek a position as a tech and why would an employer seek an EE to do a tech's job? In any case, the guiding principles in this thread were established by billvon and Trippy.
No you alone are making assumptions. I am confirming their literacy in this topic, and your illiteracy, that's all.
It has been entirely a one way exchange.

18. ### TrippyALEA IACTA ESTStaff Member

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You don't even understand capacitors.

No. DC current flows through a capacitor while the capcitor is charging or discharging, this is why they are incorporated in monostable vibrators. The capacitor has a maximum amount of charge it can store in coloumbs, you put it in series with a resistor to limit the current, the number of coloumbs per second. You then attach this circuit to the base of a transitor and you get a circuit that stays on for a period of time (determined by how long it takes the capcitor to charge) and then switches off once the capcitor is charged. Monostable vibrators work because, essentially, the second transistor provides a path to discharge the capacito that is switched on by switching off the first transistor.

No. It has nothing to do with magnetic fields. Capacitors operate on electric fields, not magnetic ones. Inducing a changing charge on plate, by charging or discharging the capacitor, or providing an AC voltage to it, induces a changing electric field between the plates, which in turn induces a change in the charge on the opposing plate.

Wrong. Capacitors will allow a DC current to pass while charging or discharging. The only requirement for a capacitor to pass a current is that the current is changing - this is also why a charged capacitor acts as an open circuit, because the electric field isn't changing, so no current is induced.

Wair, wut?

...

19. ### billvonValued Senior Member

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No, capacitors do not create magnetic fields. They create electric fields (and they are definitely not the same thing.)
AC current does indeed flow through capacitors. That is why they are used.

Trippy likes this.
20. ### TrippyALEA IACTA ESTStaff Member

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Somewhere around I have a diagram for building a tesla coil that uses oven trays to build a capacitor.

21. ### Dr_ToadIt's green!Valued Senior Member

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(Sorry, missed the 'replied to' message)

Aren't there some sites that can help with introductory electronics you could recommend, instead of this? (Even though the refresher is nice. Isn't this just something intelligent people should just know?)

22. ### Aqueous Idflat Earth skepticValued Senior Member

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You should try to learn how capacitors work. After that, you should learn how stepping motors work. After that, you would be better prepared to follow the answers you're getting.
Whoa dude. You really are confused.

23. ### Dr_ToadIt's green!Valued Senior Member

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Off-topic, but you reminded of a friend who made his own memory boards in a toaster oven. Man...