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

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

1. ### TrippyALEA IACTA ESTStaff Member

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No. I'm characterizing your argument. If I wanted to call you childish I would do so.

I'm quite willing to listen to reason, you simply haven't given me reason to change my mind.

As I've pointed out, repeatedly, an amplifier doesn't give a squarewave output from a sine wave input, when it behaves in that manner it's acting as an analog to digital converter - IE a switching circuit, not an amplifier. An amplifier preseves the original wave form without distortion. I've tried explaining this to you several times, but you keep confusing the issue by talking about how amplifiers "give square waves loving to" but that's got nothing to do with what I actually said.

This coming from the person that thinks that the orientation of the transistors is somehow important when drawn on a circuit diagram.

Best report me if you think I'm trolling you.

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3. ### LaymanTotally Internally ReflectedValued Senior Member

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I never said that it had a squarewave output from a sine wave input. I said it had a squarewave input and a squarewave output. I also gave several sources from wiki that mentioned motors running on DC circuits. You just ignored those.

There you go insulting me again. You would always rather assume that, because I don't know what I am talking about. That is where you are wrong. It has obviously gotten you thrown off completely. Why don't you try to go draw the circuit with both the transistors facing the same way and see what you get.

Reported.

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5. ### LaymanTotally Internally ReflectedValued Senior Member

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It would require a change in voltage to generate power or current. Moving an electric field close to a wire would create current, because it would change the voltage induced on the wire. Suffice it to say, no electrical circuits have inductors that vary in distance from the source to induce current. A voltage level would be induced on a circuit with a stationary magnet, but it wouldn't have any current without a pathway to ground. The key phrase here would be closed circuit.

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7. ### TrippyALEA IACTA ESTStaff Member

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I didn't say you said that - I said that was one of the points that I was making.

Then you don't understand the circuit - the output of the hall sensor varies linearly according to the strength of the magnetic field it sees. On a 16v power rail it gives a high output of at least 14.6v when it sees a magnetic field strength of -30mT and a low output of at most 0.4v when it sees a magnetic field strength of +30 mT. It's mounted to 'see' the magnetic field of the rotor which means it's going to see a varying magnetic field that varies in a sinusoidal fashion, and so will generate a sinusoidal output varying from (at a guess) between about +0.3v and 11v, which is being used to switch the coils on and off in alternating pairs. Hence, Sine qave input, square wave output.

No I didn't. I've implicitly or explicitly made two points in that regard.

1. having a DC powersource does not preclude generating an AC signal, neither does including a protecting diode in the circuit - you've been given examples of this..
2. DC motors still need rotating magnetic fields to work, and there are different ways of generating them. We're only talking about basic AC and DC theory here...

I only repeated what you yourself have said.

You get exactly the same circuit with an extra unconnected crossing in it - it just, visually, looks a little messier.

Generally telling people that you've reported them is considered to be inflammatory.

8. ### LaymanTotally Internally ReflectedValued Senior Member

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The input of the circuit is 12 volts... There is no output to the circuit besides running the fan...

Do you mean to say that a sine wave and a square wave is occupying the same lines now?

This really has gotten interesting.

Like to congratulate you on the bang up job your doing here teaching people about electronic circuitry, lol. Please note, I was commenting on the actual job your doing here, not you yourself personally.

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10. ### TrippyALEA IACTA ESTStaff Member

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Correct - and the figures I gave of 11v and 0.3 v were adjusted for a 12 volt power rail.

You're sure about that?

Not what I said.

No, you are, by your own definition, trolling.

11. ### TrippyALEA IACTA ESTStaff Member

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I'm 99% certain that that sentence does not mean what you seem to think it means. It's talking about the operating magnetic flux density, not the operational output. What it means is that it doesn't matter if you run it on 5 volts or 15 volts, it's still going to respond to magnetic fluxes in the range -30mT to 30mT. As opposed to the scenario where lower operating voltages might require higher flux densities to register.

12. ### LaymanTotally Internally ReflectedValued Senior Member

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http://en.wikipedia.org/wiki/Regulated_power_supply

"A regulated power supply is an embedded circuit; it converts unregulated AC into a constant DC. With the help of a rectifier it converts AC supply into DC. Its function is to supply a stable voltage (or less often current), to a circuit or device that must be operated within certain power supply limits. The output from the regulated power supply may be alternating or unidirectional, but is nearly always DC (Direct Current).[1]

The type of stabilization used may be restricted to ensuring that the output remains within certain limits under various load conditions, or it may also include compensation for variations in its own supply source. The latter is much more common today."

The hall sensor uses one of these.

13. ### TrippyALEA IACTA ESTStaff Member

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Take another look at the block diagram

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The output isn't connected to the regulated power supply.
Now, review what I said:
On a 16 volt power rail the hall sensor produces a voltage that varies linearly between 0.4 and 14.6 volts in response to a magnetic field varying linearly between -30 mT and +30mT. In the case of the motor circuit, because it sees a rotating magnetic field, it's going to produce a sinusoidal output varying between about 0.3 and 11 v.

14. ### TrippyALEA IACTA ESTStaff Member

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Irony - as it happens you're right, but for the wrong reasons (Oh how that must sting).

It has nothing to do with the stabilized power supply, but rather, the output of the Schmidt trigger, so yes, the output from this may in fact be a square-wave, but not because of anything to do with the nature of the stabilized power supply.

Regardless of that, it still doesn't mean that it's acting as anything other than a switching circuit.

15. ### leopoldValued Senior Member

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i was just about to comment on this.
yes, the shmidt will square the waveform into the output stage.
yes again, it has nothing to do with the regulated supply (within reason).
if the supply swings too wildly you could wind up with burned out components or a circuit that won't work because of inadequate supply voltage/current.

aside:
doesn't a hall sensor produce sharp spikes?
maybe i'm thinking of a different sensor.

edit:
there are both types, linear and "digital".
en.wikipedia.org/wiki/Hall_effect_sensor

Last edited: Nov 22, 2014
16. ### TrippyALEA IACTA ESTStaff Member

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The hall effect is the response of moving electircal charges to a static field. If you take a plate, pass a current through it, and a magnetic field through it at right angles to the current, then you get a voltage difference at right angles to the current and the magnetic field.

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All hall effect sensors produce an intrinsicly analogue signal - the stronger the magnetic field, the stronger the voltage. Some sensor packages are set up with things like schmitt triggers to effectily convert the analogue signal into a digital one.

In this case, the hall effect sensor produces an AC analogue signal, which is amplified (unless, I suppose, the amplifier is wired to act as a voltage comparitor and simply produces a signal when the input voltages are different), then passed through the schmitt trigger which filters for noise and converts the signal to digital DC signal which is in turn passed to the output stage to be expressed in terms of the voltage on the power rail - which leads us back to the question that I have asked Layman, which is, why he thinks amplification is neccessary in the first instance.

I still stand by my assertion that it is acting as a switching circuit rather than an amplifier because the transistors in the circuit are being operated with their bases above saturation - they're no longer producing an output that is proportional to the input.[/img]

17. ### TrippyALEA IACTA ESTStaff Member

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The hall effect is a linear response. Hall effect sensors have additional components in them in addition to the hall element that determine whether the output from the sensor is linear or on/off, however, the output from the hall element is always linear.

18. ### leopoldValued Senior Member

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my guess would be to isolate the sensor from the output stage.
it's been a long time and i'm not sure of a typical base resistance, but i assume it's much greater than the sensors, or vice versa.
in this instance the "amplifier" is simply an impedance matching device.

actually its the collector/emitter current that becomes saturated, and it's caused by varying the base current too far.
applying a strong enough sine wave base signal will result in an almost perfect square wave output.

19. ### TrippyALEA IACTA ESTStaff Member

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That's not amplification, that's isolation, and it's also not what Layman was referring to - he thinks that the transistors in this diagram:

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Are acting as amplifiers - my point is, and always has been that you could replace them with relays and the circuit would still function (If you could get relays able to switch fast enough).

He balked at the fact that I pointed out that it had more in common with a monostable flip-flop:

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I originally suggested that the hall sensor was in there to ensure that it starts in the correct phase, since then I have explicitly acknowledged that I may have been mistaken in that, and I now recognize that the hall sensor and the motor effectively replace the capacitor as the driver of this circuit and providing the so-called 'missing cross-connection' (which, ironically) was in essence how I regarded it originally)..

I know, I'm just lazy with my language.

Which is a point that I have been trying to make to Layman.

20. ### leopoldValued Senior Member

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not necessarily.
saturation of a high power mosfet can achieve the same thing.
they can sink large currents, and they have a high input impedance.
high power BJTs would work too, but they have a low input impedance and would need some kind of buffer (amplifier or the shmidt).
the relays trippy mentioned would work too up to a certain rpm.

the signal from the hall sensor in ALL of the above cases acts like a timing pulse, a trigger.
this in turn implies a digital, not an analog, operation.

21. ### LaymanTotally Internally ReflectedValued Senior Member

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In the circuit of the monostable vibrator, the base of the first transistor connects to the collector of the second transistor through the resistor R4. This means that when Q2 is active it will drain the base of Q1. In the computer fan circuit, the base of the first transistor is just connected to the hall sensor. This means that the monostable flip-flop would have a different type of interaction that exist which isn't present in the other circuit.

The collector from the first transistor is connected to the base of the second transistor like in the computer fan, but that isn't going to produce the same type of operation! Then both the emmiters are connected to ground, and that is nothing special. To say these are the same circuits, and produce the same type of operation is actually rather sloppy. The difference in having that one extra line would change operation completely. In now way shape or form, does the operation of the second transistor affect the operation of the first transistor. The signal has already been generated, so it would just act like an amplifier. Being a DC circuit, the capacitor connecting the collectors of the transistors would just allow for feedback from the inductor to be drained to ground while the second transistor is active. Otherwise, it would just be an open to current between the collectors.

There would be no reason for this type of operation to exist in the circuit, because the hall sensor would already produce a signal as it is already. Then it makes perfect sense why it would need a power regulator to prevent a type of feedback loop that would alter the signal being produced by the hall sensor from the magnets in the fan itself.

22. ### LaymanTotally Internally ReflectedValued Senior Member

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The output of the regulated power supply is connected to the power source of the output stage...

23. ### LaymanTotally Internally ReflectedValued Senior Member

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I don't actually believe that amplification is necessary. I just think that is the only function the transistors could serve in this arrangement. I would assume that this diagram was developed by an actual engineer, and that this circuit has actually produced fans that operate effectively. Then with a positive 12 volt power supply, it would appear to be necessary in this type of arrangement. That doesn't mean that a stronger power source couldn't be applied that would allow for it to run without them.

I am not saying that I am without fault. I just don't go out of my way to correct people when I have no idea what I am talking about (hint, hint). I haven't really done electronics in a long time. It kind of got me thinking about what Bvillion said earlier in this thread. Since, the collector of the first transistor is on the other side of the resistor of the line to the base of the second transistor, it would actually reverse the inversion effects of the amplifier. Thereby, it could create an output signal that is not inverted from the signal coming out of the first transistor. That means that both transistors could be active at the same time. That would make it more likely to coincide with the evidence we saw in the video of the experiment of making it run with 4 magnets.

A magnetic field is created when a current is running down a straight wire. Coils in that wire would make that magnetic field rotate. I believe this has become confused with having to actually rotating the wire physically (this is not necessary in any type of electronics). It really seems like the problem here comes from incorrect interpretations of definitions. Like Faraday's laws applies to a specific type of circuit (the closed circuit), and voltage can be present without the presence of power. I could agree that a closed circuit wouldn't generate power in that circumstance. If it applied to an open circuit, that would be different or the definition wouldn't specify that the circuit is closed. A voltage divider would generate current, by creating a difference in voltage.