# 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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I have started this thread to address some of Layman's misconceptions about electronics. It was originally from a thread about magnet motors, and it got bogged down because Layman seemed to disagree with everything once basic motor drives were brought up. To begin:

Of course electrons come from the ground, and of course they are negative. A negative flow from - to + is mathematically equivalent to a positive flow from + to -, which is why current is shown flowing from + to - in all electronics analysis. In semiconductor physics, physicists describe the flow of charge carriers (electrons) and holes (the opposite of electrons; a "void" where the electron normally goes.) One is positive, one is negative. P-type semiconductors have an excess of holes; N-type semiconductors have an excess of electrons.

FETs are field effect transistors. They comprise a large group of three terminal semiconductor devices. They operate on voltage. A FET is off as long as the voltage on its gate is below its threshold voltage. As the gate voltage (specifically the Vgs voltage) increases, current starts to flow through the drain. At a sufficiently high Vgs the drain current is as high as possible, and at that point current is limited only by the device's inherent resistance, called Rds_on. DC current does not flow into the gate of most FETs.

In some ways, a bipolar transistor is the opposite. It is also a three terminal device, but uses current instead of voltage to turn its base on. When the base current exceeds a threshold, then current begins to flow between the collector and emitter. The voltage during this time remains the same (around .6 volts.) It again increases up to a maximum; after a certain base current the device is said to be saturated, and no further current can flow.

The important point to remember is that bipolars are current operated devices (NOT voltage) and FETs are voltage operated devices (NOT current.)

3. ### billvonValued Senior Member

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The circuit in question is here:

In this circuit, the first transistor functions primarily as a power switch. When the Hall sensor detects the rotor is in the correct position, then its output goes open circuit. This allows the 1K resistor to apply a current to the base of the first transistor. The first transistor turns on, pulling its collector down to near ground and allowing current to flow through the first winding.
This has the effect of turning OFF the second transistor, since its base is connected to the collector of the first transistor. Thus when the first transistor is off the second is on and vice versa. Thus the secondary function of the first transistor is to function as an inverter. It is not an amplifier since the amplitude of the signal at the collector of the first transistor is the same as the signal at the collector of the second transistor.

5. ### LaymanTotally Internally ReflectedValued Senior Member

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That would just mean that the base of the second transistor would have the same signal going into it's base as the signal that was amplified from the first transistor. Often times a two stage amplifier will be arranged in that pattern. If you just forget about the motor itself, and redraw the circuit so that the motor is separate lines of a power source and the emitter is two separate lines of a ground source, you are just left with a 2 stage amplifier...

7. ### billvonValued Senior Member

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Incorrect. The signal is not the same; it is phase reversed. In other words, the first transistor is off when the second transistor is on. That is why it is connected like this - by alternating power to the coils you create the rotating magnetic field essential for motor operation. It could be done explicitly by using two Hall sensors or an inverter before the second base, but this is simpler and cheaper.

8. ### LaymanTotally Internally ReflectedValued Senior Member

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It would be more like just this diagram.

The only difference here really, is just that the collector isn't tied directly to the base of the second transistor and it is connected to the top of the resistor instead.

I guess the signal would be inverted, so it could actually act as a 2 stage amplifier and br inverted. I would assume that is so that the inductors could be activated at opposite times that are on opposite sides of each other. That would act against the permanent magnets better.

Last edited: Nov 18, 2014
9. ### billvonValued Senior Member

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It is indeed similar. (Note that the waveforms shown are NOT the same - they are phase reversed, or inverted, just as they are in the motor drive.) The big differences are:

1) In this circuit there is negative feedback provided by the base-emitter resistor. This tends to keep the device in its linear range. The motor drive circuit does not contain this resistor and thus operates in saturation. (Which means there is no amplification between the first and second stages, just inversion.)

2) In this circuit the base is DC-isolated by the capacitors. This allows the transistors to bias themselves in their linear range, which is important for an amplifier. The motor drive circuit does not use DC blocking caps because it operates in saturation. This is OK since the second stage does not operate as an amplifier.

10. ### LaymanTotally Internally ReflectedValued Senior Member

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I agree, it does invert the signal, that way the line going to two inductors would be off at the same time the other two inductors would be on. Got's me thinking that it is just 1 amplifier and 1 inverter now.

11. ### TrippyALEA IACTA ESTStaff Member

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You still haven't explained why you think amplification is neccessary. In the other thread I posted the data-sheet for the 6851 - the hall effect sensor used in this circuit.

Here it is again: http://pdf.datasheetcatalog.com/datasheet/panasonic/SPC00005CEB.pdf

It has a low voltage at 30mT of up to 0.4 v and a high voltage at -30mT of up to 14.6v (with a 16v power rail).

12. ### billvonValued Senior Member

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Layman, you seem to think (from the other thread) that you cannot get an AC waveform if there is a diode in the source. Below is an example of a motor drive (single coil) that generates an AC waveform with a DC blocking diode on its power supply.

13. ### TrippyALEA IACTA ESTStaff Member

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Here's my example form the previous thread:

And when I get home I'll modify it to include the protection diode.

14. ### LaymanTotally Internally ReflectedValued Senior Member

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You two seem to be confusing the diode with a zerner diode.

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

"A Zener diode is a diode which allows current to flow in the forward direction in the same manner as an ideal diode, but also permits it to flow in the reverse direction when the voltage is above a certain value known as the breakdown voltage, "Zener knee voltage", "Zener voltage", "avalanche point", or "peak inverse voltage"."

A normal diode doesn't allow current to flow in the opposite direction, that is they whole point or purpose to using a diode.

15. ### billvonValued Senior Member

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Nope. A protection diode on the input can either be a zener (to shunt the input in the event of an overvoltage or reverse polarity) or a standard diode (to open circuit the input in the event of a reverse polarity.) My schematic shows a standard diode used in the latter fashion.
Right. Yet amazingly, both circuits above can produce AC, and can do so with a standard diode in series with the input!

16. ### LaymanTotally Internally ReflectedValued Senior Member

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This circuit has 2 power supplies. V2 could allow for AC.

17. ### LaymanTotally Internally ReflectedValued Senior Member

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I already stated that it would be necessary in order to create a high enough voltage to create a large enough magnetic field from the inductors. The inductors only have a transistor between them and ground. The negative voltage is on the ground side, hence it would have a larger negative flow of electrons being separated by the resistors from the power supply. That way the inductors have more electrons to create the electromagnetic force needed.

18. ### LaymanTotally Internally ReflectedValued Senior Member

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Look at this circuit diagram.

Now does this look more like this...

Or do you think it looks more like this..

If you said the second one again, your obviously suffering from some sort of brain damage! Then you shouldn't let the sine waves in the amplifier circuit fool you. It can function just the same with a square wave. The diagram just doesn't have the capacitor C2, because being a DC circuit that would make it on open circuit!

19. ### TrippyALEA IACTA ESTStaff Member

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Apparently you still haven't understood my points. A second amplifier isn't needed in that circuit and the maximum output voltage of the 6851 is higher than the voltage difference between the power rail and the ground.

20. ### TrippyALEA IACTA ESTStaff Member

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The only person expecting it to look like the second diagram is you. I didn't claim it looked like the second circuit and that's not why I posted the second circuit. The second circuit was posted to illustrate a separate point - the most I said was that to me it resembled a monostable vibrator, and Billvon said yes, but it's being driven by the Hall sensor rather than a capacitor.

21. ### TrippyALEA IACTA ESTStaff Member

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This is the vibrator I originally posted:

You also still haven't addressed the point that I have made regarding the polarity of the inductors and the need to switch them alternately.

22. ### billvonValued Senior Member

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No. V2 is a signal source and is only there to provide the on/off control for the transistors. However if that confuses you here is the circuit without the signal source: