Positive or Negative Alternation of the sine wave developed across the load.

[GoldDigger]

The voltage at the top of the resistor is positive with respect to the bottom end of the resistor because electron flow is always out the top of the resistor.
That current through the resistor causes a voltage difference across the same resistor.
If you cannot see that, then I am afraid it will take some face to face hand waving to figure out just where your problem in understanding lies.

If instead of the transformer and diode bridge you had only a battery with the + pole at the top are you comfortable with electron current going up through the resistor?
With no other elements in the circuit the voltage at the top of the resistor must be positive, since it is at exactly the same potential as the battery top.

Now add the diode oriented with the "arrow" pointing at the resistor. Nothing changes except that the resistor top voltage is lower by the forward diode drop. (.1 to 1.0 volts depending on the diode type and the current)

P.S. The x-axis voltage is *arbitrarily* chosen to be zero at the bottom of the resistor. You have to choose a reference point somewhere.

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[wwhitney]

How can I tell whether the wave is going to be a positive one or a negative one simply by looking at the circuit?

So the diagrams from the OP show a sine wave next to the voltage source, and show presumptive current half-waves next to the top of the resistor.

Unspecified is the sign convention for each of these graphs. If we assume that the diagrams are for conventional charge (positive charge carriers), then one sign choice would be that positive charge carriers flow out of the top of the voltage source when the voltage is positive. And one convention for measuring current through the node at the top of the resistor is that current flowing towards the resistor is positive current.

Those two choice produce the current half-waves shown. In the second diagram, during the first half-cycle (positive voltage), current flows out of the voltage source at the top, through the diode (which is forward biased), and through the measurement node into the resistor. During the second half-cycle (negative voltage), the diode is reverse biased, and no current flows. That produces the current graph shown.

The situation in the first diagram is the opposite: no current flow during the first half-cycle, and during the second half cycle, current flows out of the bottom of the voltage source, through the resistor, through the measurement node, and through the forward-biased diode. Since the current is now flowing out of the resistor through the measurement node, the current flow is negative. This produces the current graph shown.

Now, I don't actually know if there are generally understood standards for which sign convention should be made in each case, I just chose the sign conventions that give the current graphs shown.

Cheers, Wayne

 

[GoldDigger]

You do not have any freedom or wiggle room in the sign convention for voltage, since nobody disagrees with the charge on an electron as being negative.
The sign convention of voltage by which electric field is defined in terms of the force exerted on a unit positive charge and voltage as being the scalar potential field of the vector electric field.
So that leaves us with the someway unfortunate, but usable, resolution that when you use electron flow to choose the sign of current, Ohms Law becomes E=-IR. Live with it.
The equations for magnetic field suffer the same problem, because both electric field and magnetic field were given their sign conventions at a time when everybody used conventional current.

Incidentally, the term conventional does not have any connotation of obviousness, fundamental rightness, etc.
It reflects the fact that you can define current any way you want (in terms of sign and its relation to the real majority charge carriers in any material. So the choice of the flow of positive charges (or equivalently the negative of the flow of negative charges) is an arbitrary decision, that is a convention, which was adopted in the early days.

 

[wwhitney]

You do not have any freedom or wiggle room in the sign convention for voltage

OK, but when one writes a vertical squiggle symbol for an AC voltage source, and specifies that voltage is positive during the first half-cycle (as the little graph next to it shows), is it not an arbitrary choice whether that means the positive charge carriers are flowing out of the top of the vertical squiggle symbol during the first half-cycle, as opposed to out of the bottom of the vertical squiggle symbol? Is there a convention?

Cheers, Wayne

 

[GoldDigger]

Showing the top half sine wave above the x axis means the the voltage is zero for one half wave and positive for the other half. That is how a graph works. You do not need to assume a time correlation with the full sine wave of the source to determine that.
The graph which shows a half flat line with a negative going half wave, on the other hand, means that the top of the resistor will be negative with respect to the bottom end. Nothing arbitrary about that.

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[wwhitney]

Showing the top half sine wave above the x axis means the the voltage is zero for one half wave and positive for the other half.

Hold on there a second. In the OP, each of the two diagrams has two graphs. One is a graph of sin(x) for 0 < x < 2*pi, that's next to the voltage source in each diagram. The other graph is a graph of sin(x) * Phi(x), where Phi(x) is 0 or 1, depending on whether 0 < x < pi or pi < x < 2*pi, and depending on which diagram we are looking at. This second graph is next to a node, just above the resistor.

My initial impression was that the graph of sin(x) was a voltage graph, since it was next to the voltage source, and that the second graph was a current graph, since it was next to a node. My responses so far have been based on that interpretation.

Your interpretation is that they are both voltage graphs? So for the sin(x) graph, you infer that 0 volts is defined as the bottom of the voltage source, the sin(x) graph is the voltage at the top of the voltage source, and the sin(x) * Phi(x) graph is the voltage at the node just above the resistor?

OK that works. And I agree that if both graphs are voltage graphs, then 0 volts has to be the bottom of the voltage source, in order for the other graphs to be correct voltage graphs.

But absent any other information, if I write a vertical voltage source squiggle and write a voltage sine wave graph next to it, is there any convention on whether 0 volts is defined as the bottom of the squiggle or the top of the squiggle?

Cheers, Wayne

 

[gar]

170627-2425 EDT

lbmcse:

Get the following items or equivalents:

A Simpson 260 or a Fluke or some similar instrument to use in the ohmmeter mode.

A high impedance DVM such as a Fluke 27 to used a in a DC volt mode.

A diode, almost any will do, such as a 1N4148 or 1N4004.

A 6 V battery or low voltage DC power supply.

A 1 k to 5.6 k resistor. Power rating should be greater than 2*V^2/R. For 1 k and 6 V use something greater than or equal to 1/2 W.

Determine the voltage polarity of the ohmmeter with the DVM.

Use the ohmmeter to determine the diode resistance at whatever balance occurs for both diode orientations.

Use the series combination of the battery, diode, and resistor with the battery in each of its possible orientations and measure voltages around the loop. You could also put a milliampere meter in series with the loop.

A safe milliampere meter in this circuit is an auto-ranging voltmeter across a 1 k resistor inserted in the loop. One volt across 1 k is 1 mA. You won't burn this type of meter out, except possibly the resistor for any reasonable source voltage.

Experiment.

.

 

[gar]

170627-2455 EDT

wwhitney:

In a circuit (diagram) a two terminal voltage source will usually have one of its terminals used as a voltage reference point. Lacking any definition pick what you want as the reference point.

But in the case of a symmetrical AC waveform the output will swing equally + and - relative to what is selected as the reference point. The selected reference point would only change the phasing in relation to some external constant reference.

.

 

[Smart $]

...
But absent any other information, if I write a vertical voltage source squiggle and write a voltage sine wave graph next to it, is there any convention on whether 0 volts is defined as the bottom of the squiggle or the top of the squiggle?
...

Actually, convention for [sinusoidal AC] voltage has t=0 as the positive peak: cos(ωt)


However, when graphed, it is more commonly shown as cos(ωt+φ), where φ is -pi/2 or 3pi/2, which is the equivalent of your sin(x).

 

[Strathead]

Thanks for your second reply, Wayne. Much appreciated. I understand all of what you say. Alas, my question remains. How can I tell whether the wave is going to be a positive one or a negative one simply by looking at the circuit?

It is hard to get inside someone else's head, but I am thinking you are WAY overthinking this and I mean no disrespect. I am beginning to think your question is a conundrum. Like, "Is the earth positively or negatively charged?" This is a question that has no answer because the question isn't complete. Since positive and negative are relative and require one additional piece of information. A reference point against which positive or negative is measured. Similar to, "How far is San Francisco?" To me it is about 3500 miles, to Golddigger it is about 130 miles (if I remember correctly). In your case, a diode is a half wave rectifier. It either passes that half wave above a horizontal zero reference line, or below a horizontal zero reference line, to which we assign the terms positive and negative. If I stand on my head then the side of the line it resides on is the "other" side.

Again, no disrespect meant.

 

[Carultch]

To reiterate (I'm sure you know my question by now)--what about the circuit dictates whether the diode will conduct on the positive, or negative half-cycle. Both diodes are fwd biased according to electron theory. The flow is against the arrow.

It has to do with which terminal of the diode is connected to the diode's n-type silicon, and which terminal is connected to the p-type silicon. Or other semiconductor, but usually silicon. When a diode is drawn as an arrow against a line, the line side is the n-type silicon. N-type means that the semiconductor has intentional impurities which are more electronegative than the semiconductor itself, such as silicon doped with phosphorous. P-type means that the semiconductor has intentional impurities which are less electronegative than the semiconductor itself, such as silicon doped with boron. Don't let the fact that phosphorous has a symbol P fool you, it is n-type silicon when phosphorous dopes silicon.

The diode allows current to flow from the p-type side to the n-type side, and will block current like an open circuit, if there is a voltage difference that attempts to do the opposite. An ideal diode is an open circuit when subject to a reverse bias voltage, and a short circuit when subject to a forward bias voltage. A real diode will have some voltage drop across it, and is not a perfect short circuit, in forward bias. The relationship between voltage and current for a diode is extremely non-linear, in fact, it is an exponential growth curve in forward bias. Reverse-bias isn't a perfect open circuit either in a real diode, as eventually, the diode has a breakdown voltage.

In an AC circuit, "forward bias" and "reverse bias" are temporary conditions, that alternate as the source voltage alternates. Forward bias is when the voltage applied to the diode decreases from the p-type terminal to the n-type terminal. Reverse bias is when the opposite happens.

 

[wwhitney]

Actually, convention for [sinusoidal AC] voltage has t=0 as the positive peak: cos(ωt)

That may be (and makes sense for the convenience of using complex methods), but doesn't address my question. You don't get a V(t) to graph until you have chosen a point as the 0 volt reference. And as gar put it, it is conventional to pick one side of the voltage source as 0 volts, but there is no convention on which side to pick.

Cheers, Wayne

 

[Smart $]

That may be (and makes sense for the convenience of using complex methods), but doesn't address my question. You don't get a V(t) to graph until you have chosen a point as the 0 volt reference. And as gar put it, it is conventional to pick one side of the voltage source as 0 volts, but there is no convention on which side to pick.

Cheers, Wayne

Sorry. Apparently, I thought you were talking about the other squiggle.

I believe ground is the conventional reference when any point on the secondary is grounded.

 

[wwhitney]

I believe ground is the conventional reference when any point on the secondary is grounded.

Thanks for pointing that out, I completely overlooked the ground symbol. That convention eliminates one arbitrary choice.

Cheers, Wayne

 

[Carultch]

Sorry. Apparently, I thought you were talking about the other squiggle.

I believe ground is the conventional reference when any point on the secondary is grounded.

Correct.


In any case, it is the direction of the difference in voltage across the diode terminals, and which terminal is connected to which kind of semiconductor, that makes it in either forward bias or reverse bias. Not the absolute voltage.

 

[wwhitney]

OK, now that we've cleared up those details, here's my latest/best answer to the OP. This will surely be a recapitulation of what everyone else has said in this thread, just in my own fashion:

The diagrams in the OP have several graphs, which really should be labeled. We'll assume that they are all voltage graphs (thank to Golddigger for pointing this out to me). And we'll take the 0 volt reference point to be the ground symbol in each diagram (thanks to Smart$ for pointing out this convention to me).

Each diagram has a voltage source on the left with a sine wave graph next to it. The ground symbol is connected to the bottom of the voltage source, so the bottom connection of the voltage source is 0 volts. That means the sine wave represents the voltage produced at the top connection of the voltage source, call that point the source point. The other voltage graph in the each diagram is located at the node next to the top of the resistor, call that point the measurement point. My current understanding of your question is: how do we check those measurement point voltage graphs are correct? I.e. if they were missing, how do we recreate them?

Let's assume the diodes pictured are perfect diodes. That means each diode will be in one of two possible states at any point in time: The first possibility is that the diode is forward biased, in which case it is a perfect conductor and will pass current with 0 voltage drop. This occurs when the voltage on the triangle side of the diode (labeled A in diagram 2) is higher than the voltage on the line side of the diode (labeled K in diagram 2). In this case the voltage at the measurement point will equal the voltage at the source point.

The second case is that the diode is reverse biased, in which case it acts as an open circuit. This occurs when the voltage on the A side of the diode is lower than the voltage on the K side of the diode. In this case there is no current flowing in the circuit, so the voltage drop across the resistor is 0. That means the voltage at the measurement point is the same as the voltage at the bottom of the resistor, namely ground or 0 volts.

Now we can apply these two cases to each diagram. In diagram one, during the first half-cycle with positive voltage at the source point, the diode is reverse biased. That means the voltage at the measurement point will be 0. During the second half-cycle with negative voltage at the source point, the diode is forward biased. That means the voltage at the measurement point will equal the voltage at the source point, and we get a negative voltage half-cycle. This matches the diagram shown.

In diagram two, during the first half-cycle with positive voltage at the source point, the diode is forward biased. That means the voltage at the measurement point will equal the voltage at the source point, and we get a positive voltage half-cycle. During the second half-cycle with negative voltage at the source point, the diode is reverse biased, and the voltage at the measurement point will be 0. This matches the diagram shown.

Cheers, Wayne

The OP diagrams: View attachment 17852

 

[lbmcse]

First some points to clarify:

• The voltage source on the left side of each diagram is the secondary of a transformer, let's say from 120 VAC to (shown) the 24 VAC winding.
• The sine wave immediately to the right on top and bottom only indicates an AC voltage source.
• I'm using electron theory since the program I'm engaged in, and all associated texts, assume this convention (rather than the conventional theory); so electrons are flowing from negative to positive.
• The arrow on the far left of each drawing indicates direction of flow, and nothing else.
• The resulting pulsed DC wave appears at the far top right of each figure.
• The resistor on the right indicates a load, and is in place to bring the current down to a range which will not destroy the PN junction of the diode.

Furthermore, for information only, and for the sake of folks not having to reiterate these points to teach me (although I'm CLEARLY missing something, and am open to being taught whatever you have to throw my way), here's what I do understand, or think I know.

• As a diode is fwd biased, the depletion region/layer shrinks to almost nothing, and flow is achievable.
• Conversely, when reverse biased, the depletion region widens and flow is prevented.
• The definition of fwd bias is the anode being more positive than the cathode (anode is positive with respect to the cathode.)
• The anode is P material, the cathode is N material and are represented by an arrow, and a "minus sign" turned on its side, respectively.
• Flow, when fwd biased can only occur against the arrow from negative (N material) side of the diode, to the positive/arrow (P material) side of the diode.
• In an AC circuit, with a half-wave rectifier, as illustrated, flow happens once per half-cycle, resulting in the pulsed half-waveform shown. The straight line in the waveform indicates the half-cycle where the diode is reverse-biased.
• Full wave rectifiers are available in center tapped, and "bridge" configurations, the former containing 2 diodes, and the latter, 4. In this configuration, the pulsed DC "wave" is a ripple above the x axis (ground?), twice per cycle, resulting in 120 Hz, but only in a full-wave rectifier.

Since I'm not "getting it," I think I'll have to take the advice of the poster who showed the scope shots, and take a 12V AC power source, a silicon diode, and appropriately "sized" resistor (which I'll figure from ohms law) that will limit current so the PN junction will not be damaged, and take some scope shots of my own with the circuits physically laid out just as they are in the two shots of the half-wave rectifiers I referenced in the .jpg. Only then will I see the results for myself. I'll put the diode before the resistor, scope it; then place it after the resistor and do the same. I will manipulate the secondary of the transformer, the diode, the resistor into every iteration that can be created, and see what happens.

I know full well by listening to many of you, that most have perhaps forgotten more than I'll ever know, and PLEASE! NOT TO BE TAKEN AS A CRITICISM OF THE BOARD/FORUM, but it seems you've all done your best to explain, and I'm no further from when I was when I first posted the question. To use an old worn out cliche of our high school days getting dumped for the first time, "it's not you, it's me."

I thank everyone for their help, but I feel my ability to extract an answer is, let's say; a dead short.

As recommended, I will experiment. I have the meters, resources and what I need--when I resolve my issue, I'll get back to you all and explain to you all EXACTLY what truly stupid is.

Thank you so much for your time, and trying to help me. It is sincerely appreciated. I'll report back when I get to where I need to be.

 

[GoldDigger]

Let me make one more try before you go:

The half-sine wave graph in each of the two displays represents the voltage at the top of the resistor, which is also the voltage across the resistor.
During the half-wave where that graph is flat the voltage across the diode will be non zero. It will in fact be the other half of the sine wave, since the diode is not conducting and the sum of the voltage across the resistor and the voltage across the diode must be equal to the voltage across the source transformer (by Kirchoff's Law of Voltages).

You can also just as easily consider the same half-sine graph to be represent the current through the resistor, using the sign convention that electron current up through the resistor is graphed as positive.

Same, with appropriate wording change, applies to the right hand diagram.

 

[lbmcse]

No way, Goldigger. I'm not "leaving." I'm just saying that I don't want you guys to feel like you're talking to a wall. I'll listen to anything anyone has to say. I'm here. I'm reading. I will also experiment as one suggested.

Thank you, sir.

E.T.A.:

Could we say "after" the resistor instead of top, and "before" instead of bottom? Are you saying that the shape of the wave is dependent upon whether it's upstream or downstream of the load (resistor)?

 

[GoldDigger]

I am saying that the waveform across the resistor (ground referenced) and the waveform which represents the difference in voltage between the two sides of the diode (not ground referenced) make up the the two halves of the full sine wave that you see at the top of the source.

I am OK with using "before" and "after" if you are willing to use the exact same designations independent of which way the current is flowing through the resistor.
(A casual reader might think that before the resistor is at the top on the right and at the bottom in the left. That would be too confusing.)

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