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

[lbmcse]

Hello everyone. This is my first post, and is a question that has plagued me from the days when I was a first year apprentice, until now, 13 years later. See the photo below.


Well, I thought it would be bigger, but here goes. I understand fwd. biasing of the PN junction, and how the left figure's diode is fwd biased, and the right figure is too. What I have never quite grasped, is what is the reason on these two half-wave rectifiers--that the left one develops only the negative alternation across the load, while the right figure does the opposite? I realize that in both, the cathode is negative with respect to the anode (the definition of fwd. biasing), but WHY does one create just the negative portion of the sine wave while the other renders the positive half to the resistor?

If I could get this answered, my kingdom is yours. Not much in the way of jewels and binoculars, but boy do I feel like an ass.

Help?

Thank you.
Larry.

 

[kwired]

Hello everyone. This is my first post, and is a question that has plagued me from the days when I was a first year apprentice, until now, 13 years later. See the photo below.
View attachment 17852

Well, I thought it would be bigger, but here goes. I understand fwd. biasing of the PN junction, and how the left figure's diode is fwd biased, and the right figure is too. What I have never quite grasped, is what is the reason on these two half-wave rectifiers--that the left one develops only the negative alternation across the load, while the right figure does the opposite? I realize that in both, the cathode is negative with respect to the anode (the definition of fwd. biasing), but WHY does one create just the negative portion of the sine wave while the other renders the positive half to the resistor?

If I could get this answered, my kingdom is yours. Not much in the way of jewels and binoculars, but boy do I feel like an ass.

Help?

Thank you.
Larry.

From a hydraulics analogy perspective consider the diode a check valve, you installed it the opposite direction in the second setup, it only allows flow in one direction, unless enough force is applied to overcome it's resistance to flow.

Now back to electrical- the AC current wave is blocked in one direction, turning the diode end for end reverses that direction for the entire circuit. Full AC voltage wave is applied, it just only allows current to flow one direction.

The load only sees DC (pulsing DC in a simple circuit like this) because half the cycle is essentially DC current flowing one direction, the other half cycle the diode becomes like an open circuit because it isn't allowing any current to flow the opposite direction.

 

[Carultch]

Hello everyone. This is my first post, and is a question that has plagued me from the days when I was a first year apprentice, until now, 13 years later. See the photo below.
View attachment 17852

Well, I thought it would be bigger, but here goes. I understand fwd. biasing of the PN junction, and how the left figure's diode is fwd biased, and the right figure is too. What I have never quite grasped, is what is the reason on these two half-wave rectifiers--that the left one develops only the negative alternation across the load, while the right figure does the opposite? I realize that in both, the cathode is negative with respect to the anode (the definition of fwd. biasing), but WHY does one create just the negative portion of the sine wave while the other renders the positive half to the resistor?

If I could get this answered, my kingdom is yours. Not much in the way of jewels and binoculars, but boy do I feel like an ass.

Help?

Thank you.
Larry.

Are you asking for an explanation at the atomic level of why a diode works?

Or are you asking for an explanation of your circuit results, given that a diode works the way it does?

 

[lbmcse]

Gentlemen, thanks for the replies. I understand why and how, holes and electrons, depletion regions widening, narrowing, and the implications of both--and the check valve analogy is a familiar one. I'm aware WHY current flow is only possible clockwise in one picture, and counterclockwise in the other, what I DON'T understand--is why clockwise current flow produces a NEGATIVE half-cycle pulse, while the clockwise flow yields a POSITIVE (above the X axis) pulsed DC voltage across the load. See the very small pulsed DC representation on the upper right of each diagram?

BTW, these are not "my" circuits, they are from a text, representing a simple half-wave rectifier.

To reply to Carultch, I'm asking for the latter. Chances are I'm missing something very simple. Just trying to button this up.

Thank you again.

 

[GoldDigger]

Part of your confusion, I think, is that the arrows next to the resistor are showing electron flow, not "conventional" current.
Conventional current is drawn as if positive charges were moving.
The voltage across a resistor is proportional to the conventional current through it.
In the left picture conventional current is flowing counter-clockwise. That makes the top of the resistor negative, as shown in the wave at the end of line.

When you use conventional current, the flow through a diode is in the direction of the arrowhead in the symbol.

PS: Or else the arrow direction is just a mistake!

Sent from my XT1585 using Tapatalk

 

[kwired]

Gentlemen, thanks for the replies. I understand why and how, holes and electrons, depletion regions widening, narrowing, and the implications of both--and the check valve analogy is a familiar one. I'm aware WHY current flow is only possible clockwise in one picture, and counterclockwise in the other, what I DON'T understand--is why clockwise current flow produces a NEGATIVE half-cycle pulse, while the clockwise flow yields a POSITIVE (above the X axis) pulsed DC voltage across the load. See the very small pulsed DC representation on the upper right of each diagram?

BTW, these are not "my" circuits, they are from a text, representing a simple half-wave rectifier.

To reply to Carultch, I'm asking for the latter. Chances are I'm missing something very simple. Just trying to button this up.

Thank you again.

Take the AC source away and replace it with a battery. Then keep flipping polarity of the battery in each circuit and explain why current flows when it does and when it doesent in each of the circuits you posted in OP.

Think of the battery being installed one way when the AC voltage is above the X axis, and the opposite direction when the AC voltage is below the X axis, because that is what AC does, it changes polarity every half cycle.

 

[ggunn]

Gentlemen, thanks for the replies. I understand why and how, holes and electrons, depletion regions widening, narrowing, and the implications of both--and the check valve analogy is a familiar one. I'm aware WHY current flow is only possible clockwise in one picture, and counterclockwise in the other, what I DON'T understand--is why clockwise current flow produces a NEGATIVE half-cycle pulse, while the clockwise flow yields a POSITIVE (above the X axis) pulsed DC voltage across the load. See the very small pulsed DC representation on the upper right of each diagram?

BTW, these are not "my" circuits, they are from a text, representing a simple half-wave rectifier.

To reply to Carultch, I'm asking for the latter. Chances are I'm missing something very simple. Just trying to button this up.

Thank you again.

I am taking an on line course in electron tubes (actually it's just reading text and taking an exam), and the text was written by a US Navy instructor many years ago. The convention for current flow he used is from negative to positive, and I am a little embarrassed by how much trouble that is giving me.

Why tubes? Musician, tube amps...

 

[lbmcse]

Thanks all for trying to help. Perhaps I'm not asking my question correctly. Here's what I think I DO know:

• that current will only flow when the diode is fwd biased.
• the diode in both pictures are fwd biased.
• current is flowing clockwise on the left, counter clockwise on the right.
• each picture conveys the P material is positive with respect to the N material and the depletion region all but disappears, allowing current to flow.
• a battery wouldn't produce a pulsed DC south of the X axis in any scenario.
• on the left, the "top" of the XFMR is negative and the "bottom" is positive.
• on the right, the "top" of the XFMR is "positive" and the "bottom" is negative.

So my question put a different way is, what is the reason for one picture's resultant DC pulse wave being positive for each half cycle instead of negative? And the converse as well--why is the resultant DC pulse wave of the other picture, negative for half-a-cycle.

If someone doesn't mind reading the thread, please tell me if someone has answered already and I'm just not "getting" it. Does the positive or negative aspect of the "wave" have nothing to do with any aspect of the circuit? I'm hitting the wall here.

Have mercy.

 

[mivey]

Part of your confusion, I think, is that the arrows next to the resistor are showing electron flow, not "conventional" current.
Conventional current is drawn as if positive charges were moving.
The voltage across a resistor is proportional to the conventional current through it.
In the left picture conventional current is flowing counter-clockwise. That makes the top of the resistor negative, as shown in the wave at the end of line.

When you use conventional current, the flow through a diode is in the direction of the arrowhead in the symbol.

PS: Or else the arrow direction is just a mistake!

Sent from my XT1585 using Tapatalk

The arrow may be indicating voltage rise, which would not be uncommon. The text or conventions from the source or would help clarify.

 

[mivey]

Thanks all for trying to help. Perhaps I'm not asking my question correctly. Here's what I think I DO know:

• that current will only flow when the diode is fwd biased.
• the diode in both pictures are fwd biased.
• current is flowing clockwise on the left, counter clockwise on the right.
• each picture conveys the P material is positive with respect to the N material and the depletion region all but disappears, allowing current to flow.
• a battery wouldn't produce a pulsed DC south of the X axis in any scenario.
• on the left, the "top" of the XFMR is negative and the "bottom" is positive.
• on the right, the "top" of the XFMR is "positive" and the "bottom" is negative.

So my question put a different way is, what is the reason for one picture's resultant DC pulse wave being positive for each half cycle instead of negative? And the converse as well--why is the resultant DC pulse wave of the other picture, negative for half-a-cycle.

If someone doesn't mind reading the thread, please tell me if someone has answered already and I'm just not "getting" it. Does the positive or negative aspect of the "wave" have nothing to do with any aspect of the circuit? I'm hitting the wall here.

Have mercy.

Current is counter clockwise on left and clockwise on the right.

PS: and the arrows show the voltage rise across the resistor (so going up on the left resistor is a fall, not a rise in voltage).

 

[mivey]

Gentlemen, thanks for the replies. I understand why and how, holes and electrons, depletion regions widening, narrowing, and the implications of both--and the check valve analogy is a familiar one. I'm aware WHY current flow is only possible clockwise in one picture, and counterclockwise in the other, what I DON'T understand--is why clockwise current flow produces a NEGATIVE half-cycle pulse, while the clockwise flow yields a POSITIVE (above the X axis) pulsed DC voltage across the load. See the very small pulsed DC representation on the upper right of each diagram?

BTW, these are not "my" circuits, they are from a text, representing a simple half-wave rectifier.

To reply to Carultch, I'm asking for the latter. Chances are I'm missing something very simple. Just trying to button this up.

Thank you again.

You just mis-read the diagram. You got the CW and CCW mixed up by assuming the resistor arrows meant conventional current flow when they are probably indicating voltage rise. It is just that simple.

 

[Smart $]

Thanks all for trying to help. Perhaps I'm not asking my question correctly. Here's what I think I DO know:

• ...
• on the left, the "top" of the XFMR is negative and the "bottom" is positive.
• on the right, the "top" of the XFMR is "positive" and the "bottom" is negative.

...

Assuming the inductor to the left is a transformer secondary winding, it is oriented the same in both diagrams. Note the annotation symbology immediately to the right of the inductor symbol. That's a voltage waveform indicator, not current. Would have been better if it had a conventional dot... but alas, it does not.

Anyway, it would not matter which way the winding is oriented. With respect to the circuit, the current waveforms would be the same if measurements are taken 180° later in a cycle.

 

[lbmcse]

The arrow may be indicating voltage rise, which would not be uncommon. The text or conventions from the source or would help clarify.

Sorry, mivey for the lack of information from my end. The text from which this is snipped indicates (as Smart $ has said), the coil on the left is the secondary winding of a XFMR. The text also indicates that the arrows beside the resistors/load indicate direction of current travel, nothing more.

Current is counter clockwise on left and clockwise on the right.

PS: and the arrows show the voltage rise across the resistor (so going up on the left resistor is a fall, not a rise in voltage).

Current flow according to the text indicates CW on the left, and CCW on the right. The text I'm using specifies in the beginning that while neither theory has been disproven, it uses electron flow theory, rather than conventional. This is true of the text, and the 5 year program from which it's a part.

Finally, the pulsed DC is shown at the upper right of each diagram. That is not attached to the circuit. It's included only to show the resultant pulsed DC waveform produced by each diagram. As you can see, it's indicating one full cycle, hence the rise/or fall, flat line included for the other half-cycle, while it blocks/does not conduct.

 

[mivey]

Sorry, mivey for the lack of information from my end. The text from which this is snipped indicates (as Smart $ has said), the coil on the left is the secondary winding of a XFMR. The text also indicates that the arrows beside the resistors/load indicate direction of current travel, nothing more.



Current flow according to the text indicates CW on the left, and CCW on the right. The text I'm using specifies in the beginning that while neither theory has been disproven, it uses electron flow theory, rather than conventional. This is true of the text, and the 5 year program from which it's a part.

Finally, the pulsed DC is shown at the upper right of each diagram. That is not attached to the circuit. It's included only to show the resultant pulsed DC waveform produced by each diagram. As you can see, it's indicating one full cycle, hence the rise/or fall, flat line included for the other half-cycle, while it blocks/does not conduct.

OK, electron flow it is. I thought you were confused about a sign conflict.

So you understand diodes. You understand current. You understand the circuit. Just what is it that is you don't understand? Are you saying you don't understand that current through a resistor produces a voltage with the appropriate polarity?

 

[mivey]

hence the rise/or fall, flat line included for the other half-cycle, while it blocks/does not conduct.

You mean rise/fall as increasing/decrease but not a polarity change. Is that your question? Still not sure what you are trying to clarify.

 

[mivey]

You do understand in one circuit the diode only conducts on the positive half cycle while in the other circuit the diode only conducts on the negative half cycle?

 

[mivey]

I would have sworn your post #4 would indicate a confusion with conventional notation vs. electron flow.

 

[lbmcse]

. . . Are you saying you don't understand that current through a resistor produces a voltage with the appropriate polarity?

Perhaps that's it. I don't know why when it flows one way, the pulsed DC is "positive" and when it flows the other way, the waveform created is "negative." As I alluded to earlier, this is probably something so fundamental that when someone explains it--I'll kick myself.

Thanks for sticking with me. I imagine the crowd is face-palming as we speak. At the very least, nodding in disbelief. :happysad:

 

[Smart $]

Perhaps that's it. I don't know why when it flows one way, the pulsed DC is "positive" and when it flows the other way, the waveform created is "negative." As I alluded to earlier, this is probably something so fundamental that when someone explains it--I'll kick myself.

Thanks for sticking with me. I imagine the crowd is face-palming as we speak. At the very least, nodding in disbelief. :happysad:

All voltage measurements are referenced to ground.

 

[lbmcse]

All voltage measurements are referenced to ground.

Could you elaborate? I realize when taking a voltage reading that's not phase to phase, I'm measuring a voltage, "with reference to ground." I get that. Would you mind tying that in to my OP?

Thanks for taking the time, Smart.