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

[ggunn]

You got that backwards.

I don't think so. A forward biased diode has a positive charge on the anode relative to the cathode. The anode is the plate in a tube diode, represented by the vertical line of the diode symbol. Conventional current flow is from anode to cathode, backwards from the arrows you have drawn.

 

[mivey]

I don't think so. A forward biased diode has a positive charge on the anode relative to the cathode. The anode is the plate in a tube diode, represented by the vertical line of the diode symbol. Conventional current flow is from anode to cathode, backwards from the arrows you have drawn.

Think again. Electrons do indeed flow from the cathode (heated element) to the anode (unheated element).

In a tube the cathode could be a coil and the anode a plate.

In the diode symbol, the cathode side is the bar side, not the triangle side. Any plus symbols on the bar side has to do with the high part of the rectifier circuit, not an indicator of bias as you might be thinking.

Just the way it is.

 

[GoldDigger]

Conventional current flows through a diode in the direction of the arrow.
Electrons flow from cathode to anode in a tube, so conventional current flows from anode (plate) to cathode.
That is consistent with the bar side being the cathode.

Sent from my XT1585 using Tapatalk

 

[mivey]

Think again. Electrons do indeed flow from the cathode (heated element) to the anode (unheated element).

In a tube the cathode could be a coil and the anode a plate.

In the diode symbol, the cathode side is the bar side, not the triangle side. Any plus symbols on the bar side has to do with the high part of the rectifier circuit, not an indicator of bias as you might be thinking.

Just the way it is.

If it helps you remember, just think of the triangle part and ignore the bar (which can change for other types anyway).

Think of electrons spraying from the pointy end (cathode) to wide end (anode). Picture the spray from a paint nozzle.

 

[Smart $]

If it helps you remember, just think of the triangle part and ignore the bar (which can change for other types anyway).

Think of electrons spraying from the pointy end (cathode) to wide end (anode). Picture the spray from a paint nozzle.

Easy to remember if one thinks of the now-dated picture tube... aka cathode ray tube (CRT).

 

[mivey]

Easy to remember if one thinks of the now-dated picture tube... aka cathode ray tube (CRT).

Yes. How easily things fade from active memory.

 

[Strathead]

Why tubes? Musician, tube amps...

My first thought was, I have to fix my Fender Champ. I would like to see that course.

 

[Strathead]

I wonder, could your answer be as simple as this:

This is a diagram of a full wave rectifier


Credit where credit is due, I stole it from this site. https://www.elprocus.com/full-wave-rectifier-circuit-working-theory/

 

[gar]

170626-0833 EDT

To add to my plots of yesterday.

These plots show the effect of adding some filter capacitance in parallel with the resistive load. The resistance is 5 ohms, and the capacitance is about 1100 ufd.


Plot 4. Transformer output voltage, red, and voltage across the parallel combination of resistor and capacitor after the diode, blue.

The capacitor does not fully discharge before the next positive half wave voltage starts.

At 8 milliseconds before the screen center there is a dip in the source voltage resulting from current flow starting to the capacitor-resistor and the resulting voltage drop within the transformer from its equivalent series resistance and inductance.

At abour 2.6 milliseconds before center there is a step change in voltage. This is where diode conduction turns off.



,
Plot 5. Transformer output voltage, same as input to diode, red, and diode current, blue.

The current curve is probably showing a resonant effect from the transformer leakage inductance and the filter capacitor. The step change in source voltage is clearly shown as occurring when diode current stops.




,
Plot 6. Capacitor-resistor voltage, red, and diode current, blue.



,


,


,

 

[ggunn]

Think again. Electrons do indeed flow from the cathode (heated element) to the anode (unheated element).

In a tube the cathode could be a coil and the anode a plate.

In the diode symbol, the cathode side is the bar side, not the triangle side. Any plus symbols on the bar side has to do with the high part of the rectifier circuit, not an indicator of bias as you might be thinking.

Just the way it is.

I stand corrected. Apparently the other text I was reading was incorrect, or more likely I misinterpreted it.

 

[ggunn]

My first thought was, I have to fix my Fender Champ. I would like to see that course.

I'll look it up and send you the link. A warning, though; I found it when I was researching continuing ed credits for my PE license (I didn't wait until the last minute before my license expires, but damn near), and it was listed as being worth 4 PDH hours. It sounded perfect, as I have a bunch of tube amps and I'd like to know more about how they work. I downloaded the text, and it's over 200 pages. Four hours? I don't think so.

 

[Carultch]

So, you wanna get philosophical, eh? Actually "current" itself is a convention and flows in whichever direction you define it. Electrons (also a convention) travel the same direction (from cathode to anode within a forward biased diode) whichever way you define current.

It is an artifact of history that what we speak of as "current" flows in the opposite direction as electrons actually flow.

The sign convention on charges was assigned before we even discovered the electron. The charge left on the glass by the silk was defined as positive, and the charge remaining on the initially neutral silk was defined as negative. No scales existed at the time that were precise enough to measure that the silk gained material and the glass lost material, so the arbitrary guess was that the glass gained electrical fluid.

Current is defined as an equivalent flow of positive charges, to make the same electric and magnetic field effects as the actual charges flow. Since it is usually electrons that flow, which are negative, electrons flow the opposite direction as current.

Would we be better off if we defined it the other way around? Not really. This mistake gets us to confront the problem head-on, that it doesn't necessarily need to be electrons that flow, but charges in general. Positive objects such as sodium ions could also be flowing, as they do in human nerves. If we hadn't picked the electrons as positive, we might just define current as the flow of electrons, oblivious to the possibility that it could be something else in other applications.

 

[lbmcse]

I certainly appreciate all the replies, folks. I am reading everything but right now am swamped with reading for work, as well. I'll settle the juggling soon enough. I think the scope shots were particularly useful. Again, I sincerely thank you all for the help.

 

[mivey]

OK. Please clarify any remaining questions after you get caught up.

 

[lbmcse]

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?

I'm reading over the thread, and this is a good question. Yes. I understand this, as the resultant waveform suggests only this could be the case. Where the pulsed DC wave is cup shaped, that suggests that it's only conducting on the negative half-cycle. Where dome-shaped, the diode is allowing conduction on the positive half of the cycle.

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.

 

[wwhitney]

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.

The diode is going to spend half its time forward-biased and conducting (half of each cycle) and half its time reverse-biased and not conducting (the other half of each cycle). Swapping the two wires going to the diode (the only difference between your two original diagrams) changes the half-cycle during which the diode is forward-biased, from the negative half-cycle to the positive half-cycle, or vice versa.

Cheers, Wayne

 

[lbmcse]

I wonder, could your answer be as simple as this:

This is a diagram of a full wave rectifier
View attachment 17864

Credit where credit is due, I stole it from this site. https://www.elprocus.com/full-wave-rectifier-circuit-working-theory/

Appreciate the input, Strathead, but no, I'm afraid not. I'm familiar with the full wave bridge rectifier with capacitor smoothing the pulses between half-cycles. I understand that flow is allowed through the two "parallel" (meant in the 4 sides of the square/diamond, not in the electrical sense) diodes each half cycle, resulting in a pulsed DC that is resultingly 120 Hz now instead of 60 because flow happens in this rectifier circuit twice per cycle.

Still however, I haven't a clue why the pulse is "positive" with respect to the x-axis.

 

[lbmcse]

170626-0833 EDT

To add to my plots of yesterday.

These plots show the effect of adding some filter capacitance in parallel with the resistive load. The resistance is 5 ohms, and the capacitance is about 1100 ufd.


Plot 4. Transformer output voltage, red, and voltage across the parallel combination of resistor and capacitor after the diode, blue.

The capacitor does not fully discharge before the next positive half wave voltage starts.

At 8 milliseconds before the screen center there is a dip in the source voltage resulting from current flow starting to the capacitor-resistor and the resulting voltage drop within the transformer from its equivalent series resistance and inductance.

At abour 2.6 milliseconds before center there is a step change in voltage. This is where diode conduction turns off.

View attachment 17865

,
Plot 5. Transformer output voltage, same as input to diode, red, and diode current, blue.

The current curve is probably showing a resonant effect from the transformer leakage inductance and the filter capacitor. The step change in source voltage is clearly shown as occurring when diode current stops.

View attachment 17866


,
Plot 6. Capacitor-resistor voltage, red, and diode current, blue.

View attachment 17867

,


,


,

Wow, gar! That plot 4 shows beautifully how the capacitor acts as a filter and smooths out the pulsed wave. And FWIW, if what I've learned is true, that capacitor better NOT completely discharge between pulses, that would break some RC time constant rules. Thank you for that.

 

[wwhitney]

Still however, I haven't a clue why the pulse is "positive" with respect to the x-axis.

So in your OP, is your question "why are the two current graphs the opposite sign of each other?" That's the question I've been trying to help you with. Or is your question "why is the first current graph negative, and the second graph positive, rather than vice versa?" That's just a sign convention, which others have already commented on.

Cheers, Wayne

 

[lbmcse]

The diode is going to spend half its time forward-biased and conducting (half of each cycle) and half its time reverse-biased and not conducting (the other half of each cycle). Swapping the two wires going to the diode (the only difference between your two original diagrams) changes the half-cycle during which the diode is forward-biased, from the negative half-cycle to the positive half-cycle, or vice versa.

Cheers, Wayne

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?