Sorry, GoldDigger. I should have replied to this. No, I'm going with electron flow, as that's what my text, and program "assume." I know neither has been disproven, but there you go. So electrons always flow against the arrow of a diode, in my neck of the woods, so to speak.
Positive or Negative Alternation of the sine wave developed across the load.
- Thread starter lbmcse
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Sorry, GoldDigger. I should have replied to this. No, I'm going with electron flow, as that's what my text, and program "assume." I know neither has been disproven, but there you go. So electrons always flow against the arrow of a diode, in my neck of the woods, so to speak.
Well technically, it wouldn't matter if the voltage was not referenced to ground... but it would have to be referenced to some point in the circuit. In this case, it is referenced to the lower "rail", which is bonded to ground.
What does matter is that the voltage is positive with respect to the reference for a half cycle and negative for the other half cycle. When the diode blocks current, there is no voltage measured at the upper right node, i.e. it is at the same potential as the reference. While the diode is forward biased, the voltage measured is a result of conducting current during the positive half cycle or the negative half cycle.
ggunn
PE (Electrical), NABCEP certified
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Something I just learned in that electron tube class I am taking (which started with tube diodes) is that the "arrow" in a diode symbol isn't an arrow at all. Early tube diodes had a self heating filament as a cathode and a plate as an anode. The cathode was bent at a sharp angle with the middle point oriented toward the plate, and electrons boiled off the hot cathode and jumped across to the anode during forward bias. The diode symbol is a representation of the bent cathode and the anode plate, and current flows in the direction of the "arrow" only if you use electron flow as the convention for current direction.
kwired
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does this drawing help answer your question?
ggunn
PE (Electrical), NABCEP certified
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- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Only if your current arrows follow electron flow. Conventional positive to negative current flows opposite to your arrows.
does this drawing help answer your question?
kwired
Electron manager
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I don't think OP is stuck on electron flow vs.conventional flow, so we just need to pick one or the other and stick with it while working the rest of what he is trying to figure out, you can pick conventional and I can pick electron flow and it still works out.
Actually the current flows the same no matter which method is used.
Just that conventional indicates in reverse. I'd prefer if we would call it traditional, because completely conventional it is not.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Actually the current flows the same no matter which method is used.
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.
kwired
Electron manager
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- NE Nebraska
Actually the current flows the same no matter which method is used.
Just that conventional indicates in reverse. I'd prefer if we would call it traditional, because completely conventional it is not.
exactly, the only thing that is reversed is the directional theory we use when looking at a drawing. Some more complex drawings that direction may be more critical. OP's question it doesn't matter, but you must pick one method and stick with it when dealing with OP's question.
wwhitney
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As Smart$ already pointed out, this is where you are going wrong. The XFMRs are the same in both diagrams.
The only difference between the diagrams is the polarity of the diode. That one flip results in the current waveform being multiplied by -1, so the negative half-wave becomes a positive half-wave.
Cheers, Wayne
mivey
Senior Member
You got that backwards.
mivey
Senior Member
That I can explain, but not from this mobile keyboard.
mivey
Senior Member
That covers it well. The only thing left is why the current causes a voltage to appear across the resistor, if that is what the OP wants to hear. Too much typing to cover that at the moment.
mivey
Senior Member
The cathod is the plate looking thing on the symbol. Either you misunderstood the instructor or the instructor needs instructing.
thanks
if we agree current flows using the sign convention shown
and ignore the small drop across the diode
the loop v must sum to 0 per kvl
v source + v resistor = 0 or v source = -v resistor
and from Ohm: v resistor = i loop x r
so if we have i and r we must have v drop and it is equal but opposite sign to v source
gar
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170625-1617 EDT
Some scope plots.
The circuit consists of a single transformer secondary winding with one common terminal, call it ground, the other terminal is a 60 Hz sine wave of some value Vpeak relative to ground. The instantaneous value swings from a maximum positive peak value of +Vpeak to a maximum (possibly you may want to call it a minimum) value of -Vpeak relative to the common measuring point, ground.
The load applied is a resistor with a series diode, and the two orientations of the diode.
The plots are from a scope with a suitable time base selected, and synchronized always from the AC line in exactly the same way.
Red plot is source voltage, blue is load resistor voltage, or current.
Plot 1 --- Voltage only, no load. Note the positive-negative symmetry of the wave form. Time synchronization is at the middle of the plot.
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Plot 2 --- Voltage and associated loading on only the positive half of the cycle.
The diode only conducts when its anode is more positive than its cathode. In this case the cathode is connected to the load resistor. The resistor only has a voltage drop across it when current flows. Current only flows thru the diode when the source voltage is positive because of the selected diode orientation.
No current and the resistor drop is zero.
Note: the positive voltage is slightly depressed compared to no load, and the negative peak is unaffected. This is because of the transformer internal impedance.
The waveforms are measurements relative to common (ground). The load waveform is always above ground and thus its average DC value is positive.
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Plot 3 --- Voltage and associated loading on only the negative half of the cycle.
Relative to Plot 2 this plot has the diode orientation reversed.
The diode only conducts when its anode is more positive than its cathode. In this case the anode is connected to the load resistor. The resistor only has a voltage drop across it when current flows. Current only flows thru the diode when the source voltage is negative because of the diode orientation.
No current and the resistor drop is zero.
Note: the negative source voltage is slightly depressed compared to no load, and the negative peak is unaffected. This is because of the transformer internal impedance.
The waveforms are measurements relative to common (ground). The load waveform is always below ground and thus its average DC value is negative. The average value is about 0.318 times Vpeak.
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Some scope plots.
The circuit consists of a single transformer secondary winding with one common terminal, call it ground, the other terminal is a 60 Hz sine wave of some value Vpeak relative to ground. The instantaneous value swings from a maximum positive peak value of +Vpeak to a maximum (possibly you may want to call it a minimum) value of -Vpeak relative to the common measuring point, ground.
The load applied is a resistor with a series diode, and the two orientations of the diode.
The plots are from a scope with a suitable time base selected, and synchronized always from the AC line in exactly the same way.
Red plot is source voltage, blue is load resistor voltage, or current.
Plot 1 --- Voltage only, no load. Note the positive-negative symmetry of the wave form. Time synchronization is at the middle of the plot.
.
Plot 2 --- Voltage and associated loading on only the positive half of the cycle.
The diode only conducts when its anode is more positive than its cathode. In this case the cathode is connected to the load resistor. The resistor only has a voltage drop across it when current flows. Current only flows thru the diode when the source voltage is positive because of the selected diode orientation.
No current and the resistor drop is zero.
Note: the positive voltage is slightly depressed compared to no load, and the negative peak is unaffected. This is because of the transformer internal impedance.
The waveforms are measurements relative to common (ground). The load waveform is always above ground and thus its average DC value is positive.
.
Plot 3 --- Voltage and associated loading on only the negative half of the cycle.
Relative to Plot 2 this plot has the diode orientation reversed.
The diode only conducts when its anode is more positive than its cathode. In this case the anode is connected to the load resistor. The resistor only has a voltage drop across it when current flows. Current only flows thru the diode when the source voltage is negative because of the diode orientation.
No current and the resistor drop is zero.
Note: the negative source voltage is slightly depressed compared to no load, and the negative peak is unaffected. This is because of the transformer internal impedance.
The waveforms are measurements relative to common (ground). The load waveform is always below ground and thus its average DC value is negative. The average value is about 0.318 times Vpeak.
.
.
.
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