Possible dumb question about AC-DC
This might be easier to understand if you would Google up "half wave" and "full wave" rectifier. Think of the output winding (transformer) as an "instant battery" "frozen in time." Depending on the polarity of the sine wave, "frozen" the output of a transformer is a battery, either positive at one end, or switched around, positive at the other end.
So let's start simply. In a simple half wave, the top of the transformer output is positive half the time. Current flows against the arrow of the diode, neg to pos, to the left. So current flows when the top of the transformer winding is positive. Electron flow is from the bottom of the load (neg) up through the load, through the diode to the top of the transformer, and out the bottom of the transformer, back to the load
THE ARROWS IN THESE DIAGRAMS ARE BACKWARDS FROM CONVENTIONAL 'ELECTRON' FLOW
When the top of the transformer is negative, the diode shuts off, just like a check valve in water During this half of the cycle, with no current flowing, the "gap" is produced in the pulsing output waveform. THIS IS WHY half wave rectifiers are harder to filter. This gap causes a "rough" output, more of a "buzz" so to speak. This is known as the "ripple frequency" and is half the frequency of a full wave rectifier
Below is a "full wave center tap" rectifier. Essentially, this is "two half wave back to back"
In this circuit, a negative ground supply, if you wish to think in those terms, would be grounded at point C
At a point in the sine wave, A will be positive, C will be less positive, and B will be least positive. That is if you were to put a scope across A and B, you would see a sine wave. If the transformer was perfectly made, C would be at the zero crossover.
So, with A positive, and C less positive, current flows through the top diode JUST LIKE our half wave rectifier earlier. This produces a pulse of power. Our instantaneous circuit is from the left side of the load, through the load left to right, back through the diode to the top (A) of the transformer, and out the center tap and back to the load
At this same instant, B is the most negative to C, so the bottom diode is OFF
Now the sine wave switches, and things are reversed. Now A at the top is most negative, and the TOP diode is shut off
B at the bottom is now the most positive, with C being less positive, so now the BOTTOM diode conducts and this produces the second pulse right next to the first. This time our instantaneous circuit is from the left side of the load, through the load to the right, down through the bottom diode to "B" at the bottom of the transformer, and back out through the center tap....to the left side of the load
IN A half wave rectifier, nothing would be conducting during this time, and that would produce the gap in the DC pulse waveform.
The rest of this page goes on with a bridge
http://www.electronics-tutorials.ws/diode/diode_6.html
The bridge is different than a full wave center tap because there "is no" center tap. In this diagram, again we have the transformer positive at the top, with the bottom of the load being our NEG side of the supply. With the top of the transformer positive, current flows from (incorrectly labeled in the diagram, 'we' go by electron flow) the bottom of the transformer, .........from right to left in the bottom blue diode, ...........out and up through the load, ...........from right to left in the top blue diode and back to the positive top of the transformer. THIS PRODUCES one pulse in the output DC waveform.
Now the sine wave has switched, and so now the BOTTOM of the transformer is positive. We have exactly the same situation as above. This time electrons flow from the TOP of the transformer, down right to left through the top blue diode, out and up through the load, down and to the right through the bottom right hand blue diode, and to the (now positive) bottom of the transformer.
This produces the next pulse of DC in the output, stacked right next to the first
This might be easier to understand if you would Google up "half wave" and "full wave" rectifier. Think of the output winding (transformer) as an "instant battery" "frozen in time." Depending on the polarity of the sine wave, "frozen" the output of a transformer is a battery, either positive at one end, or switched around, positive at the other end.
So let's start simply. In a simple half wave, the top of the transformer output is positive half the time. Current flows against the arrow of the diode, neg to pos, to the left. So current flows when the top of the transformer winding is positive. Electron flow is from the bottom of the load (neg) up through the load, through the diode to the top of the transformer, and out the bottom of the transformer, back to the load
THE ARROWS IN THESE DIAGRAMS ARE BACKWARDS FROM CONVENTIONAL 'ELECTRON' FLOW
When the top of the transformer is negative, the diode shuts off, just like a check valve in water During this half of the cycle, with no current flowing, the "gap" is produced in the pulsing output waveform. THIS IS WHY half wave rectifiers are harder to filter. This gap causes a "rough" output, more of a "buzz" so to speak. This is known as the "ripple frequency" and is half the frequency of a full wave rectifier
Below is a "full wave center tap" rectifier. Essentially, this is "two half wave back to back"
In this circuit, a negative ground supply, if you wish to think in those terms, would be grounded at point C
At a point in the sine wave, A will be positive, C will be less positive, and B will be least positive. That is if you were to put a scope across A and B, you would see a sine wave. If the transformer was perfectly made, C would be at the zero crossover.
So, with A positive, and C less positive, current flows through the top diode JUST LIKE our half wave rectifier earlier. This produces a pulse of power. Our instantaneous circuit is from the left side of the load, through the load left to right, back through the diode to the top (A) of the transformer, and out the center tap and back to the load
At this same instant, B is the most negative to C, so the bottom diode is OFF
Now the sine wave switches, and things are reversed. Now A at the top is most negative, and the TOP diode is shut off
B at the bottom is now the most positive, with C being less positive, so now the BOTTOM diode conducts and this produces the second pulse right next to the first. This time our instantaneous circuit is from the left side of the load, through the load to the right, down through the bottom diode to "B" at the bottom of the transformer, and back out through the center tap....to the left side of the load
IN A half wave rectifier, nothing would be conducting during this time, and that would produce the gap in the DC pulse waveform.
The rest of this page goes on with a bridge
http://www.electronics-tutorials.ws/diode/diode_6.html
The bridge is different than a full wave center tap because there "is no" center tap. In this diagram, again we have the transformer positive at the top, with the bottom of the load being our NEG side of the supply. With the top of the transformer positive, current flows from (incorrectly labeled in the diagram, 'we' go by electron flow) the bottom of the transformer, .........from right to left in the bottom blue diode, ...........out and up through the load, ...........from right to left in the top blue diode and back to the positive top of the transformer. THIS PRODUCES one pulse in the output DC waveform.
Now the sine wave has switched, and so now the BOTTOM of the transformer is positive. We have exactly the same situation as above. This time electrons flow from the TOP of the transformer, down right to left through the top blue diode, out and up through the load, down and to the right through the bottom right hand blue diode, and to the (now positive) bottom of the transformer.
This produces the next pulse of DC in the output, stacked right next to the first