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Subject - AC Ripple Voltage Component
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BillyB
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When I rectify 100 Vac with a half wave rectifier I measure 55 Vac as the AC ripple voltage. Then when I rectify 100 Vac with a full wave bridge rectifier I measure 45 Vac as the AC ripple voltage. WHY ?
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PEI
| OKAY, Let's start with some basics. Where are you getting the 100VAC and what in the world are you using to measure the output? For what you are doing and wanting to know, you need an oscilloscope. You want to see the real wave form, not just 'attempt' to quantify it. Let me be blunt, your meter is lying to you! A few meters will read the voltage peak to peak, most will read volts RMS (root mean square). Since you appear to be experimenting, you need to fully understand by seeing on an o-scope what has happened to your sine wave. Now, I will attempt in words to tell you what you should see. Essentially, the 1/2 wave bridge blocks the lower excursion of the sine wave entirely. During the second 1/2 cycle, no current flows. Thus on an o-scope you would see a 1/2 sine wave, then nothing for a 1/2 cycle then a 1/2 sine wave and repeating. Current only flows during the 1/2 cycle of sine wave, flowing in only 1 direction, then for a 1/2 cycle, there is no current flow. A full wave bridge, will 'flip' the negative 1/2 cycle of the sine wave above the zero line, such that current will flow almost the whole cycle, flowing in only 1 direction. The reason I bring up the fact it only flows in 1 direction (like DC) is, that is why you rectify AC, to get DC. The problem so far is that your voltage fluctuates up & down with the shape of the sine wave. Obviously for DC, you want to hold some voltage level, not swing. If you would introduce a good sized capacitor, you will ride out the 1/2 wave that is missing, almost until the next wave comes to create a flow all the time. Now, what the o-scope will show... Imagine a cemetery with round top tombstones next to each other, the voltage of the 1/2 wave rectifier will rise along the outline, fall to the ground, skip a stone, and rise again with the next wave and so on. The full wave will look like the tombstones are side by side, rising then falling to zero voltage then immediately rising again, etc. The addition of a capacitor will attempt to fill in the gaps as the voltage falls to zero, smoothing the waveforms. That little bit of fall that remains, in comparison to the peak, is the ripple voltage you are worried about. You can add more capacitance and even some circuitry to get that down to a very insignificant amount. Why did I ask about where you were getting the 100 vac? Because everyone calls household power 120 volts, well....it really is 120 V RMS, which means root mean square.
If you looked at that sine wave on an oscilliscope, you'd see that it is really about 170 volts peak. In other words the maximum amplitude of that sine wave is really 170 volts. Why 120, why RMS? Well, someone decided to measure is at its true heating value which happens to the peak voltage divided by 1.414, giving you 120. What I just said is a simplification for some calculus, let's not go there. This part I have glossed over, you can read up on this in a text. My point is to prove that you are not using the proper devices to measure what you are doing. I also wouldn't be surprised that your meter is 'hunting' trying to get a number to represent what it is reading and constantly changing the value on the LCD. This is due to the fact that it is interpolating the RMS value, which it can only do if it were reading a true sinusoidal waveform, and it is continually sampling the value and will not consistently hold to a value. If this is not happening, you are most likely using a VOM (or possibly an older model VTVM) with a needle and scale to read. In either case, the device is simply not designed to measure these values accurately. For your purposes, these values could be considered relative. In other words, the ripple (or rising/falling) voltage for a 1/2 wave rectifier would be greater than that of a full wave bridge and you could expect that those voltages would be near 1/2 the input voltage.
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