dKingston
05.06.2021 15:44:59
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COMPUTER SIMULATION
Schematic with SPICE node numbers:
COMPUTER SIMULATION
Schematic with SPICE node numbers: more
INSTRUCTIONS
This rectifier circuit is called full-wave because it makes use of the entire waveform, both positive and negative half-cycles, of the AC source voltage in powering the DC load. As a result, there is less “ripple” voltage seen at the load. The RMS (Root-Mean-Square) value of the rectifier’s output is also greater for this circuit than for the half-wave rectifier.
Use a voltmeter to measure both the DC and AC voltage delivered to the motor. You should notice the advantages of the full-wave rectifier immediately by the greater DC and lower AC indications as compared to the last experiment.
An experimental advantage of this circuit is the ease of which it may be “de-converted” to a half-wave rectifier: simply disconnect the short jumper wire connecting the two diodes’ cathode ends together on the terminal strip. Better yet, for quick comparison between half and full-wave rectification, you may add a switch in the circuit to open and close this connection at will: more
LEARNING OBJECTIVES
Design of a bridge rectifier circuit
Advantages and disadvantages of the bridge rectifier circuit, compared to the center-tap circuit [links]
PARTS AND MATERIALS
Low-voltage AC power supply (6 volt output)
Four 1N4001 rectifying diodes (Radio Shack catalog # 276-1101)
Small “hobby” motor, permanent-magnet type (Radio Shack catalog # 273-223 or equivalent) here
INSTRUCTIONS
This circuit provides full-wave rectification without the necessity of a center-tapped transformer. In applications where a center-tapped, or split-phase, source is unavailable, this is the only practical method of full-wave rectification.
In addition to requiring more diodes than the center-tap circuit, the full-wave bridge suffers a slight performance disadvantage as well: the additional voltage drop caused by current having to go through two diodes in each half-cycle rather than through only one. With a low-voltage source such as the one you’re using (6 volts RMS), this disadvantage is easily measured. Compare the DC voltage reading across the motor terminals with the reading obtained from the last experiment, given the same AC power supply and the same motor.
I can tell you how to hook up the bridge module, but I can't tell from your description if it will be large enough to survive for very long. I strongly advise you to try to find the correct replacements and use them instead of doing this. Google search Allied Electronics or Newark Electronics to find the correct ones. Most diodes have a number on them that begins with !N****. You will need this number to find them. Also, I strongly recommend that you replace all of them as they frequently fail in pairs.
Please see the pictures below: more
In my hurry, I didn't pay attention to how the rectifier assembly went. I need help with putting it back together correctly. There are 2 taps that go to the wire feed motor, then a transformer handles the +, and the - is connected directly to the wand. more
Negative cathode output to wire feed motor
Negative cathode diode plate
Insulator
Anode from two diodes on the negative plate
Connection from AC transformer (16v)
Fiber insulator
Anode from other two diodes on the negative plate
Cathode from two diodes on the positive plate
Cathode from the other two diodes on the positive plate
Connection from AC transformer (16v)
Insulator
Positive anode diode plate
Positive anode output to wire feed motor [links]
And here is how I put it back together in a hurry: here
If you still want to use the bridge module from Radio Shack there are 4 output connections on the bridge module. 2 of these are marked AC or have a sine wave (wavy) symbol on them. These should be connected to the output windings from the transformer. It doesn't matter which way you connect them. The other 2 connections on the bridge module are marked + and -. These should be connected to the output terminals of your welder or the leads that run in that direction. The + lead must hook to the + connection and the - lead must hook to the - connection of the module. They must be connected to the proper terminal or your welder will have incorrect polarity output.
The function of this capacitor, known as a reservoir capacitor (aka smoothing capacitor) is to lessen the variation in (or 'smooth') the rectified AC output voltage waveform from the bridge. One explanation of 'smoothing' is that the capacitor provides a low impedance path to the AC component of the output, reducing the AC voltage across, and AC current through, the resistive load. In less technical terms, any drop in the output voltage and current of the bridge tends to be cancelled by loss of charge in the capacitor. This charge flows out as additional current through the load. Thus the change of load current and voltage is reduced relative to what would occur without the capacitor. Increases of voltage correspondingly store excess charge in the capacitor, thus moderating the change in output voltage / current. Also see rectifier output smoothing. [links]
In some designs, a series resistor at the load side of the capacitor is added. The smoothing can then be improved by adding additional stages of capacitor-resistor pairs, often done only for sub-supplies to critical high-gain circuits that tend to be sensitive to supply voltage noise.
Because a bleeder sets a minimum current drain, the regulation of the circuit, defined as percentage voltage change from minimum to maximum load, is improved. However in many cases the improvement is of insignificant magnitude. [links]
The idealized waveforms shown above are seen for both voltage and current when the load on the bridge is resistive. When the load includes a smoothing capacitor, both the voltage and the current waveforms will be greatly changed. While the voltage is smoothed, as described above, current will flow through the bridge only during the time when the input voltage is greater than the capacitor voltage. For example, if the load draws an average current of n Amps, and the diodes conduct for 10% of the time, the average diode current during conduction must be 10n Amps. This non-sinusoidal current leads to harmonic distortion and a poor power factor in the AC supply.
After it was all said and done, I checked it with my digital multi-meter, and found that it put out nearly a full volt more at idle, than with the stock rectifier. more