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university:courses:alm1k:circuits1:alm-cir-diode-rectifier [23 Aug 2019 16:28] – [Questions:] Doug Mercer | university:courses:alm1k:circuits1:alm-cir-diode-rectifier [03 Nov 2021 20:29] (current) – [Activity: Diode Rectifiers] Doug Mercer | ||
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- | ======Activity: | + | ======Activity: |
=====Objective: | =====Objective: | ||
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====Directions: | ====Directions: | ||
- | Four diodes (1N914) can be arranged in a bridge configuration to provide a full-wave rectification from a single | + | Four diodes (1N914) can be arranged in a bridge configuration to provide a full-wave rectification from an AC source such as a center-taped transformer winding. The center taped winding can be represented by the two AWG outputs centered around a common node if they are 180 degrees out of phase with the center tap being the +2.5 V common rail, shown in figure 4. The load resistors R< |
{{ : | {{ : | ||
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====Hardware Setup:==== | ====Hardware Setup:==== | ||
- | The AWG CH-A waveform generator should be configured for a 100 Hz Sine wave with 0 volt Min value and 5 volt Max value. | + | The AWG CH-A waveform generator should be configured for a 100 Hz Sine wave with 0 volt Min value and 5 volt Max value. Check the B= Comp A check box as shown in figure 5 to set channel B to be the complement of channel A i.e. same p-p value but 180 degrees out of phase. Set the mode to be SVMI Split I/O to generate the sine waves on the CH A and CH B pins and measure the voltage waveforms at the bridge output on the AIN and BIN pins. |
+ | {{ : | ||
+ | |||
+ | <WRAP centeralign> | ||
+ | |||
+ | Checking the Sync AWG box means you will not need to set a trigger input source or level. | ||
====Procedure: | ====Procedure: | ||
- | Under the curves | + | Under the Curves |
- | To display the absolute value of the channel A current (in mA) enter the following Math Formula: | + | On the Math setup screen select CA-CB to display the differential voltage waveform across the load resistors at the output |
- | abs(IBuffA[t])*1000 | + | {{ : |
- | Under Math select Formula to display the calculated waveform. Set the Math-Axis to I-A. This also in effect displays the voltage across R<sub>L</sub>. The voltage is the current (in A) times the 100 Ω value of R< | + | <WRAP centeralign>Figure 6, Voltage and Current traces</WRAP> |
The disadvantage of this circuit is that now two diode drops are in series with the load and the peak value of the rectified output is less than the AC input by 1.5 Volts rather than the 0.75 V in the previous circuits. | The disadvantage of this circuit is that now two diode drops are in series with the load and the peak value of the rectified output is less than the AC input by 1.5 Volts rather than the 0.75 V in the previous circuits. | ||
+ | |||
+ | **Added activities** | ||
+ | |||
+ | Try replacing the diodes with RED (D< | ||
+ | |||
+ | Connect a 100 uF capacitor across the bridge output to filter the rectified voltages. Be sure to note the correct polarity of the electrolytic capacitor with respect to the positive and negative outputs of the bridge. How does this filter cap change the voltage and current waveforms? | ||
====Questions: | ====Questions: | ||
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**Resources: | **Resources: | ||
- | * LTSpice files: [[ https:// | + | * LTSpice files: [[downgit>education_tools/ |
- | * Fritzing files: [[ https:// | + | * Fritzing files: [[downgit>education_tools/ |
**For Further Reading:** | **For Further Reading:** | ||
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[[wp> | [[wp> | ||
- | **Return to Lab Activity [[university: | + | **Return to [[university: |
+ | **Return to [[university: | ||