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An introduction to the behavior and application of diodes in electrical circuits. It covers various diode configurations, the process of determining their state, and the impact of applied loads. The text assumes a forward resistance negligible compared to other network elements and explains how to identify 'on' and 'off' states based on voltage and current levels.
Typology: Lecture notes
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Note for future reference that the polarity of VD is the same as would result if in fact, the diode were a resistive element. The resulting voltage and current levels are the following: Reversing the diode as shown in this figure: Mentally replacing the diode with a resistive element as shown in the below figure will reveal that the resulting current direction does not match the arrow in the diode symbol. The diode is in the “off” state, resulting in the equivalent circuit as shown in the below figure: Due to the open circuit, the diode current is 0 A and the voltage across the resistor R is the following: The fact that VR = 0 V will establish E volts across the open circuit as defined by Kirchhoff’s voltage law. Always keep in mind that under any circumstances—dc, ac instantaneous values, pulses, and so on—Kirchhoff’s voltage law must be satisfied!
EXAMPLE PROBLEMS (SOLUTION WILL BE DISCUSSED IN THE ONLINE SESSION)
process. When employed in the rectification process, a diode is typically referred to as a rectifier. Its power and current ratings are typically much higher than those of diodes employed in other applications, such as computers and communication systems During the interval t = 0 -->T/2 in the above figure, the polarity of the applied voltage vi is such as to establish “pressure” in the direction indicated and turn on the diode with the polarity appearing above the diode. Substituting the short-circuit equivalence for the ideal diode will result in the equivalent circuit of below figure , where it is fairly obvious that the output signal is an exact replica of the applied signal. The two terminals defining the output voltage are connected directly to the applied signal via the short-circuit equivalence of the diode. For the period T/2-->T, the polarity of the input vi is as shown in below figure, and the resulting polarity across the ideal diode produces an “off” state with an open-circuit equivalent. The result is the absence of a path for charge to flow, and vo = iR =(0)R = 0V for the period T/2-->T. The input vi and the output vo are sketched together in below figure for comparison purposes. The output signal vo now has a net positive area above the axis over a full period and an average value determined by: The process of removing one-half the input signal to establish a dc level is called half-wave rectification. Consider the figure below to demonstrate the effect of using a silicon diode with VK = 0.7 V for the forward-bias region.
The applied signal must now be at least 0.7 V before the diode can turn “on.” For levels of vi less than 0.7 V, the diode is still in an open-circuit state and vo = 0 V, as shown in the same figure above. When conducting, the difference between vo and vi is a fixed level of VK = 0.7 V and vo=vi- VK, as shown in the figure. The net effect is a reduction in area above the axis, which reduces the resulting dc voltage level. For situations where Vm >> VK, the following equation can be applied to determine the average value with a relatively high level of accuracy. In fact, if Vm is sufficiently greater than VK , is often applied as a first approximation for Vdc.
The peak inverse voltage (PIV) [or PRV (peak reverse voltage)] rating of the diode is of primary importance in the design of rectification systems. Recall that it is the voltage rating that must not be exceeded in the reverse-bias region or the diode will enter the Zener avalanche region. Considering this circuit: The required PIV rating for this half-wave rectifier can be determined on below figure , which displays the reverse-biased diode of the circuit above with maximum applied voltage. Applying Kirchhoff’s voltage law, it is fairly obvious that the PIV rating of the diode must equal or exceed the peak value of the applied voltage. Therefore, EXAMPLE PROBLEM https://youtu.be/kVBtYfC_pgs
Since the area above the axis for one full cycle is now twice that obtained for a half-wave system, the dc level has also been doubled and If silicon rather than ideal diodes are employed as shown in Fig. 2.58 , the application of Kirchhoff’s voltage law around the conduction path results in
relatively high level of accuracy: Then again, if V m is sufficiently greater than 2 V K, then Eq. (2.10) is often applied as a first approximation for V dc.
PIV The required PIV of each diode (ideal) can be determined from Fig. 2.59 obtained at the peak of the positive region of the input signal. For the indicated loop the maximum voltage across R is V m and the PIV rating is defined by Center-Tapped Transformer A second popular full-wave rectifier appears in Fig. 2.60 with only two diodes but requiring a center- tapped (CT) transformer to establish the input signal across each section of the secondary of the transformer. During the positive portion of v i applied to the primary of the transformer, the network will appear as shown in Fig. 2.61 with a positive pulse across each section of the secondary coil. D 1 assumes the short-circuit equivalent and D 2 the open-circuit equivalent, as determined by the secondary voltages and the resulting current directions. The output voltage appears as shown in Fig. 2.61.
