CHAPTER 1 DIODE CIRCUITS Resistance levels Semiconductor act differently to DC and AC currents There are three types of resistances 1. DC or static resistance The application of DC voltage to a circuit containing a semiconductor diode will result in an operating point on the characteristics curve that will not change with time. i.e. For specified applied voltage the diode will have a specified current I d and specified resistance R D. The resistance of the diode depends on applied voltage V D and can be found by The D.C resistance levels at the knee and below will be greater than the resistance levels obtained for the vertical section of the charecterstics. The resistance levels in the reverse bias region will naturally quite high. Also higher the current through the diode the lower is the resistance level. 2. AC or dynamic resistance The varying input will move the instantaneous operating point up and down a region of characteristics. 1
Without AC signal applied to the diode the operating point Q which is fixed not moving hence it is called Quiescent point or Q point. A straight line drawn tangent to the curve through the Q point as shown in the figure will define particular change in voltage and current that can be used to determine AC or dynamic resistance for this region of diode characteristics. The steeper the slope, the lower is the value of for the same change in, and there by lowering the value of resistance. The A.C resistance in the vertical-rise region of the characteristics is quite small, Whereas A.C resistance is much higher at low current levels. The lower the Q point of operation, the higher is the A.C resistance. Mathematical expression of AC resistance We have [ ] [ ] 2
Where Hence Flipping the result we have Note: Substituting n =1 and V T = 26mV We have Observation The equation can be used directly by substituting Q point diode current I D to find the AC resistance. At lower values of I D, and the value of obtained must be multiplied by factor of 2. For small values of below the knee of the curve, the equation becomes inappropriate. Every diode has some resistance due to resistance of semiconductor material which is called body resistance and the resistance introduced by the connection between the semiconductor material and external metallic conductor called contact resistance. These can be included in the equation i.e. 3
Average AC resistance If the input signal is sufficiently large to produce a broad swing as indicated in figure the resistance associated with the device for this region is called average AC resistance. The average AC resistance is determined by a straight line drawn between the two intersections established by the maximum and minimum values of the input signal voltages. 4
Summary Type Equation Special characteristics Graphical determinations DC or static resistance Defined as a point on the characteristics AC or Dynamic resistance Defined by a tangent line at the Q point Average AC Defined by a straight line between limits of operation Diode Equivalent Circuits An equivalent circuit is combination of elements properly chosen to best represent the characteristics of a device in a particular operating region. Piecewise linear equivalent circuit The equivalent circuit for a diode can be obtained by approximating the characteristics of the diode by straight line segment which is as shown in the figure 5
The barrier potential which is equal in magnitude to the cut in voltage is represented as a battery of e.m.f V K The resistance r av offered by the forward bias diode is shown as an external resistance in series with the ideal diode. Simplified equivalent circuit For most application the resistance r av is quite small compared to the other elements in the network. So removing r av from the equivalent circuit the characteristics of the diode is as shown in the figure, along with reduced equivalent circuit. 6
Ideal equivalent circuit Let us ignore both resistance and cutoff voltage V K = 0.7V. In the case the equivalent circuit will be reduced to that of an ideal diode as shown in figure Transition and diffusion capacitance Every electronic or electrical device is frequency sensitive. In the p-n semiconductor diode, there are two capacitive effects to be considered 1. Transition capacitance 2. Diffusion capacitance The basic equation for the capacitance of a parallel plate capacitor is defined as Where Transition capacitance In the reverse bias region there is a depletion region that behaves essentially like an insulator between the layers of opposite charge. As the voltage is increased the depletion region widens and hence its capacitances decreases. 7
Diffusion capacitance In forward bias region the width of the depletion region is much reduced resulting in the increased levels of current that will result in increased level of diffusion current. Reverse Recovery time The diode conducts rapidly when forward biased and blocks conduction when reverse biased. If the applied voltage is reversed to establish a reverse bias situation we would ideally like to see the diode change instantaneously from the conduction state to non conduction state, however the diode current will simply reverse as shown in figure and stay at this level for the period of time t s (storage time) required for minority carrier to move into other region, after this reverse current decreases exponentially for a time t t (transition time). 8
Load line analysis Let us consider a diode D in series with a load resistance R. The diode characteristics are placed on the same set of axis as a straight line defined by the parameters of network. The intersection on the vertical axis is defined by applied load R. Hence the analysis is therefore called load line analysis Applying KVL to the circuit Hence When V D = 0, we have When I D = 0, we have E A straight line drawn between the two points will define load line which is as shown in the figure Change the value of R and the intersection on the vertical axis will change. The result will be a different point of intersection between the load line and device characteristics. 9
The point of operation is called Quiescent point which gives current through diode for given load and input voltage By mathematically calculate we have from equation (1) And Example 1. For the series diode configuration of fig (a) employing the diode characteristics of figure (b) Determine I) II) Fig. a Fig.b I. 10
From the resulting load line analysis we find II. 2. Repeat the above problem for Approximate equivalent model we have The level of remains same 11
3. Repeat the above problem for for ideal diode model We have Rectification Rectification is the process of converting alternating current into direct current. Rectifier Rectifier is device that converts AC current into DC current (pulsating DC) Half wave rectifier The following figure shows the half wave rectifier circuit. An alternating input voltage is applied to the diode connected in series with load resistance R During the interval t = 0 T/2 The polarity of V i makes the diode to turn ON this provides short circuit equivalence for the ideal diode. The output signal is exact replica of the applied signal which is as shown in figure 12
During the interval t = T/2 T The polarity of V i makes the diode to turn off. This provides open circuit equivalence for the ideal diode. The result is absence of path which is as shown in the figure When we sketch input V i and the outputs together then we get The D.C output waveform is expected tobe a straight line but the half wave rectifier gives output in The D.C output waveform is expected to be a straight line but the halfwave rectifier gives output in the form of positive sinusoidal pulses. The average value of D.C voltage is given by 13
Since we are using silicon diode we have a cut-in voltage V K of 0.7V. Hence the diode will be in off state when V i is less than 0.7V hence output will be zero which is as shown in figure. The net effect is reduction in area which reduces the average DC voltage hence Note: One complete cycle 0 to 2Π. To find he average value of alternating waveform, we have to determine the area under the curve one complete cycle and then dividing it by the base i.e 2 Π I m peak value of load current 2Π Π current flows only in positive half cycle [ ] [ ] 14
Average D.C load voltage Neglecting the internal resistance we have Example 1. Sketch the output V o and determine the DC level of the output for the network of figure (a) 2. Repeat part (a) if the ideal diode is replaced by a silicon diode 3. Repeat part (a) and (b) if V m is increased to 200V and compare solutions. Using equation (1) and (2). Solution 1. The diode will conduct for negative half cycle hence the output is as shonwn in figure 15
2. For silicon diode. 3. Peak inverse voltage (PIV) or Peak reverse voltage (PRV) The peak inverse voltage rating of the diode is of primary importance in the design of rectification system. It is the voltage rating that must not be exceeded in the reverse bias region. The required PIV rating can be determined from the following figure. Full wave reactivation The following figure shows the full-wave bridge rectifier circuit with four diodes in a bridge configuration. An alternating in out is applied to diodes with a load resistance R. 16
During the interval t = 0 T/2 The polarity of input V i makes the diodes D 2 and D 3 to turn ON whereas D 1 and D 3 to turn off which is as shown in figure. Since the diodes are ideal, the load voltage is V o = V i. During the interval t = T/2 T The polarity of input V i makes the diodes D 1 and D 4 to turn ON where as D 2 and D 3 to turn off which is as shown in figure. Since the diodes are ideal the polarity across the load resistor R is same as shown in figure (a) thus establishing a second positive pulse as shown in figure (b) When we sketch input V i and the outputs together we get 17
The D.C output waveform is expected to be a straight line but the full wave rectifier gives output in the form of positive sinusoidal pulses. The average value of D.C voltage is given by If we use silicon diode as shown in the figure the application of KVL around the conduction path result in The peak output voltage is therefore The DC voltage of silicon diode Peak inverse voltage (PIV) or Peak reverse voltage (PRV) The required PIV rating can be determined from the following figure 18
Center tapped transformer The following figure shows the full wave rectifier circuit with only two diodes, but requiring a center tapped transformer to establish the input signal across each section of the secondary of the transformer. During the interval t = 0 T/2 The positive portion of V i applied to primary of the transformer. The polarity of V i makes the diode D 1 to turn ON where as D 2 to turn off which is as shown in figure During the interval t = T/2 T The negative portion of V i applied to primary of the transformer. The polarity of V i makes the diode D 2 to turn ON where as D 1 to turn off which is as shown in the figure 19
When we sketch input Vi and the outputs together we get Peak inverse voltage (PIV) or Peak reverse voltage (PRV) The required PIV rating can be determined from the following figure Example Determine the output waveform for the network of figure and calculate the output DC level and the required PIV of each diode. Solution The network appears as shown in the figure below for the positive region of the input voltage 20
Redrawing the above circuit we have Applying voltage divider rule we have Hence For the negative part of the input the roles of the diodes are interchanged and V o appears as shown below 21
Clippers The circuits which are used to clip off unwanted portion of the waveform by making use of diodes without disturbing the remaining part of the waveform are called clippers. The half wave rectifier is an example of the simplest form of diode clippers which contains one resistor and one diode. Clippers are also called limiters or slicers. Depending on the orientation of the diode the positive or the negative region of the applied signal is clipped off. Clippers are classified into two types 1) Series clippers 2) Parallel clipper 1. Series clipper: In series clipper the diode is connected in series with the load as shown in figure Figure (a) shows a negative series clipper circuit, because it passes only the positive half cycle of an alternating waveform and clips off the other half cycle. A diode series clipper is simply a half wave rectifier circuit Output: Figure (b) shows a positive series clipper circuit because it passes only the negative half cycle of an alternating waveform and clips off the positive half cycle. Output: 22
The zero level output from a series clipper circuit is not exactly zero. The reverse saturation current (I R ) of the diode produce a voltage drop across resistor. This is almost so small that it can be neglected. 2. Shunt clippers: In shunt clipper the diode is connected in parallel with the load as shown in figure The figure shows positive shunt clipper circuit because it passes only the negative half cycle of an waveform and clips off the other half cycle. The figure above shows negative shunt clipper circuit because it passes only the positive half cycle of an waveform and clips off the other half cycle Examples Series clipper 1) Positive clipper or to pass negative peak above V k level 23
During Positive half cycle Anode is at ground potential. Cathode sees variable positive input voltage from o to +V m. For complete positive half cycle diode becomes reverse biased and hence V out = 0V. The circuit and the corresponding output waveform is as shown below During negative half cycle Anode is at ground potential. Cathode sees variable negative input voltage from o to -V m. For complete negative half cycle diode becomes forward biased and hence the output is given as V out = -V m + V k. The circuit and the corresponding output waveform is as shown below Hence applying K.V.L to the loop we have If V in = 10V p p and V K = 0.7V then 24
2) To pass negative peak above some level say (V R + V k ) During Positive half cycle Anode is at ground potential. Cathode sees variable positive input voltage from V R to V m + V R. For complete positive half cycle diode becomes reverse biased and hence V out = 0V. The circuit and the corresponding output waveform is as shown below During negative half cycle Anode is at ground potential. Cathode sees variable negative input voltage from V R to -V m + V R. When the diode is reverse biased and V o = 0V When the diode is forward biased. The circuit and the corresponding output waveform is as shown below 25
Hence applying K.V.L to the loop we have If V in = 10V p p, V K = 0.7V and V R = 2.3V then 3) To pass +ve peak above V K level During Positive half cycle Cathode is at ground potential. Anode sees variable positive input voltage from 0 to +V m. For complete positive half cycle diode becomes forward biased. The circuit and the corresponding output waveform is as shown below If V in = 10V p p, V K = 0.7V 26
During negative half cycle Cathode is at ground potential. Anode sees variable positive input voltage from 0 to -V m. For complete negative half cycle diode becomes reverse biased and hence V out = 0V. The circuit and the corresponding output waveform is as shown below 4) To pass positive peak above some level (V R + V k ) During Positive half cycle Cathode is at ground potential. Anode sees variable positive input voltage from -V R to +V m V R When the diode is reverse biased and V o = 0V When the diode is forward biased. The circuit and the corresponding output waveform is as shown below 27
Hence applying K.V.L to the loop we have If V in = 10V p p, V K = 0.7V and V R = 2.3V then During negative half cycle Cathode is at ground potential. Anode sees variable positive input voltage from V R to V m V R For complete negative half cycle diode becomes reverse biased and hence V out = 0V. The circuit and the corresponding output waveform is as shown below Write the output wave form for the following circuit A) Series Clippers 1) 28
2) 3) 4) 29
5) 6) 7) 30
8) 9) 10) 31
11) 12) 13) 32
14) B) Shunt Clippers 1) 2) 33
3) 4) 5) 34
6) 7) 8) 35
9) 10) 36
1) Sketch the output waveform V o to the time scale Solution Case: 1 for Diodes D1 and D2 are off therefore V o = V i. Case: 2 When Diode D1 conducts and diode D2 is open circuited, Which is as shown in the figure below Applying K.V.L to loops we have 37
Case: 3 When Diodes D1 and D2 are off therefore V o = V i. Case: 4 When Diode D2 conducts and hence Hence the input and output waveform along with transfer characteristics is as shown 2) The input voltage V i to the two level clipper circuit as shown in the figure caries linearly from 0 150V. Sketch the output voltage to time scale. Assume the diodes are ideal. 38
Solution: Case: 1 for Diode D1 and D2 are off. The circuit is as sown in figure, Hence Case: 2 for 25V Diodes D1 is off and D2 is on. The circuit is as shown below Applying K.V.L to the loop we have Substituting (1) in and (2) we have [ ] 39
Thus for 25V, From equation (3) we have At, At, Case: 3 for 100V Both the diodes becomes on. The circuit is as shown below Apply K.V.L to the above circuit Hence the input and output waveform along with transfer characteristics is as shown 40
Clamping circuit Clampers are networks or circuits that changes the input signal to a different DC level but the peak to peak swing of applied signal will remain the same. It adds a DC voltage to the A.C signal hence it is also called as DC restorer DC inserter circuits. The capacitor diode and resistance are the three basic elements of a clamper circuit The clamper circuit relies on a change in the capacitors time constant current path with changing input voltage. The magnitude of R and C are chosen so that T = RC is large enough that the voltage across the capacitor does not discharge significantly during the diodes non conduction state Design: If f = 1KHz T = 1ms Let C = 1μF 1. Positive peak clamped at reference level Capacitor charges only when diode conducts in above circuit diode gets conducted during positive half cycle. When the input voltage is +Ve (V i > 0) the diode is ON and capacitor charges to peak value of input signal as shown in figure 41
Apply KV to 1 st loop We know that, Applying K.V.L to 2 nd loop When the input voltage is ve (V i < 0) the diode is OFF as shown in figure Applying KVL Input /output waveform is as shown below 42
2) Positive peak clamped at +ve reference (+2V) When diode ON and capacitor starts charging at a very low time constant which is as shown in the figure. Applying KVL to the circuit shown in figure Applying KVL to the second loop Hence 43
When the input voltage is ve (V i < 0) the diode is OFF as shown in figure Applying KVL to the circuit shown in figure Input /output waveform is as shown below 3) Positive peak clamped at ve reference level say (-2V) 44
When diode ON and capacitor starts charging at a very low time constant Which is as shown in the figure. Applying KVL to the circuit shown in figure Applying KVL to the second loop Hence When the input voltage is ve (V i < 0) the diode is OFF as shown in figure Applying KVL to the circuit shown in figure 45
Input /output waveform is as shown below 4) Negative peak clamped level Diode starts conducting during negative half cycle. During this capacitor starts charging to the peak value of the input voltage which is as shown in the figure Apply KV to 1 st loop 46
Applying K.V.L to 2 nd loop When the input voltage is +ve (V i > 0) the diode is OFF as shown in figure Applying KVL to the circuit shown in figure Input /output waveform is as shown below 5) Negative peak clamped to +ve reference level (say +2v) 47
When input voltage is ve ( shown in figure ) the diode is ON applying KVL to the circuit as Apply KV to 1 st loop Applying KVL to second loop Hence When input is +ve ( >0V) the diode is off applying KVL to circuit in figure Applying KVL to the circuit shown in figure 48
Input /output waveform is as shown 6) Negative peak clamped at ve reference level (-2V) When the input voltage is negative (V in < 0) the diode is ON which is as shown in the figure Apply KV to 1 st loop 49
Applying KVL to second loop Hence When input is +ve ( >0V) the diode is off applying KVL to circuit in figure Applying KVL to the circuit shown in figure Input /output waveform is as shown 50
1) Determine V o for the network as shown in the figure below (ideal diode) Solution Case1: During the interval Diode is ON capacitor charges to peak value of the input signal which is ass shown in the figure Applying K.V.L to the loop1 Applying K.V.L to the loop2 Case2: During the interval Applying K.V.L to the loop 51
Case2: During the interval Applying K.V.L to the loop1 Applying K.V.L to the loop2 Input /output waveform is as shown *******END******* 52