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CHAPTER TWO POWER SEMICONDUCTOR DIODES AND CIRCUITS DESIGNED BY DR. SAMEER KHADER PPU “E-learning Project”

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Diode circuits I- (S.2.3) : Diode Characteristics: Power diode is a two-terminal pn-junction device ……….); The equivalent circuit and i-v curve are displayed as follows, where the diode conducts when a battery is connected across its terminals. Symbol P-N junction

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II- (S.2.4) : Reverse Recovery Characteristics: When the current falls to zero, the diode continues to conduct under the action of minority carriers that remain stored in the pn- junction and Bulk resistance. These carriers require a certain time to recombine with opposite charges and to be neutralized. This time is called the reverse recovery time trr. The figure shown below displayed the current falling process and the diode reverse parameters. 1- The reverse recovery time trr consist of two parameters ta, and tb: trr =ta+tb ; Irr=ta.di/dt ta- due to storage charge in the depletion region tb- due to storage charge in the bulk resistance. The softness factor Sf=tb/ta. Trr- depends on the junction temperature, di/dt, and the diode forward current 2- The reverse recovery charge Qrr : this is the charge carriers across the diode flows in the reverse direction due to changeover of the conduction state.

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Example 2.1: Given a diode circuit (p.2.1) with reverse recovery time trr=5 µS, and the rate of fall of the diode current is di/dt= 80A/ µS with softness factor Sf=0.5. Determine : 1- the storage charge Qrr; 2- the peak reverse current Irr. Solution: III- (S.2.5) : Diodes Classification : Depending on the recovery characteristics, and manufacturing techniques, there are three types: 1.General - purpose diodes: they have high trr=25 µS and with frequency < 1kHZ, applied in AC to DC circuits. The current rating up to 1000A, and up to 5kV. 2.Fast recovery diodes: they have small trr=5 µS and with frequency < 5kHZ, applied in DC to DC and DC to AC circuits. The current rating up to 100A, and up to3kV.

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3. Schottkey diode: they have approximately zero reverse recovery time with high frequency up to 10kHz, and applied in high current low voltage applications. The current rating up to 300A and 100V circuit voltage. IV- (S.2.8) : Series –Connected Diodes : In high voltage dc applications (mainly), the diodes are connected in series with purpose to increase the reverse blocking capabilities. The difference in the i-v curve in the reverse blocking condition occurs due to manufacturing errors and tolerances, therefore each diode should carry different voltage, while the leakage current is the same. The solution is to force equal sharing voltage across the diodes by connecting a sharing resistances as well shown below:

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THERE ARE TWO APPROACHES: 1. Equal sharing resistances connected across the diodes 2. Equal voltage sharing while the sharing resistances may differs. Mathematical Modeling: In general second case is the most applicable when sharing resistances are used.

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Example 2.2: Two diodes are connected in series as well shown on up mentioned circuit, where the circuit parameters are: 7kV source voltage ( DC), leakage current of first diode Is1=40mA and of second diode Is2=50mA. 1- Find the diode voltages if the resistances are equals Rs1=Rs2=R=80 kΩ. 2- Find the sharing resistances if the diode voltages are distributed equally. Solution: Two cases must be described as follow Case#1: Rs1=Rs2=R=80 kΩ

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V- (S.2.9) : Parallel –Connected Diodes : In high power applications, diodes are connected in parallel with purpose to increase the current carrying capability. Due to some differences in the Bulk resistances of both diodes, there is a different current will flow through the diodes. Therefore by connecting resistances in series with the diodes the diode voltage is shared equally as well shown below: The function of both Ls1 and Ls2 is to equally sharing the current under dynamic behaviors. When I D1 rises, the inductor voltage Ls1.d(I d1 )/dt increases, and a corresponding voltage of opposite polarity is induced across inductor Ls2. This resulting low impedance in the circuit of D2, therefore shifting the current to flow through D2 path.

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Mathematical Modeling: The following equation are derived under steady-state conditions Example 2.3: Find the value of R required for adjusting the voltage across the diodes D1 and D2, if I D1 =50A, I D = 95A, V D1 =1.8V, and V D2 =2V. Solution:

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Mode#1 D1 conducts: The current will flow from the source to the load through D1, where the current i1: IV- (S.2.12) : Freewheeling Diodes If switch S in the figure shown below is closed for time t1, a current I1 is established through the load; and then if the switch is opened for time t2 the current continues to flow through the inductor and the diode” closed path’. If there in no closed path the inductive energy induces a very high voltage and this energy is dissipated in form of heat and spark. The diode realized closed path is called usually FREEWHEELING DIODE. The operation of the proposed circuit is divided into two modes:

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Example 2.4: Determine the value of Ls and Cs for the diode circuit shown behind with L=5mH; R=100Ω; Vs=200V; and di/dt = 50A/ µS. Solution: by applying Lenz law the circuit inductance Ls is determined as follows: Mode#2 Dm conducts: The current will continue to flow through the load and the diode Dm., where the current i2:

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The excess energy due to current increasing is determined as:

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Simulation Results: By applying the simulation software “ POWERSYS” for the same circuit is applied as follows for load voltage Vp1, circuit current I1, load current I2, excess energy due to current increasing is determined as: The source current I1 at first instant rises significantly, and I2 rises also. At steady state the current I1 and I2 are stabilized at a value of 2A.

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THANK YOU FOR LISTENING QUESTIONS…..? PLEASE DON’T HESTATE TO CONTACT ME BY ON

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