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Eighteenth lecture / electronic circuit design

12/6/2021

‫المرحلة الثالثة‬/‫ تصميم دوائر االلكترونية‬.......................................................... ...................... ‫احمد عدنان المظفر‬.‫م‬.‫م‬ 2- Principle of Step Up Chopper Operation In the step-up chopper we can obtain an average output voltage V 0 greater than input voltage. Figure (a) shows the elementary form of a step-up chopper. In step-up chopper a large inductor, L is in series with the source voltage Vs. This forms a closed path as shown in the figure (b). During the time period T on the chopper is on the inductor stores energy. When the chopper is turned off the current is forced to flow through the diode and load for a time TOFF and as the inductor current cannot die 77 ‫المرحلة الثالثة‬/‫ تصميم دوائر االلكترونية‬.......................................................... ...................... ‫احمد عدنان المظفر‬.‫م‬.‫م‬ suddenly. When the current decreases the polarity of the emf induced in L is reversed. Fig (c). As a result the total voltage available across the load is given by the equation V0 = Vs + L (di/dt) . The voltage V0 exceeds the source voltage and hence the circuit acts as a step-up chopper and the energy which is stored in L is released to the load. When the chopper is turned ON the current through the inductance L will increase from I1 to I2. As the chopper is on the source voltage is applied to L that is VL = VS . When the chopper is OFF, the KVL for the figure (c) can be written as VL – V0+Vs =0 or VL =V0 -Vs where VL is the voltage across L. Variation of source voltage VS , source current IS , load voltage V0 and load current iO is sketched in the 78 ‫المرحلة الثالثة‬/‫ تصميم دوائر االلكترونية‬.......................................................... ...................... ‫احمد عدنان المظفر‬.‫م‬.‫م‬ fig (d) . Let us assume that the variation of output current is linear, the energy input to inductor from the source, during the time period T on , is Win= Vs (I1+I2/2) Ton 𝑉𝑆 = 𝐿 𝐿 𝛥𝑖 𝑇𝑂𝑁 𝑑𝑖𝑜 𝑑𝑡 = 𝑉𝑆 → 𝛥𝑖 = 𝑉𝑆 𝐿 𝑇𝑂𝑁 During the time TOFF the chopper is off, so the energy released by the inductor to the load is WOFF = (V0-Vs)(I1+I2/2).TOFF 𝑉𝑂 = 𝑉𝑆 + 𝑉𝐿 𝑉𝑂 − 𝑉𝑆 = 𝐿 𝐿 𝛥𝑖 𝑇𝑂𝐹𝐹 𝑑𝑖𝑜 𝑑𝑡 = 𝑉𝑂 − 𝑉𝑆 → 𝛥𝑖 = 𝑉𝑜 −𝑉𝑠 𝐿 𝑇𝑂𝑁 Let us assume that the system is lossless, then the two energies say W in and WOFF are equal. So equating these two we will get Vs (I1+I2/2) Ton Vs Ton = = (V0-Vs)(I1+I2/2).TOFF (V0-Vs) TOFF V0TOFF = Vs (TOFF + Ton) = Vs .T V0 = VS (T/TOFF) = VS (T/T-Ton) =VS (1/(1-K)) ………….(1) From the equation 1 we can see that the average voltage across the load can be stepped up by varying the duty cycle. If the chopper in the figure (a) is always off, K=0 and V0= Vs. If the chopper is always on, K =1 and V0 = infinity as we can see from the graph. In practical applications the chopper is turned on and off so that the required step-up average output voltage, more source voltage is obtained. 79 ‫المرحلة الثالثة‬/‫ تصميم دوائر االلكترونية‬.......................................................... ...................... ‫احمد عدنان المظفر‬.‫م‬.‫م‬ K Figure shows variation of load voltage V0 with duty cycle . Example 55: A step up chopper has an input voltage of 150V. The voltage output needed is 450V. Given, that the thyristor has a conducting time of 150μseconds. Calculate the Duty cycle and chopping frequency. Example 56: A step-up chopper supplies a load of 480 V from 230 V dc supply. If blocking period of the thyristor to be 80 microsecond, find the required pulse width. Example 57: In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor is connected across the load. The inductor current is assumed to be continuous. The average voltage across the load and the average current through the diode will respectively be 80