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Unformatted text preview: UC Berkeley, EECS Department B. E. Boser EECS 40 Lab LAB5: Boost Voltage Supply UID: Enter the names and SIDs for you and your lab partner into the boxes below. Name 1 SID 1 Name 2 SID 2 Boost Converters We have tried to use resistors (voltage dividers) to transform voltages but found that these solutions suffer from very poor efficiency: A significant fraction of the total power is dissipated in the resistors and not available for the load. Moreover, dividers are limited to lowering the voltage. This is problematic in many applications such as micro-mechanical actuators (MEMS) that often require high voltages for operation. With inductors and capacitors we can overcome both problems. Since these elements (ideally) only store but do not dissipate power, much higher efficiencies are attainable. In this laboratory we design and test a special kind of switching power supply called boost converter that boosts the input voltage to a higher value and dimension the circuit to generate 15 V from a 5 V input. Figure 1 shows the schematic diagram. The device labeled IRF510 is a transistor. Download its datasheet from the course web. The diode conducts current only in the direction of the arrow. To analyze the circuit we assume first that it is working correctly, in particular that the output voltage is 15 V. We will later verify of course that this is indeed the case. The voltage V c is a pulse train and changes between 0 V and 5 V. For V c = 5 V the transistor (IRF510) is on and behaves essentially like a short circuit. Then V boost = 0 V and V diode = V boost- V out =- 15 V. Since V diode is negative, the diode does not conduct any current, i.e. it behaves like an open circuit. With V c = 0 V the situation reverses: now the transistor is off and the diode conducts. Figure 2 on the next page illustrates the two situations....
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This note was uploaded on 01/15/2011 for the course EE 40 taught by Professor Chang-hasnain during the Fall '07 term at University of California, Berkeley.
- Fall '07