UC Berkeley EECS Department EECS 40 Lab B Boser UID LAB4 Boost Voltage Supply 1 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 Vc is a pulse train and changes between 0 V and 5 V For Vc 5 V the transistor IRF510 is on and behaves essentially like a short circuit Then Vboost 0 V and Vdiode Vboost Vout 15 V Since Vdiode is negative the diode does not conduct any current i e it behaves like an open circuit With Vc 0 V the situation reverses now the transistor is off and the diode conducts Figure 2 on the next page illustrates the two situations In situation a Vc 5 V the supply voltage Vin appears across the inductor From the differential equation for inductance we observe that inductors integrate voltage Therefore the inductor current IL is a ramp with slope determined by Vin and L In situation b the inductor again integrates the voltage Vin Vout 10 V that appears across it In steady state the current increase and decrease must be identical as otherwise the average current would continually increase or decrease Since it is negative the current through the inductor decreases as shown in Figure 3 on page 3 Since voltage is proportional to the slope of the current we note intuitively that reducing the ratio of Toff Ton results in higher output voltage Vout This is because the positive slope is proportional to Vin and the negative slope of the decreasing current is proportional to Vout Vin In the laboratory we will analyze this relationship quantitatively IL Vdiode L Vboost Vin 5V IRF510 0 5V square wave Cfilt RL Vout 15V Vc Figure 1 Boost converter 1 March 16 2009 LAB4 v1120 L L Vin 5V Vout 15V Cfilt RL Vin 5V Cfilt RL a Vc 5V Vout 15V b Vc 0V Figure 2 Boost converter operating principle with the switch transistor closed a and open b Design Let s first derive an expression for the voltage boost factor Vout Vin We start by writing expressions for IL during Ton and Toff At this point enter only the expressions Once you have determined the value of L see below you can solve for and enter the numerical answer Same for the simulation result Hint set up the differential equation for current and voltage in the inductor during the two phases Expression During Ton IL During Toff IL Simulation Calculation 1 pt 1 pt 1 pt 0 1 pt 0 1 pt 0 1 pt 1 1 1 From the timing diagram shown in the guide we know that the magnitude of IL is the same during Ton and Toff Equate the equations above and solve for the voltage boost factor Vout Vin Remarkably this result depends only on Ton and Toff and is independent of the value of the inductance Calculate Ton Toff for Vout 15 V and Vin 5 V 1 pt Ton Toff 2 For simplicity in this laboratory we will generate Ton and Toff with the pulse generator More practical implementations adjust this ratio dynamically to keep the value of Vout constant in the presence of variations of Vin and the load current Calculate the value of Ton Toff that keeps Vout constant despite varying Vin Vin 2 7 V Vin 6 3 V 1 pt Ton Toff 3 1 pt Ton Toff 4 To finalize the design of the boost converter we must determine the operating frequency f 1 T with T Ton Toff and the values of L and Cfilt We pick f 100 kHz to account for the frequency limitation of solderless breadboards1 From this we can calculate Ton and Toff and then solve for L from one of the equations for IL 6 mA Round L to the nearest available value use the resistor scale i e multiples of 10 12 15 etc 1 pt L 5 During Ton the diode is not conducting and the entire current to the load comes from Cfilt Because of this the output voltage will drop Keeping this drop to Vout 28 mV for R L 1 k determines the value of Cfilt use the next larger available value in the lab Realizing that Vout Vout we conclude that the current through the resistor is approximately constant IRL Vout R L 1 pt Cfilt 6 Verify your result with SPICE For simulation only add a 6 resistor in series with the inductor to account for the winding resistance do not add this resistor in the actual circuit you will be building Attach a transient simulation showing Vc Vboost Vdiode Vout and the current through the inductor for 3 cycles in steady state to your lab report 4 points no credit for lab reports without simulation Simulated Vboost 1 Switching 1 pt 7 power supplies are usually operated at higher frequencies to reduce the size of the inductor 2 March 16 2009 LAB4 v1120 Vc Ton Toff 5V 0V T 1 f time IL IL IL0 time Figure 3 Boost converter timing diagram Now you are ready to test the boost converter in the laboratory Although it is designed to generate only 15 V it can produce voltages in excess of 30 V e g when the input voltage is chosen higher than 5 V Exert extra caution and touch circuit nodes only after having determined e g with the oscilloscope that voltage levels agree with your simulation results and are below 30 V Also complete the entire circuit before turning on power Especially do not omit the diode and load resistor Measure Vc Vboost Vdiode Vout with the oscilloscope and compare your result to SPICE Comment on any discrepancies hint consider the assumptions made for the calculations Explain discrepancies between calculations simulations and measurements 3 pts 7 3 March 16 2009 LAB4 v1120 In SPICE and the actual circuit vary the load resistor R L from 100 to 20 k and graph your result Label the axes Vout IRL Demo the circuit to the GSI Ideally the
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