UC Berkeley, EECS Department B. BoserBoost ConvertersWe have tried to use resistors (voltage dividers) to transform voltages but found that these solutions suffer fromvery poor efficiency: A significant fraction of the total power is dissipated in the resistors and not available forthe load. Moreover, dividers are limited to lowering the voltage. This is problematic in many applications such asmicro-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 donot 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 booststhe input voltage to a higher value and dimension the circuit to generate 15 V from a 5 V input. Figure 1 shows theschematic diagram. The device labeled IRF510 is a transistor. Download its datasheet from the course web. Thediode 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. Wewill later verify of course that this is indeed the case. The voltage Vcis a pulse train and changes between 0 V and5 V. For Vc= 5 V the transistor (IRF510) is on and behaves essentially like a short circuit. Then Vboost= 0 V andVdiode= Vboost− Vout= −15 V. Since Vdiodeis negative, the diode does not conduct any current, i.e. it behaves likean open circuit. With Vc= 0 V the situation reverses: now the transistor is off and the diode conducts. Figure 2 onthe next page illustrates the two situations.In situation (a), Vc= 5 V, the supply voltage Vinappears across the inductor. From the differential equation forinductance we observe that inductors integrate voltage. Therefore the inductor current ILis a ramp with slopedetermined by Vinand L. In situation (b) the inductor again integrates the voltage Vin− Vout= −10 V that appearsacross it. In steady state the current increase and decrease must be identical as otherwise the average currentwould continually increase or decrease. Since it is negative the current through the inductor decreases, as shownin Figure 3 on page 3.Since voltage is proportional to the slope of the current, we note intuitively that reducing the ratio of Toff/Tonresults in higher output voltage Vout. This is because the positive slope is proportional to Vinand the negativeslope of the decreasing current is proportional to Vout− Vin. In the laboratory we will analyze this relationshipquantitatively.Vin5VIRF510CfiltRL0/5Vsquare waveVcVboostVout15VVdiodeLILFigure 1 Boost converter.EECS 40 Lab LAB4: Boost Voltage Supply UID:1March 16, 2009 LAB4 v11201LCfiltRLVout15VVin5VLCfiltRLVout15V(a) Vc= 5V (b) Vc= 0VVin5VFigure 2 Boost converter operating principle with the switch (transistor) closed (a) and open (b).DesignLet’s first derive an expression for the voltage boost factor, Vout/Vin. We start by writing expressions for ∆ILduringTonand Toff. At this point, enter only the expressions. Once you have determined the value of L (see below) youcan solve for and enter the numerical answer. Same for the simulation result. Hint: set up the differential equationfor current and voltage in the inductor during the two phases.Expression Simulation CalculationDuring Ton∆IL=1 pt.01 pt.01 pt.0During Toff∆IL=1 pt.11 pt.11 pt.1From the timing diagram shown in the guide we know that the magnitude of ILis the same during Tonand Toff.Equate the equations above and solve for the voltage boost factor Vout/Vin.Remarkably this result depends only on Tonand Toffand is independent of the value of the inductance. CalculateTon/Tofffor Vout=15 V and Vin=5 V.Ton/Toff=1 pt.2For simplicity, in this laboratory we will generate Tonand Toffwith the pulse generator. More practical implemen-tations adjust this ratio dynamically to keep the value of Voutconstant in the presence of variations of Vinand theload current. Calculate the value of Ton/Toffthat keeps Voutconstant despite varying Vin.Vin= V Ton/Toff=1 pt.3Vin= V Ton/Toff=1 pt.4To finalize the design of the boost converter we must determine the operating frequency f = 1/T with T =Ton+ Toffand the values of L and Cfilt. We pick f =100 kHz to account for the frequency limitation of solderlessbreadboards1. From this we can calculate Tonand Toffand 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).L =1 pt.5During Tonthe diode is not conducting and the entire current to the load comes from Cfilt. Because of this theoutput voltage will drop. Keeping this drop to ∆Vout= mV for RL=1 kΩ determines the value of Cfilt(usethe next larger available value in the lab). Realizing that ∆Vout≪ Voutwe conclude that the current through theresistor is approximately constant, IRL= Vout/RL.Cfilt=1 pt.6Verify your result with SPICE. For simulation only, add a 6 Ω resistor in series with the inductor to account for thewinding resistance (do not add this resistor in the actual circuit you will be building). Attach a transient simulationshowing Vc, Vboost, Vdiode, Voutand the current through the inductor for 3 cycles in steady state to your lab report (4points; no credit for lab reports without simulation).Simulated Vboost=1 pt.71Switching power supplies are usually operated at higher frequencies to reduce the size of the inductor.2March 16, 2009 LAB4 v11202.76.328timetimeVcIL5V0VIL0ILTonToffT=1/fFigure 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, itcan produce voltages in excess of 30 V e.g. when the input voltage is chosen higher than 5 V.Exert extra cautionand touch circuit nodes only after having determined (e.g. with the oscilloscope) that voltage levels agree withyour simulation results and are below 30 V.Also, complete the entire circuit before turning on power. Especiallydo not omit the diode and load resistor. Measure Vc, Vboost, Vdiode, Voutwith the oscilloscope and compare yourresult to SPICE. Comment on any discrepancies (hint: consider the assumptions made for the calculations).Explain discrepancies between calculations, simulations, and measurements:3 pts.73 March 16, 2009 LAB4 v1120In SPICE and the actual circuit, vary the load resistor RLfrom 100 Ω to 20 kΩ and graph your result. Label
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