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The Boost Converter

I. Introduction

The Boost Converter, also known as the step up converter, is a power converter circuit that

takes an input DC voltage and converts the output voltage to greater a DC voltage. In this

project, a 12V DC input voltage was converted to 48V DC. The switching in the circuit that is

caused by a metal–oxide–semiconductor field-effect transistor (MOSFET) gives the person

designing the circuit the ability to operate a greater voltage device than its original power input

source. Boost converters are used in battery powered devices, where the electronic circuit

requires a higher operating voltage than the battery can supply. This report introduces a design

technique for boost converter that is implanted in the laboratory and obtain experimental results.

The experimental results will consist of measurements taken with the o-scope at certain points

throughout the circuit. The experimental results are then compared with the simulated results that

were developed using PSpice code. The voltage and current ripple effects caused by passive

components, and the power losses that the boost converter produces from drawing too much

current are both analyzed in this lab project.

II. Theory

A. Operation of Ideal Boost Converter To gain a better understanding of how a boost convert functions, a single double throw

switch (SPDT) switch can be modeled in the circuit to grasp the understanding of how the MOSFET fast acting switching operates. The ideal model contains an input dc voltage source

Vin, an inductor with voltage VL and current iL, an output capacitor C with a current ic, and a power resistor R acting as the load at the output of the circuit parallel with C as shown in Figure 1 below:

Figure 1. Boost Converter Circuit with Ideal Switch

The ideal model of the boost converter is a simple way of visualizing how the MOSFET works in the circuit. Initially the switch moves to position 1 which will cause a short circuit that occurs

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right after the inductor. The inductor will then store energy by charging up until it reaches voltage Vin. Then the switch moves to position 2 where the total impedance should increase by R

and C in parallel causing the current to be less than the current in position 1, but since circuit is in a transient state, the inductor will prevent that current from decreasing and obtain a constant

current by reversing its polarity. The output voltage V across R will have a higher voltage which will be the sum of the voltages Vin and VL. When the switch moves back to position 1, this is when the capacitor C starts to discharge through R because they are isolated from the rest of the

circuit since that short circuit is there.

B. Actualization of Boost Converter The realistic model of the boost converter contains the same components of the ideal model

but instead of a SPDT switch, the realization that MOSFET Q1 and a Schottky diode D1 are

visualized in the circuit as shown in Figure 2, and the on/off topologies of the power transistor are discussed a little more in detail of what actually happens during the switches of the MOSFET

Q1.

Figure 2: Actualization of Boost Converter Circuit

In this circuit model, the boost converter, specifically at the gate of the Q1, receives the input of a square wave at a frequency of 100kHz. During this initial stage, the MOSFET switches on and starts conducting with the square wave rising, and places a short circuit at the right side of the

inductor to the negative input supply terminal. The current flows through the inductor between the positive and negative supply terminals. When the current is flowing, the inductor is storing

magnetic energy using its magnetic field. Initially, no current will be flowing through the diode, capacitor and resistor when the switch is acting like a short circuit. When the MOSFET switches off, the square wave is falling, and the amount of current starts decreasing which causes an

electromagnetic field to occur by the inductor using opposite polarity returning the energy to the circuit which was initially stored when Q1 was on. Now there is a greater voltage flowing

throught the circuit, that is Vin + VL while MOSFET is off. This forward biases the diode in order to charge the capacitor with the voltage from the input source, and the voltage from the inductor (Vin + VL). The resistor load also receives the same dc voltage the capacitor charges up to since

R and C are in parallel. When the MOSFET switches back on after the initial startup, and the load continues to be supplied with the desired dc output voltage from the charge of capacitor.

Even though the charge of the capacitor drains away through the resistor, it is recharged each time the MOSFET switches to position 2 turning off. The constant on and off switching produced by the MOSFET gives a steady output dc voltage across the load.

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C. Mathematical Concepts for Boost Converter The fundamental mathematical concepts for the operation of the boost converter are

introduced in order to gain some knowledge of how the circuit is designed and what specifications are required in order to completing the objective to convert 12V to 48 V.

The inductor that is used in the boost converter introduces the concepts of the inductor volt-

second balance and inductor current ripple. The output capacitor that is used in the boost

converter introduces the concepts of the capacitor charge balance and capacitor voltage ripple.

The MOSFET that is used in the circuit has function to produce fast turn-on and turn-off

switching to reduce power switching losses in the circuit. The duty cycle D of the boost

convert’s signal that is controlled from the function generator is needed for the boost converter to

operate. The principle of the inductor volt-second balance states that the total area (or volt-

seconds) under the inductor voltage waveform is zero whenever the converter operates in steady

state. The derivation is shown below starting with inductor defining the relation:

𝑉𝐿 (𝑡) = 𝐿𝑑𝑖𝐿

𝑑𝑡

Integrating on both sides over one complete switching period gives:

𝑖𝐿(𝑇𝑠) − 𝑖𝐿(0) =1

𝐿∫ 𝑉𝐿 (𝑡)𝑑𝑡

𝑇𝑠

0

For the periodic input and steady-state operation, the net average inductor current is zero as

shown below

0 = ∫ 𝑉𝐿(𝑡)𝑑𝑡𝑇𝑠

0

= 1

𝑇𝑠∫ 𝑉𝐿(𝑡)𝑑𝑡

𝑇𝑠

0

= < 𝑉𝐿 >

Where <VL> is The average inductor voltage is zero in steady state. Using the inductor volt-

second balance to find the gain of the boost conver, thus the duty cycle starts with the Net volt-

seconds applied to inductor over one switching period:

∫ 𝑉𝐿 (𝑡) 𝑑𝑡 = (𝑉𝑖𝑛) 𝐷𝑇𝑠 + (𝑉𝑖𝑛 – 𝑉𝑜) 𝐷′𝑇𝑠

𝑇𝑠

0

Setting the equation to zero and collect the terms:

𝑉𝑜 (𝐷 + 𝐷′) – 𝑉𝑜𝐷′ = 0

Solving for 𝑉𝑜 gives

𝑉𝑜 =𝑉𝑖𝑛

𝐷′

The voltage conversion ratio is therefore:

𝑀(𝐷) =𝑉𝑜

𝑉𝑖𝑛

=1

𝐷′=

1

1 − 𝐷

(1)

(2)

(3)

(4)

(5)

(6)

(7)

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Through this equation is where the gain can be found, and solving for the duty cycle D results in the required duy cycle the boost converter needs in order for operation. The Power MOSFET

transistors that are used in this lab contain fast switching rates that will make the output voltage greater than the input voltage tthen remain a constant. What helps this happen is when the output

capacitor charges up the voltage when the MOSFET is off and then discharges when the MOSFET is on after the initial stage. This helps achieve the desired output voltage but also creates an undesired voltage ripple in the signal. The determination of inductor current ripple is

the change in inductor current during postion 1, when the MOSFET is on, equals the slope multiplied by the length of postion 1:

2𝛥𝑖𝐿 =𝑉𝑖𝑛

𝐿𝐷𝑇𝑠

The peak current ripple from the inductor then can be solved to get:

𝛥𝑖𝐿 = 𝑉𝑖𝑛

2𝐿𝐷𝑇𝑠

The determination of capacitor voltage ripple is the change in capacitor’s voltage during postion

1, when the MOSFET is on, equals the slope multiplied by the length of postion 1:

2𝛥𝑣 = −𝑉𝑜

𝑅𝐶𝐷𝑇𝑠

The peak voltage ripple from the capacitor then can be solved to get:

𝛥𝑣 = −𝑉𝑜

2𝑅𝐶𝐷𝑇𝑠

Electrolytic capacitors that were used in the boost converter circuit contain what is called

Equivalent Series Resistance (ESR). ESR is a measure of the total lossiness of a capacitor.

Current is constantly changing in the circuit so another capacitance is to be put in parallel with

the input voltage soure Vin. The capacitor serves as an input buffer to supply energy when the

input voltage is low, so it stores energy while the dc-dc converter changes to a new energy level,

thus a greater voltage. It also prevents noise casued from the MOSFET switching to reach the

input source. The output capacitance in the boost converter acts like a filter and a sink for the

current that is produced by the inductor. The output voltage wants to be greater than the input

voltage, so the output capacitance functions as an energy buffer. The losses due to ESR will

prevent the input capacitance to quickly source charge and the output capacitance to quickly sink

charge. If ESR is increased at the input, this results in high frequency noise across thhe

capacitance which will decrease the effect of the filter, so ESR needs to be as low as possible. If

ESR is increased at the output , this results in more voltage ripple which will effect the stability

of the signal. A way to decrease ESR is to use multiple capacitors in parallel instead of just using

once capacitor. When capacitors are in parallel, they increase the capacitance by adding up, and

when the resistors are in parallel, the resisitance is reduced, thus decreasing ESR. The output and

(8)

(9)

(10)

(11)

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input capacitance that is that is needed for the boost converter can be determined by the

following equations:

𝐶 = 𝑉𝑜

2𝑅𝛥𝑉𝐷𝑇𝑠

𝐶𝑖𝑛 = 𝛥𝑖𝐿

4𝛥𝑉𝑇𝑠

A power MOSFET can be more efficient if a MOSFET driver is used at the gat of the transistor. This results in better swtitching time that is coming from the function generator. For this lab project a TC4428 IC chip was used. The TC4428 driver helps prevent powerl loss within the

circuit because of the improvement it has on the switching spped at the gate of the MOSFET. Figure 3 shows the pin diagram of the driver IC chip

Figure 3. Pin Diagram for the IC MOSFET Driver

Power losses in this Boost converter consist of the both conduction losses and switching losses.

The conduction losses come from the MOSFET, the inductor and the diode. The switching losses

come from the MOSFET. The conduction losss in the inductor is dependent on the dc and ac

inductor resistances which were calculated after the inductor was designed. The equation is

shown below:

𝑃𝐿 = 𝐼𝑖𝑛2 𝑅𝐷𝐶 + 𝐼𝐴𝐶

2 𝑅𝐴𝐶

The MOSFET experinces conduction loss because of the short ciruit it places in the circuit when

it is on, causing current to flow only during the positive half of the duty cycle so the equation is

as follows:

𝑃𝐿−𝐹𝐸𝑇 = 𝐼𝐿2𝑅𝐷𝑜𝑛 𝐷

When the gate of the MOSFET is switching on and off, the voltages rises and falls, and in that

certain time period there is power loss which can be determined by:

𝑃𝐿−𝑠𝑤 =𝑉𝑜𝐼𝐿

2(𝑇𝑜𝑛 + 𝑇𝑜𝑓𝑓)

(12)

(13)

(14)

(15)

(16)

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The Schottky diode experiences power losses almost the same way the pwer MOSFET does, but

diode dissipates power in reverse bias so it will be at the negative half of the duty cycle. The

equation is as follows:

𝑃𝐷1 = 𝑉𝐷1 𝐼𝐷1𝐷′

To calculate the power efficiency on the circuit

η =𝑃𝑜

𝑃𝑖𝑛

∗ (100)

The input power can be calculated by

𝑃𝑖𝑛 = 𝑉𝑖𝑛 − 𝐼𝑖𝑛

The power output can be calculated by

𝑃𝑜 = 𝑃𝑖𝑛 − ∑ 𝑃𝐿

Where ∑ 𝑃𝐿 is the sum of the conduction and switching losses in the circuit. It is also useful to

calculate the rise temperature of the MOSFET to makure it will not over heat

∆𝑇 = 𝑅𝜃𝐽𝐻 ∗ (𝑃𝐿−𝐹𝐸𝑇 + 𝑃𝐿−𝑠𝑤 )

III. Experimental

The objective of this experiment is to build and test a boost converter that operates at 100kHz

and with a dc input source of 12V at a given current of 2A with a output to be at a current of 0.5

and 48V. The first step in the lab experiment was to calculate all the values needed to build the

circuit and operate it. Using (7), a voltage gain of 4, and a duty cycle of 75% was used for the

function generator. The inductor L was desginesd at an earlier experiment that calculated to be

159µF. A given power resistor was given at 100Ω. A desired output capacitor that should not

exceed 100mVpp. The output capacitance was calculated to be 36µF using (12). To try to

achieve a low ripple, a 33µF and 10µF were placed in parallel to give a total output capacitance

of 43µF. A desired input capacitor that should not exceed 100Vpp as well. The input capacitance

was calculated to be 20µF using (13). To also try to achieve a low ripple by reducing ESR, the

input capacitor used two 10µF capacitors to get a total of 20µF. The power efficiency of the

boost conveter was also calculated after finding the values of the components. Conduction losses

from the inductor using (14) calculated to be 114mW. Conduction losses from the Schottky

diode using (17) was calculated to be 168mW. Conduction losses from the MOSFET using (15)

was calculated to be 121mW and the switching losses from the power MOSFET was calculated

to be 588mW. The total power losses calculated to be 0.991W to get the power efficiency of

96%. The next step was to construct the boost converter on a breadboard with the compoents

shown below in Figure 4.

(17)

(18)

(19)

(20)

(21)

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Figure 4. The Acual Experimental Circuit Schematic Diagram

A 100kHz square wave with a 75% duty cycle was connected to pin 4 of the driver IC chip

shown in Figure 3. The ouput pin 5 was conncted to the gate terminal of the MOSFET. The rest

of the connections for the circuit was simple as shown in Figure 4.

IV. Results

Figure 5. Screenshot of Gate-Source Voltage and the Drain-Source Voltage

In Figure 5, Channel one shows the waveform that is the input voltage source from the function

generator to the gate of the power MOSFET. It also shows that the boost conveter is at a 75%

duty cycle. Channel two shows the waveform of the drain-source voltage which is the voltage of

the MOSFET. When the MOSFET is on in Channel one, the drain source voltage in Channel two

drops almost to zero. Channel two, the drain source voltage rises to the output voltage which is

seen to be at 45.4V, which is close to the desired output voltage of 48V. The drain source voltage

(channel two) was measured using the cursors on the o-scope.

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Figure 6. Drain-Source Voltage and Output Voltage Ripple AC coupled

In Figure 6, channel two is the drain-source voltage of the MOSFET being compared with the

output voltage which is channel one. The output voltage ripple was measured using the cursors

on the o-scope for channel one to get a ripple voltage of about 584mV. This ripple is very high

compared to the theoretical output voltage ripple of 100mVpp. The ouput ripple voltage is

significantly high because of the ESR. More capacitors could have been put in parallel to reduce

this effect.

Figure 7. Output Voltage (Vo) is the bottom waveform V(3). Diode current (ID1) is the top

waveform I(D1)

In Figure 7, the PSpice simulation shows the output voltage Vo at the bottom. Comparing with

output voltage in Figure 6 from experimental measurement, there is a difference in the waveform

shapes because of ESR. Like stated in Figure 6, the output capacitor could have been used with

more capacitors in parallel to achieve a better looking waveform, thus a lower ripple votlage.

Time

50.000ms 50.002ms 50.004ms 50.006ms 50.008ms 50.010ms 50.012ms 50.014ms 50.016ms 50.018ms 50.020ms 50.022ms 50.024ms 50.026ms 50.028ms 50.030ms 50.032ms

V(3)

47.04V

47.08V

47.12V

47.16V

I(D1)

-10A

-5A

0A

3A

SEL>>

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Figure 8. Inductor Voltage (VL) is bottom waveform V(2). Inductor Current (IL) is top waveform

I(L).

In Figure 8, PSpice simulation shows the inductor and the inductor curren simulated. Compered

to Figure 9 below, the inductor waveforms match up which gives the desired result of the incutor

current. The cursors were used on the o-scope for channel one to measure the current ripple. The

current ripple came out to be 472mA as shown in Figure 9. This is much greater than the desired

result because of the ESR again. Not that to measure the current that was captured in Figure 9,

an AC current probe for channel one was needed to be used to get the actual measurement.

Channel two in Figure 9 is the drain source voltage which is coming from the inductor.

Comparing to the bottom waveform in PSpice simulation in Figure 8, the waveforms match up.

Figure 9. Drain- Source Voltage and Inductor Current.

Time

50.000ms 50.002ms 50.004ms 50.006ms 50.008ms 50.010ms 50.012ms 50.014ms 50.016ms 50.018ms 50.020ms 50.022ms 50.024ms 50.026ms 50.028ms 50.030ms 50.032ms

V(2)

0V

20V

40V

55V

I(L)

1.50A

1.75A

2.00A

2.25A

SEL>>

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Figure 10. Drain- Source Voltage and Diode Current AC coupled

In Figure 10, the drain source voltage is the voltage of the MOSFET is channel two and channel

one is the diode current that was AC coupled and measured using a curren probe for channel one.

Compaing the diode current to Figure 7 of the Spice simulation, the waveforms are little different

because of the ESR.

Figure 11. MOFET current (IQ) is the bottom waveform I(SFET). The Output Capacitor Current

(Icout) is the top waveform I(Cout).

In Figure 11, PSpice simulation shows the current plots of the capacitor output and the MOSFET

current. The waveforms can be seen that they are in steady state because each period of the

signal is is the same as period before and after.

Time

50.000ms 50.002ms 50.004ms 50.006ms 50.008ms 50.010ms 50.012ms 50.014ms 50.016ms 50.018ms 50.020ms 50.022ms 50.024ms 50.026ms 50.028ms 50.030ms 50.032ms

I(SFET)

0A

4A

8A

12A

I(Cout)

-12.0A

-8.0A

-4.0A

0A

2.5A

SEL>>

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Figure 12. Drain-Source Voltage and Input Voltage.

In Figure 12, the drain-source voltage is from the MOSFET is channel two. Channel one is the

input voltage being measured. Using the cursors on the o scope with channel one, the ripple

voltage was measured to be 496mV as shown in the capture. It is much greater then the input

ripple desired because of the ESR. More capacitors could have been put in the parallel at the

input to reduce this ripple voltage.

V. Conclusion

The boost converter was modeled and described as an ideal circuit with a switch instead of the

MOSFET. The switch was then replaced with the MOSFET and was described of how the boost

converter functioned with fast switching. The inductor volt-second balance was introduced in

order to find the equation to get the current ripple. The duty cycle needed for operation ended up

being 75% in order for the boost converter to function as desired. The conduction losses and the

switching losses were calculated in order to find the power efficiency of the boost converter,

which ended up being 96%. The experimental ripples that were measured in the screenshots

ended up being much higher than the theoretical ripples. This was a result due to equivalent

series resistance that occurs in electrolytic capacitors. To reduce this large ripple, capactiors for

the inout and output coulf be place in parallel adding the capacitance up and decreasing the

resistance, thus the ripple.

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Appendix

Figure 13. PSpice Code Developed in Pspice to simulate the boost converter’s signals