DC/DC Boost Converter

JoVE Science Education
Electrical Engineering
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JoVE Science Education Electrical Engineering
DC/DC Boost Converter

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12:18 min

April 30, 2023

Visão Geral

Source: Ali Bazzi, Department of Electrical Engineering, University of Connecticut, Storrs, CT.

Boost converters provide a versatile solution to stepping up DC voltages in many applications where a DC voltage needs to be increased without the need to convert it to AC, using a transformer, and then rectifying the transformer output. Boost converters are step-up converters that use an inductor as an energy storage device that supports the output with additional energy in addition to the DC input source. This causes the output voltage to boost.

The objective of this experiment is to study different characteristics of a boost converter. The step-up capability of the converter will be observed under continuous conduction mode (CCM) where the inductor current is non-zero. Open-loop operation with a manually-set duty ratio will be used. An approximation of the input-output relationship will be observed.

Princípios

A boost converter relies on energy stored in the inductor, L, to supply energy to the output side where the load is supported, in addition to a DC source being the main energy source. The main concept behind boost converter operation is that an inductor will flip its voltage polarity to maintain current flow. As shown in Fig. 1(a) for a simple boost converter circuit, when the switch is on for a duty cycle D of the switching period T, inductor voltage VL builds up. When the switch is off, the inductor current has to continue flowing and therefore the inductor's voltage polarity will flip to add to the input voltage Vde.

However, when the switch is on, the load is short-circuited and the output voltage is zero, which is not desired. Therefore, a blocking diode is added at the output side as shown in Fig. 1(b) to prevent the load from being short-circuited. This diode still does not solve the issue of the load seeing no voltage when the switch is on, so a capacitor is added as shown in Fig. 1(c) to provide the load with necessary current during the period when the switch is on. Note that the when the switch is on, the diode is off (reverse biased), and vice versa. The average output voltage is thus related to the input voltage as: <Vout>=Vde/(1-D).

Figure 1
Figure 1. Steps to building a boost converter

As this experiment proceeds, it will be shown that the average output voltage increases as the duty cycle, D, increases. This is true since the output voltage to input voltage relationship is inversely proportional to –D, and thus the output voltage and D have a positive correlation.

Note that the equation presented is for an ideal boost converter, and may seem as if a D=1 will yield infinite output voltage, but that is not true. In reality, parasitic elements and resistances in the boost converter cause D to be limited to around 70-80% after which parasitic effects start to dominate circuit operation and cause significant voltage drops. At such a point, the output voltage starts to decrease as D increases. With higher switching frequencies, the voltage ripple at the output will decrease since the voltage charging and discharging times at the capacitor become significantly shorter with a decreased switching frequency.

Procedimento

ATTENTION: This experiment is designed to limit the output voltage to be less than 50V DC. Only use duty ratios, frequencies, input voltage, or loads that are given here.

This experiment will utilize the DC-DC converter board provided by HiRel Systems. http://www.hirelsystems.com/shop/Power-Pole-Board.html

Information about the board operation can be found in this collections video “Introduction to the HiRel Board.”

The procedure shown here applies to any simple boost converter circuit that can be built on proto boards, bread boards, or printed circuit boards.

1. Board setup:

  1. Connect the ±12 signal supply at the "DIN" connector but keep "S90" OFF.
  2. Make sure that the PWM control selector is in the open-loop position.
  3. Set the DC power supply at 10 V.
    1. Keep the output disconnected from the board for now.
  4. Before connecting the load resistor, adjust it to 20 Ω.
  5. Build the circuit shown in Fig. 2 by using the lower MOSFET, upper diode, and BB magnetic board.
    1. Note the inductance value shown on the board.
  6. Connect "RL"across "V1+" and "COM."
    1. Note that the input and output connections are flipped compared to those in the buck converter experiment.
    2. NEVER Disconnect the load during the experiment as the boost converter can become unstable and cause damage to the board.
    3. Make sure the switch array for MOSFET selection (lower MOSFET), PWM selection, and other settings are correct to achieve a functional circuit as in Fig. 2.

Figure 2
Figure 2. Boost converter circuit

2. Adjusting the Duty Ratio and Switching Frequency

  1. Connect the differential probe across the gate-to-source of the lower MOSFET.
  2. Turn ON "S90." A switching signal should appear on the scope screen.
    1. Adjust the signal time axis to see two or three periods.
    2. Adjust the frequency potentiometer to achieve a frequency of 100 kHz (period of 10 µs).
  3. Adjust the duty ratio potentiometer to achieve a 10% duty ratio (on-time of 1 µs).

3. Boost Converter Testing for Variable Input

  1. Connect the input DC power supply, which is already set at 10 V, to "V2+" and "COM."
  2. Connect the differential probe to measure the inductor current at "CS5."
    1. Connect the other probe across the load. Make sure the ground connector is connected to "COM."
    2. Capture the waveforms and measure the output voltage mean, inductor current ripple, and inductor current mean.
    3. Record the input current and voltage readings on the DC power supply.
  3. Adjust the input voltage to 8 V, 12 V and 14 V, and repeat the above steps for each of these voltages.
  4. Disconnect the input DC supply and adjust its output to 10 V.

4. Boost Converter Testing for Variable Duty Ratio

  1. Connect the differential probe across the gate to source of the lower MOSFET.
    1. Connect the other probe across the load. Make sure the ground connector is connected to "COM."
    2. Connect the input DC supply to "V2+" and "COM."
    3. Capture the waveforms and measure the output voltage mean and on-time of the gate-to-source voltage (also the duty ratio).
    4. Record the input current and voltage readings on the DC power supply.
  2. Adjust the duty ratio to 20%, 40%, and 60%. Repeat the above steps for each of these three duty ratios.
  3. Reset the duty ratio to 10%.
  4. Disconnect the input DC supply.

5. Boost Converter Testing for Variable Switching Frequency

  1. Connect the differential probe across the gate to source of the lower MOSFET.
  2. Connect the other probe across the load with the ground connector connected to "COM."
  3. Connect the input DC supply to "V2+" and "COM."
  4. Adjust the switching frequency to 70 kHz.
  5. Capture the waveforms and measure the output voltage mean and on-time of the gate-to-source voltage (also the duty ratio).
  6. Record the input current and voltage reading on the DC power supply.
  7. Adjust the switching frequency to 40 kHz, 20 kHz, and 10 kHz (or minimum possible if 10 kHz cannot be reached).
  8. Repeat the above steps for each of these three switching frequencies.
  9. Turn OFF the input DC supply and "S90," and then disassemble the circuit.

Boost converters are used in electronics to generate a DC output voltage that is greater than the DC input, therefore boosting up the supply voltage. Boost converters are often used in power supplies for white LEDs, battery packs for electric automobiles, and many other applications. A boost converter stores energy in an inductor’s magnetic field and transfers it to a load with a switching circuit. The transfer of energy from the inductor’s magnetic field enables the increase in DC output in a single stage. This video will illustrate the construction of a boost converter and investigate how changing the converter’s operating condition affects its output voltage.

This simple boost converter circuit consists of an input DC voltage source connected to an inductor and a switch. The switch may be a bipolar transistor, a MOSFET or, other similar electronic device that alternately connects and disconnects the inductor from the common line of the power supply. A blocking diode connects the inductor to a capacitor which filters the ripple in the output voltage. Increasing the capacitance decreases the ripple. For a sufficiently large capacitance the output becomes a steady DC voltage. A digital pulse train opens or closes the switch. The pulse has a duty ratio which is the ratio of the on time to the period. The duty ratio may vary from zero or increase up to one with more and more on time. When the pulse is on, the switch closes and the inductor is connected across the supply voltage. In this state, the inductor terminal connected to the output of the power supply has the higher potential and the terminal connected to the common has the lower potential. Now current flows through the inductor increasing linearly with time for sufficiently high switching frequencies. During this time the inductor voltage is defined to be positive because the slope of the current versus time is positive. The inductor stores energy proportional to the square of the current in its magnetic field. The longer the inductor is connected to the power supply, the more current increases and the more energy it stores. When the switch opens, current through the inductor must continue flowing in the same direction. This current also decreases because the inductor now gives up energy to the load. The inductor voltage becomes negative because the slope of current versus time is negative. As a result, the inductor’s polarity flips and now adds to input voltage “V in” producing a higher potential at the output. The circuit in this state, forward biases the diode and the inductor discharges current, some going to the load, and some going to the capacitor which then stores the charge. When the switch closes again the diode becomes reverse biased disconnecting the inductor from the output and preventing a short circuit of the load. During this time the inductor recharges and in its place the capacitor provides current to the load. This cycle of capacitor charging and discharging produces an average output voltage with some amount of ripple. At sufficiently high switching frequencies, the capacitor’s charge and discharge times are short and the output reaches a steady state voltage with relatively little ripple. This switching cycle repeats indefinitely and is the basis of boost converter operation. Ideally the average output voltage increases as the duty ratio increases and a duty ratio of one generates infinite voltage. However parasitic elements and resistances in the boost converter limit useful values of D to a maximum of about 0.7 or 0.8. If D is sufficiently large, parasitic effects dominate circuit operation and output voltage decreases even as D continues to increase. In the following experiments we will study how a boost converter steps up voltage in continuous conduction mode, also called CCM, a condition when the inductor operates at all times with non zero current.

The output voltage in this experiment is limited to 50 volts DC or less. Use only the specified duty cycles, frequencies, input voltages, and loads. These experiments utilize the HiRel Systems Power Pole Board which is designed for experimentation with different DC to DC converter circuit topologies. With signal supply switch S90 turned off, plug the +/- 12 volt signal supply into den connector J90. Set the PWM control selection jumpers J62 and J63 to the open loop position. Adjust the DC power supply to positive 10 volts but do not connect the power supply output to the board. Next build the circuit as shown with the lower MOSFET, the upper diode, and the BB magnetic board. Record the value of the inductor on the BB magnetic board. The load resistor is a power potentiometer. Use a multi meter to measure it’s resistance while adjusting it to 20 ohms. Then connect the potentiometer between terminals V1+ and COM. Set switch selector bank S30 as follows: PWM to bottom MOSFET, use onboard PWM, and switched load off. Connect the oscilloscope’s differential probe between terminal 16 which is the gate of the lower MOSFET and terminal 12 which is the source. Turn on switch S90. The pulse train that drives the MOSFET should appear on the scope’s screen. Select the scope’s time axis to display several periods of this wave form. Set frequency adjustment potentiometer RV60 to produce a switching frequency of 100 kilohertz. Set duty ratio potentiometer RV64 so the pulses have an on time of one microsecond which corresponds to a duty ratio of 0.1.

Connect the DC power supply to input terminals V2+ and COM. To measure the inductor current connect the differential scope probe between terminals CS5 and COM. To measure the voltage across load resistor RL, connect the other differential probe between terminals V1+ and COM. The output voltage should be a triangle wave. The upward ramps occur when the boost convertor switch is open and the inductor is transferring energy to the load. The downward ramps occur when the switch is closed, the inductor is disconnected from the output, and the capacitor is supplying energy to the load. The inductor current is a triangle wave which ramps up linearly during the on time of the pulse train, then ramps down linearly during the off time. The offset is the average current. Using the scope’s built in measurement functions, measure the mean value of the output voltage and the mean value of the inductor current. Repeat these steps with the input DC power supply set to eight, 12, and 14 volts. For a fixed duty ratio as the input voltage increases the output voltage of an ideal boost converter should increase proportionally.

This part of the experiment measures the duty ratio of the pulse train instead of the inductor current. Connect the scope probes between terminals 16 and 12 which are the gate and source of the lower MOSFET respectively. Connect the input DC power supply to terminals V2+ and COM. Like before, the output voltage is a triangle wave resulting from the inductor and capacitor alternately supplying current to the load. The gate source voltage of the MOSFET is a digital pulse train with a frequency of 100 kilohertz, a period of 10 microseconds, and an on time of one microsecond. Measure the mean value of the output voltage and the on time of the gate to source voltage along with the input current and voltage readings from the DC power supply. Repeat this test after adjusting duty ratio potentiometer RV64 so the pulse stream has an on time of two, four, and six microseconds, which correspond to duty ratios of 0.2, 0.4, and 0.6 respectively.

As duty ratio D increases, the output voltage of the boost converter also increases. Ideally if D has a value of 0.2 then an input of 10 volts generates an output of about 12.5 volts. If D is 0.4 then the output would be about 16.6 volts. If D is 0.6 then the output would be about 25 volts. In general, the output voltage is less than expected from the ideal relationship because parasitic elements create non ideal voltage drops and unaccounted energy loss. As the duty ratio approaches one, the theoretical output voltage becomes infinitely large. In reality, the output voltage is limited to about three or four times the input voltage and the influence of parasitic and non ideal components causes the output voltage to decrease after D becomes sufficiently high.

Boost converters generate an output voltage that is greater than the input voltage and many applications incorporate them to increase flexibility in the choice of power sources. The voltage from a solar panel changes with the position of the sun, weather conditions, and shade. Boost converters are commonly used to step up the variable output of a solar panel array to provide a consistent voltage to feed into an electrical grid. Battery powered systems are frequently used to power devices without the use of a power cord. In order to achieve the necessary higher output voltage, battery cells are often stacked. This can take up a lot of space if many cells are needed to reach the desired output. Instead, boost converters are used to step up the voltage while conserving space.

You have just watched Jove’s Introduction to Boost Converters. You should now understand how boost converters work and how adjusting input voltage, duty ratio, and frequency affects the output voltage. Thanks for watching.

Resultados

The boost converter output-input voltage relationship is proportional to the duty cycle in the sense that higher D will yield higher output voltages for a given input voltage. If the input voltage is Vdeand the output voltage is Vout, Vout/Vde= 1/(1-D), where 0≤D≤ 100%. Therefore, for an input voltage of 10 V, Vout≈ 12.5 V for D = 20%, Vout 16.67 V for D= 40%, and Vout≈ 25 V for D = 60%.

Nevertheless, the output voltage will be lower than expected from the ideal relationship, which is linear with the duty ratio. The main reason is that the ideal converter model from which the Vout/Vde relationship can be derived does not account for non-idealities and voltage drops in the converter. Theoretically, as D→100%, Vout→∞; practically, a theoretical limit on the boosting capability is around 3-4x the input voltage, and after a certain level of D, the output voltage of the converter starts to drop rather than being boosted due to parasitic and non-ideal elements in a real converter.

Applications and Summary

Boost converters are very common in solar photovoltaic applications where the input voltage from the solar panel varies with weather conditions and available solar energy, and a boost converter can always boost from the PV panel voltage. Power factor correction to improve power quality as seen from the utility grid with power electronic loads which may require significant reactive power, e.g. motors, is another major application of boost converters.

Transcrição

Boost converters are used in electronics to generate a DC output voltage that is greater than the DC input, therefore boosting up the supply voltage. Boost converters are often used in power supplies for white LEDs, battery packs for electric automobiles, and many other applications. A boost converter stores energy in an inductor’s magnetic field and transfers it to a load with a switching circuit. The transfer of energy from the inductor’s magnetic field enables the increase in DC output in a single stage. This video will illustrate the construction of a boost converter and investigate how changing the converter’s operating condition affects its output voltage.

This simple boost converter circuit consists of an input DC voltage source connected to an inductor and a switch. The switch may be a bipolar transistor, a MOSFET or, other similar electronic device that alternately connects and disconnects the inductor from the common line of the power supply. A blocking diode connects the inductor to a capacitor which filters the ripple in the output voltage. Increasing the capacitance decreases the ripple. For a sufficiently large capacitance the output becomes a steady DC voltage. A digital pulse train opens or closes the switch. The pulse has a duty ratio which is the ratio of the on time to the period. The duty ratio may vary from zero or increase up to one with more and more on time. When the pulse is on, the switch closes and the inductor is connected across the supply voltage. In this state, the inductor terminal connected to the output of the power supply has the higher potential and the terminal connected to the common has the lower potential. Now current flows through the inductor increasing linearly with time for sufficiently high switching frequencies. During this time the inductor voltage is defined to be positive because the slope of the current versus time is positive. The inductor stores energy proportional to the square of the current in its magnetic field. The longer the inductor is connected to the power supply, the more current increases and the more energy it stores. When the switch opens, current through the inductor must continue flowing in the same direction. This current also decreases because the inductor now gives up energy to the load. The inductor voltage becomes negative because the slope of current versus time is negative. As a result, the inductor’s polarity flips and now adds to input voltage “V in” producing a higher potential at the output. The circuit in this state, forward biases the diode and the inductor discharges current, some going to the load, and some going to the capacitor which then stores the charge. When the switch closes again the diode becomes reverse biased disconnecting the inductor from the output and preventing a short circuit of the load. During this time the inductor recharges and in its place the capacitor provides current to the load. This cycle of capacitor charging and discharging produces an average output voltage with some amount of ripple. At sufficiently high switching frequencies, the capacitor’s charge and discharge times are short and the output reaches a steady state voltage with relatively little ripple. This switching cycle repeats indefinitely and is the basis of boost converter operation. Ideally the average output voltage increases as the duty ratio increases and a duty ratio of one generates infinite voltage. However parasitic elements and resistances in the boost converter limit useful values of D to a maximum of about 0.7 or 0.8. If D is sufficiently large, parasitic effects dominate circuit operation and output voltage decreases even as D continues to increase. In the following experiments we will study how a boost converter steps up voltage in continuous conduction mode, also called CCM, a condition when the inductor operates at all times with non zero current.

The output voltage in this experiment is limited to 50 volts DC or less. Use only the specified duty cycles, frequencies, input voltages, and loads. These experiments utilize the HiRel Systems Power Pole Board which is designed for experimentation with different DC to DC converter circuit topologies. With signal supply switch S90 turned off, plug the +/- 12 volt signal supply into den connector J90. Set the PWM control selection jumpers J62 and J63 to the open loop position. Adjust the DC power supply to positive 10 volts but do not connect the power supply output to the board. Next build the circuit as shown with the lower MOSFET, the upper diode, and the BB magnetic board. Record the value of the inductor on the BB magnetic board. The load resistor is a power potentiometer. Use a multi meter to measure it’s resistance while adjusting it to 20 ohms. Then connect the potentiometer between terminals V1+ and COM. Set switch selector bank S30 as follows: PWM to bottom MOSFET, use onboard PWM, and switched load off. Connect the oscilloscope’s differential probe between terminal 16 which is the gate of the lower MOSFET and terminal 12 which is the source. Turn on switch S90. The pulse train that drives the MOSFET should appear on the scope’s screen. Select the scope’s time axis to display several periods of this wave form. Set frequency adjustment potentiometer RV60 to produce a switching frequency of 100 kilohertz. Set duty ratio potentiometer RV64 so the pulses have an on time of one microsecond which corresponds to a duty ratio of 0.1.

Connect the DC power supply to input terminals V2+ and COM. To measure the inductor current connect the differential scope probe between terminals CS5 and COM. To measure the voltage across load resistor RL, connect the other differential probe between terminals V1+ and COM. The output voltage should be a triangle wave. The upward ramps occur when the boost convertor switch is open and the inductor is transferring energy to the load. The downward ramps occur when the switch is closed, the inductor is disconnected from the output, and the capacitor is supplying energy to the load. The inductor current is a triangle wave which ramps up linearly during the on time of the pulse train, then ramps down linearly during the off time. The offset is the average current. Using the scope’s built in measurement functions, measure the mean value of the output voltage and the mean value of the inductor current. Repeat these steps with the input DC power supply set to eight, 12, and 14 volts. For a fixed duty ratio as the input voltage increases the output voltage of an ideal boost converter should increase proportionally.

This part of the experiment measures the duty ratio of the pulse train instead of the inductor current. Connect the scope probes between terminals 16 and 12 which are the gate and source of the lower MOSFET respectively. Connect the input DC power supply to terminals V2+ and COM. Like before, the output voltage is a triangle wave resulting from the inductor and capacitor alternately supplying current to the load. The gate source voltage of the MOSFET is a digital pulse train with a frequency of 100 kilohertz, a period of 10 microseconds, and an on time of one microsecond. Measure the mean value of the output voltage and the on time of the gate to source voltage along with the input current and voltage readings from the DC power supply. Repeat this test after adjusting duty ratio potentiometer RV64 so the pulse stream has an on time of two, four, and six microseconds, which correspond to duty ratios of 0.2, 0.4, and 0.6 respectively.

As duty ratio D increases, the output voltage of the boost converter also increases. Ideally if D has a value of 0.2 then an input of 10 volts generates an output of about 12.5 volts. If D is 0.4 then the output would be about 16.6 volts. If D is 0.6 then the output would be about 25 volts. In general, the output voltage is less than expected from the ideal relationship because parasitic elements create non ideal voltage drops and unaccounted energy loss. As the duty ratio approaches one, the theoretical output voltage becomes infinitely large. In reality, the output voltage is limited to about three or four times the input voltage and the influence of parasitic and non ideal components causes the output voltage to decrease after D becomes sufficiently high.

Boost converters generate an output voltage that is greater than the input voltage and many applications incorporate them to increase flexibility in the choice of power sources. The voltage from a solar panel changes with the position of the sun, weather conditions, and shade. Boost converters are commonly used to step up the variable output of a solar panel array to provide a consistent voltage to feed into an electrical grid. Battery powered systems are frequently used to power devices without the use of a power cord. In order to achieve the necessary higher output voltage, battery cells are often stacked. This can take up a lot of space if many cells are needed to reach the desired output. Instead, boost converters are used to step up the voltage while conserving space.

You have just watched Jove’s Introduction to Boost Converters. You should now understand how boost converters work and how adjusting input voltage, duty ratio, and frequency affects the output voltage. Thanks for watching.