Designing Closed Loop Non-Isolated Boost Converter SMPS (Part 2/12)

In the previous tutorial, SMPS open loop boost converter was designed. In this series, the following SMPS circuits are designed –

1. Boost Converters –

a) Open loop boost converter

b) Closed-loop boost converter

c) Open Loop Boost Converter with adjustable output

d) Closed Loop Boost Converter with Adjustable Output

2. Buck Converters –

a) Open Loop Buck Converter

b) Closed-loop Buck Converter

c) Open Loop Buck Converter with Adjustable Output

d) Closed-loop buck converter with adjustable output

3. Buck-Boost Converters

a) Buck Open Loop Inverter – Boost Converter

b) Buck Open Loop Inverter – Boost Converter with Adjustable Output

4. Flyback Converter

5. Push-pull converter

The open-loop boost converter designed in the previous tutorial did not have a constant, regulated voltage at the output. For stable output, a Boost converter with error detection circuit needs to be designed. Error detection can be done by adding a feedback circuit that continuously checks the error in the output voltage and provides a regulated voltage at the output.

Therefore, in this tutorial, a closed-loop non-isolated boost converter is designed. The boost converter can be designed in two ways-

Open Loop Boost Converter – In open loop boost converter, there is no feedback from the output to the input, unlike the closed loop which has a feedback circuit. Therefore, the output of an open-loop boost converter is not regulated.

Closed-loop boost converter – In closed-loop boost converter, there is feedback from the output to the input. Therefore, the output of a closed-loop boost converter is regulated.

There are certain design parameters involved in the design of the boost converter. It is important to understand these design parameters. Any boost converter can operate in any of the two possible operating modes. These modes of operation are as follows –

• Continuous Conduction Mode (CCM)- In CCM, the current in the inductor is continuous throughout the entire switching period cycle. Thus, a regulated voltage at the output is obtained, but the output is regulated only if the current is drawn within the limits of the CCM.

• Discontinuous conduction mode (DCM)- In this mode, the current in the inductor is pulsating and becomes zero during a part of the switching time. Therefore, a regulated voltage is not received at the DCM. However, the voltage can be regulated by connecting a feedback circuit from the output to the input.

In this tutorial, a non-isolated boost converter is designed, which means the input and output share the same ground. The boost converter designed in this project will step up from 12V DC to 24V DC with a tolerance limit of +/-0.5V. Once the circuit has been designed and assembled, the value of the output voltage and current will be observed using a multimeter. These values ​​will indicate the efficiency of the boost converter designed in the project.

Fig. 1: List of components required for the Closed Loop Boost converter

Fig. 2: Closed-loop Boost converter block diagram

In this experiment, a closed-loop boost converter operating in DCM mode is designed and component values ​​according to standard DCM equations are calculated for the desired output.

The boost converter designed in this tutorial will have the following design parameters –

Input voltage, Vin –A 12V lead-acid battery will be used as the source. The battery voltage will be the input voltage

Output voltage, Vout – The desired output voltage is 24V.

Maximum output current, Iout (max) – The maximum output current limit will be 100 mA.

Output ripple voltage (dV) – The maximum output voltage ripple assumed at the output will be 100mV

Load resistance – In this circuit, a resistance will be connected to the output that will act as a load for the circuit. The maximum resistance value can be calculated using Ohm’s law, which is as follows –

Vout = Iout(max)*RL(max)

RL(max) = Vout/Iout(max)

Putting all the values,

RL = 240E

Now the nominal power of the resistance can be calculated as follows – P = (Vout)2/(RL(max))

Putting all the values,

Pout = 2.4W

Therefore, a resistance with a value of 240E and a power equivalent to or greater than 2.4W will be used as a load at the output for maximum efficiency.

Frequency (Fs) – The frequency of the PWM signal generated by the microcontroller should not be too high or low, therefore a frequency of 10 KHz is selected to operate the switching components of the circuit. The frequency value is assumed.

The boost converter has the following circuit blocks –

A 12V battery is used as the input power source in the circuit. A 12V lead-acid battery is used in the project. The battery voltage itself is the input voltage to the circuit.

An oscillator is used to generate a pulse width modulated (PWM) signal of the desired frequency. In this boost converter, Arduino UNO is used to generate the PWM signal, so the Arduino board acts as an oscillator. The PWM signal is a pulse train used to turn the MOSFET on and off. MOSFET is used as switching transistor in the circuit.

For switching purposes, a transistor and a diode are used as the switching component. For transistor selection, MOSFET is chosen as FETs are known for their fast switching speed and low RDS (ON) (drain to source resistance in ON state). Therefore, an N-channel P30NF10 MOSFET (shown as Q1 in the circuit diagram) is connected in parallel to the input DC source which acts as a switch in the circuit as its threshold voltage is very low, around 4V. Therefore, it can be driven by a 5V PWM signal from the microcontroller. In the ON state, the Vds of the P30NF10 MOSFET is also very low, which reduces the power dissipation of our circuit.

To turn the MOSFET on and off, a pulse train must be applied to its gate. To do this, the controller board generates a 10kHz pulse width modulated signal. This PWM signal is used to turn the MOSFET on and off. To generate the controller's PWM signal, an Arduino sketch was written to the board. This Arduino sketch can be downloaded from the code section.

It should be noted that the switching time of the MOSFET and diode must be less than the rise and fall time of the PWM wave. A gate-to-source resistance must be used to prevent any unwanted triggering of the MOSFET by external noise. It also helps to quickly turn off the MOSFET by discharging its parasitic capacitance. A low resistor value (10E to 500E) must be connected to the MOSFET gate. This will solve the ringing (eddy oscillations) and spike current problem in the MOSFET. The PWM signal voltage level must be greater than the MOSFET threshold voltage. So that the MOSFET can be turned ON completely with minimum RDS (ON).

Another switching component used in the circuit is a diode. The diode switching time must be less than the rise and fall time of the PWM wave. The Arduino board generates a PWM wave with a rise time of 110ns and a fall time of 90ns. The forward voltage drop of the diode must also be very low, otherwise it will dissipate power, which will further reduce the efficiency of the circuit. The diode must offer low voltage drop in forward bias and the RDS (ON) of the MOSFET must be low. Therefore, in this experiment, a BY399 diode that best suits the circuit design is selected.

Before generating the PWM signal, the switching frequency of the circuit needs to be decided. For this boost converter, a switching frequency of 10kHz is selected, which will work well in this converter design.

The duty cycle of the generated PWM signal is another important consideration as it will decide the active state of the MOSFET. The duty cycle can be calculated as follows –

D% = 1- (Vin/Vo)*100

Vo=Desired output voltage, 24V

Vin =Input voltage, 12V

Putting all the values ​​in the above equation, the desired duty cycle is –

D = 50%

A capacitor and resistor of appropriate value must be used to generate the 10 kHz frequency and 50% duty cycle. The higher the frequency selected for the switching components, the greater the switching losses. This decreases the efficiency of the SMPS. But the high switching frequency reduces the size of the energy storage element and improves the transient response of the output.

An inductor is used to store electrical energy in the form of a magnetic field. Therefore, the inductor acts as an energy storage element. An inductor of value 11.5mH is used in the circuit. For an inductor, a secondary or primary coil of a transformer, a relay coil, or any standard inductor that has the desired inductance value can be used.

As a filtering element, a capacitor (shown as C1 in the circuit diagram) is used at the output of the circuit. In normal operation of the Boost circuit, transistor Q1 turns on and off according to the frequency of the oscillator circuit. This generates a pulse train in inductor L1 and capacitor C3, as well as transistor Q1. Since the capacitor is connected to the inductor only in the negative cycle of the PWM signal, this forms an LC filter that filters the pulse train to produce a smooth DC at the output. The value of the capacitor can be calculated using the following DCM equation –

C1>= (Io(max) * (1- (2*L1*Fs/RL)1/2))/ (Fs*dV)

Putting all the values ​​in the above equation,

C1 >= (0.1*(1-(2*0.0115*10,000/240) 1/2))/ (10,000*0.1)

C1 >= 3uF

As a capacitor can be larger than the calculated value, any capacitor with a value equivalent to 3uF or greater than 3uF can be used.

As it is the minimum value of capacitor required, so a standard value capacitor is used in the circuit which can be easily available and therefore a 47uF capacitor is used.

In the previous tutorial, it was seen that without any feedback circuit, a high voltage was received at the output when there was no load connected. When any load on the output was connected, the output voltage dropped below the desired output voltage. therefore, there was a need to regulate the output voltage by connecting a feedback circuit.

Therefore, in this circuit, feedback is provided from the output terminal to the input terminal which regulates the output voltage even when there is no load present at the output. The Feedback circuit must calculate the voltage error and then the microcontroller automatically adjusts the output voltage according to the desired voltage. This error voltage adjustment is managed by the Arduino sketch running on the board.

To provide feedback, a ladder of resistors is used at the output (as shown in the circuit diagram). To calculate the value of the resistor network, the maximum output voltage of the open loop circuit must be known when no load is connected to the output. This will be the desired output voltage that should feed from the resistor divider network to the analog pin of the controller (as shown in the circuit diagram). The output voltage must be equal to 5.2 V as the microcontroller provides 5.2 V as a reference voltage. The value of the resistor network can be calculated as follows –

Fig. 3: Circuit diagram of the resistive ladder used as feedback in the closed-loop boost converter

According to the resistor divider rule –

Vr = Vo*(R4/R3+R4)

Where,

Vr = desired output for analog pin

Vo = output voltage of the open-loop boost converter without load.

It was observed that the values ​​of Vr and Vo are –

Vo = 110V

Vr = 5.2V

Let's assume R3 = 10K now R4 is

5.2= 100*(R4/10K+R4)

R4 = 500E (approx.)

So R3 = 10K and R4 = 500E

Now, the rated power of resistor R3 and R4 can be calculated as follows –

(For R3),P3 = (Vo-Vr)2/(R3)

Putting all the values,

P3 = 1W (approx.)

(For R4), P4 = (Vr)2/(R4)

Putting all the values,

P = 54mW (approx.)

How the circuit works –

Any SMPS has some switching components that turn on and off at high frequency and has some storage component that stores the electrical energy while the switching components are in the conducting state and discharges the stored energy to the output device while the switching components are in state no. -driving status.

A simple boost converter consists of the inductor (L), a diode (D), a capacitor (C) and a transistor where the transistor acts as a switch. In the boost circuit, when the switch is closed, i.e., the switching component is in a conducting state, the inductor starts generating a magnetic field and stores energy. The energy stored in the inductor increases the output voltage compared to the input voltage.

When current begins to flow through the switching component, as its path is less resistive compared to the parallel path containing the capacitor and output load, the inductor generates a positive polarity at its left terminal and negative at its right terminal. . Due to the change in polarity, the diode becomes reverse biased. In this condition, the capacitor, which was charged in the previous cycle, supplies current to the load while the switching component goes into a non-conducting state or opens between ground.

When the switch is open, the current is reduced as the impedance increases, so the magnetic field generated in the inductor begins to collapse and the polarity of the inductor reverses. This makes the diode forward biased and the capacitor now starts charging with a voltage higher than the input voltage. Since the input now has two sources in series, one is the inductor and the other is the battery. Therefore, the output voltage is always greater than the input voltage.

Fig. 5: Circuit diagram showing the OFF state of the switching component in the Boost converter

Therefore, in the ON state, the Diode was in Blocking Mode (OFF) and the Transistor was ON. In the OFF state, the Diode was in conduction mode (ON) and the Transistor was OFF.

So, it can be said that the Boost Converter has two switching components – one is the transistor and the other is the diode. At a time, only one of the switching components conducts, that is, it is in the ON state, while the other enters the non-conducting state, that is, it enters the OFF state.

Testing the circuit –

The purpose of making a closed-loop boost converter is to increase the energy efficiency of the circuit and improve the stability of the output.

In this circuit, Input Voltage, Vin = 12V

The output voltage when no load is connected is 24.3V

By measuring voltage and current values ​​with different loads at the output, the following observations were made –

Fig. 6: Table listing the output voltage and current of the Closed Loop Boost Converter for different loads

Thus, it can be seen that a current of 100 mA can be consumed at the 24.3V output with a tolerance limit of +/-0.5V.

The energy efficiency of the circuit can be calculated as follows –

Efficiency% = (Pout/Pin)*100

(Output power) Pout = Vout*Iout

(Output voltage)Vout = 24.3V

(Output current) Iout = 100mA

Pout = 2430 mW (approx.)

(Input power) Pin = Vin*Iin

(Input voltage) Vin = 12V

(Input current) Iin = 220mA (measure input current using ammeter)

Pin = 2640mW

Putting all the values,

Efficiency% = 92%

It can be seen that when using a feedback circuit, the efficiency of the boost converter increases from 88% (as calculated in the previous tutorial) to 92%.

This is a closed-loop boost converter with non-isolated output and operating in DCM mode. It can be used as a switching regulator for LED drivers and as a regulated DC power supply. It can be used to supply power to low-power portable electronic devices. In battery powered applications, when there is space constraint to stack number of batteries in series to achieve high voltage, this boost converter can be used with less number of batteries to supply DC power.

This boost converter is simple to design and uses inexpensive components. It can be easily assembled in a short time. Furthermore, there is no need for control circuits for PWM signal generation in this boost converter design.

Circuit-Diagram-Closed-Loop-Boost-Converter