DC-DC Buck Converter Design – Critical Inductance 98 μH

A comprehensive design solution using a roll-number determined constraint

electronic circuit components on printed circuit board

Highlights

Critical Parameter: The inductor value is fixed at 98 μH, directly influencing other converter parameters.
Step-by-Step Computation: Detailed calculations determine duty cycle, ripple current, output capacitance, and switching frequency adjustments.
Circuit Operation & Waveforms: Explanation includes converter circuit diagram, switching signals, inductor current ripple, and voltage waveforms.

1. Converter Circuit Topology and Critical Parameter

The assignment requires designing a familiar DC-DC converter by setting all parameters of our choice except one design constraint: the inductor value must be equal to your roll number (98 μH). For this design, a Buck Converter (step-down converter) is chosen due to its wide usage in power electronics for voltage reduction applications. In a buck converter, the transistor (switch) periodically connects the input voltage to the inductor which then supplies power to the load through an output filter.

Buck Converter Basic Circuit Diagram

The typical buck converter consists of:

Input Voltage (Vin): The DC supply voltage.
Switch (SW): A controlled switch (MOSFET) that chops the DC input.
Diode (D): Provides a freewheeling path when the switch is off.
Inductor (L): Stores energy during the switch on period and releases it during the off time. Here, its value is fixed at 98 μH.
Capacitor (C): Filters the output voltage to smooth out ripples.
Load Resistor (R_load): Represents the load which consumes power from the converter.

Circuit Schematic

The schematic for a buck converter is illustrated below:


         Vin
          │
          ├─────────────● SW (MOSFET)
          │             │
          │             ├───── L = 98 μH
          │             │        │
          │             │        ├────● Vout
          │             │        │       │
          │             │        │       │
          │             │        │       ▼
          │             │         └─ C  (Output capacitor)
          │             │                │       
          │             └────── D (Diode)
          │                       │
          └───────────────────────┘

2. Parameter Selection and Computation

This section details the steps required to compute the parameters controlled by a fixed inductor value. In our design, aside from the critical inductor value of 98 μH, other parameters can be chosen to tailor converter performance. Here, two examples are provided with different sets of input and output conditions to illustrate their impact on the converter design.

Design Example 1

Assumed Parameters

Input Voltage (Vin): 24 V
Output Voltage (Vout): 12 V
Desired Load Current (Iout): 1 A
Initial Switching Frequency (fs): 100 kHz
Ripple Current (ΔI): Initially 20% of Iout (i.e., 0.2 A) as a design target

Step-by-Step Calculations

(a) Duty Cycle (D):

The basic relation for a buck converter operating in continuous conduction mode (CCM) is:

$$ D = \\frac{V_{out}}{V_{in}} = \\frac{12\\,V}{24\\,V} = 0.50 $$

(b) Calculating the Inductor Current Ripple (ΔIL):

The relation governing the peak-to-peak inductor ripple current is:

$$ \\Delta I_L = \\frac{(V_{in} - V_{out}) \\times D}{L \\times f_s} $$

Substituting in the known values:

$$ \\Delta I_L = \\frac{(24V - 12V) \\times 0.5}{98 \\times 10^{-6}\\,H \\times 100\\,kHz } = \\frac{12V \\times 0.5}{98 \\times 10^{-6}\\,H \\times 100\\times10^{3}} $$

$$ \\Delta I_L = \\frac{6}{9.8} \\,A \\approx 0.612\\,A $$

This value is higher than our initial target of 0.2 A. To meet the design target, we need to adjust a parameter that is not fixed. Since the inductor value is the critical parameter, we can modify the switching frequency.

(c) Adjusting the Switching Frequency:

To achieve the desired ripple current (0.2 A), we rearrange the ripple current equation to solve for the switching frequency (fs):

$$ f_s = \\frac{(V_{in} - V_{out}) \\times D}{L \\times \\Delta I_L} = \\frac{12V \\times 0.5}{98 \\times 10^{-6}\\,H \\times 0.2A} $$

$$ f_s = \\frac{6}{19.6\\times10^{-6}} \\approx 306.12\\,kHz $$

For practical purposes, we select a switching frequency of 300 kHz.

(d) Recalculating the Ripple Current:

With the new switching frequency, the ripple current becomes:

$$ \\Delta I_L = \\frac{(V_{in} - V_{out}) \\times D}{L \\times f_s} = \\frac{12V \\times 0.5}{98\\times10^{-6}\\,H \\times 300\\,kHz} \\approx 0.204\\,A $$

This is nearly equal to the target ripple current.

(e) Output Capacitor (Cout) Calculation:

The output capacitor is chosen to limit the voltage ripple (ΔVout). Assume a desired peak-to-peak voltage ripple of 1% of Vout. For Vout = 12 V, 1% gives:

$$ \\Delta V_{out} = 0.12\\,V $$

A common approximate relation to determine the capacitor value is:

$$ C_{out} = \\frac{I_{out} \\times D}{f_s \\times \\Delta V_{out}} $$

Substituting our values:

$$ C_{out} = \\frac{1A \\times 0.5}{300\\,kHz \\times 0.12V} \\approx 13.89\\,\\mu F $$

(f) Load Resistor (R_load) Calculation:

If the output current is 1 A with an output voltage of 12 V, the load resistance is given by Ohm's law:

$$ R_{load} = \\frac{V_{out}}{I_{out}} = \\frac{12V}{1A} = 12\\,\\Omega $$

Summary Table – Design Example 1

Parameter	Value	Unit
Input Voltage (Vin)	24	V
Output Voltage (Vout)	12	V
Inductor (L)	98	μH
Duty Cycle (D)	0.50	-
Original Switching Frequency	100	kHz (initial)
Adjusted Switching Frequency (fs)	300	kHz
Inductor Ripple Current (ΔIL)	≈0.204	A
Output Capacitor (Cout)	≈13.89	μF
Load Resistance (R_load)	12	Ω

Design Example 2

To explore how design choices vary while constrained by the fixed inductor value, consider another example with different operating conditions:

Assumed Parameters

Input Voltage (Vin): 100 V
Output Voltage (Vout): 50 V
Load Current (Iout): 5 A
Switching Frequency (fs): 50 kHz

Step-by-Step Calculations

(a) Duty Cycle (D):

$$ D = \\frac{V_{out}}{V_{in}} = \\frac{50V}{100V} = 0.5 $$

(b) Inductor Ripple Current (ΔIL):

$$ \\Delta I_L = \\frac{(V_{in} - V_{out}) \\times D}{L \\times f_s} = \\frac{(100V - 50V) \\times 0.5}{98 \\times 10^{-6}\\,H \\times 50\\,kHz} $$

$$ \\Delta I_L = \\frac{25}{4.9} \\approx 5.10\\,A $$

Here, the ripple current is approximately the same order as the average load current (5 A). This level of ripple might be acceptable in certain designs with large filter components, but it also indicates a design trade-off – a high ripple may cause increased losses and could necessitate a higher switching frequency or additional filtering.

(c) Inductor Current Peak and Valley:

Since the average inductor current (I_L,avg) in steady state equals the load current (Iout = 5 A), the peak-to-peak ripple divides evenly about 5 A:

$$ I_{L,peak} = I_{L,avg} + \\frac{\\Delta I_L}{2} = 5 + 2.55 \\approx 7.55\\,A $$

$$ I_{L,min} = I_{L,avg} - \\frac{\\Delta I_L}{2} = 5 - 2.55 \\approx 2.45\\,A $$

(d) Output Capacitor (Cout) Calculation:

Assume a desired peak-to-peak output voltage ripple (ΔVo) of 1 V for a 50 V output:

$$ C_{out} = \\frac{\\Delta I_L}{8 \\, f_s \\, \\Delta V_{out}} = \\frac{5.10}{8 \\times 50\\,kHz \\times 1V} \\approx 12.75\\,\\mu F $$

In practice, a higher actual capacitance (such as 47 μF) might be selected to account for equivalent series resistance (ESR) and component tolerances.

(e) Load Resistor (R_load):

$$ R_{load} = \\frac{V_{out}}{I_{out}} = \\frac{50V}{5A} = 10\\,\\Omega $$

Summary Table – Design Example 2

Parameter	Value	Unit
Input Voltage (Vin)	100	V
Output Voltage (Vout)	50	V
Inductor (L)	98	μH
Duty Cycle (D)	0.5	-
Switching Frequency (fs)	50	kHz
Inductor Ripple Current (ΔIL)	5.10	A
Inductor Peak Current	≈7.55	A
Inductor Valley Current	≈2.45	A
Output Capacitor (Cout)	≈12.75	μF (minimum)
Load Resistance (R_load)	10	Ω

3. Signal Waveforms and Dynamic Behavior

Practical design documentation requires illustrating the typical waveforms across various nodes of the buck converter. Once the operating parameters are computed, the key signals to be drawn include:

Switching Signal

This is a square wave that controls the MOSFET switch. In the first design example (24 V to 12 V at 300 kHz), the duty cycle is 50%. Thus, the switch remains on for half the period and off for the other half. The corresponding switching period T is:

$$ T = \\frac{1}{f_s} = \\frac{1}{300\\,kHz} \\approx 3.33\\,\\mu s $$

Inductor Current

The inductor current exhibits a triangular or sawtooth waveform due to the continuous energy storage and release during each switching cycle. For Design Example 1, the current ramps upward when the switch is on and downward when the switch is off, oscillating between approximately 0.90 A and 1.10 A (centered around the average of 1 A with a ripple of about 0.204 A peak-to-peak).

Output Voltage

The output voltage ideally remains close to the desired DC level (12 V for Example 1 or 50 V for Example 2). Although small ripples occur due to the switching action, the filter capacitor smooths them, resulting in a nearly constant DC voltage with ripple defined by the capacitor design.

Diode Current

During the off period of the switch, the diode conducts the freewheeling current. Its waveform closely follows the inductor’s current when the switch is off.

Illustrative Timing Diagram

A simplified timing diagram for Design Example 1 can be summarized as follows:

t = 0: Start of switching cycle
t_ON: Duration corresponding to the duty cycle (approximately 1.67 μs for fs = 300 kHz and D = 0.5)
t_OFF: Remaining time of the cycle

These signals can be simulated using software tools such as SPICE to verify dynamic behavior and waveform integrity.

Conclusion

In this assignment, a DC-DC buck converter design was explored in depth with the given constraint that the inductor value must equal 98 μH (derived from the roll number). Two distinct design examples were detailed:

The first example considered a 24 V to 12 V conversion with refined switching frequency adjustments to achieve a target ripple current of approximately 0.2 A.
The second example employed higher voltage and current levels (100 V to 50 V at 5 A), resulting in larger ripple currents and emphasizing the limitations imposed by the fixed inductance value.

By calculating the duty cycle, ripple current, output capacitor value, and other relevant parameters, we demonstrated how the fixed inductor of 98 μH directly influences the performance and design trade-offs of the converter. Furthermore, waveform considerations and schematic representations have been provided with detailed simulations in mind. This design exercise underscores the importance of parameter interdependence in power electronics and illustrates the necessity to adjust non-critical parameters (like switching frequency) to obtain optimal performance while meeting design specifications.

DC-DC Buck Converter Design – Critical Inductance 98 μH

A comprehensive design solution using a roll-number determined constraint

Highlights

1. Converter Circuit Topology and Critical Parameter

Buck Converter Basic Circuit Diagram

Circuit Schematic

2. Parameter Selection and Computation

Design Example 1

Assumed Parameters

Step-by-Step Calculations

Summary Table – Design Example 1

Design Example 2

Assumed Parameters

Step-by-Step Calculations

Summary Table – Design Example 2

3. Signal Waveforms and Dynamic Behavior

Switching Signal

Inductor Current

Output Voltage

Diode Current

Illustrative Timing Diagram

Conclusion

References

Recommended