Electrical Engineering - Buck Converter, Prevent Saturation

In summary, the question is asking to determine the inductance needed to prevent saturation in a buck regulator with a 12VDC input and 5VDC output at 4A with a frequency of 100 kHz. The answer to part a is correct, but the answer to part b is incorrect. The formula used for lambda in part b is correct, but the calculated value for L is incorrect. The minimum inductance needed to average 4A without exceeding 10A should be the same as L_CCM. The waveform of the current in the inductor will rise and fall as the PWM changes phase.
  • #1
YoshiMoshi
226
8

Homework Statement



I need help with part b-

A buck regulator is supplied from a 12VDC source and outputs 5VDC at 4 A with a PWM switching frequency is 100 kHz. Answer the following then:
a- Determine the continuous conduction mode inductance (L_CCM).
b- Compute the inductance for the converter in with the requirement that the inductor current not exceed 10A to prevent saturation.

Homework Equations

The Attempt at a Solution



a.
V_s = 12 V, V_o = 5 V, I_o = 4 A, f = 100*10^3 Hz

D = V_o/V_s = 5/12
R = V_o/I_o = 5/4 ohm
L_CCM = ( (1 - D)R ) / ( 2f ) = ( (1 - 5/12)( 5/4 ) ) / ( 2*100*10^3 ) ~ 3.646*10^(-6) H

b.
R = V_o / I_o = 5/10 = .5 ohm
lambda = I_o / ( I_max - I_o ) = 4/(10 - 4) = 2/3
lambda = L / L_CCM, L = lambda * L_CCM = ( 2/3 )( 3.646*10^(-6) ) ~ 2.431*10^(-6) H

So this was my answers to the assignment. Apparently my answer to question a. is correct, and my answer to question b is wrong. I'm not sure what I'm doing wrong. My textbook provides me the lambda formula that I used in question b.

Since the value I found for L is less than L_CCM, and L has to be greater than L_CCM (not so sure), my answer to question b should be the same value as L_CCM upon realizing that the value I found for L is less than L_CCM?

Thanks for any help, much appreciated.
 
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  • #2
The average output current will flow through the inductor. The current will rise and fall as the PWM changes phase, between the switch and the freewheel diode. What will the waveform of the current in the inductor be? What is the minimum inductance that will average 4A without exceeding 10A?
 

1. What is a buck converter?

A buck converter is a type of electrical circuit that is used to step down or decrease the input voltage to a lower output voltage. It is commonly used in power supplies and electronic devices to regulate the voltage level.

2. How does a buck converter work?

A buck converter works by using a switch (usually a transistor) that is turned on and off at a high frequency. When the switch is on, the input voltage is connected to an inductor, which stores energy. When the switch is turned off, the energy stored in the inductor is released and transferred to the output, resulting in a lower output voltage.

3. What is saturation in a buck converter?

Saturation in a buck converter refers to when the inductor becomes fully charged and can no longer store any more energy. This can lead to a sudden increase in current, which can cause damage to the circuit components.

4. How can saturation be prevented in a buck converter?

To prevent saturation in a buck converter, the duty cycle (ratio of on-time to off-time for the switch) must be carefully controlled. This can be achieved by using a feedback control system that adjusts the duty cycle based on the output voltage. Additionally, using a larger inductor or increasing the switching frequency can also help prevent saturation.

5. What are the benefits of using a buck converter?

There are several benefits of using a buck converter, including high efficiency, compact size, and cost-effectiveness. Additionally, it allows for easy regulation of the output voltage, making it suitable for a wide range of applications in electronics and power systems.

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