SLUAB16 March 2025 LMR51403 , TPS629203

2 Input and Output Ripple of Buck Converter in DCM

Figure 2-1 shows the current loop of the buck converter in DCM. The high-side FET current I_{HS_FET} and the inductor current I_L are both discontinuous waves. The input current (I_IN) and load current (I_O) can be regarded as DC components of I_{HS_FET} and I_L respectively. Therefore, the input capacitor current I_CIN and the output capacitor current I_COUT are the AC components. Based on the charge balance, the input and output voltage ripples can be know by the charge change ∆Q on CIN and COUT in the pulse if ESR is ignored.

Figure 2-1 Current Loop of The Buck Converter in DCM

Input Voltage Ripple Calculation

Figure 2-1 shows the relationship of input voltage ripple ΔVIN with I_{HS_FET} and I_IN. The yellow area represents I_IN and the red line represents I_{HS_FET}. For the period when I_{IHS_FET} exceeds I_IN, CIN discharges into high-side FET, leading to ΔVIN decrease. The charge of ∆Q_{HS_FET} of the red area can by calculated by using Equation 1.

Figure 2-2 Input Voltage Ripple of Buck Converter in DCM

Equation 1.

{∆ Q}_{H S_F E T} = \frac{1}{2} \times I_{p e a k} \times T o n

According to the geometric relationship, ∆Q_CIN and can be calculated by using Equation 2. And ΔVIN can be calculated by using Equation 3.

Equation 2.

{{∆ Q}_{C I N} = {∆ Q}_{H S_F E T} \times (\frac{I_{p e a k} - I_{I N}}{I_{p e a k}})}^{2}

Equation 3.

∆ V_{I N} = \frac{1}{2 C_{I N}} \times I_{p e a k} \times T o n \times {(\frac{I_{p e a k} - I_{I N}}{I_{p e a k}})}^{2}

Output Voltage Ripple Calculation

Figure 2-3 shows the relationship ΔVOUT of with I_L and I_O in a pulse. The yellow area represents I_O and red line represents I_L. For the period when I_L exceeds I_O, COUT discharges into load, leading ΔVOUT to decrease. The charge of ∆Q_L of the red area can by calculated by using Equation 4.

Equation 4.

{∆ Q}_{L} = \frac{1}{2} \times I_{p e a k} \times (T o n + T o f f)

Figure 2-3 Output Voltage Ripple of Buck Converter in DCM

According to the inductor current in DCM of Equation 5, the charge of ∆Q_L can be further represented as Equation 6.

Equation 5.

I_{p e a k} = \frac{V_{I N} - V_{O U T}}{L} \times T o n = \frac{V_{O U T}}{L} \times T o f f

Equation 6.

{∆ Q}_{L} = \frac{1}{2} \times I_{p e a k} \times \frac{V_{I N}}{V_{O U T}} \times T o n

According to the geometric relationship, ∆Q_COUT can be calculated by Equation 7. And ΔVOUT can be calculated by Equation 8.

Equation 7.

{{∆ Q}_{C O U T} = {∆ Q}_{L} \times (\frac{I_{p e a k} - I_{L O A D}}{I_{p e a k}})}^{2}

Equation 8.

∆ V_{O U T} = \frac{I_{p e a k} \times T o n}{2 C_{O U T}} \times \frac{V_{I N}}{V_{O U T}} \times {(\frac{I_{p e a k} - I_{L O A D}}{I_{p e a k}})}^{2}