Stepping into Transfer Functions of a Buck in DCM: Part 4 of 6

Welcome to the intricate world of power electronics, where the buck converter's performance can be finely tuned by understanding its different conduction modes. In this segment, we will examine the detailed steps involved in deriving the transfer function for a buck converter operating in DCM, utilizing various methods. This journey will not only reveal the complex relationship between the input and output voltages but also confirm that the buck converter in DCM still performs its essential step-down function. When we create a transfer function, the load current (and other parameters) will come into play. 

The transfer function Vout/Vin is more complex to establish.  

Transfer Function of a Buck in DCM Step 1: In and Out Power Balance 

Figure 1: Buck Transfer Function – Step 1: Power Balancing Principle 

To determine the transfer function, which tells us how the output voltage (Vo) relates to the input voltage (Vin) and other factors such as inductance (L), resistance (R), capacitance (C), time period (T), and duty cycle (d), the effect of the load needs to be considered. We’ll use a method known as "power balance" to solve this. This method is based on the principle that at a steady state, the power going into the circuit (Pin) should be the same as the power coming out (Pout), assuming there's no loss in all the components. So, the first step in this method is to set up the equation: Pin = Pout. 

Transfer Function of a Buck in DCM Step 2: Input Power Injected 

Figure 2:  Buck Transfer Function in DCM – Step 2: Input Power Injected 

The power going in at a steady state is simply Pin = Vin(t) * IN(t). Vin(t) is the DC voltage at the input. It is also relatively constant at steady state, where Vin(t) = Vin. In this diagram, the current from Vin is shown in blue. When the switch is on, the current in the inductor IL(t) matches the input current IN(t). This current increases linearly from zero to a peak value (INpeak or ILpeak). During Ton, IN(t) and IL(t) are identical. That is, IN(t) = IL(t), and thus the peak current INpeak = ILpeak. 

The voltage across the inductor is Vin – Vout. Therefore, the current IN(t) = IL(t) = (Vin-Vout)/L, leading to a peak current, INpeak or ILpeak, at the end of Ton. INpeak = Ton * (Vin-Vout)/L = d*T*(Vin-Vout)/L (equation a). For the total power at Vin, we have d*T*Vin* INpeak/2during the time the switch is on (Ton) and zero during the time the switch is off (Toff), averaging to d*Vin*INpeak/2 (equation b). Finally, combining equations a and b, we find that the power input, Pin, equals d²*T*Vin(Vin-Vout)/2L (equation c).  

Transfer Function of a Buck in DCM Step 3: Output Power Collected 

 Figure 3:  Buck Transfer Function in DCM – Step 3: Output Power Collected 

On the output side of a circuit, in a steady state, the power is simply Vout*Iout = Vout²/R. (assuming a standard resistive load). At this steady state, the average current on the output capacitor must be zero. If it weren’t in a steady state, there would be an accumulation of charge on the capacitor. 

Transfer Function of a Buck in DCM Step 4: In and Out Power Equalization 

Figure 4:  Buck Transfer Function in DCM – Step 4:  Power Equalization 

When we balance the input power (Pin) and the output power (Pout), we arrive at an expression for the output voltage (Vout) that is influenced by several factors: the input voltage (Vin), the duty cycle (d), the switching period (T), the resistance of the load (R), and the inductance (L). This expression is more complex than the one given for CCM, where Vout =d*Vin. Despite the complexity, this relationship still describes a step-down function, meaning that Vout will be less than or equal to Vin for any combination of R, d, T, Vin, and L values.  But can we confirm it’s still a buck?  

Transfer Function of a Buck in DCM Step 5: Is a Buck in DCM Still a Buck? 

Let’s examine the Vout expression found earlier: 

Vout = kVin/2 * [-1 + (1+2/k)1/2] 

 To prove it is a buck, Vout/Vin is <1  

Vout/Vin = (k/2)*[-1 + (1+2/k)1/2] < 1 

This implies: 

 (1+2/k)1/2 > 1 

Or 2/k > 0, which is always the case regardless of R, L, T. 

We are OK! 

Conclusion 

By carefully deriving the transfer function using the various methods, we have seen that the output voltage (Vout) in DCM is influenced by a combination of input voltage (Vin), duty cycle (d), switching period (T), load resistance (R), and inductance (L). Despite the added complexity, the relationship still ensures that Vout is less than or equal to Vin, confirming that the buck converter in DCM remains a true step-down converter. Understanding these modes and their transfer functions is essential for optimizing the performance and efficiency of your buck converters. Armed with this knowledge, you can make informed decisions to ensure your power electronics systems operate optimally. 

 See blogs in the DCM & CCM in SMPS  series.