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Basic SMPS structures Refresh Part 2 of 4

Introduction 

After understanding the inductive switched mode power supply’s (SMPS) position within the vast landscape of energy converters, we embark on the next leg of our journey toward SEPIC. Along this path, we encounter the fundamental SMPS schemes that serve as the building blocks of more complex topologies: the buck, boost, and inverter. Those foundational structures, familiar to many, are presented in a manner that equips readers with the knowledge necessary for the subsequent exploration of SEPIC architectures. 

  Fundamental SMPS schemes: buck, boost, and inverter

  Figure 1: Fundamental SMPS schemes: buck, boost, and inverter 

(Source: self-made) 

 

The basic topologies are built by connecting three simple elements: switch (S), inductor (L), and diode (D). The various ways these elements are arranged and oriented determine whether the circuit functions as a buck, boost, or inverter converter. Note that buck, boost, and inverter have other names: step-down, step-up, and buck-boost, respectively. 

The value of Vout primarily depends on Vin and on the duty cycle of the switching process. If you're not already familiar with these different setups, it can be a bit challenging to remember them. 

One helpful "trick" is to think of the vertical branch: D, S, and L correspond to the three schemes—buck, boost, and inverter, respectively. 

The step-Down converter can be recognized because the vertical branch starts with "D" like diode. 

The booSt converter is identified by the switch, which is like the "S" in the word "boost." 

As for the Inverter, it's the one left after recognizing the other two, or use the first letter of "inverter" to relate to the inductor in the vertical branch. 

To control the output voltage, a PWM (pulse-width modulation) or PFM (pulse frequency modulation) generator is used. It adjusts the duty cycle or pulse widths based on the comparison between the actual Vout and a reference voltage. 

 

Step-Down (Buck) Converter Operation 

   Basic buck operation

 Basic buck operation

 Figure 2: Basic buck operation 

(Source: self-made) 

 

To determine the relationship between Vout, Vin, and the duty cycle (d), the simplest approach is to create an equation that balances the current flowing through the inductor within one cycle (T). The nulling of current balancing is crucial to reflect—in a stable circuit, there is no accumulation of charge in the coil. 

When the switch is ON, the inductor (L) is charging during Ton using Vin, which happens at a rate of (Vin-Vout)/L. When the switch is OFF, the inductor discharges at a rate of Vout/L. 

By solving this equation, we find that Vout equals d times Vin. Since d falls between 0 and 1, we have a step-down function. This means it's not possible to obtain a Vout value higher than Vin. 

 

Step-Up (Boost) Converter Operation 

  Basic boost operation

 Figure 3: Basic boost operation 

(Source: self-made) 

 

We're applying the same basic principle here: within one cycle (T), we consider the circuit is stabilized and therefore that the current in the inductor (L) is balanced; average current is zero. 

When the switch is ON, the inductor charges up using Vin. During the OFF period (Toff), the inductor discharges, and this happens at a rate of (Vin-Vout)/L. 

By solving the equation, we find that Vout equals Vin multiplied by (1-d). It's worth noting that Vout is consistently higher than Vin, and it can even approach infinity, but it's impossible for Vout to be lower than Vin. 

However, this design has a significant drawback that its two brothers (buck and inverter) do not have: if Vout is accidentally shortened, this will also lead to Vin collapsing. This occurs because the path created by the coil and the diode becomes fully conductive. 

 

Inverter (Buck-Boost) Converter Operation 

  Basic inverter operation

 Basic inverter operation    

Figure 4: Basic inverter operation 

(Source: self-made) 

 

We use a similar approach for the inverter, we aim to balance the current in the inductor (L) within each cycle (T). When the switch is ON, the inductor charges up using Vin. When the switch is OFF, the discharge of L happens through Vout. By solving the equation, we find that Vout equals (-Vin) times d divided by (1-d).  

Since the duty cycle varies between 0 and 1, Vout can either be greater or less than Vin, allowing this structure to work as a step-down or step-up converter: that is a significant advantage versus the buck and the boost.  

Unlike the boost structure, we don't encounter the issue of Vout accidentally being shortened. This is because the diode (D) acts as a barrier separating Vin from Vout. One can argue if Vin disappears, then the circuit offers a DC path from Vout to Vin. That’s correct, but this will only discharge the output capacitor since Vout is not producing any voltage. However, there's one drawback: Vout consistently has the opposite polarity compared to Vin; hence the name “inverter”. 

 Conclusion 

As we delve deeper into our exploration, we discover that of the three fundamental structures, only the inverter (or buck-boost) offers the ability for Vout to be either lower or higher than Vin. This is particularly advantageous in applications where the energy sources are constrained, such as batteries or harvested sources, where maximizing every ounce of available energy is crucial.  

In the next part of the series, we continue our journey and take another significant step by transforming the familiar buck-boost structure into the flyback topology. This transition will naturally lead us to the SEPIC family, where we’ll uncover even more intriguing possibilities. 

Click here to see the previous post.

Click here to see all the posts in this series.

  • Hello KCC, thanks for the article!
    Could you clarify on the sentence: "One can argue if Vin disappears, then the circuit offers a DC path from Vout to Vin."? Do you mean when S is turned on and Vin is shorted to ground?
    Also, could you clarify on why average current in the inductor is zero in your sentence: "the current in the inductor (L) is balanced; average current is zero."? Do you mean to say average current is constant?