There's a lot of interest in DC-DC Converters these days with all the new Single IC converter ICs. But the app notes for these will rarely explain the theory very well or give many equations. To keep the blog reasonably short, I'll limit the discussion here to the power stage of a buck converter, which take a higher DC input voltage and reduce it down to a lower DC output voltage with high efficiency using switching techniques and PWM modulation. We'll be considering only Continuous Conduction Mode (CCM) here. This means that in steady-state (equilibrium), the inductor current never goes to zero. This is generally true except at very light loads. The inductor current has a triangular wave shape on top of a DC value and this is the Inductor ripple current which causes the output ripple voltage. I've drawn three schematics below. The first one is a Buck Converter, less the FET drive and regulation IC, just the basic power stage. The second is the equivalent circuit for the first switching interval which I'm calling Interval-1, inductor charging interval. And the last schematic is the equivalent circuit of the second switching interval, Interval-2, inductor discharging interval.

Although there's only one actual buck converter circuit (see first schematic, below), there are effectively two separate circuits to analyze. We must split the switching interval into two sections for analysis. First, is Interval-1 for the part when L is charging (increasing inductor current) and second, Interval-2, is for the part when L is discharging (decreasing inductor current). Then, the two results can be combined. The entire switching time period, T can be written as T= Ton + Toff. Also, the duty cycle, which is simply the percentage of time the main switch is on with respect to the entire switching period, called D, can be written as: D = Ton/T, or D*T = Ton. And similarly, for the off time, (1-D) = Toff/T, or (1-D)*T = Toff. Note that D has a maximum value of 1 and a minimum value of 0. For the first part of the switching cycle, the inductor is across Vin to Vout.

Please refer to the second schematic, below. Note that the freewheeling diode is missing. Again, this equivalent circuit is for Interval-1 when the inductor is charging up with current. The diode is effectively out of the circuit and doesn't conduct except for a very small leakage current that we can ignore. Similarly, in the third schematic, for Interval-2, both Vin (input voltage) and the MOSFET are gone as the inductor's current continues to flow in the same direction as in Interval-1, except that it's discharging. The current flows thru the freewheeling diode in this interval and inductor current discharges thru the load.

We'll start the analysis with the basic inductor property: E=-L*di/dt, where: E= Voltage across the inductor, L = Inductance value in Henry, di = change in Inductor current, and dt = change in time. Note the negative sign.in front of L We can get an equation for Interval-1, the first part of the switching cycle (L charging) as follows: di = -(Vin-Vout)*dt/L. Without getting too deep into integral calculus, by integrating both sides of that equation from t=0 to t=Ton, we get an equation for I1(t), with I1(t)=[(Vout-Vin)*Ton/L]. I should also point out that Vin, Vout, and L are constants for the purpose of the integration, which makes integrating this pretty easy. We integrate from t=Ton to t=Toff. And since it's a definite integral where we subtract the value at the lower limit from the value at the upper limit, there's no constant of integration to deal with as it cancels out.

For the second part of the switching cycle, the inductor is across Vout to Gnd (neglecting the voltage drop of the freewheeling diode or synchronous rectifier FET). So, we get a similar formula here after integrating from t=Ton to t=Toff of I2(t) = [-Vout * Toff/L].

These two equations can be rewritten in terms of D instead of in terms of Ton and Toff as shown in an earlier paragraph as follows:

1. Inductor Charging Time Period: I1(t) = D*T*(Vout-Vin)/L

2. Inductor Discharging Time Period: I2(t) = (D-1) *T*Vout/L

In CCM, these two equations are equal, so we can write the following: D*T*(Vout-Vin)/L=(D-1)*T*Vout/L and the Ts and the Ls cancel, leaving us with: D*(Vout-Vin) = (D-1)*Vout, which simplifies to: Vout = Vin.* D.

This equation, Vout = Vin * D is probably the single most important equation for a buck converter. And even though it's almost self-evident, we've proved it here using electrical engineering theory and mathematics. There are other ways to analyze the buck converter, such as noting that the energy gained by the inductor equals the energy lost over one full cycle (Interval-1 + Interval-2).

I should also mention that the diode in the buck converter power stage is often replaced by another MOSFET which performs the same switching function as the diode, but without the typical 1V or more drop in the diode. This is called Synchronous Rectification and it needs a control circuit (usually an IC) to switch on and off at the proper times.

Here's a brief description of the timing and operation of the buck converter power stage, starting from an off condition and not from a steady-state condition. Please refer to the first schematic, below. At first, FET M1 is off and initially, cap C1 is discharged as is inductor L1. Imagine a control circuit with a FET driver that turns FET M1 on for a time period Interval-1. During this time, the current in L1 increases linearly (with a linear ramp shape) from zero to a peak value Ipk. During this interval, both the C1 and load voltage increase but probably won't get to their steady-state value on the first switching interval. Next, FET M1 is switched off. Now, the current in L1 continues to flow in the same direction toward the load, but it's decreasing. Instead of current flowing thru M1, in Interval-2 it flows thru D1. Note that in Inverval-1, the voltage polarity on L1 is positive on the D1 side. But on Interval-2, the voltage polarity across L1 reverses since L1 becomes a power source. Since the polarities of V1 and L1 are opposite, they get subtracted, or "bucked-out" and this is why it's called a Buck Converter. Capacitor C1 stores energy and releases it, thereby smoothing the output voltage.

One bad point about buck converters is that if FET M1 shorted, the full input voltage would appear across the load and depending on the particulars, may damage the load. This is especially true for high-voltage input buck converters. For example, a 100 VDC input with a 3.3V output. Most any 3.3V IC loads would fail dramatically if the 100VDC input suddenly appeared across the output. But with well-designed and well protected buck converters, these kinds of failures are rare.

I think I'll end today's blog here. I'll probably do similar blogs for the boost converter and buck-boost, or inverting converter.