不要忘记直流传递功能中的电压降

元件

开关转换器结合了无源元件，如电阻器，电感器，电容器，还有有源器件，如电源开关。
当您研究功率转换器 - 例如降压转换器 - 大多数这些组件都被认为是理想的：当开关关闭时，它们不会降低电压，电感器没有欧姆损耗等等。
实际上，所有这些元素，无论是被动还是主动，都远非完美。
图1显示了这些寄生术语的简化表示。
图1：我们在功率转换中使用的元件不是完美的主机寄生术语
这些不同的下降会影响转换器的交流和直流传递功能。
其他复杂现象也起作用（磁损耗，恢复效应等），但这里不予考虑。
如果这些液滴的影响在高压应用中不那么重要，例如24 V应用中的1 V Vf，那么在低压电路中就不能忽视它们，例如在便携式电池供电应用中遇到的那些。
在考虑或不考虑这些寄生项的情况下计算降压转换器的输出电压可以以不同的方式执行。
最简单的选择是使用所谓的伏秒平衡定律计算电感两端的平均电压。
它表示，在稳态（意味着转换器达到其输出目标并且稳定）时，电感两端的平均电压为0 V.在数学上，您可以这样写：
（1）
图形上，如图2所示，表示导通状态（串联开关导通时）和关断状态（二极管续流时）的电感电压。
然后通过将矩形高度乘以其基数来计算开启和关闭状态下的区域。
计算面积实际上与在开启或关闭时间内对变量（此处为vL（t））进行积分相同。
随着电感电压随时间积分（伏特 - 秒，V-s）具有磁通尺寸，您可以计算电感在导通时间和关断时间内的磁通偏移。
在平衡时，由于平均值必须为零（正通量偏移返回其起始点，否则可能发生饱和），两个区域必须相等。
图2：电感中的磁通平衡意味着0以上和0以下的区域相等
让我们现在进行练习，同时考虑完美的元素，没有欧姆损失和下降。
在降压转换器中，当开关在ton期间闭合时，在稳态下，一个电感器端子接收Vin而第二个电感器端子处于Vout。
导通时间V-s计算如下：
（2）
在该表达式中，D是占空比，Tsw是开关周期。
在关断期间，电感器电流以与ton期间相同的方向循环，但现在找到通过导电二极管的路径。
先前偏置为Vin的电感器端子降至0 V，因为二极管被认为是完美的。
瞬时电感电压反转，我们可以写：
（3）
在平衡时，（2）和（3）的总和必须返回0：
（4）
在上面的等式中求解D给出了经典的直流传输值，用于标记为M的完美降压转换器：
（5）
这是“un cas parfait”（请原谅我的法语），不考虑寄生虫。
现在让我们通过添加rds（on），电感器欧姆损耗rL和二极管正向压降Vf来使电路复杂化。
在开启状态期间，我们有图3中的电路：
图3：在导通期间，电流通过MOSFET和其他欧姆路径循环
在接通时间期间的电感器电压不再由（2）描述并且需要更新。
导通期间的循环电流为Iout等于。
因此
（6）
在关断期间，电感器电流通过现在续流的二极管保持在相同方向（仍为Iout）的循环。
电感电压反转。
图4显示了功率MOSFET关闭时更新的电流路径：
图4：在关断期间，二极管导通并将电感左端拉至-Vf。
我们可以通过考虑电感器右端子偏置在Vout而其左端子处于，来计算关断期间的电感器伏秒。
因此，我们有：
（7）
如果我们求和（6）和（7）然后求解M得到0，我们得到：
（8）
在该表达式中，我们可以看到rDS（on）平均贡献由占空比D加权，而二极管正向压降Vf取决于（1-D）。
在以低占空比（例如12到1.2 V转换）工作的CCM转换器中，因此最好将精力集中在二极管特性（D'很大）上，并且可能通过选择低Vf肖特基或实施来最小化其影响。
同步整流。
由于D很小，rDS（on）贡献将不那么重要。
相反，对于更大的占空比，rDS（on）对效率的贡献将更大。
无论占空比如何，电感器欧姆损耗rL在开启和关闭时间期间都存在，并且必须保持在其最低值。
从中，我们可以提取将由控制回路调整的占空比值，以使Vout保持在目标上：
（9）
假设一个降压转换器由12 V电源供电，必须在5V输出电流（R =1Ω）下精确提供5 V电压。
MOSFET rDS（on）为56mΩ，此电流下的二极管正向压降为787 mV，电感ESR为70mΩ。
准确输出5 V的占空比值是多少？
应用（9）我们有
（10）
在这个例子中，（5）将返回0.417，一个较低的值。
我们可以使用诸如[1]中描述的有损平均模型来测试（10）。
原理图如图5所示。工作偏置点显示在原理图中（1 V = 100％），并确认（10）传递的结果。
图5：有损平均模型考虑了各种欧姆路径带来的影响

以上来自于谷歌翻译

以下为原文

Switching converters combine passive elements such as resistors, inductors, capacitors but also active devices like power switches. When you study a power converter – a buck converter for instance – most of these components are considered ideal: when switches close they don’t drop voltage, inductors have no ohmic losses and so on. In reality, all these elements, either passive or active, are far from being perfect. Figure 1 shows a simplified representation of these parasitic terms.

Figure 1: Components we use in power conversion are not perfect and host parasitic terms

These various drops affect the ac and dc transfer functions of a converter. Other complex phenomena are also at work (magnetic losses, recovery effects and so on) but will not be considered here. If the impact of these drops is less important in high-voltage applications, for instance a 1-V Vf in a 24-V application, you cannot neglect them anymore in low-voltage circuits such as those encountered in portable battery-powered applications.
Calculating the output voltage of a buck converter with or without consideration for these parasitic terms can be performed in different manners. The simplest option is to calculate the average voltage across the inductor using the so-called volt-second balance law. It says that, at steady-state (meaning the converter has reached its output target and is stabilized), the average voltage across the inductor is 0 V. Mathematically, you write it this way:
(1)

Graphically, as shown in Figure 2, you represent the inductor voltage during the on-state (while the series switch is turned on) and during the off-state (when the diode freewheels). Then you compute the area under the on- and off-state lines by multiplying the rectangle height by its base. Calculating the area is actually the same as integrating the variable - here vL(t) - during the on- or off-times. As inductor voltage integrated over time (volts-seconds, V-s) has a dimension of a flux, you calculate the flux excursion in the inductor during the on-time and the off-time. At equilibrium, as the average value must be zero (the positive flux excursion returns to its starting point, otherwise saturation may occur), both areas must be equal.

Figure 2: Flux balance in the inductor implies that area above and below 0 are equal

Let’s now run the exercise while considering perfect elements, no ohmic losses and drops. In a buck converter, when the switch closes during ton, at steady-state, one inductor terminal receives Vin while the second is at Vout. The on-time V-s are computed as:
(2)
In this expression, D is the duty ratio and Tsw the switching period. During the off time, the inductor current circulates in the same direction as during ton but now finds a path through the conducting diode. The inductor terminal previously biased to Vin drops to 0 V as the diode is considered perfect. The instantaneous inductor voltage reverses and we can write:
(3)
At equilibrium, the sum of (2) and (3) must return 0:
(4)
Solving for D in the above equation gives the classical dc transfer value for a perfect buck converter noted M:
(5)
This is “un cas parfait” (pardon my French) where no parasitics are considered.
Let’s now complicate the circuitry by adding the rds(on), the inductor ohmic loss rL and the diode forward drop Vf. During the on-state, we have the circuit in figure 3:

Figure 3: During the on-time, current circulates through the MOSFET and other ohmic paths

The inductor voltage during the on-time is no longer described by (2) and needs an update. The circulating current during the on-time is Iout equal to . Therefore
(6)
During the off-time, the inductor current keeps circulating in the same direction (still Iout) through the diode that is now freewheeling. The inductor voltage reverses. Figure 4 shows the updated current path while the power MOSFET is turned off:

Figure 4: During the off-time, the diode conducts and pulls the inductor left terminal to –Vf.

We can calculate the inductor volt-seconds during the off-time by considering the inductor right terminal biased at Vout while its left terminal is at . Therefore, we have:
(7)
If we sum (6) and (7) then solve for M to obtain 0, we have:
(8)
In this expression we can see that the rDS(on) average contribution is weighted by the duty ratio D while the diode forward drop Vf depends on (1-D). In CCM converters operating at low duty ratios (12 to 1.2 V conversion for instance), it is therefore preferable to concentrate the effort on the diode characteristics (D’ is large) and minimize its effects perhaps by selecting a low-Vf Schottky or implementing synchronous rectification. As D is small, the rDS(on) contribution will be less important. On the contrary, for larger duty ratios, the rDS(on) contribution to the efficiency will be greater. Regardless of the duty ratio, the inductor ohmic loss rL is present during the on and off times and must be kept at its lowest value.
From , we can extract the duty ratio value that will be adjusted by the control loop to keep Vout on target:
(9)
Assume a buck converter supplied from a 12-V source that must deliver 5 V precisely at a 5-A output current (R = 1 Ω). The MOSFET rDS(on) is 56 mΩ, the diode forward drop at this current is 787 mV and the inductor ESR is 70 mΩ. What is the duty ratio value to exactly deliver 5 V? Applying (9) we have
(10)
In this example, (5) would return 0.417, a lower value. We can test (10) by using a lossy averaged model such as the one described in [1]. The schematic appears in figure 5. The operating bias points are displayed in the schematic (1 V = 100%) and confirm the result delivered by (10).

Figure 5: A lossy average model accounts for effects brought by the various ohmic paths

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