原文地址：/blog/articles/h-bridge-secrets/asynchronous-sign-magnitude-drive/
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Asynchronous Sign-magnitude Drive
异步符号幅度驱动器
1.Introduction
引言
The following article will go though a drive mode where the catch diodes are conducting for a significant amount of time in each cycle. If you haven’t read theintroductory pageof the series and you’re not familiar with H-bridge concepts, I suggest you do and only than continue reading this article. I also strongly suggest you read the page onsign-magnitude drivebecause much of the discussion here will build on the material covered over there.
接下来的文章讨论一种驱动模式，泄流二极管在每个周期内会有大量的时间在工作。如果您还没有阅读本系列的介绍，而且您不熟悉H桥的概念，那么我建议到之前的页面阅读。我还强烈建议你阅读符号-幅值驱动的那一篇文章，因为这里的大部分讨论将建立之前的基础上。
If however you did read that introductory article, you might be a little surprised. There I put together a neat little taxonomy of available drive modes, and concluded that there’s only two reasonable variants. Those, thesign-magnitude driveand the lock anti-phase drive’ we’ve already covered in previous chapters. So what’s all this about than?
如果你阅读了那篇文章，你可能会有点惊讶。在那里，我对可用的驱动模式进行了简洁的分类，并得出结论: 只有两种合理的驱动模式。符号-幅值驱动和锁定-反相驱动我们已经在前面的章节中讨论过。那么，下面要讨论的会是什么？
Well, as it turns out, I’ve lied. To be more precise, I didn’t tell the whole truth. Before we talk about the details and the reasons however, let’s review the basics! Our motor control circuit is an H-bridge, with the following configuration:
事实证明，我撒谎了。更确切地说，我没有说出全部真相。然而，在我们讨论细节和原因之前，让我们回顾一下基础知识！我们的电机控制电路是一个 H 桥，具有以下配置:
The load, the DC motor is a complex beast, but we will mostly use just a very simple model for it:
电机的负载，以及直流电机是一个复杂的东西，但我们大多情况下只使用一个非常简单的模型:
For this discussion we won’t be able to ignore the internal resistance for the most part.
在下列讨论中，我们不会忽略电机内阻。
Finally, a warning: while the previously discussed drive modes are in wide use, I haven’t found too many resources that describe what I’m about to here. That means that many of the terms I’m using (including the name actually) I came up with and may not correspond to the industry-standard nomenclature. If you find that misleading or know the official term for something in here, please let me know so I can make the appropriate corrections.
最后，需要注意的是: 虽然前面讨论的驱动模式得到了广泛的应用，但是我还没有找到太多的资源来描述我在这里要讲的内容。这意味着我使用的许多术语(实际上包括名字)可能不符合行业标准的命名法。如果您发现有错误或知道官方术语，请让我知道，以便我可以作出适当的纠正。
2.Motivation
动机
The principle assumption leading to the original analysis and the resulting two possible drive modes was that we wanted the switches to conduct during both the on-time and the off-time. Our intention was to minimize heat dissipation on the switches and since in most cases the voltage drop on the switch would be less than that on the diode, it’s better to use the switches for conducting current.
产生的两种可能的驱动模式的前提假设是，我们想要在“工作状态”和“停机状态”下切换。我们的目的是尽量减少开关元件的散热，因为在大多数情况下，开关管上的电压降会小于二极管上的电压降，所以最好使用开关管来切换电流。
As it turns out, both the lock anti-drive drive and the sign-magnitude drive suffer from a serious problem: regenerative braking. I say problem, because this phenomena, while sounds cool at first – putting energy back into the battery – complicates system design quite a bit and pretty much forces closed-loop control of the bridge with additional sensing circuitry.
事实证明，锁定-反相驱动和符号-幅度驱动都存在一个严重的问题:再生制动。我之所以说是问题，是因为这种现象，乍看起来很酷——把能量放回电池——然而会使系统设计相当复杂，而且几乎是用附加的感应电路来实现闭环控制。
If we look back at what causes regeneration, we’ll see that it happens whenever Vmot_avgand Vgare of the same polarity but Vgis the greater of the two. Under those circumstances, the average current starts flowing in the reverse direction, recharging the battery. The current could reverse direction, because our switches allow current to flow both ways. If there was a way to block the current from reversing – at least through the battery – we could potentially make an H-bridge which doesn’t have any state with regeneration. Turns out, such reverse-current protected elements are diodes, and we already have four of them in the bridge. The question is, can we make use of them. And we’ll investigate the answer in the next chapter.
如果我们回顾一下是什么导致了再生电流，我们会发现：Vmot_avg和Vg具有相同的极性，但Vg > Vmot_avg。在这种情况下，平均电流开始向相反的方向流动，给电池充电。电流可以反向，因为我们的开关管允许电流向两个方向流动。如果有一种方法可以阻止电流倒流——至少是通过电池——我们就有可能制造出一个没有任何再生状态的H桥。事实证明，这种反向电流保护元件是二极管，我们已经在桥上安装了四个二极管。问题是，我们能否利用它们。我们将在下一章研究这个问题。
3.Basic operation
基本操作
The switching patterns of this drive mode are a modification of the sign-magnitude drive. The idea is that during the off-time we’ll leave only one of the four switches closed, instead of two as with the sign-magnitude drive. This will force the off-time current to flow through one of the catch-diodes. There are four reasonable ways of doing that:
这种驱动模式的切换模式是对符号-幅值驱动的修改。它指，在“非工作”时间，我们将只关闭四个开关管中的一个，而不是像符号-幅值驱动那样关闭两个管子。这将迫使“非工作时间”的电流流过其中一个寄生二极管。有四种合理的方法可以做到:
(If you are pedantic, you’ll notice that there are four other possibilities, bur they are effectively identical to the four I have listed at least as far as the discussion here is concerned.)
(如果你仔细看，你会注意到还有其他四种可能性，但是至少就这里的讨论而言，它们实际上与我列出的四种完全相同。)
The four combinations are the four possible answers to the two binary questions:
这四种组合是两个二进制问题的四种可能的答案:
Is the on-time voltage applied in the forward or reverse direction? 导通电压是正向施加还是反向施加？Does the high- or low-side switch stay closed during the off-time?高或低侧开关在关闭时间是否保持关闭？
There usually is a control signal deciding the answer to the first question, while the second one is usually answered during the design of the bridge (there are exceptions we’ll talk about in a later article).
通常有一个控制信号决定第一个问题的答案，而第二个问题通常在H桥设计期间得到回答(有些情况我们将在后面的文章中讨论，这与元器件有关）
For the general discussion it doesn’t matter too much which direction the on-time voltage is applied, so I’m going to stick with my old habit of choosing one where the motor is driven in the forward direction (mapping A or B). For the most part, I’ll assume that we keep the high-side switch closed, so I’ll use ‘mapping A’ from now on.
对于一般性的讨论，实时电压施加在哪个方向上并不太重要，所以我将坚持我的旧习惯，选择一个电机在正向方向运行的情况(映射 a 或 b)。在大多数情况下，我假设我们保持高侧开关关闭，所以从现在开始我将使用“映射 a”。
Let’s take a look at the switching diagram first:
让我们先来看一下交换图:
You see that during the off-time, I drew the ‘b-side’ motor potential being slightly above Vbat, and at the same time, the motor voltage being slightly negative. To understand the reasons for that, let’s look at the current flow diagrams! During the on-time, the current does what it always does: flows through the two closed switching elements.
你可以看到，在“非工作时间”，我画的“ b侧”电动机电势略高于 Vbat，同时，电动机电压为负值。为了理解其中的原因，让我们看看当前电流的流向！在”工作1状态“下，电流流过两个打开的开关管元件，如下图所示，电流通过了Q1，流经电机，最终通过Q4流向地：
When the switch-over happens to the off-time, Q4 opens and only Q1 stays closed. At that point the motor current (being an inductive load) will have to continue flowing and the only way it can is through forward-biasing D3. So, the off-time current flow will be the following:
当切换到”非工作时间“时，Q4管子关闭，只有 Q1保持打开状态。此时，电机电流(作为一个感应负载)将不得不继续流动，而唯一的途径是流过D3。因此，关闭时间的电流路径如下:
In order for current to flow through D3 however, the b-side lead of the motor needs to be at a slightly higher potential than Vbat. This voltage is the forward-bias voltage of the diode, and is in the range of 0.2 and 1V depending on construction, current, temperature among other things. In the following discussions I will disregard this little voltage difference though, and assume that the diode drop is 0. Just keep in mind, that for more precise calculations, you will have to take that forward drop into consideration as well.
然而，为了使电流流过 D3，电动机的 B侧的电位需要比 Vbatl略高.这个电压是二极管的正向偏置电压，根据结构、电流、温度等的不同，这个值在0.2到1V 的范围内。在接下来的讨论中，我会忽略这个小的电压差，并假设二极管压降为0。请记住，为了得到更精确的计算结果，你必须考虑到这个正向导通压降。
In other drive modes (mapping B for example) it’s the low-side diode and switch that conducts the current:
在其他驱动模式(例如映射 b)中，低侧二极管和开关导通电流:
In both cases however, essentially the motor is short-circuited, the same way as in sign-magnitude drive. Consequently the current diagram of the motor doesn’t change much either.
然而，在这两种情况下，本质上电机都是短路，和符号-幅度驱动的情况一样。因此，电动机的电流回路也没有太大的变化。
4.Continuous and Discontinuous Current
连续和不连续电流
Now, before we go any further, we’ll have to discuss the topic of current flow a little deeper. What happens, when (during the off-time) the current decreases to 0? In both of the sign-magnitude and lock anti-phase drive modes the current is free to turn negative, as the conducting elements are bi-directional. Not the case here! The conducting diode (D3 for example) will close as soon as the current reaches 0. If that happens, there’s no current flow in the circuit for the rest of the cycle. So, depending on the slopes of the on-time and off-time currents (and the times spent in each state) the current might flow continuously or discontinuously in the system:
现在，在我们进一步讨论之前，我们必须更深入地讨论电流流向的问题。当电流减小到0时会发生什么？在符号-幅值和锁-反相驱动两种模式中，由于导电元件是双向的，电流可以自由地转为负值。但在这里不是这样的！一旦电流达到0，导电二极管(例如 D3)就会关闭。如果发生这种情况，在剩下的周期中，电路中就没有电流了。因此，根据”工作状态“和”非工作状态“下电流的斜率(以及在每种状态下所持续的时间) ，电流可能在系统中连续或不连续地流动:
This phenomena has some nice benefits as we’ll see later, but unfortunately means that our equations will become messier as their form changes depending on which of the two distinct operating mode the bridge is in.
这种现象有一些好处，我们将在后面看到，但不幸的是，这意味着我们的方程将变得更加混乱，因为它们的形式变化取决于两个不同的运行模式桥是在哪一个。
You can see that I introduced three time-periods:
你可以看到我引入了三个时间段:
tonis the on-timeton就是”工作时间“toff_conducts the portion of the off-time when the bridge is conducting current 、toff_conduct是电桥在”停机时间“电流仍在流动的时间toff_zerois the portion of the time when the current is 0.toff_zero是当电流为0时的部分时间
There are some obvious relationships between these:
这两者之间有一些明显的关系:
toff_conduct+ toff_zero= toffton+ toff= tcycle
Finally there’s another time metric, worth introducing, which is the conduction time:
最后还有一个值得介绍的时间指标，即传导时间:
tconduct= ton+ toff_conduct传到时间 = 导通时间 + 关断时传导时间
Using this terminology, the difference between continuous and discontinuous current modes is if toff_zerois zero or not.这两个术语，连续和不连续的电流模式的区别在于toff_zero是否为零。
While many things are different, some of the knowledge that we’ve gained by analyzing the simpler drive modes, carries over. For example, the slope of the current change during the on- and off-time (as long as it conducts) can be found by the following equations:
虽然很多事情是不同的，但我们通过分析更简单的驱动模式所获得的一些知识仍然有效。例如，通断时间（只要导通）期间电流变化的斜率可以通过以下等式求得：
on_time:dI/dt = (Vbat– Vg– Imot_avg*Rm)/Lm
导通时间内，等效在电机电感上的电压为电池电压-反电势-电机内阻压降，而电机电感器上的电流与Lm微分关系。电流是指数变化的。IL = L*di/dt
off_time: dI/dt = (– Vg– Imot_avg*Rm)/Lm
关断时间内，电池不再给电机供电，此时只有电机本身产生的反电势，同样电机内阻上也有压降，这些共同产生了电感器上的电流变化。
As I promised, I am already involving the voltage drop on the internal resistance in these equations: we will need to make use of them pretty soon.
正如我前面所说的，我已经在这些方程中包含了内阻的电压降: 我们很快将会用到它们。
5.Average motor voltage
平均电机电压
In order to be able to calculate the average motor voltage, there’s one last thing we’ll have to discuss: what is the voltage on the motor terminals in discontinuous current mode, when the current is 0? The answer is fairly simple: Since the current doesn’t change, the inductor of the motor should see 0 voltage difference. The internal resistance drops 0 as well, since the current is 0. This means that the motor terminal voltage must cancel out the generator voltage, Vg. In other words the motor voltage during this period should be Vg. Now, using that knowledge, we can calculate the average motor voltage:
为了能够计算出电机的平均电压，我们还要讨论最后一件事:当电流为0时，电机终端在不连续电流模式下的电压是多少？答案相当简单: 由于电流不变，电机的电感上的压降为0。内阻上的压降也为0，因为电流是0。这意味着电机端电压必须抵消发电机电压。换句话说，这段时间的电动机电压应该是Vg。现在，利用这些知识，我们可以计算出平均电机电压:
Vmot_avg= (Vbat*ton+ Vg*toff_zero)/tcycle
如上面所言，电流为不连续模式时，在ton阶段，电机两端电压为Vbat，而在toff_zero阶段，电机电压与发电机电压相抵消，即为Vg。诚然，这里是近似计算，配合之前的图容易理解:
It’s also beneficial to introduce another voltage, which is the average voltage the motor sees during the time the bridge conducts current:
这里还要引入另一个有益的电压，这个电压是电桥传导电流时电机上的平均电压:
Vavg_conduct= Vbat*ton/tconduct
Note, that if the bridge is in continuous current mode, the two values are equal as tcycle= tconductand toff_zerois 0.
注意，如果电桥处于连续电流模式，tcycle = tconduct，toff_zero为0。
6.Steady state
稳态
Our steady-state is defined when the parameters of the bridge stay constant from cycle-to-cycle. This means both electrical and mechanical characteristics: currents, voltages, torque, acceleration etc. If that’s true, the current-change during the on-time and off-time must chancel each other out, and the change (the ripple-current) is the following:
我们的稳态定义为，在一个周期内，电桥的参数为常数。这意味着电气和机械特性: 电流，电压，扭矩，加速度等都是常数。如果这是真的，那么在“工作时间”和“停机时间”的电流变化必须互相推翻，变化(纹波电流)如下:
Iripple= (Vbat– Vg– Imot_avg*Rm)/Lm*ton= (Vg+ Imot_avg*Rm)/Lm*toff_conduct
Expressing Vgfrom this, we get:
求解 Vg，我们得到:
Vg= Vbat* ton/tconduct– Imot_avg*Rm
which is the same as:
也就是:
Vg= Vavg_conduct– Imot_avg*Rm
And solving it for the average motor current, we get:
然后求出平均电机电流，我们得到:
Imot_avg= (Vavg_conduct– Vg) / Rm
Now, let’s put this back into the ripple-current equation:
现在，让我们把个带入文波电流方程得到:
Iripple= 1/Lm* Vavg_conduct* toff_conduct
7.Finding the operating mode
寻找工作模式
Let’s see if we can figure out if the bridge at a given operating point is in continuous or discontinuous mode.The external conditions that define the operating point are the following:
让我们看看我们是否可以计算出电桥在某个状态是在连续还是不连续的工作模式。定义操作点的外部条件如下:
The operating frequency represented by t用 t cycle表示的工作频率The duty-cycle, represented by D占空比由D表示The battery voltage Vbat电池电压表示为vbatThe speed of the motor – or the generator voltage, represented by Vg 电机的速度——或发电机的电压，用Vg表示The motor parameters Rm and Lm电机参数Rm 及Lm
Before we can attack this problem, we’ll have to do one more thing. It turns out, we can arrive at the average current for discontinuous current mode in a different way as well. For this operating mode, that is when the current starts from 0 in each cycle, the area under the current ‘triangle’ should be the same as the area under the average current rectangle:
在解决这个问题之前，我们还要做一件事。结果表明，我们也可以用不同的方法得到不连续电流模式的平均电流。对于这种操作模式，也就是当电流在每个周期从0开始时，当前“三角形”下的面积应该与当前平均矩形下的面积相同:
Note, that this approach doesn’t work for continuous current mode, as the current doesn’t reach zero during the cycle.
请注意，这种方法不适用于连续电流模式，因为在周期内电流不会达到零。
From this insight, we get this:
从这个观点，我们可以得出这样的结论:
Imot_avg* tcycle= 1/2 * Iripple* tconduct
电机平均电流 * 持续时间 = 1/2 * 峰值 * 传导时间，tconduct = toff_conduct + ton,即如图所示，三角形的面积
Putting the expressions for Imot_avgand Irippleinto this and expressing Vg, we get:
把 imot_avg 和 Iripple 的表达式放到这里求解Vg，我们得到:
Vg= Vavg_conduct* (1 – 1/2 * Rm/Lm* tconduct* toff_conduct/ tcycle)
∵Imot_avg= (Vavg_conduct– Vg) / Rm 以及Iripple= 1/Lm* Vavg_conduct* toff_conduct
∴(Vavg_conduct– Vg) / Rm * tcycle = 1/2 *1/Lm* Vavg_conduct* toff_conduct * tconduct
Then Vg =Vavg_conduct -1/2 *1/Lm* Vavg_conduct* toff_conduct * tconduct * Rm / tcycle
So Vg =Vavg_conduct * （1-1/2 *1/Lm* Vavg_conduct* toff_conduct * tconduct * Rm / tcycle）
The final insight is this: there’s a critical point where the continuous and discontinuous current modes meet: at this point the off-time current just reaches zero at the very end of the off-time. So in that case, the above equation is correct, whiletoff_zerois zero as well:
最后的结论是: 有一个临界状态，连续和不连续的电流模式相交: 在这一点上，关断时间的电流刚好在关断时间的最后达到零。因此，在这种情况下，上面的等式是正确的，而 toff_zero 也是零:
For this special-case we can do some simplifications to the previous equation, and after substituting Dcritical= ton_critical/tcycle, we get:
对于这个特殊情况，我们可以对前面的方程进行一些简化，用 Dcritical = ton _ critical/tcycle替换，我们得到:
Vbat* Dcritical* (1 – 1/2*Rm/Lm*tcycle*(1-Dcritical)) = Vg
This is a quadratic equation for Dcritical:
这是关于Dcritical 的一元二次方程:
Dcritical^2– Dcritical*(1+2 * Lm/Rm* 1/tcycle) + 2 * Lm/Rm* 1/tcycle* Vg/Vbat= 0
This equation has at most two roots. However, as long as Vghas the same polarity as Vbat(we’ll talk about what happens if it’s not in a minute), only one of the roots is between 0 and 1, leaving us with exactly one physically valid root:
这个方程最多有两个根。然而，只要 Vg 具有与 Vbat 具有相同的极性(我们会在稍后讨论) ，其中只有一个根介于0和1之间，这样我们就只剩下一个物理上有效的根:
Dcritical= 1/2 +Lm/Rm* 1/tcycle– sqrt[ (1/2 +Lm/Rm* 1/tcycle)^2+ 2 *Lm/Rm* 1/tcycle*Vg/Vbat]
Not a nice equation, but at least a solution.Finally we’re at the point where we can figure out in what operating mode the bridge is in: With the operational parameters, we calculate the critical duty-cycle. If the actual duty-cycle is above this critical value, the bridge operates in continuous current mode. If it is below that, the bridge is in discontinuous current mode.
这不是一个好的方程式，但至少是一个解决方案。最后，我们可以计算出电桥在什么样的工作模式下: 通过运行参数，我们计算出临界占空比。如果实际占空比高于这个临界值，则电桥工作在连续电流模式下。如果低于这个值，则电桥处于不连续电流模式。
If you look carefully at the solution, you’ll see that it depends on a few ratios only:
如果你仔细观察解决方案，你会发现它只取决于几个比率:
First, it’s the ratio of Vg/Vbat, in other words the relative speed of the motor. 首先，它取决于Vg/Vbat，它取决于电机的相对速度The second is Lm/Rm * 1/tcycle. Since Lm/Rm is the electrical time constant of the motor, this ratio is the relative time constant of the motor to the bridge cycle time第二个是是电动机的电气时间常数Lm/Rm * 1/tcycle.，这个比率是电动机的相对时间常数与电桥周期的比值
Usually a bridge operates at lower cycle time as the motor time constant and even if it didn’t the calculations above assumed that; We assumed the current change to be linear during the on-time and the off-time and that is only true if the bridge operates much faster than the motor time constant. So Lm/Rm* 1/tcyclemust always be – for our purposes at least – bigger than 1.
通常一个电桥的周期时间低于电机时间常数，即使它没有上面的计算假定; 我们假定电流的变化在开时间和关时间内是线性变化的，这是真实的，所以电桥工作频率必须比电机时间常数快得多。因此Lm/Rm * 1/tcycle必须始终大于1——至少对于我们的目的是如此。
I’ve plotted a few of these curves in the following graph:
我在下面的图表中绘制了其中的一些曲线:
You can see that each of the curves cut the square which represents all possible duty-cycle and relative motor speed combinations (in the forward direction) into two. The area above the curves is the land of continuous current mode operation, the area below them is for discontinuous current modes. You can see that the curves don’t deviate too much from the diagonal, so a good first-order approximation is that if the relative motor speed is less than the duty cycle, the bridge operates in discontinuous current mode. This approximation has a roughly 15% error, but the error is less and less if you increase the operating frequency of the bridge.
你可以看到，每个曲线切割的方块代表所有可能的占空比和相对电机速度的组合(假设电机正转)。曲线上面的区域是连续电流模式运行的区域，曲线下面的区域是不连续电流模式运行区域。你可以看到，曲线不会偏离对角线太多，它是一阶近似曲线，如果相对电机转速小于占空比，电桥在不连续电流模式下工作。这个近似值有大约15% 的误差，但是如果你增加电桥的工作频率，误差会越来越小。
Finally, the promised explanation for negative Vg: If Vgis negative (that is opposing Vbat), the off-time current will never be able to close the diode as Vgbiases the diode in the forward direction. So in that case, the bridge is always in continuous current mode.
最后，关于负的 Vg 的解释是: 如果 Vg 是负的(即相对于Vbat反向) ，关断时的电流将永远不能关闭二极管，因为 Vg 向前偏置二极管。所以在这种情况下，电桥总是处于连续电流模式。
8.Finding tconduct
The last topic to discuss is this: let’s say your bridge operates in discontinuous current mode. What is the portion of the time it spend conducting?
最后要讨论的话题是: 假设你的电桥在不连续电流模式下工作。它花在传导的时间有多少？
For the answer, we’ll have to go back to the equation in the previous chapter:
为了找到答案，我们不得不回到前一章的等式:
Vg= Vavg_conduct* (1 – 1/2 * Rm/Lm* tconduct* toff_conduct/ tcycle)
From this, we get (after expressing Vavg_conductfrom Vbatand rearranging things):
由此，我们得出(使用Vbat 替换了Vavg_conduct) :
Vg/Vbat*tconduct=ton* (1 – 1/2 * Rm/Lm* tconduct* (tconduct-ton) / tcycle)
This is a quadratic equation for tconduct. I’m not going to give you the closed-form solution, because it’s an eye-sore and it’s actually easier to substitute numbers into this equation and solve if for a particular case, than to derive the parametric solution. What is important to note here is that just as before, from the two possible solutions, only one will be physically meaningful, that is between 0 and tcycle.
这是一个关于tconduct的一元二次方程。我不打算给你们显式解，因为这是一个棘手的问题，实际上，对于一个特定的情况，用数字代替方程并求解 ，比导出参数解要容易得多。这里需要注意的是，就像以前一样，在两个可能的解中，只有一个在物理上是有意义的，它位于在0和 tcycle 之间。
9.Transient states and input capacitor
暂态和输入电容
So far it seems it’s impossible for the current to flow in the reverse direction, and we’ve really covered quite a bit of ground. The previous discussion not only applies to steady-speed operation but to acceleration, deceleration as well as long as the cycle-to-cycle averages don’t change.
到目前为止，电流似乎不可能逆向流动，而且我们已经涉及了相当多的领域。前面的讨论不仅适用于稳速操作，而且适用于加速、减速以及只要周期到周期的平均值不变化的情况。
When one of the parameters (mainly D or Vg) changes, it takes time for the system to settle in the new steady-state. In that transient, the current can flow in the reverse direction during the on-time. Let’s take a quick example to understand why!
当其中一个参数(主要是 周期D 或 反电势Vg)发生变化时，系统需要一段时间才能达到新的稳态状态。在这种暂态过程中，电流可以在通电期间以相反的方向流动。让我们举一个简单的例子来理解为什么！
Let’s say we operate the bridge in the reverse direction at 50% duty-cycle and under relatively heavy loads. Let’s also assume the motor has a really low Rm. In this case Vgis close to -Vbat/2 and the average motor current is a relatively high positive number. Because of the bridge operates in the reverse direction, the motor current flows through the battery in the positive direction as well, discharging the battery.
假设我们在相对较重的负载下以50% 的占空比反向操作电桥。让我们也假设电机内阻 Rm很低。在这种情况下，Vg 接近-Vbat/2，平均电机电流是一个相对较大的正数。由于电桥朝反方向运行，电机电流也朝正方向流过电池，使电池放电。
Now, let’s imagine that we abruptly change the bridge top operate in the forward direction, still with 50% duty-cycle. The first cycle with the changed duty-cycle will see the motor current to be opposite to the battery voltage, just as Vg. This current will start increasing (decrease in absolute value) during the on-time, as Vbatis applied in the opposite direction to the motor current now, but for the moment it still is flowing into the battery, recharging it.
现在，让我们想象一下，我们突然改变桥的操作方向，仍然使用50% 的占空比。改变占空比的第一个循环将看到电动机电流与电池电压相反，等于Vg。这个电流将开始增加(绝对值减少)在导通时间，由于Vbat的作用方向与电机电流方向相反，但目前它仍在流入电池，给电池充电。
When the off-time comes along, something strange happens: As Q4 closes but Q1 stays open, the motor current (still flowing in the negative direction) will have to find a way to continue flowing. It can’t flow through D3, since that diode can’t conduct in the reverse direction. The only way the current can flow is through D4:
当停机时间到来时，奇怪的事情发生了: 当 Q4关闭，但 Q1保持开放，电机电流(仍然流向负方向)将不得不找到一种方法继续流动。它不能通过 D3，因为这个二极管不能在相反的方向传导。电流通过 D4的唯一途径是:
What that means, is that the bridge is essentially in the same mode for both the on- and the off-time. The only difference is that it’s the diode and not the switch that conducts the current, but it flows through the same leg in the same direction.
这意味着，桥梁在开关时间上，基本上处于相同的模式。唯一的区别是，导电的是二极管，而不是开关，但它沿着同一方向流过同一桥臂。
This also means that the current will continue increasing (decrease in absolute value) during the off-time as well.
这也意味着，在关闭时间内电流将继续增加(绝对值下降)。
The next on-time will continue in the same way just as the next off-time and so on. This process continues as long as the motor current reaches 0 (coming from the negative side) at which point the normal operation of the bridge will resume: the off-time current will start flowing through D3 as usual.
下一个工作时间将以同样的方式持续到下一个停机时间。只要电机电流达到0(来自反方向) ，这个过程就会继续，此时电桥将恢复正常运行: 关断期间电流将像往常一样开始通过 D3。
From this example, the following can be derived: in the transient, the motor current increases through potentially several cycles from it’s initial negative value to 0. The increase will not depend on the timing of the bridge as the current flow and voltages stay the same during the on- and off-times. The current change will follow an exponential curve (we can no longer assume linear process as the current change is too large to assume a constant voltage drop on Rm), according the following equation:
从这个例子中，可以得出以下结论：在瞬态过程中，电机电流可能通过几个周期从初始负值增加到0。由于电流和电压在通断时间内保持不变，因此增加值不取决于电桥的时间。电流变化将遵循指数曲线（我们不再假设线性过程，因为电流变化太大，无法假设Rm上的电压降恒定），根据以下等式：
Imot(t) = Imot(0) + (Vbat/Rm– Imot(0)) * (1 – e^(-t/(Lm/Rm)))
Note that Imot(0), the initial current through the motor when the change in the duty-cycle happens is negative, so is Imot(t). Of course only the portion of the curve where the current is negative is valid:
注意，当占空比发生变化时，通过电机的初始电流为负值，Imot (0)也为负值。当然，只有曲线中电流为负的部分才有效:
As we’ve discussed it several times, current flowing back to the battery is a problem in most implementations. The main goal with using this switching pattern to begin with was to get rid of this reverse-current! Well, apparently we weren’t completely successful.
正如我们多次讨论的那样，电流回流到电池是大多数实现中的一个问题。使用这种开关模式开始的主要目标是摆脱这种反向电流！显然我们没有完全成功。
However, as long as we can put a limit on the charge that this reverse current pushes back into the battery, we can handle it by putting a capacitor on the input on the bridge that can temporarily soak up the extra charge.
然而，只要我们能够限制电池的充电量，我们就可以通过在电桥的输入端加一个电容器来暂时吸收额外的电量。
So, let’s figure out if there’s an upper bound to this charge? The total charge delivered is the integral of the current over time, so the first order of business is to figure out how long does the current flow. It stops flowing when the motor current reaches 0, so:
那么，让我们来看看这个电荷是否有一个上界？交付的电荷总量是电流随时间变化的积分，因此，业务的第一要务是计算出电流流动的时间。当电机电流达到0时，它就停止流动，所以:
Imot(0) + (Vbat/Rm– Imot(0)) * (1 – e^(-tstop/(Lm/Rm)))= 0
Now, solving it for tstop, we get:
现在，为了解决这个问题，我们得到:
tstop= – Lm/Rm* ln(1 – Imot(0) / (Vbat/Rm– Imot(0)) )
Next, to do the integral, we have to find the primitive function of the current curve. It is the following:
接下来，要做这个积分，我们必须找到当前曲线的原函数:
Vbat/Rm* t + (Lm/Rm) * (Vbat/Rm– Imot(0)) * e^(-t/(Lm/Rm)) + C
With these two, we can calculate the integral. After some simplifications, the total charge released is the following:
利用这两种方法，我们可以计算积分，经过一些简化，总的电荷释放如下:
Qreverse= – Lm/Rm* (Imot(0) – (Vbat/Rm) * ln(1 – Imot(0) / (Vbat/Rm– Imot(0)) )
As expected, the sign is negative, so the charge is coming out of the bridge, into the battery (or the capacitor we’re trying to size).
正如预期的那样，这个符号是负的，所以电荷是从电桥出来的，进入电池(或者我们试图计算的电容)。
So what is the maximum value of this charge? It is when Imot(0) is maximal (in absolute value). So how big this current can be? We can take two approaches, just as we did in the lock anti-phase drive. We can take either the short-circuit current of the motor or the current-limit of the bridge into account. The latter is still a very conservative estimation and is usually much lower than the short-circuit current of the motor, so let’s use that. In the following equations, I’ve changed the signs so that we substitute the positive bridge current:
那么这个电荷的最大值是多少？当Imot（0）为绝对值时。那么这个电流有多大？我们可以采取两种方法估算，就像我们在锁-反相位驱动中所做的那样。我们可以考虑电动机的短路电流或电桥的电流限制。后者仍然是一个非常保守的估计，通常比电机的短路电流低得多，所以我们用这个。在下面的方程式中，我改变了符号，这样我们就可以用正桥电流来代替：
Qreverse_max= Lm/Rm* (Imax+ (Vbat/Rm) * ln(1 + Imax/ (Vbat/Rm+ Imax) )
This charge changes the capacitor voltage following the familiar formula: Vripple_max= Qrelease_max/C. If we know how much ripple voltage the circuit can tolerate, we can calculate the size of the capacitor needed:
这种电荷会按照我们熟悉的公式改变电容器的电压：Vripple_max=Qrelease_max/C。如果我们知道电路能承受多大的纹波电压，我们就可以计算出所需电容器的尺寸：
C= Lm/Rm* (Imax+ (Vbat/Rm) * ln(1 + Imax/ (Vbat/Rm+ Imax) ) / Vripple_max
Let’s see an example! Let’s say our motor has a 30uH inductance, and a 1Ω resistance. Let’s say we operate from a 20V supply and the bridge can sustain a 10A current. Finally let’s assume that we can tolerate 5% ripple current. The result is a 473µF.
让我们看一个例子！假设我们的马达有30uH电感，和1Ω个电阻。假设我们用20V 电源操作，这座桥可以承受10A 的电流。最后，让我们假设我们可以容忍5% 的纹波电流。结果是473uF。
计算过程：C = 0.00003*（10+20/1）*ln（1+10/（20/1+10））/21= 0.000473 F = 473uF
If we want to build a bridge with a 100A current delivery capability, the input capacitance needed would jump to almost 10x, or around 3500µF.
如果我们想要建立一个具有100A 电流传输能力的电桥，需要的输入电容将跳到10倍，大约为3500uf。
If we assume a much lower resistance, let’s say 100mΩ only (and still 10A current limit), the capacitor value jumps to over 5000µF. That is a trend: for low resistances, the capacitance required is more or less proportional to 1/Rm.
如果我们假设一个小得多的电阻，假设只有100mΩ(仍然是10A 电流限制) ，电容值就会跳到5000uF以上。这是一种趋势: 对于低电阻电机，所需的电容大致与1/Rm 成正比。
You can also play around with the equation to get a feel to how the various parameters affect the solution.
您还可修改各种参数以了解它们如何影响解。
Another interesting experiment is this: what if we want to calculate the capacitance needed for the stall-current of the motor? The stall current is Vbat/Rm, so if we put that in the place of Imaxin the previous equation we get some nice simplifications:
另一个有趣的实验是: 如果我们想计算电动机失速电流所需的电容呢？失速电流是 Vbat/Rm，所以如果我们把它放在 Imax 的位置，我们得到了一些很好的简化:
C = 1.405 * Lm/Rm^2* Vbat/Vripple_max
计算过程：C = 1.405*0.00003/1*20/21= 0.0000401 = 40uF
This is much easier to comprehend, but as you can see it has an inverse quadratic relationship to Rm. That is understandable since in the previous calculations we’ve assumed Imaxto be independent of Rm, while in here it’s not the case anymore.
这很容易理解，但是正如你所看到的，它与 Rm 有一个反二次方的关系。这是可以理解的，因为在以前的计算中，我们假设 Imax 是独立于 Rm 的，而在这里情况不再是这样了。
Summary
摘要
We started this article with the idea to construct an H-Bridge drive mechanism that doesn’t suffer from regenerative braking. After a lot of ugly math I hope I convinced you that it is in deed possible. There are of course down-sides to this approach:
我们从构造一个不受再生制动影响的 h 桥驱动机构的想法开始这篇文章。在做了很多难懂的数学运算之后，我希望我能让你相信这确实是可能的。当然，这种做法也有不利的一面:
The catch diodes conduct for a significant portion of the time and that is usually going to generate more heat than if the switches conduct the current.捕获二极管在相当长的一段时间内导通，这会比电流通过开关管时产生更多的热量The bridge has two operating modes: continuous and discontinuous current mode and the operating parameters, including the motor torque (average motor current) behaves differently in those two modes. This change in behavior complicates closed-loop speed-control design.电桥有两种工作模式：连续电流模式和不连续电流模式，在这两种模式下，包括电机转矩（电机平均电流）在内的运行参数表现不同。这种行为变化使闭环速度控制设计复杂化We still need to employ an input capacitor to handle transient states, and the capacitor value can be rather large, especially for high-current bridges.我们仍然需要使用一个输入电容来处理瞬态，而且电容值可能相当大，特别是对于大电流电桥。The analysis of the bridge is significantly more complicated due to the non-linear behavior of the diodes. 由于二极管的非线性特性，对电桥的分析要复杂得多。
With all that, this drive mode has one huge advantage, one that in many cases outweighs the negatives: it is inherently safe, as it can’t over-charge the battery during braking.
尽管如此，这种驱动方式有一个巨大的优势，在很多情况下，它的优点大于它的缺点：它本质上是安全的，因为它不会在刹车时给电池过度充电。
Where to go from here?
接下来该怎么办？
If you’re really pedantic, you’ll notice that there’s another possible drive mode we haven’t discussed so far: you can take the lock anti-phase drive and modify it to work in an asynchronous fashion just as we did with the sign-magnitude drive here. I will leave that operating mode alone at least for now as I don’t see too much practical value to it.
如果你真的很较真，你会注意到还有另一种可能的驱动模式，我们到目前为止还没有讨论过的: 你可以采用锁定-反相驱动，并以异步方式工作，就像我们在这里使用符号-幅值驱动一样。至少现在我不会去管那种操作模式，因为我认为它没有太多的实际价值。
So now that we’ve finished with the detailed analysis of all interesting drive modes, we’re almost ready to move on and discuss drive circuitry and component selection. However before we jump into that topic, the next installment will start with a comparison of the drive modes we’ve discussed here to be able to see the differences and the similarities side-by-side.
现在我们已经完成了所有有趣的驱动模式的详细分析，接下来准备好继续讨论驱动电路和组件选择。然而，在我们开始讨论这个话题之前，下一期文章将首先将会比较一下我们在这里讨论的驱动模式，以便能够同时看到它们的区别和相似之处。
