在没有控制器的情况下怎么驾驶STEVAL-IHM023V2

再见
 
 也许我不明白你的问题，但如果你想保持一条腿（以及逆变器的任何一个开关）打开你需要保持相对输入不活动是正常的。这意味着在STEVAL-IHM023V2的情况下J5引脚：
 3,7,11到GND（这些输入为高电平有效）
 5,9,13到Vdd（这些输入为低电平有效）
 
 当然，如果让它们保持浮动状态，它们将通过内部上拉或下拉电阻进入非活动状态。
 
 再见
 梁咏琪



以下为原文





Ciao

Maybe I don't understand your question but it is normal that if you want to keep a leg (and also any of the switches of the inverter) open you need to keep inactive the relative inputs. This means in case of  STEVAL-IHM023V2 the J5 pin:
3,7,11 to GND (these inputs are active high)
5,9,13 to Vdd (these inputs are active low)

Of course if you keep them floating they will be bring to the inactive state by means of internal pull up or pull down resistor.

Ciao
Gigi

你好吉吉，
 
谢谢你的答案，但我们已从数据表中推断出。
 
让我重新解释一下我的问题：
采取任何3相负载（例如3个电阻：R_A，R_B和R_C）。我们想要一个通过R_A和R_B的电流，所以I_A = -I_B和I_C = 0。因此，必须关闭阶段A中的上部开关，必须关闭阶段B中的下部开关，并且必须保持所有其他开关保持打开。
 
我的问题：三相电桥可以实现吗？
 
我们知道，如果我们将阶段A中的上部和下部开关发送5V，将阶段B中的上部和下部开关发送到0V，并且将阶段C中的上部和下部开关发送到不同的电压（例如0V到上部），这应该工作和5V到更低）。
 
你看到我们的错误，因为这不适用于我们的设置吗？
 
提前致谢！
弗洛里安



以下为原文





Hello Gigi,

thank you for your answer but we deduced that already from the datasheet.

Let me rephrase my question:
take any 3 phase load (3 resistors for example: R_A, R_B and R_C). We want a current through R_A and R_B, so I_A=-I_B and I_C=0. Therefor the upper switch in phase A has to be closed, the lower switch in phase B has to be closed and all the other switches have to be left open.  

My question: is this possible with the 3 phase bridge?

We know that this should work if we send 5V to the upper and lower switch in phase A, 0V to the upper and lower switch in phase B and a different voltage to the upper and lower switch in phase C (for example 0V to the upper and 5V to the lower).

Do you see our mistake because this isn't working on our setup?  

Thanks in advance!
Florian

有没有其他人有这个问题或试图自己驾驶STEVAL？
 
 
对于ST：有没有办法可以获得更多关于如何从微控制器驱动STEVAL的信息？



以下为原文





Is there anyone else who had this problem or tried to drive the STEVAL by themself?


For ST: is there a way we can get some more information on how you drive the STEVAL from the microcontroller?

Ciao Florian，
 
 您提到的模态是6步驱动的基本配置之一，应由3相桥支持，我不认为STEVAL-IHM023V2有任何问题。
 当您按照描述提供输入时，您获得的行为是什么？
 
 相位A的上下开关为5V，
 0V到B相的上下开关，
 相位为0V，高压为5V，相位为5V
 
 如果您打开所有开关，可能还需要注意J5的pin1。紧急停止通常用于过电流保护。这应该是拉起来否则所有的swicthes被强行打开。
 
 再见



以下为原文





Ciao Florian,

The modality that you mention is one of the basic configuration of the 6-step driving that shall be supported by a 3 phase bridge and I don't see any problem to be applied with the STEVAL-IHM023V2.
Which is the behaviour that you get when you provide the input as you describe?

5V to the upper and lower switch in phase A,
0V to the upper and lower switch in phase B,
0V to the upper and 5V to the lower in phase C

If you get all switches open probably you have to take care also to the pin1 of the J5. It is the Emergency Stop usally used for over current protection. This should be pull up otherwise all swicthes are forced open.

Ciao

你好吉吉，
 
 
我做了你提出的建议并将所有信号测量到他们进入芯片的位置。换句话说：我确定你所说的输入信号。
 
输出是什么：A相中的两个开关保持断开，B相中的上部开关保持断开，B相中的下部开关闭合，C相的开关保持断开。我还测量了SD引脚（连接到J5的引脚1）并且它一直保持高电平，所以这不是问题所在。结论：B相和C相的正常行为，而不是上部开关应该闭合的A相。
 
知道这可能是什么？我使用3个电阻作为负载，每个7W，以星形连接，R = 100欧姆
 
亲切的问候，
弗洛里安



以下为原文





Hello Gigi,


I did what you proposed and measured all the signals down to where they enter the chip. In other words: I'm certain of the input signals as you stated them.

What was the output: both switches in phase A remain open, the upper switch in phase B remains open, the lower switch in phase B closed and the switches of phase C remained open. I also measured the SD pin (connected to pin 1 of J5) and it stayed high the whole time so that's not the problem. Conclusion: normal behavior for phase B and C, not for phase A where the upper switch should have closed.

Any idea what this could be? I use 3 resistors as load, each 7W, connected in star, R=100 Ohm

Kind regards,
Florian

Ciao Florian，
 
 在打开高压侧开关之前，您可能需要为启动电容充电。
 这意味着您必须先应用三个低侧开关闭合一段时间（例如10ms），然后再进行配置。一段时间后，引导盖将被放电，因此最好为高边应用PWM。
 
 再见
 梁咏琪



以下为原文





Ciao Florian,

Probably you need to charge the boot capacitors before to turn on the high side switches.
This mean that you have to apply first a configuration with the three low side switched closed for a while (10ms for instance) and then your configuration. After a while the boot cap will be discharged so is better to apply a PWM for the high sides.

Ciao
Gigi

你好吉吉，
 
当我这样做时，我得到一个33V的栅极电压，它下降（线性）到30V然后立即下降到0V。所有这些都发生在18ms，换句话说，上部开关关闭18ms。这是你所期望的吗？
 
之前我使用了一些PWM信号来测试桥接器并且没有工作，所以我将记录下来并在明天发送它。
 
问候，
弗洛里安



以下为原文





Hello Gigi,

when I do that I get a gate voltage of 33V which goes down (linear) to 30V and then instantlydrops towards 0V. All this takes place in 18ms, in other words the upper switch is closed for 18ms. Is this what you expected?

I used some PWM signals before to test the bridge and that didn't work so I'll document that and send it through tomorrow.

Regards,
Florian

Ciao Florian，
 
 这是合理的，因为当高压侧开启时，启动电容将放电，因此需要应用PWM，因为在关闭期间，启动盖重新充电。
 
 再见



以下为原文





Ciao Florian,

it is reasonable because the boot capacitor will be discharged when the high side is turned on for this reasion you need to apply a PWM because during the off time the boot cap are re-charged.

Ciao

你好吉吉，
 
我一直在考虑你说的话，这是一个说明我的想法的例子：
 
（S_1是上限，S_2是下限...... ......）
 
I_A = + 1，I_B = -1，I_C = 0
S_1关闭，S_2打开，S_3打开，S_4关闭，S_5打开，S_6打开
I_A = + 1，I_B = 0，I_C = -1
S_1关闭，S_2打开，S_3打开，S_4打开，S_5打开，S_6关闭
I_A = 0，I_B = + 1，I_C = -1
S_1打开，S_2打开，S_3关闭，S_4打开，S_5打开，S_6关闭
I_A = -1，I_B = + 1，I_C = 0
S_1打开，S_2关闭，S_3关闭，S_4打开，S_5打开，S_6打开
I_A = -1，I_B = 0，I_C = + 1
S_1打开，S_2关闭，S_3打开，S_4打开，S_5关闭，S_6打开
I_A = 0，I_B = -1，I_C = + 1
S_1打开，S_2打开，S_3打开，S_4关闭，S_5关闭，S_6打开
 
如果我们从步骤2到3，那么S_3必须关闭，但是在步骤2中，S_3和S_4打开...换句话说，冷凝器将不会被充电而S_3不能关闭...除非下一步（在6中）步行驱动器）在18ms内......
 
PWM不能解决这个问题，因为B阶段必须在步骤2中保持开路（因此没有PWM）。
 
测量结果也表明了这一点......如果我的信号频率足够高，那么电流会按照预期的那样流动。如果我降低频率（电机的速度较低）并且相位必须保持打开超过18ms，则电流不再流动，因为我们想要它。原因是因为上部开关不想再关闭。



以下为原文











Hello Gigi,

I have been thinking about what you said, here's an example to state what I think:

(S_1 is upper, S_2 is lower, ... ... ...)

I_A=+1, I_B=-1, I_C=0
S_1 closed, S_2 open, S_3 open, S_4 closed, S_5 open, S_6 open
I_A=+1, I_B=0, I_C=-1
S_1 closed, S_2 open, S_3 open, S_4 open, S_5 open, S_6 closed
I_A=0, I_B=+1, I_C=-1
S_1 open, S_2 open, S_3 closed, S_4 open, S_5 open, S_6 closed
I_A=-1, I_B=+1, I_C=0
S_1 open, S_2 closed, S_3 closed, S_4 open, S_5 open, S_6 open
I_A=-1, I_B=0, I_C=+1
S_1 open, S_2 closed, S_3 open, S_4 open, S_5 closed, S_6 open
I_A=0, I_B=-1, I_C=+1
S_1 open, S_2 open, S_3 open, S_4 closed, S_5 closed, S_6 open

If we go from step 2 to 3, then S_3 has to close but in step 2, S_3 and S_4 were open... In other words the condensator will not be charged and S_3 cannot close... UNLESS the next step (in 6 step drive) is within 18ms...

PWM will not solve this problem because phase B has to stay open in step 2 (so no PWM).

Measurements also show this... If the frequency of my signal is high enough then current flows as it is suppose to. If I lower the frequency (lower speed of the motor) and the phase has to stay open for more then 18ms, the current doesn't flow anymore as we want it. The reason for this is because the upper switch doesn't want to close anymore.

Ciao Florian，
 
 这是6步驱动的典型问题。通常，为了向施加到电动机的电压调制施加到高侧开关的PWM信号si，这解决了启动电容器的放电问题。
 如果你需要施加100％的占空比，这意味着最高电压，因此高端关闭两个连续的步骤，那么你有两个步骤的最小持续时间的限制，必须小于放电时间，这给了你是旋转场最小速度的限制。
 为了增加引导盖的放电持续时间，一个选项可以是增加电容值。例如，将它加倍，你应该将每个高侧开关的最大Ton时间增加到36ms。
 
 再见
 梁咏琪
 
 附：分析中缺少的是当存在感应电动机时（如电动机的情况），LS开关的自由弯曲二极管允许对启动电容器充电，因此仅在HS上施加PWM就足够了。如果你只有阻性负载，那么你需要在高侧和低侧都应用互补PWM来保护行李箱盖。



以下为原文





Ciao Florian,

this is a typical problem of 6-step drive. Usually to modulate the voltage applied to the motor a PWM signal si applied to the high side switches and this solve the problem of discharging of the boot capacitor.
If you need to apply 100% of duty, this means maximum voltage and so the high side closed for two consecutive steps, then you have a limitation in the minimum duration of the two step that must be less than the discharging time, and this give you a limitation in the minimum speed of the rotating field.
To increase the duration of the discharging of the boot cap one option can be to increase the capacintance values. For instance doubling it you shall increase to 36ms the maximum Ton time of each high side switches.

Ciao
Gigi

P.S. What is missing in the analysis is that when a inductive motor is present (like in case of motor) the free weeling diodes of LS switches allows to charge the boot capacitor so just applying the PWM on the HS is sufficient. While if you have only resistive load then you need to apply a complementary PWM in both high side and low side to cherge the boot cap.

你好吉吉，
 
这可能是问题的解释......但是如果你问我，那应该是数据表中的一段文字;）到本周末，我将在电机上测试电路板。
 
感谢您的帮助，如果一切正常，我会通知您。
 
亲切的问候，
弗洛里安



以下为原文





Hello Gigi,

that could be an explanation for the problem... But if you ask me, that should be a paragraph of text in the datasheet ;) By the end of the week I will test the board on the motor.

Thank you for your help and I'll let you know if everything works fine.

Kind regards,
Florian