L298 question

[wakoko79]

I've got a question about L298.. I attached a picture of the logic of L298
View attachment 60213

Now focus on one H-bridge(the left one)

Assuming the enable are always high, and motor terminals(brushed DC motor) are connected at out1 and out2, the logic would be:

in1 | in2 | logic
0 | 0 | slow decay (the current path would be motor>LOWrightbjt>LOWleftbjt>motor)
0 | 1 | forward (current path is HIGHrightbjt>motor>LOWleftbjt)
1 | 0 | reverse (current path is HIGHleftbjt>motor>LOWrightbjt)
1 | 1 | slow decay (current path is motor>HIGHrightbjt>HIGHleftbjt>motor)

My question is, is this logic correct?
I input a pwm signal on in1 and grounded in2.
when pwm goes LOW, a slow decay will ensue correct?
The mode is called slow decay because the current path when the PWM goes from HIGH to LOW is just going in circles right? And so, basically the motor terminals are grounded, this leads the DYNAMIC breaking correct?
Now, if its that the case, then why can I read a voltage from the current sense pin? I mean no current flows through the sense resistor and yet I can still read voltage from it.

View attachment 60215
See? Actually, the rising part of the wave is the point where the pwm signal goes HIGH, and the peak is when the PWM signal goes low.


I'm really concerned about this since I'm basically replicating a driver unit composed of L298's.

Enlighten me please.

Thanks!!

 

[koolguy]

Hi,

I have used 298 chip it was working fine, but your question is different i am not getting you?

 

[wakoko79]

Uhhm. Basically, I'm concerned about the collapse path of the current.
See, in driving H-bridges, a current path must be made for the load to receive power. By controlling how the elements of an H-bridge switch, we can alter the path of the current flowing through a load. In brushed DC motors, this means we can effectively change the direction of its rotation. Old stuff.

When we turn off the bridge, we are basically cutting out the power supply from the load. But in this case, the load is a motor, and motors are inductive which means even if we cut out the its connection to the supply, the current WILL find a way to circulate. And this is what I'm talking about.

Basically there are two ways for the current to collapse: slow decay and fast decay. Just google it if you're not familliar with it.

Like what I've said on my previous post, the collapse current circulates that way. And with the supplied schematic of the L298, I'm pretty sure it shows that a slow decay will ensue if the current collapsed, which is evident in the above truth table.
When that happens, the current will recirculate to the motor for a relatively longer time because there wouldn't be a resistance that will dissipate that current (well, everything in the real world is lossy, so the collapse current will eventually die out). And from what I described the SUPPOSED to be path of the collapse current, the current sensing resistor won't be getting any current flowing through it (look at the schematic). V=IR. no current means no voltage.
So I'm asking, why is there a voltage reading on the oscilloscope? I mean, the rising part is understandable because the current really flows at that time since the PWM signal is ON then. But I'm clueless about the falling part, it shouldn't be an exponential decay, it should instead go down to 0V instantly.
When we

 

[koolguy]

Can you just tell about that software or that simulation u are using.

 

[ronv]

Do you have the clamp diodes installed?

 

[wakoko79]

Do you have the clamp diodes installed?

sir, do you mean catch diodes? yes they are installed. do you have any internal structure of the L298? I really want to know how it could sense the falling part so I can somehow do it on my circuit.

One more thing is that I don't think this current is just mirrored from the actual current flowing through the load. This is because when you don't connect a sense resistor OR ground the current sense pin, there won't be any current flowing, i.e. motor won't rotate. I think this is the actual current that flowed through the load.