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Anik , 01-20-2024, 06:16 PM

For H bridge motor driver we need two N channel mosfet and two P cannel mosfet. But some of the diagram I can see there are 4 mosfet but they are same. Its also on a polulu motor driver board they also have used same 4 mosfet. I dont understand this.

Anik , 01-21-2024, 03:24 PM

I have designed like this...

QDrives , 01-21-2024, 05:00 PM

The L6388 does not work with P-Channel mosfets. Only N-Channel. That is why you use a driver like the L6388 is that you only need N-Channel. The trick is in the bootstrap...

Anik , 01-21-2024, 05:02 PM

ok. But What about the 4 same mosfet design?? How does a N channel mosfet switch positive side??

QDrives , 01-21-2024, 05:07 PM

Reread my last sentence again and then check that word in the datasheet. Or google it to find a ton of information about it.

Anik , 01-22-2024, 09:33 AM

I have searched a lot for last two days. But I could not find information what I need to know. I know how H bridge motor driver works. But how the boostrap helping the mosfet to turn the motor on. I am confused about this table its from IR2104 driver. How the VB nad VS working. Also I dont understand the functional diagram how they are working VB HO VS and VCC LO GND
I always first try to find on google as well as chatgpt. In many ways I dont know what should I search for my problem.

Anik , 01-22-2024, 09:44 AM

All of the diodes are SS34 and I need to drive 24V and 5A motor

QDrives , 01-22-2024, 10:10 PM

https://duckduckgo.com/?q=bootstrap+mosfet+driver&atb=v314-1&iax=videos&ia=videos

SS34 - why do you pick such bulky diode that do not allow for the current you want? Then again, what current do you need? When does the current go through the diode?

Anik , 01-23-2024, 03:49 AM

I have seen some video through this link.
I found the D22 and D21 is for charging the Boostrap capacitor for switching the HO. Here do I need Higher current supply?
I need 5A current for the motor. Here D17,18,19,20 are for back emf protection. I dont understand about diode current rating.

QDrives , 01-23-2024, 04:54 PM

For D21 and D22 the way to reduce the current peak is to add a series resistor, but this was probably also mentioned in one of the video's.

"Back EMF protection"? What needs protection and why?

The only reason to add Schottky diodes as you have done, is to reduce/eliminate reverse recovery of the MOSFET's intrinsic diode.
Btw, all 'silicon' diodes have a reverse recovery.

Anik , 01-23-2024, 08:08 PM

i quite understand the video.
"For D21 and D22 the way to reduce the current peak is to add a series resistor,"
Do I need to reduce the current to the capacitor??
Here my 12V 150mA supply is from a 7812 voltage regulator. You can see on the diagram.
I am really very confused what should I do with the circuit. This topic getting very hard to me. I found two reference note from the video. I will read them.

Anik , 01-23-2024, 08:08 PM

And mt PWM supply is from esp32 its 3.3V

Anik , 01-23-2024, 08:11 PM

""The only reason to add Schottky diodes as you have done, is to reduce/eliminate reverse recovery of the MOSFET's intrinsic diode.
Btw, all 'silicon' diodes have a reverse recovery.""
Ok sir thank you, I got it. I thought they are for the motor. Cause as far I know when the brush motor turned of it through voltage in reverse voltage .

Mini , 01-23-2024, 08:15 PM

Just a side note about your power supply. If you are using linear power supplies like 7812 you must understand how they work. You have to calculate thermal properties. How much current your 12V takes? If it is 150 mA and it means almost 2W of power dissipated on 7812, which means your 7812 can reach 100+ degrees celcius depending on which package you use and do you have airflow etc.

Anik , 01-23-2024, 08:47 PM

How can I calculate the current of the boostrap capacitor need?? I tried something from internet. But I think its not the right ...

QDrives , 01-23-2024, 11:08 PM

"Do I need to reduce the current to the capacitor?" -- Lets assume the cap is charged to 10V, Vf=1V and the 12V is well 12.0V. How much current would flow through the diode? Mathematical it would be infinite. Now if you add a 2.2 ohm resistor, the peak would be ~5A.

I see 78L12, usually the L is for low power. What are the loads of the 12V regulator? I assume that the 150mA is the current limit?

"Cause as far I know when the brush motor turned of it through voltage in reverse voltage ." ??? A DC brush motor will also act as a generator, is that also what you try to say here?

Theoretical the DIScharge current could reach 7A (assuming your numbers and calculations are correct).
However, the formula for tswitch is not (entirely) correct. R*C is a time constant. 1RC = ~63%, 3RC=~95%.
This means that the absolute peak current is simply 12V / Rg = 12V / 1.3 = 9.2A.
However, the driver too has resistance. And you are advised to also add an additional gate resistor, a typical value for something like this is 10...47 ohm.

Now if you take that dV=50mV, and the 2.2 ohm resistor in the beginning, you get a peak of ~25mA. No problem for small diodes such as BAS40.

https://www.youtube.com/watch?v=of_v2N5f788&pp=ygUZZ2F0ZSBkcml2ZSBzYW0gYmVuIHlhYWtvdg%3D%3D
https://www.youtube.com/watch?v=Aq1Iw6ByXAw&pp=ygUZZ2F0ZSBkcml2ZSBzYW0gYmVuIHlhYWtvdg%3D%3D

Anik , 01-24-2024, 08:02 PM

thank you so much sir. I have learned many things by this time. I have completed the schematic. If you can review It willbe proper clear for me what I have designed. I have used esp32 as microcontroller and here most of the device provide analog value output. There is no specific reason to use esp32 but I have done a lot project on it and it has a lot of analog IO. every pin has PWM capabilities. If you can take a look for a little time and make me aware about my faults in the design.

QDrives , 01-24-2024, 09:30 PM

Why don't you draw the mosfet as a mosfet?
What about current limitation/protection? Shortcircuit protection?

Anik , 01-24-2024, 09:38 PM

EasyEDA and LCSC parts supplier from JLCPCB combined them. So that all components of LCSC has its own symbol and footprint created and its can be found on the easyEDA software. Easy to use so that I didn’t change them.

I never have use efuse on circuit before. But I will add a fuse on it.

QDrives , 01-25-2024, 09:56 PM

I did not mention e-fuse.
Resistor, op-amp, optionally comparator...

Anik , 01-26-2024, 02:19 AM

Ok sir. But I have not sure how much current that can consume from the board. Do I keep a pot for controlling it? Like this

QDrives , 01-26-2024, 04:10 PM

The comparator goes to the SD# pin of your gate driver. Or better, to a flip-flop and then to the gate driver and also to the MCU.

Anik , 01-26-2024, 05:21 PM

I got it sir, thank you.
but for 24V 5A supply the shunt resistor willbe very big??

QDrives , 01-27-2024, 08:07 PM

I am using 4x 0508 resistors in a 48V 25Arms drive. They are 4m ohm each, so 1m ohm total. P = I² * R.
For your design a single 1206 10m ohm would do. Just make sure that the op-amp used has a low input offset voltage.

Anik , 01-27-2024, 08:10 PM

If I dont use protection and just keep a 6A fuse on the motor terminal will it be ok?? cause sir the this circuitary going costly and complex. j

QDrives , 01-27-2024, 08:14 PM

I cannot answer that question as I do not know the application.
I designed a motor controller for trains and (non-resettable) fuses were out of the question.
Besides, electronic protection is hardly and bigger.
In your design you are unable to measure anything from the driver, neither current nor voltage. How do you want to control the motor?

Anik , 01-27-2024, 09:00 PM

its from another person project, he only gave me he has 24V and 5A motor.

QDrives , 01-28-2024, 04:37 PM

Yes, but does not explain how you want to control the motor?

Anik , 01-28-2024, 04:39 PM

he told me he want to controll the motor speed 0 to 100% speed . Its probably a grear motor.

Anik , 01-28-2024, 04:39 PM

I have asked again for more details,I am waiting for his response.

QDrives , 01-29-2024, 08:00 PM

Speed is linear to the supplied voltage. Torque is linear to the current through the motor.
So you could 'control' the speed by varying the voltage (duty cycle).
However, if the motor needs to provide more torque, then the current will go up. I*R is the voltage you lose.