Hello, I made an all NPN type H-Bridge using IRL22303.

Things works well aside from the these two things:

1. When one of the upper bridge is turned on, the side of the upper bridge also turns on.
2. Same scenario with the lower bridge.

Here is my schematic diagram:

 

Even if there is no control drive to either Q1 or Q3 aren't these biased on by the presence of R5 & R8 tied to the 12V rail and feeding same to the gates of Q1 & Q3?

 

I see... so what should I modify? Can I just remove R5 and R8? or do I have to attach it somewhere?

 

You might consider tying them to ground. But you might also need a voltage greater than 12V to drive the gates sufficiently high (to get a high enough Vgs) when you actually want to turn the Mosfets [Q1 or Q3] on via the opto-isolators.

Have you tried Googling "all N-channel mosfet H-bridge" to see how others have done this?

 

Thank you!

There is no problem driving my IRL2203.

My hardware worked when I attached my R5 and R8 to the GND. xD

Though there is almost a 3V voltage drop when I measured the voltage on my motor. Will that be okay? I am running on a 12V/10A battery. Do I have to put a heat sink then?

 

Voltage drop where?
You need at least 10-15 Vgs to turn the high side on, otherwise you will burn the mosfets. That means you have to supply the high side mosfets with about 24V total on the gate to turn them on.

 

My IRL2203 turned on. using the schematic I posted having R5 and R8 as pull down resistors instead of pull up. I can successfully drive a motor forward, reverse, and even with PWM.

I am using a 12V/9A battery, and when I measure the voltage on the motor itself, it is 9V. So the 3V will become heat I guess. My IRL2203s are acting normal, they do not have any heat even turned on for 15 minutes.

 

You need to be a bit more thorough with the measurement. Does it work the same for both directions? What is the battery voltage on load? What is the motor current?
You should measure the voltage drop on the transistors directly, rather than substracting.

 

Yes the speed is similar may it be forward or backward. The transistors give minimal heat when we turned on the motor for 15 minutes. We measured the motor voltage, it is 9.xxV whereas our supply is 12V. The current on the motor is 7A, our supply is 9A. How was that?

 

You mean you measured different current on the supply? It could be shoot-through, so try measuring the current through the off transistors.

 

I had a similar discussion many moons ago, some of it might help...

High Side MOSFET Drivers

BTW, referring to those components as NPN marks you as a beginner. NPN refers to a BJT (bipolar junction transistor). No big deal, we like beginners here.

The core concept is there is a capacitive voltage multiplication going on for the high side drivers, to boost the voltage over 10V from the source. The voltage on the gates is around 20-24VDC. This is absolutely critical, as has been mentioned.

The other thing is use of a specialized chip to prevent shoot through. The idea is the top one is turned off before the bottom MOSFET is turned on, and visa versa. It would take a lot of extra circuitry to do the same thing, which makes the IC economical.

 

I have to disagree sir
If it is driving a DC motor, the upper and lower transistors don´t alternate fast, so shoot through can happen only if the transistors are wrongly biased or driven.

 

I have checked the transistors, they are doing what they supposed to. Q1=Q3 and Q2=Q4.

 

The IRL2203N's are logic level MOSFETs. In order to turn them on, you need to have Vgs (voltage on the gate using the source terminal as a reference) at anywhere from 4.5v to 10v, or they will dissipate power as heat, and you will not get full voltage to the motor.

The trouble is that you just have the 12v battery supply. Without a voltage source higher than 12v, you won't be able to turn on the upper MOSFETs completely.

 

Exactly. I have a 3V loss, someone knows how to compute about power to heat conversion?

 

Power in Watts = Voltage x Current.
You have a 3v drop.
What's the current?
hint: it's going to change considerably from when the motor is in a stall condition (rotor locked; high current) to when there is no load on the motor and it is at speed (lowest current).

 

I have 3V drop and approximately 1A-2A drop. How was that? Do I have to put some heat sink? a fan?

 

3v loss using a mosfet with an Rds-on of 0.007 ohm...
1/4 of your power is lost on the mosfet as heat, if this is not an indication of improper driving of the mosfet then I wonder what is.

I have explain very clearly the difference of wrong driving (in your edaboard forum post) where you get 3v voltage drop


and correct driving where you get 0.05v voltage drop

You can either do it with an isolated power supply, an available power supply with higher voltage from the one that the mosfet uses or using bootstrap technique.

And your answer was

There is no problem activating the IRL2203

and then your final comment

Well, I thought I was receiving pieces of information to help me with my circuit. But it turned out to be NOT

You are welcome
Alex

 

It is because you always give me theoretical elaboration on things where in fact, the error I was made on the schematic was I placed the resistors on the upper bridge as pull up instead of pull down.

I have 1V+ voltage drop directly on my transistors. The 3V drop is included upon implementing the other components like the resistors.

I was looking for a straightforward answers merely before I study things.

Anyway, we turned the h-bridge on the breadboard for 30 minutes. Still, cold MOSFETs are showing.

 

3v*1A = 3 Watts of power wasted.
3v*2A = 6 Watts of power wasted.