Hello,

I'm using the Ardumoto from SparkFun to control a stepper motor.
It have a L298 dual full-bridge driver and protection diodes on board.

Somehow I manage to break one transitor from the L298 full-bridge driver. The transistor that fails is marked in red.

I'm not sure what the cause is, I can think of 2 possibilities:
- 1.7A current DC operation
- Manual trying to move the rotor in the reverse direction of its movement

What is more likely the problem?

Thank you.

 

First things that come to mind are not having some sort of snubber circuitry on the device it's driving, second is that it's switching far too fast.

 

The protective diodes are the snubber circuitry right?

I'm driving the stepper motor in 4 steps, the fastest switching that I have tried is 250Hz (1ms per step).

 

What is the part number of the diodes that you are using?
They must be either fast-recovery or Schottky diodes.

The SparkFun board doesn't seem to have much in the way of heat sinking. If you're trying to put 1.7A current through the L298, you may very well have overheated the IC.

 

I'm not very sure about the diodes, but I guess they are the right one for the job. On the body of the diode is printed "S403", with a bit of Google-ing I think it's the High Voltage Switching Applications Diodes from Thoshiba.

Overhitting could be the problem, I did feel the case with my finger while working, it's indeed hot, but I can stay in contact with it the whole time. The total power absolute rating is 25W. Vin for the stepper motor is 12V. Vss for the driver is 5V.


Is "manual trying to move the rotor in the reverse direction of its movement"
is not really a problem then?

 

The 1SS403 diodes are fast-recovery type, but are only rated for 100mA average current. I would not be surprised if you found the protection diode for that transistor is burned open.

You really need a diode rated for 1A or more.

What is the steppers' rated voltage? The L198 will drop roughly 4v across itself with 2A current flowing through it.

 

I'm sorry, I'm not really sure what type of diode it's. Testing with the multimeter I know that all the diodes are still good. And the forward voltage is 0.17V .

The steppers' ratings is (for one coil):
4.1V and 1.1A
3.6ohm
1.8 deg/step

I'm powering 2 coils per step, bipolar stepping.

 

So, what are you using to limit the current?

The L298 will drop about 4v across itself; since you're using 12v for the supply, there's 8v left across the motor. 8v/3.6 Ohms = 2.2 Amperes, or about double the motors' rating, and exceeding the L298's rating by 10%.

Holding the motor still or turning it in reverse causes the motor to be operating at stall; maximum current.

 

Just being curious...is post one a picture of the insides of an L298, and how can a person see into an L298 to see exactly which transistor is bad?

 

Did you stop the motor while it was running and try to turn it backward?

Stalled motors draw a great deal of current.

I'm curious how that transistor was isolated as well. A new L298 is needed no matter which part(s) failed.

What is the load on the motors? Is there a heatsink on the driver IC?

 

Just being curious...is post one a picture of the insides of an L298, and how can a person see into an L298 to see exactly which transistor is bad?

If OUT2 can source current but not sink current, that would be a good indication that the lower transistor or it's driver circuit has failed.

 

So, what are you using to limit the current?

The L298 will drop about 4v across itself; since you're using 12v for the supply, there's 8v left across the motor. 8v/3.6 Ohms = 2.2 Amperes, or about double the motors' rating, and exceeding the L298's rating by 10%.

Holding the motor still or turning it in reverse causes the motor to be operating at stall; maximum current.

Since I'm powering over 2 coils, thus the voltage is over 7.2Ohm, the current is then 12/7.2 = 1.7A . Yes, that's still over the recommended rating for the motor. But it's the transistor that fails, not the motor.

Just being curious...is post one a picture of the insides of an L298, and how can a person see into an L298 to see exactly which transistor is bad?

Yes, that's the block diagram of the L298. You can turn on a transistor with the inputs In1, In2, EnA. After turning on the transistor you can measure the voltage at out1 and out2 with respect to ground.

Did you stop the motor while it was running and try to turn it backward?

Stalled motors draw a great deal of current.

I'm curious how that transistor was isolated as well. A new L298 is needed no matter which part(s) failed.

What is the load on the motors? Is there a heatsink on the driver IC?

Yes, I tried to stop the motor while it's running and turn it backward , and it does turn backward. But I didn't do it for to long.

So when the motor is stalled the current increase? It's not just Ohm law anymore?

There is also no load on the motor.

There is no heatsink. Maximum temperature rating of the L298 case is 75 Celsius.

 

The die of the L298 will heat more quickly than the package.

8v/7.2 Ohms = 1.111... Amperes. How did you arrive at 1.7A?

At 1A, the total L298 saturation voltage will be somewhere between 1.8v and 3.2v. Even if it was ~2v, leaving 10v/7.2 Ohms, you'd only have 1.39A current.

 

Sorry, I was very wrong. I thought all the supply voltage Vs is for the motor, because the driver already have the Logic Supply Voltage Vss for it's working. I forget that there is also voltage drop Vce at the 2 transistors.

I just did some measurements:
Vs : 11.45V
Vout1_out2 no load : 10.30V
Vout1_out2 stepper load : 8.62V

 

So, if Vs holds at 11.45v, then you're dropping 2.83v across the L298. 8.62v/7.2 Ohms = ~1.2A (rounded up a tad). 2.83v * 1.2A = 3.4 Watts, x2 = 6.8W power dissipation in the L298.

From the datasheet:

Top (junction operating temperature) = -25° to 130°C
Rth j-amb in the PowerSO20 is 13°C/Watt, IF mounted on an aluminum substrate.
So, 6.8x13 = 88.4
130-88.4 = 41.6°C (~107°F) maximum ambient temp IF the package has adequate heat-sinking.

However, it's probably more like 35°C/W since I don't see a heat sink anywhere.
35*6.8 = 238
130 - 238 = -108°C maximum ambient temp, or -164.2°F. That's mighty chilly.

 

So it's the heat? :-D

Thank you for all the calculations, I really appreciate it.
But why x2 = 6.8W ? out1 and out2 is enable only when out3 and out4 is disable and vica verca.

 

OK, if you're using full step, then you'll only have one channel on at a time. If you're using half-step, then you will alternate having both channels and one channel on at a time.

So, power dissipation will be either 3.4W for full step, or alternating between 3.4W and 6.8W when half-stepping.

Go through the formulas with the alternate calculations and see what you get. I get 11°C maximum ambient temp for 3.4W with no heat sink; about 52°F - but that's with a MultiWatt 15 package. It appears that yours uses the PowerSO20 package, and I can't tell if it's mounted on an aluminum substrate like the datasheet calls for. If not, I'd be willing to bet your IC will perform worse than the MultiWatt 15 package - the datasheet doesn't go into that.

 

Not to mention the fact that an inductive load can appear as a dead short for the first few milliseconds regardless of the DC resistance it measures out as.

 

Oke. So if I respect the recommended values and provide adequate cooling, manual moving the rotor in the reverse direction of its' movement is not really a probleem?

 

Why do you want to interfere with the shaft on a running motor?

What purpose could it be serving?