Help me analyze this stepper circuit - Boss machines

Ok I hate to beat this issue to death, but I want to figure out why transistor failure is so common with the Bridgeport boss machines. Their has to be some major flaw in the circuit design. You almost never hear about it in any other machine.

Here is the circuit:

http://i264.photobucket.com/albums/i...027773crop.jpg

If I understand back emf correctly, the voltage will be opposite of the applied voltage. So while the coil is active, if the voltage at line 31 is 60v, and 0v at X1, then when the transistor is shut off for the leftmost coil, the voltage at X1 will be more positive than 31, say 120v. It appears to me that their is no protection for this. The diodes D33 and D29 would only be effective if the voltage at X1 were less than ground (line 39). Is this a flaw in their logic?

Most diagrams I could find on the net show a diode across the motor from what is normally the negative side to the positive. This would make sense as it would keep the voltage that the transistor would see at the collector capped at 60v. I modified the above diagram to reflect this:

http://i264.photobucket.com/albums/i..._resistors.jpg

But, wouldn't these diodes be a dead short across the coils when freewheeling? Effectively braking the motor? And probably frying the diodes? I added resistors inline with the diodes on the following diagram, but I have yet to find a stepper driver diagram with these.

http://i264.photobucket.com/albums/i..._resistors.jpg

Can someone tell me if I am in the wrong? I am having a hard time finding information on exactly what goes on with voltages during back emf.
This also assumes that the transistors are dying because of too high voltage across collector to emitter, and not base to emitter. I need to look at the SMD board to see what kind of voltage the base could be seeing.

Thanks!
Joe

I'm starting to think its not the collector-emitter voltage that is causing the failure, I think it is spikes in the base-emitter voltage.

I added part of the SMD board that controls the failing transistor to the drawing. This just seems like a bad idea. It appears when the Q9 transistor is off, the final drive transistor is on via the 24v supply. When the Q9 turns on, it turns the final drive transistor off by bringing the base voltage up to equal the collector voltage. So the base, which has a max voltage rating of 9 volts is exposed to the wild voltage from the stepper.
Anyone else follow?

http://i264.photobucket.com/albums/i...773cropSMD.jpg

Quote:

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Originally Posted by jderou

Ok I hate to beat this issue to death, but I want to figure out why transistor failure is so common with the Bridgeport boss machines. Their has to be some major flaw in the circuit design. You almost never hear about it in any other machine.

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The most common reason for the failure of power transistors in an H-bridge is shoot-through. This occurs when the switching times between the high side and low side transistors allow both transistors to be on at the same time. This is essentially a short to ground and it will blow at least one of the transistors.

The operating temperature of the transistor effects how quickly it will turn off. As they heat up, they turn off slower. The logic that determines when to turn on which transistors needs to allow for this.

The electronics available in the '80s were no where close to what is available now. It is likely this contributes to the problem also.

Shoot-through is a very likely cause.
The diodes you refer to being used on standard motor circuits cannot be used on a stepper as you are reversing the polarity while stepping. The usual protection arrangement with diodes for stepper needs two diodes per motor connection: one diode to positive supply(cathode on positive rail) and one to negative-ground(anode to ground). This arrangement clamps any voltages on the motor lines to stay within the ground to supply voltage window. (search forum or google for 297-298 driver examples to see how diodes are used)

Well, I decided to just cut the amperage to the z-axis (the main problem) to 6 amps, and lower my max feedrate on the axis. I also stopped using the manual switch for turning the spindle on, It seems that many of the transistors I have lost was while turning the spindle on manually. Don't know why. Possibly because I start jogging the machine too soon after I turn the spindle on manually (big voltage drop). Their is a 3 second dwell if I turn it on through the control.

Because English cannot understand it, I use an interpreter.
I cannot express it definitely.

(82) leads to the transistor of both groups in your figure and cannot distinguish coil tupple.
In the case of the first figure.
The Class motor coil plays to (39) by one resistance through two one set of transistors.
By my experience, I produce the voltage of the plus in the case of a uni-polar motor by cutting it.
It is considerably big, but I produce the voltage of the minus to the coil of the other side, and it is drawn up through resistance by a power supply.
The double power supply voltage occurs to (109) then.
Therefore, the breakdown voltage of the transistor needs more than 6 times of the power supply voltage.
（The normal breakdown voltage multiplies it by three）.
In the case of the second figure.
(109) The の voltage goes along the diode, and it is sent back by a power supply.
In this case only some voltage get nervous.
However, it occurs by torque to flow long time, and it is not it on brakes.
However, time to change the direction of the electric current hangs when I did a reverse transistor to be next to enter.
（As a result, movement becomes slow）.
When the voltage occurs for resistance in the third circuit when I correct an electric current, and it did it depending on resistance value when it is it to double, zero voltage appears in a thing of a reverse coil.
This becomes equal with the first circuit.

Wow, that's a lot of information. It may take me a bit to digest it :)
Thanks!

I can drive uni-Paula stepping by two circuits.
I hope to replace the inside of the dotted line with MPU.
There are three modes.
One red -NchFET-ON, electric current is early and increases.
Two magenta -F Di,PchFET-ON, electric currents decrease slowly.
Three blue -b Di-ON, electric currents are early and decrease.
The connection diagram does not consider dead time.
When there is a drive signal of SW1; one mode.
A case beyond the analog voltage, two modes.
When there is not a signal of SW1 and SW2; three modes.

The web page is Japanese.
I added some circuit simulation.
http://den-nekonoko.blog.so-net.ne.jp/2009-04-19