How can any TB6560 driver design NOT violate the chips spec's?

[lazzymonk]

has any one tried the single axis tb6560 boards? Im new to cnc and dont want to spend more than I have to but cant find anyone that has used, or even seen these before.

3A TB6560 DC Stepper Motor Single Axis Drive Controller | eBay

[James Newton]

Reading over the TB6560 datasheet again, I noticed something else:

When you look at the actual power dissipation, (page 6 on the data sheet)
http://www.toshiba-components.com/mo...3_20080407.pdf
it handles LESS than 43 watts (43 watts with an infinitely good heat-sink, which doesn't exist) which is only a bit over 1 amp at the max 40 volts. Even with a 24 volt supply, it will fry above 43/24=1.8 amps. Only using 19 volt power brick? It will fry at 43/19=2.25 amps. Actually, it will fry before that, even with a really good heat sink.

[doorknob]

James, just a question about interpreting the spec sheet with respect to the max 43 watts dissipation figure. I'm not sure that I agree with your calculations, but then I haven't completely scoped out the fine print in the spec sheet, and I may very well be missing something (and, as I mention below, I think that there may even be a problem with one of the equations in the spec sheet).

From page 28:

Power Dissipation

The power dissipation of the IC can be calculated by the following equation:

P = VDD × IDD + IOUT × Ron × 2 drivers

The higher the ambient temperature, the smaller the power dissipation.
Examine the PD-Ta characteristic curve to determine if there is a sufficient margin in the thermal design.


My interpretation of VDD and IDD is that it represents the power to the control section of the chip (at 5 volts and perhaps 5 mA), whereas the power dissipation related to the load would be based on IOUT times Ron times 2 (but then, I would have expected that power to be calculated as IOUT-squared times Ron times 2, so I'm wondering whether that formula is wrong).

As far as Ron is concerned, I would expect that to be fairly small, as I would expect it to be operating in a switched mode, not a linear mode - page 8 shows that as max of 0.4 ohms.

But I don't completely understand how the chip implements microstepping in the switched mode - does it act as a switch throughout all of the portions of the sine/cosine waveform generation? Probably yes, but I haven't yet figured out how it does that.

(edit: I guess that it does current limiting to generate the sine/cosine waveforms, which it obviously can do in switchmode)

If power dissipation inside the chip is really dependent on IOUT-squared times Ron times 2, then at 3.5A for I and 0.4 ohms for Ron gives a very small number which is not close to 43 watts.

(edit: I calculate that as approximately 5 watts times 2, for a grand total of 10 watts at maximum rated load current)

So, what am I missing here?

I can't stick around to discuss this right now, but maybe later tonight I can revisit it.

[H500]

Doorknob, I think their equation is wrong. Ignoring switching losses, the heat produced by the chip can be estimated as 2 x .5 ohm x 3.5amp x 3.5 amp x .7 = 8.5w

[doorknob]

Doorknob, I think their equation is wrong. Ignoring switching losses, the heat produced by the chip can be estimated as 2 x .5 ohm x 3.5amp x 3.5 amp x .7 = 8.5w

Is your derating factor of 0.7 because you expect that both windings will not be fully energized at the same time, or due to something else?

[neilw20]

I squared R. that's physics.
And the switching losses can be considerable and there is little information the spec sheet.
And if the flywheel diodes are conducting a high current occurs for the reverse recovery time of the diode. That's physics.
No specs on the diodes there I notices either.

There is also a section in the specs about a 'fast' fuse.
Considerable derating is required for a 'fast fuse'.
The term fast, is not really a good term.
The I squared T rating of a 'fast' fuse is much lower, so the fault current during the arcing phase as the fuse blows is lower.
A fast fuse once arcing can keep the fault current within possibly 10% of the rated value, but a 'normal/slow' fuse may arc at 200-500%.
The time the fuse arcs for is dependent on many things.

The spec sheet says use a 'fast', and looking at fuse specs, it is quite possible for the silicon to act as a fuse as many have discovered.

And then there is leakage inductance/capacitance in the steppers that will increase dissipation during switching. significant at chopping frequencies.

3.5 A is the absolute max, and considerable derating for normal operation is needed. As James says, you need a good heatsink.

You can't just multiply by 0.7 because of the times the windings are on.
That makes little difference. It is all about losses because the thing runs in switch mode, and both channels are conducting ALL the time.

The phasing of the switching can mean both conduct at the same time.
The phasing is a function of the current limiting at a given instance.

Test in real application, and you will see how big the heatsink needs to be!
I wonder why the spec sheet says 43W?

[H500]

The .7 is for the RMS current, since the waveform is sinusoidal.

[James Newton]

I believe that is the ability of the chip when used without a heatsink. Notice graph top of page 27, where the Pd-Ta line 3 for "Chip without heatsink" is about 5watts. I was talking about the maximum possible wattage with an infinite heatsink, which is line 1 on that graph. Line 2, which is for a 3.5'C per Watt heatsink (that's a pretty good heatsink) and which shows about 25 watts max. So these things are really designed for 1 amp at 24 volts or maybe 2 amps at 12 volts WITH a reasonable heatsink. Half an amp at 40 volts. It's a wonder they hold up as well as they do.

[doorknob]

My interpretation is that at a gross level the power dissipation calculation depends on the current through the output stage and the "on" resistance of the motor driver, independent of the voltage at which it is being driven, and independent of the heat sink. (The other factors that neilw20 mentioned are presumably operating as well.) Based on that calculation, the heat sink requirements can be estimated from the graphs.

But then I'm not an EE, just an annoying hobbyist, so YMMV...

[H500]

So these things are really designed for 1 amp at 24 volts or maybe 2 amps at 12 volts WITH a reasonable heatsink. Half an amp at 40 volts.

Voltage doesn't affect the chip's power dissipation much. It's a pwm driver. It will handle 3.5 amps peak, which is equal to 2.45 amps RMS per coil.

[htrantx]

Doorknob, I think their equation is wrong. Ignoring switching losses, the heat produced by the chip can be estimated as 2 x .5 ohm x 3.5amp x 3.5 amp x .7 = 8.5w

I would say it is a little presumptuous for jumping into that conclusion.

It is what it is, an equation to estimate the chip's PD, I would say. Don't stomping down too much onto an empirical formula provided in a spec sheet.

htrantx

[H500]

I would say it is a little presumptuous for jumping into that conclusion.

It is what it is, an equation to estimate the chip's PD, I would say. Don't stomping down too much onto an empirical formula provided in a spec sheet.

htrantx

What makes you accept it without question, other than the fact that it's written in the datasheet? Power is normally equal to I^2 *R.

[htrantx]

H500, you were right.

The equation in 11/13/2007 datasheet was a typo, and the spec was updated 03/24/09 and it is:

for 2-phase excitation:

P = VDD × IDD + (Ron(U + L) × Io × Io) × 2

[H500]

I simply glossed over the second X Iout.

But you're still placing too much faith in supposed authorities.

[RomanLini]

The "2 x .5 ohm x 3.5amp x 3.5 amp x .7" from the datasheet is a simplification of the "Rds ON" losses, which is one of the switching losses of a switchmode driver;
2 (two phases in a motor)
0.5 ohm (Rds ON resistance of the switching FETs)
3.5amp x3.5amp (current squared)
0.7 (average current in one sine driving "phase")

However this does not include the dynamic switching losses, these can often be as high as the static Rds ON losses. The dynamic losses are because the IC does not switch instantly from OFF to ON, it can take a microsecond or more which adds up a lot when
the IC is chopping at 20 to 50kHz.

Also, the Rds ON value of 0.5 ohm is likely to be optimistic and will rise a LOT if the chip is hot, ie under heatsunk. I think 0.5 ohm sounds very optimistic for a bipolar driver (where current must go through 2 FETs and their skinny little SMD power leads).

Don't be surprised if that "8.5W" ends up being quite a bit more...

[doorknob]

However this does not include the dynamic switching losses, these can often be as high as the static Rds ON losses. The dynamic losses are because the IC does not switch instantly from OFF to ON, it can take a microsecond or more which adds up a lot when
the IC is chopping at 20 to 50kHz.

But, assuming that the dynamic switching losses are comparable in magnitude to the static Rds ON losses, wouldn't the fact that those two losses do not overlap in time, but rather at any instant in time the switch is effectively operating in one regime or the other, mean that the average power dissipation might indeed still be comparable to the calculation of the static Rds ON losses?

(that assumes that the dynamic switching losses would be constant over the duration when they occurred, which might not be a defensible simplification, but then maybe the net result would be less, not more, than the Rds ON loss value, depending on the behavior of the dynamic losses with respect to time)

[H500]

Romanlini, I would not expect switching losses to be that high at 20khz. Modern switching regulars commonly go up to the megahertzs. The datasheet didn't list switching losses as being significant.

Doorknob, heat is dissipated slowly. The microseconds of time difference won't make a difference.

[doorknob]

Doorknob, heat is dissipated slowly. The microseconds of time difference won't make a difference.

Yes, but that is a different issue from determining the average electrical power dissipated in the chip.

For example, if there were instantaneous on/off switching with a 50% duty cycle, the average power dissipated in the chip would be different from a 25% duty cycle or a 75% duty cycle.

[H500]

The current waveform is sinusoidal when the motor is turning. This waveform is generated by the duty cycle of the FET. This is made possible because the motor inductance prevents the current from changing instantly.

The bottom line is that the power dissipated by each fet is proportional to the current squared.

[Hommersimpson]

So after reading all this stuff on the TB6560.. Is the simplest solution to cut the traces from the main power input and run a sep 5V supply to the logic and just make sure to power on the 5v first and power the 5v off last..I am talking about the 5 axis TB6560 ..blue ebay boards