Toshiba TB6600

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Originally Posted by lucas

- The diodes on the my THB6064 design didn't have much effect, this chip is similar so I don't think you will gain much, maybe a few degrees temperature reduction, not worthwhile to install them.

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In the datasheet TB6560 or TB6600:
I do not see, if the IC has included (buildin) the diodes.
THB6064 included buildin the diodes?

Why would anyone in their h-bridge design include diodes?

Why desiden others that are not important?

Using them Prolongs transistor ?

Do not use them, gradually deteriorates transistors?
soft deterior?

Do not include a way to slowly deteriorate the circuit, so that at some point the customer having to buy another?

Exactly what happens and we're talking about?

Well I do not know much about electronics,
I read a bit, and then hise some calculations:

With 3Amp motor and 7 mH.

Time off transistor, switching transistor Output characteristics (time tf)
TB6560 is 1.0 ms = 0.000001 sec.
TB6600 is 0.5 ms = 0.0000005 sec.

The spike volt = H * A / sec.

With TB6560 = 0.007 * 3 / 0.000001 = 21,000 Volts
With TB6600 = 0.007 * 3 / 0.0000005 = 42,000 Volts

Please review these calculations, it could be a mistake.
I do not know whether that voltage is enough to make a arc.

I do not know how to calculate the duration of the peak,
I would like to know how to calculate it.
Also like to know how to calculate the peak current (ampere).
Maybe then I could choose a diode or decide not diode.
What is the best diode for this?
How as you do for deciding not to install diodes?
you, as you do?

I wish I knew.

My doubts are multiplied exponentially. -LOL-

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Originally Posted by riphet

Why would anyone in their h-bridge design include diodes?

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Have a look in the datasheet at the current paths through the components in all modes.
You will notice that in some cases there's current flowing through the internal diodes integrated in the mosfets's. Their spec's are not really good for some chip's: not fast enough or to high VF.
This creates additional power to be dissipated.

This current will flow through the external diodes when good and fast ones are used and this will reduce the chip's heatdissipation.
There are other reasons also but heat reduction is the most important.

Quote:

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Originally Posted by lucas

Have a look in the datasheet at the current paths through the components in all modes.
You will notice that in some cases there's current flowing through the internal diodes integrated in the mosfets's. Their spec's are not really good for some chip's: not fast enough or to high VF.
This creates additional power to be dissipated.

This current will flow through the external diodes when good and fast ones are used and this will reduce the chip's heatdissipation.
There are other reasons also but heat reduction is the most important.

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Indeed in the datasheet is written:

"Note: Parasitic diodes are indicated on the designed lines. However, these are not normally used in Mixed Decay mode. "

"As shown in the figure above, an output transistor has parasitic diodes.
Normally, when the energy of the coil is drawn out, each transistor is turned ON and the power flows in the
opposite-to-normal direction; as a result, the parasitic diode is not used. However, when all the output
transistors are forced OFF, the coil energy is drawn out via the parasitic diode."

Only when transistors are forced OFF...
That's like a car makes a journey very well, only heats when turned off.

I understand that: Tturn off the last step of each half cycle,
As off twice per cycle. This is in any mode.

Or is it the case that when it passes through zero, some transistor is turned on?

My idea is not to dissipate heat, but that energy regenerating prosecute. I understand that even the existence of regenerating ohm causes heat. But at least most regenerated.

That's my idea, use the correct diodes.
But how fast is fast?
So, I'm looking to calculate the duration of the peak. And also current.

I appreciate any related information.

Quote:

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Originally Posted by James Newton

That TB6064AH board, chip and a kit of all parts is also available in the USA from:
Stepper Motors

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At this time, no more eBay tb6600 drives for sale.

If I needed a driver now,
without hesitation,
this is the buy.

I would order only sold me with "diodes includeded" option :) ,
for induction peaks 7 ~ 9mH. 3 ~ 4 Amp.

No current will flow through the external diodes unless you're in the slow decade mode with the proper transistor turned off. That is not normally a desirable mode of operation. Unless your chip supports that mode, the diodes won't do much.

I think you are trying to calculate irrelevant parameters.

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Originally Posted by H500

No current will flow through the external diodes unless you're in the slow decade mode with the proper transistor turned off. That is not normally a desirable mode of operation. Unless your chip supports that mode, the diodes won't do much.

I think you are trying to calculate irrelevant parameters.

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I can I respect you, that:
You think anything is irrelevant and insignificant,
I do not approve but I respect it, so be it for you.

You should be powerful, capable of despising anything.
You are ignorant of the answer to my questions, then decides to belittle. I understand that you are not obligated to answer my questions. Anyway, I appreciate your time and attention.

I am interested, because I think you referred to earlier warming, should not be solved only with heat sinks. And that energy, it also damages the chip, can be recycled.

More, I can save the energy needed for the fan heatsink.

Then,
How I can calculate the correct mode of operation of the chip?
How to calculate the heat of the chip?
What is the origin of this warming?

I'm not belittling anything. I'm just telling you that you are not calculating the right things. Most of the heat comes from the I^2 *R losses through the mosfets. The datasheet tells you how to calculate that.

Read up on decay modes. The external diodes won't carry any current unless you turn off the proper fets during decay.

Quote:

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Originally Posted by H500

I'm not belittling anything. I'm just telling you that you are not calculating the right things. Most of the heat comes from the I^2 *R losses through the mosfets. The datasheet tells you how to calculate that.

Read up on decay modes. The external diodes won't carry any current unless you turn off the proper fets during decay.

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Yes, it is that you only refers to the heat generated by the IC.
Ignoring kickback , which is to cut the power to the coil.

As explained above, this can reach over 42Kv. the example presented.

The energy stored in the inductor is: W = (LI^2)/2.

Data above example: 0.007*9/2 = 0.0315 Joules
Energy that is returned to the mosfet in just only one kickback.

The heat generated by the kickback (internal arc), must be multiplied by the frequency step *2/8 (1/2 step).

Of course we have to add the heat generated by the IC itself.

All these calculations have not comprovado could be wrong.
But I find it more than 26W minute @ 500RPM, 400PPR, half step, 3Amp, 7MH. Only kicks back. You need to add the heating of the chip.

From the tb6560 datasheet:
"Power Dissipation
The power dissipation of the IC can be calculated by the following equation:
P = VDD × IDD + IOUT × IOUT × Ron × 2 phases"

But I have not completely clear this matter, I am no expert in electronics, I would like to know more about this.

For all the above, I think that energy dissipated as extra heat can be recycled and stop damaging the ICs. What do you think?

I appreciate any related information.
(I also appreciate any kind of scholarship or sponsorship. May be offered give me:)

The question is:

On "step zero", What is the state of transitors?

How I can check?

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Originally Posted by riphet

The energy stored in the inductor is: W = (LI^2)/2.

Data above example: 0.007*9/2 = 0.0315 Joules
Energy that is returned to the mosfet in just only one kickback.

The heat generated by the kickback (internal arc), must be multiplied by the frequency step *2/8 (1/2 step).

Of course we have to add the heat generated by the IC itself.

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Power = 1/2 *L* I^2 gives the instantaneous energy stored in an inductor as the current ramps up to the set value. You are incorrectly assuming that this energy is discharged into the chip when the transistor turns off. The current only drops slightly during the PWM off time. It shows up as ripple on the waveform. The heat energy dissipated in the chip continues to be I^2 *R, even during a step cycle when the current changed direction.

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For all the above, I think that energy dissipated as extra heat can be recycled and stop damaging the ICs. What do you think?

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It cannot be recycled. The current needs to flow through the mosfets or diodes. The loss can be reduced by using lower resistance fets.

Quote:

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Originally Posted by riphet

The question is:

On "step zero", What is the state of transitors?

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Quote:

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How I can check?

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Measure the voltage across the coil.

Quote:

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Originally Posted by H500

Measure the voltage across the coil.

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Excellent,

I have no instrument to measure peaks.

Anyone have any examples of measurements in IC toshiba drives,
with and without diodes diodes?

(TB6560 and TB6600)

The data sheets do not indicate if they have diodes inside.
Less information on the voltage of the diode.