DRV8825: DRV8825 absolute current rating

Hi Elysaf,

1. In my circuit i designed " i chop" to 3.5A. I saw on every phase 2.5A. Dose the reason is that in full step every phase current is 71% so 3.5A*0.71=2.5A?

Yes. Full step operates at 71% of the full scale current setting (Ifs).


2. The absolute current rating is 2.5A continues. In "full step mode" it's means that "I chop" should not pass 3.5A so on every phase would be 2.5A (3.5*0.71=2.5) , or " I chop" should not pass 2.5A so on every phase would be 1.775A (2.5*0.71=1.775).

The absolute continuous current rating is at 25C ambient and with proper heatsinking. As the ambient temperature increases, the continuous current decreases.

3. My profile operation is 1.5 SEc ,one time operation, in 4KHz step . The absolute current rating is 2.5A is for continues working. I suppose that 2.5A absolute current rating is for continues working . Is it possible to operate with more current if the operating time is only 1.5 sec one time operation?

It may be possible, but it will depend on the ambient temperature and heatsinking. Also, the overcurrent protection can activate at 3A.

If you need more current, please consider the 2x DRV8840 or 2x DRV8842. These devices can drive up to 5A, but require two devices.

4. If so how to calculate how much current could i use?

This is a thermal transient event. Please refer to www.ti.com/.../sloa104.pdf to determine how to measure this. The value will depend on the heatsinking (ground plane) and solder coverage to the thermal pad.

Hi
Thanks about your answer.

I didn't understand the answer about"2". in data sheet its written that the absolute current is 2.5 A.
do you mean 2.5A "i chop" or do you mean 2.5A on the phase. (71% from I chop)

also about "3" you wrote that "Also, the overcurrent protection can activate at 3A." 3A "ichop" or on phase
Elyasaf

"also about "3" you wrote that "Also, the overcurrent protection can activate at 3A." 3A "ichop" or on phase

>>> If Ifs is set such that the current could exceed 3A, there is an internal overcurrent protection circuit that could activate to protect the FETs as low as 3A. Please refer to Figure 6, and note the OCP paths."

Without referring to the problem of the  power dissipation (short activation) .I understand that  my " I chop"  could be 4A , the current on phases will be 2.85A. (71% FULL STEP MODE)', and the " internal overcurrent protection" would not activate?

Hi Elyasaf,

Yes. If power dissipation is ignored and Ifs is to 4A with full step mode, the device will attempt to regulate the current at 2.85A.

But power dissipation at 2.85A is at least 8W during the time the outputs are enabled. The results will depend on the heatsinking connection, the ambient temperature, and the RDSon variation of the FETs.