DRV8412: Power Calculation to estimate IC temperature

Hi Kevan,

Power dissipation of the IC can simply be given by I_rms^2 * R_DSon, since most of the power would be dissipated in the switches. (neglecting switching losses)

Assuming that you're using it for motor drive application, your torque/ acceleration would be proportional to the I_rms and you can directly estimate the losses when accelerating.

Also assuming that you're braking dynamically, the power dissipated at that time would simply be the energy lost by your motor flywheel in the braking process.

Once you have a rough estimate of the power dissipation, the temperature estimation becomes fairly simple since the thermal resistances are provided in the datasheet.

Thanks and Regards,

Avish Gupta

Hi Avish Gupta,

Thanks for your prompt reply.

My RMS current is low 0.3A, and with one H-Bridge active I used 2 * RDS(on) as resistance. This gives power as 20mW and temperature rise as 0.5C (at 25C/W) but I get much higher IC temperature rise (+45C after 300s). My applications isn't motor drive - I am driving a 1.6mH inductor from 24V with a 0.08 duty cycle at a repetition rate of 333Hz.

Are there other IC losses that I need to consider?

Regards

Kevan 

Hi Kevan,

Avish is correct. The datasheet typically suggest using  (I_rms^2) * (2 *R_DSon) as a power estimate. This is generally a good starting point when the FETs are the primary heat source.

When the RMS current is low, other sources and factors may become more significant.

The GVDD_x and VDD currents.
The quality of the connection from the thermal pad to the PCB
The size of the ground plane
Switching losses of the FETs (at 333Hz, this is probably negligible).

Are you using the second H-bridge?
If not, please consider using the device in parallel mode to reduce the RDSon further. This will require additional components to connect the two bridges (ferrite beads as recommended in the datasheet).

Hi Rick,

Thanks for your help.

We are using the second H-bridge but it is inactive during this measurement.

The temperature rise is orders of magnitude larger than the I*I*RDS(on) calculation. How can this be explained?

What are the heating effects from the GVDD_x and VDD currents?  

Any other factors that can increase RDS(on) significantly?

Is there a Spice model available for the part?

Regards

Kevan

Hi Kevan,

In your application, the many sources of power are the PVDD, GVDD_x, and VDD currents.

What is the PVDD voltage?

Quiescent losses for PVDD can be calculated as (PVDD * (4 * Ipvdd))

The GVDD_x and VDD currents have an impact on the heating, as do the PVDD_x currents when idle.

Quiescent losses for VDD can be calculated as VDD * Ivdd

Quiescent losses for GVDD can be calculated as (GVDD * 1.7mA * 4 ) +  (GVDD * 6.3mA * (2 * number_of_RESET_xx_pins_high)).

The 1.7mA is the reset mode current, and the 6.3mA is the difference between reset and operating mode.

Assuming PVDD = 24V and 1 RESET_x pin is high, the power due to the PVDD, GVDD_x, and VDD currents is ~426mW. This would be roughly 10C increase.

There is no Spice model available for the part.

Hi Kevan,

Kevan Johnston said:

Hi Avish Gupta,

Thanks for your prompt reply.

My RMS current is low 0.3A, and with one H-Bridge active I used 2 * RDS(on) as resistance. This gives power as 20mW and temperature rise as 0.5C (at 25C/W) but I get much higher IC temperature rise (+45C after 300s). My applications isn't motor drive - I am driving a 1.6mH inductor from 24V with a 0.08 duty cycle at a repetition rate of 333Hz.

Are there other IC losses that I need to consider?

Regards

Kevan 

Can you please elaborate your application.

What happens during the off time?

Do you de-energize the inductor through the freewheeling path?

Thanks and Regards

Avish Gupta

Hi Avish Gupta,

Thanks - how would the inductor be de-energized through the freewheeling path?

Regards

Kevan

Hi Kevan,

If you freewheel long enough, you'll end up dissipating all the stored energy through the solenoid resistance and switch resistance.

Thanks and Regards,

Avish

Hi Kevan,

Was your question answered?

Hi Rick,

Perhaps, need to take some more measurements on my test rig. Still interested to know what factors will effect the RDS(on) value.

Regards

Kevan

Hi Kevan,

Can you share your schematic with the component values. I suspect the bootstrap circuitry is causing excessive heating.

Thanks,

Avish

Hi Avish,

Thanks. 

Each BST pin has a 220nF, 25V  capacitor connected to it's OUT pin. PVDD is +24V, GVDD is +12V, M1 = VREG, M2,M3 = 0V (AGND)  

Regards

Kevan