DRV5013-Q1: Output Low level voltage

Hi Cafain,

I will need to check internally with our team on this.  I expect to be able to provide a response by Thursday.

Regards,

Mekre

Hi Cafain,

There is still one more thing we have to verify.  I expect to have an answer Monday or Tuesday at the latest.  Sorry for the delay.

Regards,
Mekre

Hi, Mekre san

Thank you for checking.

If you refer to your answer, you will get the following results.
When Vcc=5V and Io=10mA, Vol=50mV, so Ron=5Ω.
When Vcc=24V and Io=10mA, Vol=240mV, so Ron=24Ω
Is it OK to understand this calculation?

However, from the datasheet(page6), when Vcc=3.3V and Io=10mA, Ron=50Ω, so Vol=500mV!?
Is this idea wrong?

Should I consider Vol=50mV at all temperatures if Vcc=5V and Io=10mA or less, regardless of the formula?

best regards
cafain

Hi Cafain,

To elaborate, R_on should not vary too much across current unless the current goes above 10 mA since that would start activating the overcurrent protection feature on the device. I am not saying to design with 10 mA in mind.

Since R_on shouldn’t vary that much if the current is less than 10 mA, you could use the VCC/(R_pullup + R_on) x R_on formula that you mentioned in your first post.

I’m not sure I understand the equations in your first paragraph. Could you provide more details on your method of calculation?

Regards,

Mekre

Hell Mekre san

Thank you for your support.

In the previous question, I referred to e2e below.
e2e.ti.com/.../603195

When Ron=50Ω at 5V, (Pullup=1kΩ) Vol=50mV is answered.

It does not reach 50mV when calculated.
Calculated as 5V/(1kΩ + 50Ω) x 50Ω = 238mV !

It can be calculated as Vol=238mV.
But the answer is 50mV.

Back-calculate Ron to get Vol=50mV.
Assuming Io=10mA
It can be calculated as Vol=10mA x 5Ω=50mV.

※To calculate 50mV, the pullup resistance value and power supply voltage cannot be considered.

Please tell me how 50mV was calculated.
(The same applies when V = 24V Vol = 240mV)

Regards
cafain