cancel
Showing results for 
Search instead for 
Did you mean: 

Clarification in datasheet of VNL5300S5TR-E

Bitesh
Associate II

I'm planning to use 24V 0.5A Motor in application. It will be controlled from MCU which is taking 3.3V input and it's GPIO will give 3.3V when output pin is logic high. I'm thinking to interface the Motor to MCU via driver VNL5300S5TR-E.

Now I have doubts in this driver VNL5300S5TR-E.

  1. This VNL5300S5TR-E IC is running on 3.5V to 5V. so I'm giving Vcc pin to 5V but input pin of this IC will be driven from MCU which will make input pin of IC to 3.3V.

so 3.3V voltage is sufficient to turn on the Internal MOSFET ?

2.There is a truth table 14 in the datasheet which says that STATUS Pin will get low for condition Output voltage < VOL. What is this condition, can you explain in detail.

3.Also what is this Openload detection table 10. in the datasheet. Can you tell more about this.

Regards,

Bitesh

1 ACCEPTED SOLUTION

Accepted Solutions

Sorry, I had mistyped point 3, but had already corrected it above.

There is a small leakage current that can flow from the output (drain) to GND, which can be compared to a high-impedance resistor. This "resistor" pulls the output to GND level when there is no load on the output or it has been disconnected and the MOSFET is in OFF state (INPUT = L). For details on open-load detection, see AN4798, section 1.6.

Regards

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

View solution in original post

6 REPLIES 6
Peter BENSCH
ST Employee

Welcome, @Bitesh​, to the community!

  1. VNL5300S5-E is not only a MOSFET, but also contains a complete MOSFET driver including the diagnostic function in addition to protection functions. The pin INPUT can therefore be driven in the entire range of Vsupply (3.5-5.5V) with the levels mentioned in table 9: VIL ≤0.9V, VIH ≥2.1V, which means that INPUT can be driven with a 3.3V MCU, even with TTL compatible levels.
  2. Table 14, Output voltage < VOL means: STATUS = active (L) when INPUT = L and no load connected or interrupted.
  3. Table 10, VOL: this is the limit below which a missing load (open/disconnected, interrupted) can be detected if the INPUT is simultaneously switched active inactive (L).

Does it answer your question?

Regards

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

Hi Peter,

Thanks for your response. Point 1 is clear to me now. but I still have confusion in Point 3.

Do you mean Output Voltage as Voltage on the Drain of MOSFET ?

If yes then I expect that when Input Voltage is LOW and No Load is connected then the Drain of MOSFET will be floating and reading some random noise. In this case Drain Voltage will be less than VOL and STATUS Pin will go LOW.

Is my understanding correct ?

Sorry, I had mistyped point 3, but had already corrected it above.

There is a small leakage current that can flow from the output (drain) to GND, which can be compared to a high-impedance resistor. This "resistor" pulls the output to GND level when there is no load on the output or it has been disconnected and the MOSFET is in OFF state (INPUT = L). For details on open-load detection, see AN4798, section 1.6.

Regards

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

Hi,

Thanks for the answer, now it is clear to me.

Regards,

Bitesh

Hi,

Can you please tell me what is the maximum voltage that can be applied between Drain and Source of MOSFET when Input is LOW ?

Regards,

Bitesh

This is listed in the data sheet as Vclamp (section 2.3) and has the value 41V (min), 46V (typ) and 52V (max). I would therefore recommend not applying a voltage greater than 41V there.

Regards

/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.