TLV61070A: TLV61070A

Hi Singh,

The output capability can be calculated as:

If 3.6A Iout is applied, the device will act like:

Best Regards,

Travis

Hello Travis,

Thanks for ur replies.

2) I calculated the Iout based on the equations in section 7.3.4 of the datasheet. It turned out that this regulator does not supply enough Iout. The max. value is 1.05A for our smallest motor (with 2.4V Vin and 5.5V Vout). So, I searched and found another regulator TPS61023, which has higher Valley Current limit of 3.7A. Based on similar calculations, the max. Iout is 1.65A. I can use this for our smallest motor which has rated current of 0.4A, and Stall Current (Inrush start current) of 1.6A - 2A range. So, it is possible that Inrush current may slightly exceed the 1.65A calculated value.

U had pointed out to section 7.3.7 of the datasheet as the answer. But that is not applicable here. That section is for Short-to-Ground. In case of smallest DC motor, Vout (from regulator) = 5.5V, and max. Stall Current is 2A. That means the resistance of the stator winding =   2.75Ohms. This is not a short. The entire 5.5V will appear across the winding initially. As the motor starts to rotate, the back-emf will build up and the motor current will decrease to 0.3 - 0.4A.

However, the section 7.3.4 is again applicable here. It states:

"When the load current is increased such that the inductor current is above the current limit within the whole
switching cycle time, the off-time is increased to allow the inductor current to decrease to this threshold before
the next on-time begins (so called frequency foldback mechanism).

Please Note, it says "frequency foldback mechanism". As I understand, the regulator will limit the current to its max. value of 1.65A by increasing the Off time.  1.65A is acceptable to us. Within few milliseconds, after the motor rotates, the current will decrease to the rated current of 0.3 - 0.4A anyway.

Is my interpretation correct? Since u r an expert at TI, please reply. Thanks a lot for ur advice and time. AB

Hi AB,

1. All of our boost converters has current limit operation. This means when you apply 2A on TPS61023, the input current will be limited at 3.7A and Vout will drop because of insufficient input power. I don't know about your motor driver so you'll need to confirm whether you can accept the voltage drop during the inrush.

2. About short circuit protection feature: The function is activated when Vout is dragged below Vin. Our IC does not know or care how the Vout gets below Vin. So there'll be no difference whether you short or apply heavy load which the IC cannot support.

3. The current limit operation actually looks like this (the waveform is not captured on TPS61023 but they shares the same behavior):

When Iolad increases, the inductor current will increase to maintain Vout at the target Voltage(5.5V for your application). But if the Iload keep rising and the inductor current reaches the current limit , the current will be limited at the limit (3.7A for TPS61023) and Vout will drop because input power did not match the increasing output power. So the thing is whether you can accept the voltage drop (drop to about 4V for your 2A condition) during the motor start up.

Best Regards,

Travis 

Thanks for ur detailed reply. I will contact the Motor manufacturer(s). They have to custom make motors for us, as their standard N20/N30 motors don't meet our requirements. So, I can make them aware of this fact. Hopefully, they can give me satisfactory answer. I will let u know. Thanks again. AB