TPS2596: Analog Load Current Monitor equation

user443387

Part Number: TPS2596

Section 8.3.4 of the TPS2596 datasheet explains how to determine the load current based on the voltage across R_ilm.

I'm trying to deduce the voltage range seen across this resistor, but do not seem to get to realistic results.

For example, suppose: R_ilm = 511 Ohm, which limits I_out to around 1.78 A according to equation (5).

If I would rework equation (6) to V_ilm = I_out * G_imon * R_ilm, then with an average value for G_imon of 656 uA/A and at maximum current, I would get a V_ilm of a little less than 597000 V, which is obviously incorrect.

I've checked the values several times throughout the datasheet, but cannot seem to find the source of my error. Can anybody shed some light on this?

over 5 years ago

0 Praveen GD over 5 years ago

TI__Guru 57737 points

Hi,

I am looking into this and will get back to you by tomorrow.

0 Praveen GD over 5 years ago

TI__Guru 57737 points

Hi,

Consider Iout = 1.78A, Rilm = 511 Ω and from datasheet we know Gimon = 656 uA/A

Now as per equation (6),

Vilm = Iout * Gimon * Rilm

⇒ Vilm = 1.78 A * 656 uA/A * 511 Ω

⇒ Vilm = 597000 uV = 0.597 V

For more understanding on how the current limit circuit works please refer to section 4 of the below document.

http://www.ti.com/lit/an/slvaec2/slvaec2.pdf

0 user443387 over 5 years ago in reply to Praveen GD

Prodigy 20 points

Praveen,

Thanks for the example, it helped to spot my mistake! Also thanks for the reference additional documentation.

Regards,

Jan Willem

Power management

Power management forum

TPS2596: Analog Load Current Monitor equation