Understanding cable drop compensation factor and tolerances: SC1225K

Posted by: wgebbiainvue on 06/03/2015

I'm trying to understand tolerances around cable loss compensation in a design using SC1225K. Our partner is working off an Engineering report from the ShenZhen branch that places an external resistor in parallel with the IS pin. I see how that allows for a higher operating current, but I don't see how they are able to provide the tolerances they are claiming. Our application is not a standard adaptor: we have some specialized requirements.

The claim is:

0A voltage min 5.0V nom 5.3V max 5.5V

2.3A voltage min 5.4V nom 5.6V max 5.8V

Overcurrent maximum 2.9A

I posed the following questions to our partner engineer and recieved the following answers:

Questions

· The IC is made for 2A max and they’re getting more out of it by putting an external resistor in parallel with the internal current sense. That will also reduce the cable drop compensation factor since the IC won’t actually know how much current is in the circuit.
· I can’t find the value of the internal current sense resistor: how is the external part calculated?

Answers

1) The cable lost compensation requirement is being met, we fixed the compensation value of one of the output pin.

2) IS Pin=33mV and internal current is 2A, therefore computed internal resistance R=16.5M Ohm. We can add resistor in series with this to increase resistance, in order to raise output voltage.

I've uploaded an excerpt from the engineering report they're working from. Can you help me make sense of this?

Thanks

Files

Engineering report (SC1225K for 5V A2.4A Chargeer)-Rev1 page3 (100.35 KB)

Comments

Thanks for Posting your question into forum. The Clarification for your questions are below:

(1) The SC1225K part is designed for 2A CC which means the part goes into CC mode beyond the 2A load around 2.1 A including tolerances. The Cable compensation is 6% for 2A Load. The Part goes into CC mode when the drop across the IS pin and GND pin is 33mv ( for 2A), the part measures the drop across the internal bond resistor which is connected between IS pin and GND Pin. To increase the over current of the part to deliver beyond 2A is done by connecting a resistor across the Is pin and GND pin such rhat the drop across it 33mV.

For Eg :

It's 2 A CC Part, the drop across Is pin and GND pin is 33mV as per the datasheet.

The Internal Bond resistor is = 33mV/ 2A =16.5 milli ohm

To Make the Part to work 3 A CC , 33mV/3 A = 11 m Ohm.

The Effective Resistance between IS pin and GND pin Should be 11 milli ohm but the internal resistor is only 16.5 mOhm.

To make it 11 mOhm , we need to Connect an External resistor between Is pin and GND so that the effective parallel combination of Internal bon resistor (16.5 mOhm) and Parallel resistor is 11 m Ohm. if we connect a 33mohm resistor across the Is pin and GND , the total effective resistance is 11mOhm and the part goes into CC when you inject 3 A Current as the drop across IS pin and GND is 3A * 11mohm which is equal to 33mV.

Please let us know if the above explaination clarifies your question , let us know any other questions.

Thanks,

PI-KSAN

Thank you for the information. I understand the basics of the technique, what I'm having trouble understanding is how to calculate the minimum and maximum slope based on the information given. I'm trying to establish minimum and maximum voltages at full load, maximum voltage at overcurrent, and minumum and maximum overcurrent.

According to the spec sheet CC occurs at 2A Minimum and the voltage at Is at CC is 33mV Typical. That means the effective resistance is around 16.5mOhm most of the time .Since the two numbers used to come up with 16.5 ohm were not under the same conditions, the typical value is unknown. Both parameters are specified at Tj=25C so I don't know what happens over temperature.

I assume that there is litterally a current sense resistor on the die. Monolithic resistors are known for having poor tolerances, so if the typical resistor value is 16.5 ohm the min value could easily be 14.9 ohm (10%), then there are errors involved with detecting the voltage differential. The "standard usage" spec sheet limits will account for these tolerances but we're adding new error with the shunting resistor.

Yes , there will be tolerance involved . The part actually doesn't go exactly into CC at 2 A Load but CC trip point might happens around 2.1 A Load considering tolerance.