Author Topic: ltc4006 - dual pullup ?  (Read 1088 times)

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Offline gionag

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ltc4006 - dual pullup ?
« on: June 09, 2016, 03:20:56 pm »
Hello folks, it's my first post here... and i was too shy to post that to other, more advanced "section".

I am starting to design a battery-pack charger around the ltc4006 ic from linear...

i was reading the datasheet of the ltc and i've found this section :

When the charge cycle starts, the CHG pin is pulled down
to ground by an internal N-channel MOSFET that can drive
more than 100µA. When the charge current drops to 10%
of the full-scale current (C/10), the N-channel MOSFET is
turned off and a weak 25µA current source to ground is
connected to the CHG pin
. After a time out occurs, the pin
will go into a high impedance state. By using two different
value pull-up resistors, a microprocessor can detect three
states from this pin (charging, C/10 and stop charging).

See Figure 6.

How does it works ? i don't understand how can i detect three state by using two pullups... can you explain me ? The topic interest me because i am planning to implement an mcu to monitor the process (in conjunction with a supervisor ic from ti).

thanks a lot !

Offline newbrain

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Re: ltc4006 - dual pullup ?
« Reply #1 on: June 09, 2016, 05:11:51 pm »
Ciao Michele, and welcome!
As you can see from the picture you attached, there are two pull-ups attached to the /CHG pin:
  • The fixed one, 200kohm
  • A second one, that can be driven by the micro output.

Now, let's consider the situations described in the datasheet.

To be of some use the OUT pin from the µP must be set either to High or to HiZ (e.g. by setting it as an input), using it as a push-pull output will not help.

First case: the internal N-chan MOSFET in the ltc4006 is turned on, and able to sink more than 100µA.
In this case, regardless of the status of the OUT pin the level at the IN pin will be 0 or close to 0:
Worst case is when OUT is driven High: VIN = 3.3V - (33kohm//200kohm)*100µA = 0.47V still readable as a Low level.

Second case: the 25µA source is connected.
In this case, the IN will read Low if the OUT pin is HiZ: Imax = 3.3V/200kohm = 16,5µA < 25µA
and High if the OUT pin is High: VIN = 3.3V - (33kohm//200kohm)*25µA = 2,6V

Third case: the /CHG pin is HiZ.
In this last case, IN will read High regardless of the state of the µP OUT pin, as only pull-ups are involved.

So you can see that, by driving the OUT pin between High and HiZ and reading the IN pin, it's possible to detect the three states signaled  by the /CHG pin.

I hope that's clear enough!

Edit: on second thought, using a suitable resistor and an analog input could be an easier way...e.g. with just a 68kohm pull up you'll have:
100µA: about 0V
25µA: 3.3-68k*25µ = 1.6V, square in the middle of the range.
HiZ: 3.3V
« Last Edit: June 09, 2016, 05:22:24 pm by newbrain »
Nandemo wa shiranai wa yo, shitteru koto dake.

Offline gionag

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Re: ltc4006 - dual pullup ?
« Reply #2 on: June 10, 2016, 09:04:27 pm »
Fantastic !
thanks, make perfectly sense.

when investigating by myself I considered using an adc channel to detect different states, so you have confirmed me that is possible. I have to evalute if i prefer "wasting" two gpio or one adc channel.

Thanks again for your support.


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