Help with ICL7665

Thread Starter

rndm_noob

Joined May 21, 2022
2
Hello, Im new to this forum and I'm having trouble with my little project. Please bear with me as I'm just new at this and trying to learn.

I'm trying to create a simple system to notify my esp32 when my LiPoly/LiIon battery is low at voltage of either 3.1/3.2 to shut it down. But I'm having a problem on connecting or testing the ICL7665 I purchased online. At first I thought connecting a led at OUT2 would light up my led but it isn't. On further inspection I noticed that when checking continuity on my DMM, GND shorts at OUT2 but when removing the IC on my breadboard it isn't shorted anymore.

  • My first question is that did I fry my IC? I read somewhere in the datasheet this part but I don't know how to interpret this:

Due to the SCR structure inherent in the CMOS process used to fabricate these devices, connecting any terminal to voltages greater than (V+ + 0.3V) or less than (GND - 0.3V) may cause destructive latchup. For this reason, we recommend that inputs from external sources that are not operating from the same power supply not be applied to the device before its supply is established, and that in multiple supply systems, the supply to the ICL7665 be turned on first. If this is not possible, currents into inputs and/or outputs must be limited to ±0.5mA and voltages must not exceed those defined above.
  • My second question what's the difference between on/off and hi/low in the truth table?
1653162703010.png

  • My third question is what voltage does the OUT2 outputs when it detects the undervoltage? Is it the same as VIN?

The schematic that I recreated is the same as this one but only on the undervoltage side, then on the OUT2 I connected my led to test:
1653162659557.png
 

Papabravo

Joined Feb 24, 2006
18,400
It is hard to say what is going on. We need to have schematic diagram of how the chip is connected, and showing component values. Also show the LED and the current limiting resistor that you used. The chip might still be OK, and you may or may not have correctly interpreted what your eyes are telling you.

If you don't have a schematic, tell us what values you assigned to each resistor in Figure 5., from the datasheet. That way we can check your design to see if there is a mistake there.

EDIT: I just noticed that OUT2* (can't use an overbar, so a '*' symbol is used to denote an active low logic signal) is an open-drain output and it REQUIRES a pullup resistor to function correctly. See Figure 2., in the datasheet for the functional block diagram.
 
Last edited:

Thread Starter

rndm_noob

Joined May 21, 2022
2
IMG_20220522_131102.jpg

Sorry for the ugly schematic but this is how I connected my ICL7665,
The led I used is red led
I also added R1 as the pullup resistor
V+ has a fixed 3.5v for the Vtrip
R1 is 1k resistor
R22 11.8k resistor
R12 7k resistor

is my connection correct sir?
 

Papabravo

Joined Feb 24, 2006
18,400
View attachment 267793

Sorry for the ugly schematic but this is how I connected my ICL7665,
The led I used is red led
I also added R1 as the pullup resistor
V+ has a fixed 3.5v for the Vtrip
R1 is 1k resistor
R22 11.8k resistor
R12 7k resistor

is my connection correct sir?
Since out2 is an open drain output and we don't know the forward voltage drop of the RED LED, the 1K resistor might not be able to allow sufficient current to drive the LED. An open-drain output is intended to pull current through an LED connected to a power rail. For example, if the LED has a forward voltage of 2.2 Volts there would not be much available current to light the LED in your configuration.
 
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