ICL7665 Low Voltage Cutoff Help

Thread Starter

circnow123

Joined Aug 26, 2021
7
Hello All,

I'm trying to make an ICL7665 based low voltage cutoff board work.

Here is the datasheet: https://datasheets.maximintegrated.com/en/ds/ICL7665.pdf

I'm using it to prevent undervoltage discharge on 18V lithium ion battery packs (20V fully charged). I want it to cut the load supply at 15V (3V per cell) and have about 1-2v hysteresis.

I'm using an IRFZ44N mosfet as the power transistor: https://www.infineon.com/dgdl/irfz44npbf.pdf?fileId=5546d462533600a40153563b3a9f220d

I'm using an LP2950 (3V fixed) for the LDO regulator (since the max voltage for the ICL7665 is 16V).
https://www.onsemi.com/pdf/datasheet/lp2950-d.pdf

I have potentiometers on the SET and HYST pins to "tune" the voltage manually as I couldn't make sense of the formulas in the datasheet. The resistor values on the schematic make the MOSFET turn off at 15V and turn back on at about 16.5V (hysteresis). I found these with trial and error.

The whole system is currently using about 0.6mA which is very high. The ICL7665 only uses 3uA, so the rest must be the LP2950 and resistors.

Here is my circuit:
LowVoltCut2.jpg

Here are the problems I am having:

  1. Output voltage from the regulator is being pushed to 18V when the input voltage is 20V. If I remove the ICL7665, the regulator output drops to 3.0V. I suspect voltage from the HYST pin (18V is present here) is being fed internally to the Vin. I don't know why.
  2. Gate voltage of MOSFET is very low (0.6V) when the system is in normal operation (battery voltage above 15V). Is R3 resistance too high?
  3. High current use of the system.

Thanks in advance for any help.
 

crutschow

Joined Mar 14, 2008
28,204
Output voltage from the regulator is being pushed to 18V when the input voltage is 20V.
No input voltage to the ICL7445 should be greater than its supply voltage (3V here).
You are greatly exceeding that at the HYST inputs.

Why did you not connect them as the data sheet shows?
You can't expect an IC to work properly if you don't follow the data sheet's information.
Gate voltage of MOSFET is very low (0.6V) when the system is in normal operation
Note that Q1 is a standard type MOSFET and needs a Vgs of 10V to fully turn on (20V maximum).
You need to connect Q2 to the battery voltage instead of 3V and add a resistor to ground to keep the maximum Vgs at about 15V

I see no purpose for R2 and R4.

The value of R3 is fine.
High current use of the system
The LM2950 will use about 0.1mA bias current.
R5, R6, and R7 take about 33µA.
R2 takes 60µA.
The high voltage on the HYST inputs may be taking a lot of current.
 

LowQCab

Joined Nov 6, 2012
1,343
Do You require automatic power-up ?
This Circuit may do what You want,
but would need a few more parts to be fully automatic ............
.
.
.
UVLO 1 FLAT .png
 

Thread Starter

circnow123

Joined Aug 26, 2021
7
Thank you for your reply

No input voltage to the ICL7445 should be greater than its supply voltage (3V here).
You are greatly exceeding that at the HYST inputs.

Why did you not connect them as the data sheet shows?
You can't expect an IC to work properly if you don't follow the data sheet's information.
All of the examples in the datasheet show an unregulated input voltage of 5v or 9v. My input voltage was 18v so I thought I could regulate the Vin input to 5v or 3v and feed the unregulated voltage to the other pins (obviously not).

None of the examples fit my application, so I guess I'll need to use a voltage divider to drop the unregulated supply to a level with max 3v, then use more resistors to get the hysteresis and SET voltage.
Or just feed the whole circuit from a voltage divider (and get rid of 3v regulator) since the ICL7665 uses so little current.

I’ll make changes to the mosfet gate so it’s about 15v.

I’ll get rid of R2, R4 was to invert the function of the output (didn’t need use in the end)
 

Papabravo

Joined Feb 24, 2006
17,299
Thank you for your reply



All of the examples in the datasheet show an unregulated input voltage of 5v or 9v. My input voltage was 18v so I thought I could regulate the Vin input to 5v or 3v and feed the unregulated voltage to the other pins (obviously not).

None of the examples fit my application, so I guess I'll need to use a voltage divider to drop the unregulated supply to a level with max 3v, then use more resistors to get the hysteresis and SET voltage.
Or just feed the whole circuit from a voltage divider (and get rid of 3v regulator) since the ICL7665 uses so little current.

I’ll make changes to the mosfet gate so it’s about 15v.

I’ll get rid of R2, R4 was to invert the function of the output (didn’t need use in the end)
You should NOT use a voltage divider in a POWER circuit. You only use them for signals. You can however use a pre-regulator.
 

Thread Starter

circnow123

Joined Aug 26, 2021
7
You should NOT use a voltage divider in a POWER circuit. You only use them for signals. You can however use a pre-regulator.
The ICL7665 only uses 3uA and has an internal voltage reference. Why couldn’t you power it through a divider?
Currently the LDO regulator is using 30x more (100uA) than the ICL7665 chip alone.
 

Papabravo

Joined Feb 24, 2006
17,299
The ICL7665 only uses 3uA and has an internal voltage reference. Why couldn’t you power it through a divider?
Currently the LDO regulator is using 30x more (100uA) than the ICL7665 chip alone.
Because the divider itself will consume power and you have to account for the current drawn when you compute the values for the divider. A chip should be powered from a constant voltage source so that it's operation is consistent with the datasheet and does not introduce artifacts of changing supply voltage on the output. I have to wonder where and how you got the idea of using a voltage divider to power an IC. It is not uncommon among those just starting to learn electronics and it is one of those habits and ideas that you need to let go of quickly if you want to succeed in electronics.
 

Thread Starter

circnow123

Joined Aug 26, 2021
7
Because the divider itself will consume power and you have to account for the current drawn when you compute the values for the divider. A chip should be powered from a constant voltage source so that it's operation is consistent with the datasheet and does not introduce artifacts of changing supply voltage on the output. I have to wonder where and how you got the idea of using a voltage divider to power an IC. It is not uncommon among those just starting to learn electronics and it is one of those habits and ideas that you need to let go of quickly if you want to succeed in electronics.
I’d consider myself semi experienced in electronics, but very green on ultra low power design. Most of the time burning 50mA quiescent current wouldn’t be an issue. Now I’m trying to get things below 50uA.
It feels wrong wasting so much power on the input regulator, when the main chip uses so little.
 

LowQCab

Joined Nov 6, 2012
1,343
It seems that this Thread is going off on a "Wild-Goose-Chase".

If circnow123 would provide a list of specifications,
rather than trying to figure-out how to make a possibly overly-complex
concept work, then maybe this problem could be greatly simplified for him.

If all You want to do is Power a Micro-Controller, and also provide
Battery-Under-Voltage-Protection,
the solutions can be elegantly simple.

I've provided a simple UVLO Battery Protection Circuit,
now it seems that the goal is Regulation down to ~5 or ~3-Volts.

What is the overall goal of this project ?
.
.
.
 

Thread Starter

circnow123

Joined Aug 26, 2021
7
It seems that this Thread is going off on a "Wild-Goose-Chase".

If circnow123 would provide a list of specifications,
rather than trying to figure-out how to make a possibly overly-complex
concept work, then maybe this problem could be greatly simplified for him.

If all You want to do is Power a Micro-Controller, and also provide
Battery-Under-Voltage-Protection,
the solutions can be elegantly simple.

I've provided a simple UVLO Battery Protection Circuit,
now it seems that the goal is Regulation down to ~5 or ~3-Volts.

What is the overall goal of this project ?
.
.
.
The use of this device is to provide under voltage protection when retro fitting lithium batteries to a power tool that previously accepted Ni-cad batteries. The tool has no under voltage protection so drained batteries to zero. This would wreck lithium packs.
The device needs low quiescent current as it will always be connected to the battery “ready to go”. I’d rather not have on/off buttons on the battery.
 

LowQCab

Joined Nov 6, 2012
1,343
Build this Circuit inside your Drill-Motor-Housing, NOT in the Battery-Packs.
Make sure it gets some Air-Flow when the Motor is running for extra insurance.

This Circuit has a ~2-second delay before it will shut-down because of Under-Voltage.

If the Circuit shuts-down because of severe Motor-overload
causing the Battery Voltage to drop below the preset,
then ........
un-plug the Battery,
wait 30-seconds for the Circuit to reset its self,
plug the Battery back in, and use the remaining Charge in that Battery Pack,
hopefully without as much Motor abuse.

The Voltage-Adjustment-Pot can be replaced with appropriately sized
resistors after the desired setting is found.

This Circuit will run the battery dead in about ~30-days, (maybe longer),
after which time the Battery will be permanently damaged.
Don't leave a Battery permanently Plugged into the Drill.
.
.
.
UVLO for Drill Motor 1 FLAT .png
 
Last edited:

Thread Starter

circnow123

Joined Aug 26, 2021
7
I got all of the changes done and increased most resistor values to save power.
26CEC240-4A67-4EEC-9B4D-1A1ED0B0DD0D.jpeg

I’m also now getting 200uA current draw which isn’t too bad. This should last between 6 months and a few years before draining battery depending on how charged it is.
3E192F45-1E65-4A38-A774-491D18FD5A79.jpeg

MOSFET gate voltage is now between 12-15V which turns it on properly.

I’ll post the final schematic later.

Thank you all for the help and suggestions.
 
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