Using Sensed battery voltage to pull EN/UVLO pin low

Thread Starter

jim0000

Joined Oct 28, 2020
130
My rough idea is below. I would like to pull the EN/UVLO pin low when the battery voltage reaches 13.5V. I am unsure if this design would work, I am worried I will always just be reading the output of my switching regulator which is 14.6V. How could I make sure that I am reading the battery voltage and not the switching regulator output, and then compare that battery voltage with a reference so that I can automatically pull the EN/UVLO pin low when Vbattery is 13.5v?

View attachment 282278
 

Thread Starter

jim0000

Joined Oct 28, 2020
130
I would power the LM339 and the LM431 off the battery.
Connect the reference voltage to the + input and the battery to the - input.
The LM338 is an open collector output, may not need a pullup resistor on the output.
View attachment 282279
Thank you,
I would power the LM339 and the LM431 off the battery.
Connect the reference voltage to the + input and the battery to the - input.
The LM338 is an open collector output, may not need a pullup resistor on the output.
View attachment 282279
Thank you, why did the inputs need to be switched?
Also, I am confusing myself with the output of the switching regulator and the positive battery terminal. What would be going in to the LM431 and Comparator, is it the battery voltage or switching regulator voltage in this configuration?
 

sghioto

Joined Dec 31, 2017
5,380
I just deleted my previous post. To clarify the battery is charging and you need the pin to go low at 13.5 volts, correct?
Is the switching regulator to charge the battery or is the battery powering the switching regulator?
 
Last edited:

Thread Starter

jim0000

Joined Oct 28, 2020
130
I just deleted my previous post. To clarify the battery is charging and you need the pin to go low at 13.5 volts, correct?
Is the switching regulator to charge the battery or is the battery powering the switching regulator?
Yes battery is charging and I'd like to make the EN/UVLO pin go low right when the battery reaches a charge of 13.5V. The switching regulator output goes to the battery to charge it.
 

sghioto

Joined Dec 31, 2017
5,380
Is there a load on the battery as well or is the regulator to just maintain a charge on the battery?
What is that 0-15 volt sweep circuit for?
 

Thread Starter

jim0000

Joined Oct 28, 2020
130
Is there a load on the battery as well or is the regulator to just maintain a charge on the battery?
What is that 0-15 volt sweep circuit for?
There is no load on the battery, it will just be charging. The 0-15volt sweep is to mimic a changing input (solar panels..although the change is extreme it will show via test and simulation how the EN/UVLO pin changes and then how it can go from high to low)
 

sghioto

Joined Dec 31, 2017
5,380
You wouldn't need much current to maintain the charge on the battery with no load.
Back to your original question I would think you could install a resistor between the output of the regulator and the battery.
That will separate the two voltages. You will also need to wire the comparator for some hysteresis to set the low voltage required to turn the regulator/charger back ON.
1670295535685.png
 

WBahn

Joined Mar 31, 2012
29,978
I suggested 1 ohm as shown on the drawing. Assuming the charger is activated at 12.2 volts that would provide 2.4 amps of initial current.
So what would be the point of having a charger that can charge at 25 A if it is going to be throttled to never supply more than 10% of that?

I know that if I'm picking out a charger, I'm not going to spend the extra money on something that can supply ten times the current that I need -- at least not without a good reason. I'm not familiar with switch-mode battery chargers that can be enabled programmatically, so perhaps there's a good reason that I'm not aware of. I just know that when I bought my last old-fashioned charger many years ago, there was a huge cost/size difference between a 2 A charger and a 20 A charger.
 

Thread Starter

jim0000

Joined Oct 28, 2020
130
So you could limit the current to just 5 amps if only to maintain the battery charge.
Is this a typical 12 volt lead acid battery?
It isn't desirable to have a maintenance charge for my application. I would like to isolate the battery from the charger once full charge is reached.
 

Thread Starter

jim0000

Joined Oct 28, 2020
130
So the question becomes how much current does your application NEED in order to be acceptable?
I updated my design with the comparator to a more detailed design. Would this work in the lab? I was expecting to need to switch a npn or something at the output of the comparator. I was also expecting to need a pull up or pull down resistor.

There is one problem I am seeing with my design. I tried to make it to where the output would go low right when the input reaches 13.5V. I remapped the inverting voltage using R3 and R4 attempting for V- to reach 5V right when Vbattery=13.5V which would trigger the output of the comparator to change. That seems to be delayed slightly as shown in the simulation. Why is this happening?
View attachment 282320
View attachment 282321
View attachment 282322
 
Top