High Low logic

[Xenobius]

Hi all,

Application: I have a circuit which is powered from a lipo battery which needs to shutdown when the battery hits 3v for safety reasons. I am making use of an LT1763CS8 LDO voltage regulator which coincidentally has a shutdown pin that must be pulled up to function or LOW to shutdown power.

Question: What can I use to output HIGH when the battery voltage is 3.01v+ and LOW when its 3v or lower?

Solution: I was thinking perhaps a transistor to output low whenever the battery is 3v but I'm not sure how to connect it or calculate needed values.

Thanks a lot

 

[Xenobius]

Maybe I could make use of a 3.1v zener diode connected somehow?

 

[BobK]

A single transitor cannot turn on and off shaply based on a voltage. You need a comparator to do this well. And you need a voltage reference to compare to.

On the other hand, you should be using a battery with a built in protection circuit that prevents over charge and over discharge.

Bob

 

[Xenobius]

The battery charger IC that I will be using protects from Overcharging however nothing stops me from Overdischarging. What I need is a way to trigger a LOW when there is 3V or maybe 3.1v at most.

Maybe a zener diode can be used in a way to do this?

 

[BobK]

A comparator is the only thing that will do this sharply.

There might be charging ICs that output a low voltage cutoff signal.

Bob

 

[dorke]

You can build a 2 transistor schmidt trigger comparator circuit.
Like Bob said, a single transistor circuit isn't good enough.

Alternatively,
You can use a single very small(sot-5) comparator like the TL331.

Note:
You have a very accurate Vref(1.22V) on the Adj. pin(pin-2) of the LT1763CS8 that can help you.

BTW,
Which charging IC are you using?

 

[Xenobius]

Hi,

Charging IC - MCP73831
Regarding 1.22v on ADJ how does it work when I use the ADJ version of the voltage regulator?
I have a potential divider from the OUT to ADJ to GND according to the datasheet in order to control the output of the adjustable voltage regulator.
#
Thanks

 

[duke37]

The regulator keeps 1.22V between the output and the ADJ terminal. If this terminal is connected to ground. then the output will be 1.22V
If the ADJ terminal is raised by a certain voltage, then the output will be the sum of the added voltage and 1.22V. The raised voltage is derived from the regulated output with a potential divider, the voltage across the 'top' resistor will be 1.22V and the voltage across the bottom resistor will be R2/R1 * 1.22.

 

[Xenobius]

The regulator keeps 1.22V between the output and the ADJ terminal. If this terminal is connected to ground. then the output will be 1.22V
If the ADJ terminal is raised by a certain voltage, then the output will be the sum of the added voltage and 1.22V. The raised voltage is derived from the regulated output with a potential divider, the voltage across the 'top' resistor will be 1.22V and the voltage across the bottom resistor will be R2/R1 * 1.22.

Ohhh right now it makes a lot of sense. I read this on the datasheet but you explained it much better.

 

[dorke]

The regulator keeps 1.22V between the output and the ADJ terminal. If this terminal is connected to ground. then the output will be 1.22V
If the ADJ terminal is raised by a certain voltage, then the output will be the sum of the added voltage and 1.22V. The raised voltage is derived from the regulated output with a potential divider, the voltage across the 'top' resistor will be 1.22V and the voltage across the bottom resistor will be R2/R1 * 1.22.

Well not in this case!
The Adj. pin voltage of the LT1763 is referenced to ground not to the output!
It is a fixed voltage at 1.22v ,the band gap internal voltage reference(that is why I said it is easy to use)!
It is clearly stated in several places in the data sheet.

Note also that the output voltage is set by :
Vo=1.22(1+ R2/R1) ,in which R2 is the "top" resistor in the drawing.

The explanation you gave is valid for other types of adjustable linear regulators like the LM138/338 and others but not the LT1763.
There is a difference between VADj in both cases.

Here is a circuit(classic linear regulator) that explain how the the structure of the LT1763 .

 

[BobK]

What you are saying makes it sound like the adjust pin is an output. It is not, it is an input driven by a voltage divider off rhe regulated output. Saying that it will always be 1.22V is really saying that the output is at the regulated voltage.

If the input is too low so that it drops out, or the load is too high so that ot cannot supply the current, the adj pin will not be at 1.22V.

Bob

 

[dorke]

Well,
It is an input ,no doubt.
It will be fixed at Vref. ,as long as there is a closed feedback loop.
i.e the op. amp in the diagram isn't saturated.

 

[duke37]

OK, I got it wrong, it has happened before.

The difference is that 1.22V is across the 'bottom' resistor rather than the 'top' resistor. I wonder whether this is due to the fact that it is a low dropout regulator.

 

[Xenobius]

I am sincerely learning a lot and appreciate your time too however I am a bit lost now.
Should I try to look for TL331 and try to understand how it works?

Thanks again

 

[duke37]

The TL331 is a comparator and does not have a voltage reference.

 

[AnalogKid]

I wonder whether this is due to the fact that it is a low dropout regulator.

Yes. In a floating regulator like the LM317, the voltage source that powers the internal regulator circuit is the input-output differential voltage. That is why it has a minimum value that is over 1 V. An LDO can have a Vdiff of much less than 1 V, and less than one Vbe drop. That is too little to run the guts, so the regulator is configured more like the 7805 and other early regulator ICs. In these parts, the regulator circuit is powered by the voltage between the input and GND pins.

ak

 

[dorke]

The TL331 is a comparator and does not have a voltage reference.

That is correct.
That is why I suggested using the reference voltage of the LT1763.
There are compactors with internal voltage reference, like the TLV3011-2 , that can be used.

But in general,
What you really need is a battery charger IC with integrated PowerPath control.

 

[BobK]

Well,
It is an input ,no doubt.
It will be fixed at Vref. ,as long as there is a closed feedback loop.
i.e the op. amp in the diagram isn't saturated.

No it won't. The voltage on the pin is set by a voltage divider from the output. If the output is below the regulated voltage, the adj pin will be below 1.22V by the same ratio.

Bob

 

[(*steve*)]

The voltage between the adj and output pin is held constant with a very low current flowing through the adj pin. This the current through the resistor between the output and ground is constant allowing the output voltage to be easily changed by altering the resistor between adj and gnd.

So, when operating normally, there is a known voltage between adj and output, but I've never seen it used separately as a reference.

 

[dorke]

No it won't. The voltage on the pin is set by a voltage divider from the output. If the output is below the regulated voltage, the adj pin will be below 1.22V by the same ratio.

Bob

Bob,
we are saying the same thing.
The voltage on the Adj. pin will be equal to Vref=1.22v while in regulation mode.
That means that the op. amp.(error amp.) isn't saturated,and the negative feedback loop is closed(active).
i.e V(+)=V(-)=Vref at the inputs of the op.amp.