Low Battery Indicator Question

Link to Max1674 data sheet

I've got a Max1674 working on a breadboard, but I don't really understand what I can do with the LBI / LBO part of the circuit for monitoring battery condition.

1. How much current can I use to make something happen when the battery voltage drops below the threshold?

2. Can I make a red led light up or is the current flow so low that all I could do is make something on an LCD display turn on?

Originally posted by jeff1500:
Link to Max1674 data sheet

I've got a Max1674 working on a breadboard, but I don't really understand what I can do with the LBI / LBO part of the circuit for monitoring battery condition.

1. How much current can I use to make something happen when the battery voltage drops below the threshold?

2. Can I make a red led light up or is the current flow so low that all I could do is make something on an LCD display turn on?

Click to expand...

<font size="2" face="Verdana, Arial">The drive capabilities are adequate to drive an LED at a couple of mA. Remember, though, that you need to set the LBI threshold high enough to still have adequate voltage to light a red LED. Once the voltage is below 1.6-1.7V it will not be enough to light the LED. Alternately, you can power the indicator LED off of the output voltage. In this case you can set the threshold to any value you wish.

I'm not clear on where the current to run the indicator comes from. There's a 100k resistor from Vout to LBO. As I read the data sheets, before trip, LBO sinks current to ground, after the trip LBO switches to high impedence.

Does that mean before trip, current goes from Out to LBO and then after trip it goes from Out, past LBO, to the indicator? Is that what you mean by driving the indicator from the output? Would I adjust the size of the 100k resistor to get enough current to drive the indicator. If so does that mean I'm always dumping current, either I dump it to ground from Out to LBO before the trip, or from Out to the indicator after the trip?

Originally posted by jeff1500:
I'm not clear on where the current to run the indicator comes from. There's a 100k resistor from Vout to LBO. As I read the data sheets, before trip, LBO sinks current to ground, after the trip LBO switches to high impedence.

Does that mean before trip, current goes from Out to LBO and then after trip it goes from Out, past LBO, to the indicator? Is that what you mean by driving the indicator from the output? Would I adjust the size of the 100k resistor to get enough current to drive the indicator. If so does that mean I'm always dumping current, either I dump it to ground from Vout to LBO before the trip, or from Out to the indicator after the trip?

Click to expand...

<font size="2" face="Verdana, Arial">Ignore that page 7 "simplified functional diagram". Consult the page 9 description of the LB function. The LBO is simply an open drain FET that is turned on when the LBI is low. Wire your indicator LED from Vout [thru a resistor to limit current if desired] to the LBO terminal. You can probably skip the resistor if your detection threshold is below 3V.

LB0 is an active low output, it sources current whan active. The 100K is just a pullup resistor.

Originally posted by lux0:
LB0 is an active low output, it sources current whan active. The 100K is just a pullup resistor.

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<font size="2" face="Verdana, Arial">The LBO sinks current. It does not source current.

Originally posted by Doug S:
</font><blockquote><font size="1" face="Verdana, Arial">quote:</font><hr /><font size="2" face="Verdana, Arial">Originally posted by lux0:
LB0 is an active low output, it sources current whan active. The 100K is just a pullup resistor.

Click to expand...

<font size="2" face="Verdana, Arial">The LBO sinks current. It does not source current.</font><hr /></blockquote><font size="2" face="Verdana, Arial">depends on weather you use conventional current flow or not.

Originally posted by lux0:
</font><blockquote><font size="1" face="Verdana, Arial">quote:</font><hr /><font size="2" face="Verdana, Arial">Originally posted by Doug S:
</font><blockquote><font size="1" face="Verdana, Arial">quote:</font><hr /><font size="2" face="Verdana, Arial">Originally posted by lux0:
LB0 is an active low output, it sources current whan active. The 100K is just a pullup resistor.

Click to expand...

<font size="2" face="Verdana, Arial">The LBO sinks current. It does not source current.</font><hr /></blockquote><font size="2" face="Verdana, Arial">depends on weather you use conventional current flow or not.</font><hr /></blockquote><font size="2" face="Verdana, Arial">Maxim does use conventional [positive] current flow on their datasheets. I am unaware of any major IC makers who do not.

Sorry, still confused.

Quotes from the data sheet:

Quote 1: "If the voltage at LBI falls below the internal reference voltage LBO (an open drain output) sinks current to ground."

Quote 2: "When LBI is above the threshold, the LBO output is high impedance."

Question 1: In quote 1, does "sinks current to ground" mean current comes out of the chip? ( or does it mean absorbs current and sends it to ground inside the chip?)

Question 2: In quote 2, does high impedance mean no current can come in or go out of LBO?

Question 3: If current comes out of LBO after the low battery trip, how much is there, and at what voltage?

Question 4: After trip, does current start to flow through the 100k pull up resistor? What makes it start?

Originally posted by jeff1500:
Sorry, still confused.

Quotes from the data sheet:

Quote 1: "If the voltage at LBI falls below the internal reference voltage LBO (an open drain output) sinks current to ground."

Quote 2: "When LBI is above the threshold, the LBO output is high impedance."

Question 1: In quote 1, does "sinks current to ground" mean current comes out of the chip? ( or does it mean absorbs current and sends it to ground inside the chip?)

Question 2: In quote 2, does high impedance mean no current can come in or go out of LBO?

Question 3: If current comes out of LBO after the low battery trip, how much is there, and at what voltage?

Question 4: After trip, does current start to flow through the 100k pull up resistor? What makes it start?

Click to expand...

<font size="2" face="Verdana, Arial">Questions 1 and 2. Think of a switch inside the IC that connects LBO to ground. This switch is closed when LBI is low and open when LBI is high.
Question 3: does not.
Question 4: Please note that the 100K resistor is not part of the IC. See questions 1 and 2.

I think I'm starting to understand. Can you agree with what follows?

1. LBO is just a connection to ground inside the IC. When the battery is new, LBO isn't connected to anything. When the battery is warn down, LBO connects to ground.

2. So, if I hook a red led to Vout with an in-line resistor to get the voltage just right (or maybe Vin), and then hook the ground connection from the red led to LBO, the red led will light up when the voltage at LBI drops below the threshold.

3. The data sheet says to use the 100k pull up resistor for driving CMOS circuits. I'm not, so I don't need it. What's it for?

1. Yes
2. Yes
3. Don't need it. For driving CMOS circuits.

Very good. Thanks.

Originally posted by jeff1500:

3. The data sheet says to use the 100k pull up resistor for driving CMOS circuits. I'm not, so I don't need it. What's it for?

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<font size="2" face="Verdana, Arial">Yes you dont need it. But if you were driving CMOS circuits, you shouldn't let anything float(high impedance) so you have a small resistor to make it go high (V+) when the 'swich' is open. When the 'switch 'is closed and connected to ground (V-) the resistor can't supply much currnet and it is held at ground (V-). You need to do this in CMOS circuits because the CMOS circuits inputs dont draw much current, so the slightest source of current(dirt on the circuit board) can make it change from low (V-) to high (V+). The pull up ( or sometimes a pull down)resistor makes it so you need a little more current to change state( high to low or low to high).