I have assembled this circuit:http://forum.allaboutcircuits.com/showthread.php?t=16232 the original drawing from Sgt.Wookie and added it to another circuit, before I added the low battery disconnect to my other circuit the current drawn from a fresh 9V battery is 625uA, when I add the low battery disconnect circuit it draws 5.77mA. For the FET I am using a 2n7000.

Any suggestion on what I need to do to get the current down.

Thanks

 

Hmm... I think that's the current through the Zener. Isn't it?

Do you have this breadboarded? Can you add a 100 Ohm in series with the Zener and see if that's the location of the somewhat large current?

 

thanks for the reply, do you mean to add a 100 ohm resister in series from the drain of the fet to the anode of the zener, or from the cathode of the zener to the junction of R2, R3?


Thanks for your assistance

 

I don't know your pricepoint or space available for components, but there are voltage references such as ISL60002 that operate on < 1 uA, and also comparators that use < 1 uA. Those, plus a MOSFET and a voltage divider in the range of several MegaOhms should allow you to implement a low voltage disconnect with a total current draw of under 10 uA.

For curiosity's sake, here is an article describing the problem of Zener diode current consumption, and something akin to a solution using ALD precision MOSFETs, although I would myself prefer to use an ISL60002 reference and MCP6041 op amp as I mentioned above with a 10 MegaOhm divider & 1 nF capacitor. I have found the ALD precision MOSFETs to be very interesting but also difficult to work with due to easy static damage as well as drift in the threshold voltage (possibly due to said static damage).

 

thanks for the reply, do you mean to add a 100 ohm resister in series from the drain of the fet to the anode of the zener, or from the cathode of the zener to the junction of R2, R3?

I mean from the top of the Zener to the place where it was previously connected -- just to be able to sense the current passing through the Zener as represented by the voltage over the 100 Ohm resistor that you insert in series. Say you read 0.2V over the 100 Ohm resistor -- that means you're spilling about 2 mA over the Zener.

I'm guessing you'll find about 0.3V over the 100 Ohm, for 3 mA. Zeners are not a very power efficient voltage reference.

 

I tried it on the anode and cathode side and yes the current draw goes down. All I am trying to accomplish is since I have a voltage regulator, LP2950ACZ-5.0 and a pic 16f628a processor on my board and even though I have brown out feature on the processor, as the 9v battery starts to get low I want to "disconnect" the battery from the circuit.

Thanks

 

I tried it on the anode and cathode side and yes the current draw goes down. All I am trying to accomplish is since I have a voltage regulator, LP2950ACZ-5.0 and a pic 16f628a processor on my board and even though I have brown out feature on the processor, as the 9v battery starts to get low I want to "disconnect" the battery from the circuit.

What did you read over the 100 Ohm resistor?

What is your actual load besides the PIC? There is another option that you might like. The PIC itself can run at very very low power, if you program it correctly. You could have the PIC check the voltage of the 9V battery every so often, and be the supervisor. It could shut off the load if the battery voltage gets too low, and the it will only be drawing around 1 uW to just stay asleep and wake up every second to check the voltage again.

What's the actual application?

 

However, scratch that, because your 5V regulator has a quiescent bias current of 75 uA. That makes it impossible to run the PIC on 1 uW in your circuit, even though the PIC is capable of it.

 

add the resistor in line dropped the current draw by close to 200uA. My entire circuit at full power, removing all the the low battery disconnect components on draws 28mA thats with things running, when the processor is asleep about 600uA, and the 600uA is partly due because it is also waiting for an interrupt, part of that is a divider circuit but anyhow once at at the battery disconnect it jumps up to over 5mA the 600uA I can live with but 5mA won't last long...

 

Well, I think that the current loss is over the Zener. I was asking what was the voltage drop over the 100 Ohm resistor, which you didn't mention. However, it is clear that if the overall current dropped by 200 uA when you added the 100 Ohms in series to the Zener, then a lot of the current is flowing over the Zener, and adding the 100 Ohms just reduced that a bit. My goal with the 100 Ohms was to use it as a current sense to know approx how much is going over the Zener.

I don't know what the actual application is, because you didn't mention, but even so, for a PIC to be consuming 600 uA in a sleep state is very high, even with a crystal running. Your voltage regulator uses up to 75 uA but still the PIC should be able to sleep on a few uA. Maybe you don't have it powered down to the minimum power you could. Then you also mention you have a divider somewhere. You may be able to lower that current draw as well. You probably could reduce the overall current consumption of your circuit by a healthy factor, through some design changes.

 

the majority of the current draw without the disconnect circuit attached is due to two seperate voltage dividers, one is pull from the 5v leg of the lp295 through a 680 ohm to a 1.5k to ground to give me around 3.4 volts the other is 22k from the 9v battery to a 10k to ground for about 2.81v. The first divider I mention pulled around 2mA but have reconfigured it with a 3.9k to the 5v and a 10k to ground, really cut back on the current draw I am waiting on receiving my lp2950 3.3v regulator to do away with the voltage divider. The processor goes to sleep, but wakes up on portb on-change, then runs it routine, back to sleep, without the processor out of circuit and still connected to the battery disconnect there is very little if no change in current draw.

Anyhow, thanks again