3V3 power comes from 9V standard battery. It's a temperature measurement tool, shaped as a pistol, with a trigger. So the hardware is always powered down, when the trigger is pressed the 9-3V3 regulator is kept on by a MSP keepalive signalto this regulator shutdown pin.
Ok, the problem is we need a low bat check. There's no ADC, the comparator inputis already in use. This level check should draw 5nA top, but preferably be inshutdown mode and turned on only when the MSP wants to inquire it if everythingis shining with the battery level.I'm wondering about using a commercial vco to do this. Since this is far from a new problem, I'd ask if someone could please share his/her thoughts about this?
Thanks in advance,Xezi.
I think 3V3 power comes from 9V standard battery is not a good choice. Take a look at size/Joule, weight/Joule, and cost/Joule. 9V standard battery is mostly used where you need the voltage and do not want to bother with a DC-DC converter and do not care about energy.
Well, yes, we were aware of this, but the 9V bat could not be changed in this particulardesign. Though we could improve the energy efiiciency, that's not of concern. The difficulty really stands in knowing the battery condition - we don't even need to knowits voltage, only if it's still 'good' or 'bad', if needs to be replaced.
Thank you for your reply.
Bruno Marchesithe comparator input is already in use
could you add an analogue mux/switch to it...?
Yes, an analog switch could be used, as a bit of an expensivesolution, I believe. We could do: 9V - R1 - AS input - R2 - Qs - GNDAS stands for analog switch.
R1-R2 provides the voltage drop from 9V to, say, Vcc/2,to be read from the MSP compare pin.
Qs base is tied to a MSP I/O pin. This way, wheneverthe MSP powers off, additional current drawing from the9V R1-R2 net is minimized.
We need an analog switch with nA quiescent current,or a shutdown pin that gives us that condition. Two ADG701appear to do the job, although I sometimes still got puzzledwith parameters like "1nA typ/1uA max quiescent current",in the sense this is a 1,000 fold range. If it stays around 1nA,I believe this could be done.
I still wonder if a 2x $3 part would be a good solution in this case,meaning if there isn't a better, more straightforward approach,which I really can't see right now.
Are you sure of this 5nA requirement?
A standard 9V battery has 400mAh capacity.
At 5nA it will drain in 100 years (0.4 / 5e-9 *24*365 =9132 years)
no, I'm not sure. The current drain and battery lifetime calculations I've madeare not precise. I've measured the on and off state current drawings with an Agilent 34410,and I'd conduct tests once the product is first assembled to get better aproximations.
The last data I've got is 120mAh for local market 9V batteries, andwith around 50-100nA continuous current consumption plus theon state usage - reaches 20-40mA, a lifetime of around 2 years. We were looking to something around 5 years, which is the usualbattery expire time, considering you've buyed it soon enough.
Not sure if I'm overreacting with this battery draining thing.
Thank you for your reply.
Bruno MarchesiWe could do: 9V - R1 - AS input - R2 - Qs - GND
Why not using 9V - R1 - FET - MSPin - R2 - GND?Teh FET is controlled by an MSP pin. Since the FET gate is a pure capacitive load, the port pin won't drive any current except during state change.When you raise the gate of the FET to VCC, the MSP input pin will raise from GND (through R2) to (R1*R2)/(R1+R2). If R1 and R2 are carefully chosen, this voltag eiwll be above teh high-going threshold level when the battery is still good and stay below if the battery is bad. You might want to replace R1 by an adjustable resistor, so you can trim the threshold to narrow down the threshold voltage. But if you don't need an exact trigger point, carefully chosen fixed resistors do the job fine.
However, pick a FET with a very low gate threshold voltage, as the gate will be risen to VCC, btu the FETs drain will have to also rise above the input threshold. So the difference between drain and gate will be small. It's a tight calculation. But will drop your 'comparator' load to zero if the FET gate is pulled low.
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Not sure if I kept up with your approach, but using 9V - R1- FET - MSP input pin - R2 - GND,where FET: BSH103, Rds 0.5R, vth 0.4V,
if we initially set 3V3 on R2 with Vbat 9V, neglecting Rds, this gives us, for example,
R1=173k and R2=100k. This leads to a minimum 'good' battery voltage of 7.21V,because of Vih min = 0.8Vcc
But, if we stretch the MSP input pin voltage to something near the absolute maximum rating of 3.6V,using 3.5V, we get R1=157k, R2=100k, and a minimum 'good' battery voltage of 6.79V.
That's a good example of a tight calculation, but it could work nonetheless. I'll try it out and come back.
I didn't catch up the importance of a low vth. This 0.4V was the lowest one I could find, anyway.
If this was not what you meant, if you find some time and could please elaborate on it, it'd be appreciated of course.
One more time,thanks for your reply.
Bruno MarchesiThat's a good example of a tight calculation, but it could work nonetheless
Bruno MarchesiI didn't catch up the importance of a low vth. This 0.4V was the lowest one I could find, anyway.
The problem is that Vg needs to be by Vth above Vd. Vd, however, rises when the FET begins to switch on (as Vd is also the MSP input voltage). So Vd must be above the threshold voltage, but Vd+Vth must still be reachable by the MSP output driver (<=Vcc). If Vth is too large, it won't work. Vth must be less than the difference between positive-going threshold voltage and VCC.
To use the switched voltage divider you need 1 MSP430 output and 1 ADC input..
You also need 3 resistors and BSH103 ( low vth nmos transistor).
To monitor a 9V battery I would use 90kohm resistor connected to the drain of BSH103.
The BSH103 source is connected to 10kohm resistor to ground.
the BSH103 gate is connected to 10kohm resistor to ground.
The MSP430 output is connected to the gate and the ADC input to the source.
With the output set to 0V you have just the BSH103 leakage flowing through the divider. ( < 100nA spec = ~ 5nA typ )
With the output set to 3V you have a 10 to 1 divider. ( BSH103 = < 1 ohm )
So to measure the 9V battery activate the gate circuit, ADC reference, use long ADC sampling period and convert, this should take about 100us.
The measured voltage should be 0.9V with fresh battery.
Peter DvorakTo use the switched voltage divider you need 1 MSP430 output and 1 ADC input..
Jens-Michael: The problem is that Vg needs to be by Vth above Vd.
Well, that kinda invalidate my previous calculations. -sigh-
Still, maybe something like this could do the trick... The mosfet pair is a single sot-563, $0.53 part.
With logic low at the 'shutdown' pin, ie the N mosfet gate, P mosfet gate should be at 9V,thus off. 3V3 at N mos should bring P mos gate to ~0V, and for battery voltages from 9Vtill something around 6.6V, we should have 3V3 at MSP input pin.
With almost the same amount of parts, it could be used a $1.40 ADC with 5nA-1uA standby current,I2C interface, sot-23-5 package. Maybe the circuit above still has the advantage of a near-zero standbycurrent, it seems.
Hi Bruno, what we know:
1> a Gun shaped device that measure some way temperature (no data on method)
2> an MSP is inside (no device model)
3> no data about how data is handled by
4> no data about which peripheral are onboard and why 9V battery instead of 3.6 Lithium batt so it can leave MSP live forever.
5> commercial battery, the low capacity in first line last before end date and leakage occur damaging your instrument.
6> power cycle duty, power drawn when on
The circuit you drawn test when battery voltage drop below threshold minus 3.3 V from Zener reference, this is an instable condition that appear when voltage dropped under 1.8 plus 3.3 so threshold drop with the supply and battery is too low voltage. Raise the upper zener to almost 5.6V.
Upper P channel mosfet can be overdriven over max VGS with a punch through risk, better on low side part remove N channel and substitute with a 10-100nF capacitor using it as a boostrap driver.
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SOme trick can be done using double integration converter done just with some resistor and some free cpu pin.
Comparator is needed, so is this pin available or shared?
If yes two resistive divider can charge/discharge a capacitor from cpu voltage level and then evaluate time to reach threshold, this method work comparing unknown voltage to VCC of MSP, this is as precise as VCC is stable, so it can be used till battery is not fully discharged.
If not available and spare pin are is possible to compare unknown voltage to input gate hysteresis too, this method require continuous calibration but it can work reading few bit than from precision comparator but enough to evaluate battery voltage to one decimal.
Problem on how to timely switch on instrument remain unsolved.
Bruno Marchesi Maybe the circuit above still has the advantage of a near-zero standby current, it seems.
And the two-stage FETs are a good idea. However, the zeners won't work. First, zeners require a relatively large current for their nominal voltage. Below, the voltage drops. Besides the nonlinearity of this influence, this doesn't gove you any advantage over a plain resistor divider. And raises the costs.
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