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[SOLVED] Voltage detector issue

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Veketti

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Dear All,

I used STM1061 IC with voltage treshold 3.1V with below shown circuitry to detect low voltage. I needed it to trigger when voltage was ~5.3V. This IC was only available up to 3.1V so I had to decide some circuitry to lower the voltage. One was using voltage divider, but as this is constantly powered by battery I didn't want to waste any energy so I decided to use diodes in series to drop the voltage. Datasheet for these 1N4148 say only, max forward voltage being 1V. Measured by multimeter and I got 0.56V voltage drop, so calculated I need 4 to drop enough voltage for the 3.1V IC to trigger. I only had SMD diodes and didn't bother to solder them in series to test complete circuit and now that I got the PCB's each only drop around 0.36V! So this only works when the voltage drop down to around 4.6V which is way too low.

So my question is, is there anything to do to save this design and pcb to make it work? Eg. replace any other type 1206 SMD diode or anything?

voltage_monitor.jpg


Thank you in advance.
 

betwixt

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A diode only drops a voltage across its ends when a current flows through it. There is no flow of current in your schematic unless U6.1 is low.
Look at the data sheet for the 1N4148 and you will find a Vf/If graph showing the drop at different currents. To ensure enough drop you need at least the minimum current flowing to achieve it.

Brian.
 

    Veketti

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KlausST

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Hi

The problem here is that the "pullup" creates a relatively large hysteresis.
The current through the diodes differ if OUT is LOW or HIGH.

I don't have a good solution.

Btw: I don't like the the datasheet naming "OUT" .... because it is meaningless.
Even more meaningless: Low_active .... High_active "OUT".

Klaus
 

    Veketti

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danadakk

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There are LV Zeners one could use, some 5% tolerance. With fairly low Tempco,
on the order of 3 - 5 mV / C.



Regards, Dana.
 

    Veketti

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00kam

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hello
from the data sheet there is 3uA leakage current at 25C this current increase with high temperature environment say 35C . if this current can forward those diodes you can get VCC voltage
(VCC=5.3-3*(1 diode droop voltage )) .but you take in consideration there is no device without energy losing

kamal
 

    Veketti

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Veketti

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Darn. Not the first time to face the no current no resistance -issue. Last time it was with zeners not working due to no current. Hopefully next time I remember.


So my noble quest for almost zero current consumption circuitry was flushed down the drain. I was able to reproduce that what you have said that when the U6 was low diodes dropped the correct voltage and reset happened around 5.3x V.

So my dirty approach to fix this was solder 18kohm resistor to U6 VCC leg and jumper wire it to ground. Now it draws ~270uA all the time but I quess just have to accept that. That was the highest value resistor for it to work. So now, not happy but relieved that it works.

Below picture of the ugly fix. Never mind about the ugly lonely zener which I tried whether it has more voltage drop.. Need to clean those flux residue and hotglue that mess.. :rolleyes:


Thank you all for helping.

pcb_mod.jpg


edit: corrected typos
 
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KlausST

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Hi,

You surely need no 270uA when the IC just draws 3uA.
And if you use a fixed load resistor, then replace all the diodes (with their high temperature drift) and use a resistor instead. The resistor does not need to draw more current than the diodes need...but the accuracy and precision is much higher.

Klaus
 

Veketti

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Hmm, ok so back to the voltage divider. If I were to change the series diodes to 160kohm resistor and grounding resistor 226kohm resistor, I'd get at 5.3V, 3.103V to the U6 pin 3. Would this work in that sense that would U6 be able to pull down out pin 1 and overcome 10kohm pulldown resistor?

This would have current draw of 14uA at 5.3V

Edit:
Tested it by 193kohm in place of diodes and 270kohm resistor to ground and it gave only 0.71V to pin3.. Why is it that the voltage divider calculation doesn't match?
 
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00kam

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i have an idea didn't do it before why don't remove the resistance 18kohm(183)and connect instead of it a capacitor of 0.5uf or 1uf .the capacitor at the start (an charged) appears low impedance which make it draws a limited charger current passes through diodes. after a short period when Cap.is fully charged it appears very high impedance (open circuit) and the charger current backs very small or you can say leakage current . my point is here the rapid charger current keeps the didoes in "ON STATE" even if the current backs to (3uA + Cap.leakage current uA)(which is the case no STM1062 load ).


kamal
 

danadakk

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Something like this. The IN4148 could be a schottky to reduce its T Vf dependence
on translated voltage.

As you see current falls dramatically after C1 charged.

1609416894031.png



Regards, Dana.
 

Veketti

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I replaced the 18kohm resistor with 1uf capacitor but that didn't unfortunately have any effect. U6 pulled at 4.5V. Added 100kohm resistor in parallel like in Dana's example and then U6 pulled at 4.9V.
 

KlausST

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Hi,

Edit:
Tested it by 193kohm in place of diodes and 270kohm resistor to ground and it gave only 0.71V to pin3..
I guess you did not disconnect R6....if so it surely shifts the node voltage down as it draws a lot of current.
You don't have a true voltage divider..

Klaus
 

Veketti

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You're right, I did not disconnect R6... But I quess it makes no differences to the end result as according to the STM1061 datasheet it needs pullup resistor..
 

KlausST

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Hi,

For sure you need a pull up resistor. But not to the center of a "mesurement type" voltage divider.
Usually one connects the pull up to a power supply rail.

Klaus
 

    Veketti

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Veketti

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Wohooo, that's it! So once again desoldered all of the diodes (I'm getting good at (de)soldering) and found out that even with 1% precision resistors had to choose 180kohm and 270kohm instead of 193 - 270 to get the 3.1V from 5.3V. Maybe some multi turn trimmer would have been perfect so that one could easily tweak the correct voltage to get all the juice out of batteries. Then using through hole 10k resistor from regulated 5V supply jumped the pullup. Now it works and draws only ~11uA. I got confused on the STM1061 datasheet as it said:
Connect a pull-up resistor from OUT to any supply voltage up to 6V
even though in example schematic the pullup were connected to regulated 3V so different place than the sensing VCC.

Thank you everybody once again for your help. It is so relieving to get help from you professionals that hobbyist like me can overcome issues which I can't figure out by myself.

edit: too bad the STM doesn't consider 5V users and make these STM1061 over 5V values. That would have saved a lot of hassle and components..
 

KlausST

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Hi,

Instead of a power supply monitoring circuit I'd use a comparator. The benefit is that there is a power supply input and separate signal inputs.
Or use the ADC of a microcontroller ... maybe you could use microcontroller internal comparator.

Klaus
 

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