Hi,

I need to help modify this circuit for battery box 3x 18650 (total is 11.1V)
https://www.circuitstoday.com/wire-loop-alarm
https://www.amazon.com/Sackorange-Battery-Storage-Plastic-3x18650/dp/B071ZQ2X2T

[Original components for 9V]
1x condenser: 0.1 uF
1x resistor: 100K
1x resistor: 4.7M
1x MOSFET: 2N700

my buzzer is at 5-12V I would like to power it 11.1V (box 3x 18650)
can you please advise me which electronic components should i use? (I'm not sure how to calculate it) what components do i need for 11.1V?

I want the batteries to last as long as possible in idle mode.


Thanks.

 

Welcome to AAC!

That circuit will work fine on 11.1v, no changes needed.

In idle mode there is virtually no usage of the batteries, a set of 18650 cells will last months (they will self discharge about 2% per month anyway, much faster than this circuit will discharge them).

 

I'm not at all familiar with it, so I wanted to make sure that this mosfet is suitable for a circuit 11.1V
https://www.onsemi.com/pdf/datasheet/nds7002a-d.pdf

I'm a complete novice,
there is some difference when I use it (1x resistor: 4.7M) or (3x connected in series with the resulting resistance 4.7M)?

 

That MOSFET is fine.

What is the difference you encounter between 1 x 4.7M and 3 series resistors... or were you asking if there is a difference? No, there is no difference and none of those values are particularly critical.

 

yes, I don't have 1x resistor 4.7M
So I thought of using other resistors and connect them in a series that will have the resulting resistance 4.7M

PS: Thank you for your advice

 

You're welcome.

The resistance here is not critical, but I'd go lower rather than higher; anything above 100k will work fine - actually you could go much lower but too low and you will impact battery life on idle - 100k would discharge standard 1500mA cells in a couple of years, about the same time they might self-discharge in (for cheap 18650 li-po rechargables).

 

hmm,
do you mean to replace resistance of 100k for 10k,
and leave the second 4.7M resistor unchanged?

do i understand that correctly

 

No, I was refering to the 4.7M. I'd forgotten there was R2, 100k, also in circuit. Ignore what I said above.

The important thing is , when the alarm loop is complete, the voltage at the junction of R1 and R2, a proportion of the battery voltage (Vbatt) must be less than the minimum gate-source threshold voltage (Vgs(th)) for the MOSFET, to ensure it is switched off, ie:

R2/(R1 + R2) x Vbatt < Vgs(th)​

and rearranging the formula

R1 > R2 x (Vbatt/Vgs(th) - 1)​

From the 2N7000 datasheet Vgs(th) = 0.8v minimum, so if R2 = 100k and Vbatt = 11.1v, R1 must be >1.3M. However if you are using rechargable 18650s their voltage after being fully charged is nearer 4.2v, or 12.6v for 3 so I'd use R1 >= 1.48M to be sure.

So, as I said previously, you can go lower than 4.7M just not as low as I said if R2 is 100k. However the values of R1 and R2 are not critical, its their relative values that is important. For example, if R2 was 10k, then R1 could be as small as 150k, but then battery life would be reduced as per the earlier discussion, so I wouldn't recommend reducing it that far.

 

@Irving Hi, I'm trying to use your formula to calculate the resistance for 12V from the original 11.1V


Vgs(th) = 0.8v minimum
R1 = 1300k (1.3M)
R2 =100k
Vbatt = 11.1v

R2/(R1 + R2) x Vbatt < Vgs(th)

100/(1300+100)=14x11.1=155.4 < 0.8v minimum


Vgs(th) = 0.8v minimum
R1 = 130k
R2 = 1300k (1.3M)
Vbatt = 12v

??



I'm still counting it wrong :-D

can you try to break it down more that formula? and show me the resistance calculation for 12V, thank you

 

I think you're using the formula incorrectly..

Vgs(th) = 0.8v minimum
R1 = 1300k (1.3M)
R2 =100k
Vbatt = 11.1v

R2/(R1 + R2) x Vbatt < Vgs(th)

100/(1300+100)=14x11.1=155.4 < 0.8v minimum

100/(1300 + 100) = 0.0714 x 11.1 = 0.79 which is < 0.8 so 1.3M is the lowest value on a 11.1v supply

Using the second formula I showed is easier to do:

R1 > R2 x (Vbatt/Vgs(th) - 1)

so for 11.1v R1 > 100k x (11.1/0.8-1) or R1 > 100k x 12.875 ie 1287.5k (1.3M)

or, for 12.0v R1 > 100k x (12.0/0.8-1) or R1 > 100k x 14 ie 1400k (1.4M)

Basically any value over 1.5M will work...

 

@Irving Hi Hi,

I'm wondering what capacitor I should use and how to connect it:

http://imgur.com/fcdPBOh

I'm confused how to connect it :-D

++ = --
+- = +-



PS: For a ceramic capacitor, it doesn't matter?

 

If it's an Electrolytic capacitor they are polarised and have a white stripe on one side marking the Negative.
A ceramic capacitor can be put in either way.

 

If it's an Electrolytic capacitor they are polarised and have a white stripe on one side marking the Negative.
A ceramic capacitor can be put in either way.

Yes, I know, but I don't know how to connect it :-D

Should it be (- | - = + | +) or (- | + = - | +)

 

Should it be (- | - = + | +) or (- | + = - | +)

Not following what you're trying to say. Electrolytic caps are straight forward. The white striped side (negative lead) is connected to the negative battery line. The positive side is connected to R2. This way (below). Negative lead on the bottom wire.

 

C1 is a non-polarised capacitor, eg ceramic, polyethylene, polypropylene, mylar, etc. - it doesn't matter which way round you connect it.. The one in your picture is an electrolytic and is also a much higher value - you don't need or want to use that.

@Tonyr1084 There is no C2 in this circuit, just C1. The TS was just overlaying 2 pictures to illustrate their confusion over which way round!

 

Irving thanks. I was confused by R2 flying to what I thought was the cap. So where does R2 go? Wait a minute - never mind. I don't want to be accused of hijacking another thread. Forget I asked.