Determining values for zener diode

Hi guys,

I am quite noob in electronics.

I would like to know how to determine resistors and zener diodes values for a battery life indicator similar to this :

And if I want to use a battery source of 1.5V, how will the values change?


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Orngrimm4 years ago
The problem is, that you need around 1.5-1.6V for your LED alone... So if your battery dies @ 0.7V, you cannot light the LED at all...
Also most transistors have a voltage-drop (Since they also are a diode) of around 0.5-0.6V, leaving only 0.1V or so for your LED.
IF you, however, go for a double-sourced circuit (Where the Measuring and lighting is provided by a 9V-Battery and the AA-Battery does NOT have to supply the "Batteryindicator") then it is totaly doable with some massive mods.

See, the basic prinziple of the circuit you linked is the question of the voltage is enough to cross the zener-level or not.
But with normal zener-diodes only going down to 1.8V (like the MMSZ4678 or DDZ9678) you have to do something with the voltage to being able to compare with a zener-diode. Like double it to have a switchlevel of 1.8V which becomes 0.9V at the AA-Battery-Side.

Attention: At this point it becomes purely educational since the introduced mods are WAY more complicated than if we do it directly with an operational amplifier.

Looking at gives you a simple non-inverting amplifier. Most OpAmps run with 9V quite well... Since the Gain is 1+(R1/R2) and you want a result of 2, you have to choose your 2 resistors that same value:
Gain = 1+(10k/10k)=1+1=2 if i would take 10k... But you can take whatever you want... But in the kilo-region you are good... And you already have those 10k's around... so... :)

So you simple replace the original 6.2V-Zener with a 1.8V one. Now you "split" the supplys: The AA-Battery + goes over the amplifier (factor 2) to the Zener diode. The 9V + goes to the two LEDs.
Presto. Your circuit should work.

However, here is what i would do: and save some parts:
First, i want to use it as less energy as possible thus eliminating the green LED.

Part1: Theory
Every diode has a forward-voltage. Normal silicone is around 0.6V, Shottky around 0.2V.
If we could COMPARE those added voltages (0.6+0.2=0.8V) to the voltage of the battery, we would be golden!

Part2: The making
we follow the basic principle of 2-part: One part is the Measuring and signaling and one part is the thing to be measured. Thus we need the 9V-battery in our measuring-equipment and only look at the AA-Battery without taking any (well... a few pico-Amps are to be forgotten. Your battery has a higher selfdischarge than that!) energy from the 1.5V-AA-Battery.
Both grounds from both battery have to be connected together also.

First we create the voltage to compare it against:
We wire up a (from +9V) 10k-Resistor followed by a 1N4148 (Silicone-diode) followed by a shottky diode like a BAT85 to ground. Both diodes have to conduct. They have to face downwards to Ground...
Now you can measure 0.8V across the two diodes. --> Thats what we compare against.

Now to the measuring-part:
We use a simple OperationalAmplifier to create a comperator to compare the 2 voltages:
The upper resistor would be the 10k and the lower resistor would be the 2 diodes in series.
Now if the voltage drops below the voltage we created with the diodes, the output goes... Hm... DOWN. Darn. Not what er wanted... But luckily, thats very easy fixed!
We simply switch the 2 inputs of the opAmp in the circuit. Our Reference goes to the inverting input of the opAmp and the voltage from the AA-Battery goes to the non-inverting.
Now the question goes like this:
If the reference-voltage is higher than the Voltage from the battery (like when the battery dies), the output goes HIGH.
THATS what we wanted! :)
Now add the LED and 330Ohms at the output of the Opamp and you are ready to Rumble...
kurtselva (author)  Orngrimm4 years ago
Hey thanks for your time for the detailed explanation.

After much thought, I have decided to do away with the 1.5V battery source and instead, use a 9V battery source to monitor its own battery life. So the schematic I posted in the main topic should work right?

But I want to make a few changes:

1) As you said, I want to do away with the green LED. So I would only need 1 red LED which lights up when the Voltage becomes low

2) And the Voltage when the red LED lights up I want to be at 7.5V.

So with these 2 changes in mind, what changes do I need to make with the original schematics. As in how many diodes/resistors/transistors would I need and at what values?

Sorry if the qn is too basic but as I said, I am noob to electronics.

Would greatly appreciate your help : )
No question is noobish if you keep in mind that it is better to aim for a simpler target and not for a high-tech-one you probably never will reach.

To 1):
Wise. Cuts the discharge by factor 44! :)

To 2):
The trigger-voltage is around 7.8V already with a 6.2V-Zener if we make some changes. The trigger-voltage will be zenervoltage+LEDvoltage which equals in 6.2V+1.6V=7.8V
If you want you can go for a 1N5233B with 6.0V zener-voltage which will end in 7.6V

Now to the changing-parts-game... :
I now could explain how to rebuild the complete circuit and make a new one for this simpler tasks since the original circuit is a bit overkill for the tasks at hand now and also quite nonflexible in terms of trigger-level... Or i could simply link to a schematic i made just for you...

This circuit uses a cascade of 2 transistors to switch and a simple voltage-divider to adjust when (@ what voltage) to switch.

With the given values and a normal LED it will NOT light the led till around 7.48V (Thats when the LED gets a dim 1mA of current) and will be bright @ 10mA when it reaches 7.2V.
You can adjust those values yourself by replacing the 2 R_adjust with a potentiometer of 520k or so and turn till you have the exact triggerlevel you want.
But i choose common resistor-values you can use and which will land in the ballpark you are looking for (7.5V).

In "waiting-mode" (No LED is lit) the circuit consumes around 107uA which is 0.107mA which is really lower than the self-discharge of the battery itself... :)

The highes current-use will be @ around 6.5V when the LED will be supplied with around 14.8mA.
So no current-limiting-resistor in series to the LED is needed here... Saves a part. :)

Edit: Oh: And if you want to fiddle around with the simulation yourself, get LTSpice (It is free) and go for it. :)
All you've done there is a kind of cascaded common emitter amplifier. Its not a comparator. You could knock one up, for this voltage supply, with a CMOS schmitt trigger
Thats true and the whole point: with a comfigurable amplifier like we have here, we can adjust the voltage of switching very precisely and simple without the need to rely on zener-voltages.

Also you are right: A schmitttrigger would also work... And if you talk about that: It may even work better since it has a really sharp toggle-point.
I was simply too focused on the 2-transistor-type-circuit when i whipped up the circuit above.
Mea culpa maxima! :)

Here is what steveastrouk suggested:
We use an inverting schmitt-trigger (74HC14) a "sensor" and simply adjust the input (The battery-voltage) to trigger it at the correct voltage.
Since i wanted to keep it simple, i used only single resistors and took those from the E12-series which are normally common.
Because of that, we miss our 7.5V-Target a little and go for 7.37V

Here it is...
Freshly drawn for you:

Have fun!

Edit: Oh: Dont forget to connect the 74HC14 to the supply of the battery as well. Thats not shown in the schematics because it is internally wired for this symbol...
Oh... Just forgot: If you want to cut down on the power used by the circuit, replace the 10k with a 1meg and the 5.6k with a 560k.
This cuts down the current in standby to a few microamps...
kurtselva (author)  Orngrimm4 years ago
Thats really helpful guys, appreciate it..

Just two more doubts:

I am connecting the circuit on a breadboard and need one which uses the minimum number of tie points (As I am using the same breadboard for other circuits). So of the 2 schematics, which one takes up lesser space?

And as this is for a school project, cost is also considered. So which is the cheaper circuit?
If i click the items used for the 6V-version with the 74HC14 i end up with 0.33 swiss francs a piece if i take material for 10 pieces. Add the cost of the breadboard to that.

Site? depends on how messy you want it... I could make the thing for sure fit in between the legs of the 74HC14. Doable with trough-hole-components and very easy with SMD-components (which would be a LOT cheaper by the way). So in theory no breadboard is needed at all. But the result wouldnt be pretty... :)
If you want to fit it on a breadboard, i would say the needed size is just the overlay-size:
5x7 holes should do: 4x7 for the 74HC14 and one line alongside with the 3 resistors (placed high-ways) and the LED.

Also you could (for the cost of maybe 2 cents) raise the 74HC14 on 2 connector-rows and fit the 4 parts underneath there. Letting the LED stick out on one end would look nice. :)
Also this would reduce the needed space to 4x7 holes.

If i check the price at for those breadboards, i get 5.93 Euros for one with 2318 holes.
That makes 0.0026 Euros per hole.

So the price will be around 0.43 Swiss francs a piece. if you go for 5x7 (non-messy).
kurtselva (author)  Orngrimm4 years ago
Thanks man!

I have bought the components. Now need to connect all of them to the breadboard.

So for the 74HC14 and CD40106, there are 14 pins,which ones do I need to connect to where?
For the connections of the 74HC14, see a datasheet like on page 2.
So if you are building the 6V-one like i posted here @ then you can use any of the following pairs:
where the first pin-# is always the imput and the second is output.

To save energy, you should tie the inputs of unused gates to a defined potential. So either GND or VCC. If not sone, the unused gates may pick up random noise and switch erratically --> Switching uses power.

Also here is the 3V-Clamped version with the CD40106 for the 6V-Battery which switches @ around 5.009V:
NOTE: The VDD is the supply for the CD40106! DONT supply the CD40106 with the 6V directly...
The Zenerdiode used for the supply of the CD40106 is a 4.7V-Type.
With only one resistor on each side the levels were too far off with E12-Series resistors... So i made them more exact with the circuit above.

Also the version for your 9V-Battery:
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