Trying to understand a low voltage warning circuit

[Ch33f]

Hi there everyone,

Introduction:
I'm kind of a newby in electronics but I am eager to learn and want to actually understand how things work instead of just (re-)using them. This is why I'm posting here. I hope this is the right form to post it on. In order to accomplish this goal I read Electronics for Dummies, so you guys might know what kind of a knowledgebase you are dealing with.

The question:
I am currently trying to understand a circuit that is supposed to warn you when your battery is getting low. In particular the top right i.e. the first circuit on this page: http://homepages.paradise.net.nz/bhabbott/lvw.html
I basically want to understand why current flows the way it does and how to calulate the voltagedrops that occur and the nessecary resistors.

To save you guys some googling, here is the datasheet for the TL431: https://www.fairchildsemi.com/datasheets/TL/TL431.pdf

My Thoughts:
I am going to assume this circuit is configured to light up when the battery voltage drops below 7.2V which results in R3 being chosen as a 2.5 kΩ resistor.

What I understand (I hope):
The resistors R2 and R3, wich act like a voltage devider, are used to configure the voltage at which the LEDs are supposed to light up. With R2 as 4.7 kΩ and R3 as 2.5 kΩ the Vref (pin 1) for the active zener at 7.2V battery results in 7.2 * 2500 / (4700 + 2500) = 2.5V which is the breakingpoint of the active zener. That means as long as the battery voltage is above 7.2V the zener is conductive from pin 3 to pin 2. I'll call this the ON-State from now on.

So in the OFF-State the current can only run through R1 and the LEDs (?). Since regular LEDs (when on) draw about 15mA of current the battery would need 2200 * 0.015 * 2 = 33V in order to light the LEDs up => they are off. I am actually not at all sure that my calculations and my conclusions in this paragraph are correct, so please correct me if I'm wrong.

Now for what I can't wrap my head around: In the ON-State current can flow from pin 3 to pin 2. But what does that help the LEDs? How to calculate the currents and voltages now?

Source of the problem (I guess):
I guess my problem is that I haven't yet grasped exactly enough how the TL431 works and a general problem of mine is filtering out the values that I need from a datasheet. I am kind of overwelmed by all the information in datasheets and I have trouble finding what I need, or even knowing what I need in the first place. In the datasheet of the TL431 I read something about the "output voltage" but I am not quite shure what that referes to. Applying the calculation in the datasheet the output voltage should be (1 + 4700/2500) * 2.5 = 7.2V but again I don't really know what that referes to or where that voltage is applied.


I hope this is not too broad a question and you guys can help me figure this out.
If you know any helpfull tutorials or books on this topic I'd be glad to hear your recomendations.

Thanks in advance,
A eagerly learning newby

 

[Arouse1973]

Hi there.

Well you do have a good grasp of how the circuit works but you voltage calculation is not correct. The other thing is the author is wrong about the battery cut-off voltages. 8 cell NICD cut-off is not 7.2 Volts the cut-off for a 6 cell pack is 7.2 Volts. That doesn't really matter for what you are trying to understand. The LEDs will be on when the supply is approx. 7.2 Volts. So say the LEDs have a forward voltage drop of 1.5 Volts which equals 3 Volts because you have 2 of them. So you take 7.2 Volts and take way 3 Volts for the LEDs. This leaves 4.2 Volts across the resistor. Now divide this by 2200 and you will have your current through the LEDs which is approx. 1.9 mA.

When the device is off the current through the resistor that is connected to the LEDs will be what ever supply voltage you have minus 2 Volts for the TL431, now divide this by 2200 and you have the current through the resistor when the LEDs are off which will be approx. 2.4 mA if the supply is 8.1 Volts. This will reduce as the voltage falls during discharge up until 7.2 Volts which it will be 2.14 mA just before it switches off. There will always be a small current through the LEDs when the LEDs are off in this case around 170 uA which is not enough for them to light so they appear to be off.

Thanks
Adam

 

[Ch33f]

Ok, so when the active zener is off the current can only go through the LEDs and they are provided with approximately 2 mA which is enough to light them up. So ... I got that now but now I'm confused by why they are off when the active zener is on...

So... you said when the supply voltage is 8.1 V the active zener drops 2 V.
Now from my knowledge the voltage drop across each branch of a parallel circuit is the same, so since 2 V is droped across the 2 LEDs they are not going to light up and even are only barely conductive.
But here I am struggling:
How can I determine what the voltage drop actualy is in a parallel circuit? Why is it determined by the active zener and not the LEDs?
And a little side question: Where did you find the reverse voltage drop in the datasheet? Is that delta Vref / delta Vka ?

 

[Arouse1973]

When the active zener is on it hogs all but a very very small amount of the current from the LEDs. It's like placing a low value resistor across the LEDs.
Thanks
Adam

 

[Ch33f]

Ooookay... why? Or how do I know that / figure that out without measuring it?

I read something along the lines of I1/I2 = R2/R1 in parallel circuits but since those are diodes and (to my knowledge) that equation only applies to ohmic components. So how can I figure out how much current flows where?

 

[Ch33f]

Btw. thanks a lot so far! You already helped me get a better understanding of electric current!

 

[Arouse1973]

Ooookay... why? Or how do I know that / figure that out without measuring it?

I read something along the lines of I1/I2 = R2/R1 in parallel circuits but since those are diodes and (to my knowledge) that equation only applies to ohmic components. So how can I figure out how much current flows where?

The current in the LEDs when the active zener is on is so low it's not worth worrying about. To work out the resistor value for the LED, You have to look at the datasheet for the LED and choose a current that is anywhere up to the maximum operating current for the LED. You then can work out your resistor value by looking at the voltage drop for that particular current. But then how much current do you need to use? Well it depends on the application and in most cases it's try it and see.

If you don't have the datasheet then you have to pass a known current through the LED and measure the voltage across it. Bit of a faff, And then of course different colours produce different voltage drops at the same current. So you see it's not that straight forward. Standard RED LEDs can range from 1.5-1.7 Volts for a few mA's, higher output RED LEDs could be as high as 1.9 Volts. Always check the data sheet, it's the sure way of getting it as near as you can.

Adam

 

[Ch33f]

Ok but which way do the dependencies go? I was looking at some datasheets for LEDs to get an idea and it seems like the voltage drop depends on the current running through it. That would mean the resistance is dependent on the current. But how does that work? I always thought the current depended on voltage and resistance.

This seems to me like an unsolvable circular dependency.

 

[Arouse1973]

Ok but which way do the dependencies go? I was looking at some datasheets for LEDs to get an idea and it seems like the voltage drop depends on the current running through it. That would mean the resistance is dependent on the current. But how does that work? I always thought the current depended on voltage and resistance.

This seems to me like an unsolvable circular dependency.

Diodes are different to resistor in the fact that diodes are non linear and resistors are linear devices. Diodes have a dynamic resistance which changes with current. If you try to work out the resistance of the diode by using V/I then the answer you get will be way off, maybe 10 times. If you need to work out the diodes dynamic resistance then you need to work this out from the VI curve of the datasheet or work backwards or forwards from a given dynamic resistance that the manufacturer has quoted on the data sheet. A simple approach to this is to draw a line tangent to the point on the VI curve you are interested in and workout the value of RD using delta V / delta I (change in Voltage / change in Current).

Lets do an example:
We know we have to decide on one parameter first to be able to work this out. Well because LEDs give off light which is roughly proportional to current, I think the current might be a good place to start. So the LED I have chosen is http://www.mouser.com/ds/2/216/WP7113ID-48891.pdf a standard HI Red LED.


You can see the intensity is proportional with respect to current.



But you can see the voltage across the diode is not proportional. So for this LED I have just chosen 12 mA, no reason for this. So the dashed line is the tangent line and is best that I could do in Microsoft Word. the dotted line intersects the current @12 mA. The difference in voltage between the bottom of the two lines is approx. 1.94V -1.8 V and the current we know is 12 mA. So the dynamic resistance of the diode at that point is 1.94-1.8 / 0.012 = 11.6 Ohms. But if you did it using V/I then it turns out to be 161 Ohms.

But I hear you say 12 mA * 161 Ohms is 1.94 Volts so it works...........! But now what if you want to workout what the voltage change across the diode is from one current to the other. Lets say from 12 mA to 14 mA we do this by increasing the supply voltage, Looking at the graph it should be 1.95 Volts but if we try and think of keeping the resistance the same because we just worked it out at 161 Ohms, like you would with a resistor divider for example, then the result would be 14mA * 161 Ohms = 2.25 Volts. We are way off. You can find all this information on the graph, but if you wanted to work it out you would have to do another graph for 14 mA and you should find only a small increase in dynamic resistance from 12 mA to 14 mA.

Thanks
Adam

 

[Ch33f]

Ok well, now I know how to calculate the dynamic resistance of LEDs and I am sure that will come in handy at some point but it doesn't seem to help me understand why the current mostly flows through the zener. If I understand it correctly the dynamic resistance is a function of current. As in R(I) = ....
But I don't know how much current flows through the LED because (as far as I know) the current in a parallel circuit is a function of the resistances and the total current over all parallel paths. This is that circular dependency I was talking about. How do I solve that?

This is how I feel right now (imagine a baby trying to force that round bit through that triangular hole without realizing that it can never fit). Apart from the feeling that I am still missing some pieces, I feel like can't even fit the ones I've got.

For now I'd be content with "just" learning how to predict the behaviour of currentflow in parallel circuits. Less as in "Exactly how much amperes flow where?" but more as in "This will hog most of the ampere and the rest is negligible.". Is that maybe just a thing that zeners do?

 

[Arouse1973]

You seemed confused earlier when you said "That would mean the resistance is dependent on the current. But how does that work?" I answered that for you and now your saying, you didn't need to know that! So I guess just stick with the zener hogs most of the current, it's good enough.
Adam

 

[Ch33f]

Ok, so when I see a zener of sorts in parallel with something else I will assume it consumes most of the current.

Just one followup question: For what Voltage should I configure the circuit when I want to use it for a 9 V NiMh battery with 250 mAh?