NEWB: Voltage divider question

[wireburn]

Please be kind, first post!

I have a design for a low fuel circuit for my motorcycle that involves a voltage comparator, a reference voltage and a fuel level sending unit (variable resistor). The sender goes from about 10 Ohms when full to 90 Ohms at empty. I want to make a voltage divider circuit with the sender, but I'm having problems limiting the current sunk by the sender and the resistor on the other side of the divider. My question is, how do I limit the current going into this divider circuit so I'm not smoking the resistors? No problem with the reference voltage because I'm using 10k+ values on the divider network. I can post a schematic if it helps. I understand about Ohm's law and why I'm getting so much heat, but I don't know how to avoid it!

Thanks in advance for your kind assistance!

-Mike

[Drew1]

A schematic would help, what is your other resistor value? If you are using the battery voltage to provide current to the sender, and it is 10 ohms, you could be seeing .6A to 1.2A through it depending on your battery voltage. You need to find out what the maximum wattage is that the sender is rated at, then decide what the divider reistor should be, then whether you can sense it with the comparator.

[wireburn]

Thanks for the reply, Drew.

The other side of the divider is 33 Ohm. I only picked that because the resulting voltage values are convenient (off the top of my head something like 3-8V). I could go way higher on that value, but of course, my readings would go way down and the range would narrow significantly. I don't know anything about the sender's rating, but something doesn't feel right about putting that much load through there and then sticking it in my gas tank!

So, really, there's no other way to limit the dissipation through that circuit? I can post a schematic, but a hand-drawn scanned sketch will have to do. I'll work on that today.

Thanks,
-Mike

[Drew1]

The most straight forward approach to limiting the current is just what you did, a current limiting resistor/voltage divider.
The problem is that the current is too high and heating the Sender and 33 ohm resistor. Assuming a 12V Battery, 10 ohms sender +33 ohms resistor = 43 ohms which is about 3.3W! This is huge for resistors. Your current limiting resistor should be on the order of 1200 ohms, this will give you about 10mA current through the circuit. You can expect about .837V when Tank is Empty and about 0.099V when tank is full. This should be well within the comparators ability to detect. You just need to set your threshold to about 0.8V
The basic circuit is the easy part. The more complicated aspects are designing for the environmental effects on the circuit such as temperature extremes, vibration, and electrical noise.

Hope this helps,

Drew

[wireburn]

Yes, that does help. I ended up padding both sides of the divider, giving a .5V (6.4-6.9V) swing with only .05w dissipation. When V-ref at the + input is set to 6.8V, when triggered the hysteresis will pull it down to 6.5V. See any problems with this arrangement?

Thanks,
-Mike

[Drew1]

Mike,
I took a look at the circuit and I agree with the Voltage from the sender when full and empty; however, I think there is a problem with the Vref and Hysteresis portion. The schematic you provided had a net error I think. I am assuming the 4.7K/100K/10K-pot junction are common and tied to the V+ input right?
The first issue would be the 10K pull-up on the output. This would be an indicator that this is an open collector output from the comparator am I right? There really isn’t a need to pull it up except to provide leakage current into the output vs your LED providing the leakage. This would be an okay thing to do except you will also feed current through the 100K resistor and set your threshold well above 6.8V at the V+ Input. Your LED will also conduct current through the 2.2K and through the 100K as well. The result will be that your circuit will never trip because the threshold is too high.
I also analyzed the circuit using a blocking diode placed in series with the 100K feedback resistor and did not get 6.5V when tripped. Without doing all the math to figure out superposition current s etc…The best case would be 6.64V. This is based on a parallel 100K/4.44K (the 4.4K is the calculated value of the 10K Pot). The actual value will be a little more than 6.64V.
Might try a 49.3K feedback instead of the 100K and place a diode in series with the 49.3K with the Cathode toward the output of the comparator. I think you can loose the 10K pull-up on the output.
The point of the blocking diode is to block any conduction path through the LED, any current supplied by the comparator output should it not be an open collector, or the 10K pull up should you decide to keep it. If I am interpreting this wrong or if you disagree please let me know as I am not perfect and could be off in the weeds.

[pebe]

Hi Mike,
I would like to add a couple of points to the discussion.

Drew is right about the pull-up resistor; it is not required unless you are using a comparator (which doesn’t have a pull-up).

But your circuit shows that the sensor is 10ohms when empty, whereas your first posting shows 10 ohms when full. If your first posting was correct then you will need to move the sensor to the lower arm of the divider.

My calculations show that your circuit should work with the values you have selected. The mid-point voltage at the –ve input is 6.55V, and in order to get that as a reference at the +ve input [i]without[/i] the 100K feedback resistor connected, the pot will need to be set at 4.13K.

At the +ve input, the 4.7K and the 4.13K are in parallel as far as feedback is concerned, giving effectively 2.2K. This forms a potential divider with the 100K to give .301V hysteresis.

So if you set the pot for the LED to turn on with 6.92V at the input, then it will turn off when the tank is half filled, ie. at 6.62V

(edited 21/12)

[wireburn]

Yes, pebe, the values were calculated using the spreadsheet I worked out based on hysteresis calculations, and yes, the circuit works as drawn. I had to add a 47uF smoothing cap to the sender input because I think at some points along the way the wiper loses contact and "spikes" the LED for a second. The result is better than I had hoped, works very well. Now to see if I can ride in the dead of winter and the heat of summer and still get reliable readings!

BTW, this is a comparator, so I need the pull-up.

Again, thanks for the help!

-Mike

[Drew1]

Wow was I off base. I guess I anlalyzed it following the currents and had totally different results. I cant argue with an operating circuit but I dont understand it. I'll have to take another look and see where I went wrong.......

Drew

[Drew1]

Been busy past few days but have taken a second look. My calculation for the mid-point at –V is 6.65V as follows: Assuming the sender is linear; the resistance would decline 20Ω’s per ¼ of a tank of gas used. If we assume the low fuel comes on at ¼ tank, we can say the resistance declined 60Ω’s to give us 30Ω’s. (470/(470+470+30))*14V= 6.78V. Using the same assumptions but with a tank at the halfway point, the voltage would be (470/(470+470+50))*14=6.65V. An Empty tank would be (470/(470+470+10))*14=6.92V
Since Mike’s original idea was to make the threshold 6.8V’s I am going to assume he had the number that I had.
Where I went wrong was that I had calculated the pot value based on the 100K not being in the circuit and then adding it back in assuming the 10K pull-up was there. The result was the threshold at +V going much higher. I then set about trying to fix that problem with a diode and setting the hysteresis to 0.5V forgetting the fact that all that had to be done was tweek the pot down until the threshold was 6.8V effectively negating the effect of the feed-back and pull-up resistor.
So with this new revelation; assuming the 10K pull-up and 100K feedback are in the circuit, and that the current contributed by the LED is negligible so not counted, the pot value would be 4.255K which I calculated as follows: 4.7K in parallel with (100K+10K)= 4.507K, (14V-6.8V)= 7.2V, (7.2V/4.507K)=1.598mA, (6.8V/1.598mA)=4.255K.
When the 6.8V Threshold is reached and the comparator output switches to 0 (ground), the voltage at the +V input would be 6.507V which I calculated as follows: (4.255K in parallel with 100K)=4.081K, ((4.081K/(4.081K+4.7K))*14V=6.506V. This still equates to 0.294V of hysteresis and the LED will turn off when the sender is at 71.37Ω’s, or a fuel level of a little over ¾ tank.
To be honest pebe; it has been many years since I actually did a textbook analysis of this type of circuit. I am confused about the statement of the 100K and 4.7K being in parallel for calculating the hysteresis since the 4.7K is tied to 14V and the 100K tied to ground when the circuit is tripped (LED ON). Please let me know if you agree or disagree with my analysis as I am quit curious to know if I have lost my edge

[wireburn]

Your numbers are very, very close to what I measure in the actual circuit, so I'd say you're right on with your analysis. I found the calculations on the 'net for the hysteresis function and just plugged in my numbers to get the resistor values. Sounds like you actually know what you're doing (unlike me!).

-Mike

[Drew1]

Good feedback Mike; nice to know it worked out

[pebe]

Hi Drew,
I fully agree with your analysis and calculations. I was just doing an approximation to check that Mike’s circuit would work, so your calculations are to a higher degree of accuracy than mine were.
[quote] ….I am confused about the statement of the 100K and 4.7K being in parallel for calculating the hysteresis since the 4.7K is tied to 14V and the 100K tied to ground when the circuit is tripped (LED ON).[/quote]
I think you may have misread what I said. I was saying that for the purposes of calculating feedback, the 4.7K and the 4.13K resistances could be considered to be in parallel. Here is how I calculate feedback in a case like this:

Thevenin’s theorem states that two resistors in a network can be replaced with a single resistor equal to the two in parallel, and fed from a voltage source equal to the open-circuit voltage at their junction. So if we ignore the 100k for a moment, the +ve input of the comparator is fed with an equivalent single resistor of 2.2k from a virtual voltage equal to 14*4.13/(4.7+4.13).

As the virtual voltage won't vary (because operation of the circuit will not affect the 14V supply), then the feedback voltage at the +ve input of the comparator will result from a potential divider made up of 2.2k and 100k, and its value will be 14*2.2/100+2.2 = 0.301V. The result is not as accurate as yours because I only approximated the 4.13k.

I hope that clarifies things.

[Drew1]

Thevenins! I knew there was something along those lines but could not recall the specifics and is why I questioned my results. I remember doing circuit analysis just as you described by taking the resistors to a common point, calculating and then flipping it 180 and doing the same thing. I have come to rely on the brute force method of following the currents rather than the formula's though because for me anyways it is faster than finding the ap notes, scrolling the data to find the formula etc....... Merry Christmas

Drew

[pebe]

And a Merry Christmas to you and to all members of the forum.