# Can this voltage divider be made?

Discussion in 'Homework Help' started by marshallf3, Sep 25, 2010.

1. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
202
I'm down with the flu and can't for the life of me figure this simple equation out, may not even be a solvable problem.

Two resistors in a simple series voltage divider circuit.

Top (R1) is connected to +5V
Bottom (R2) is connected to GND
Center of the two resistors is Vout

Now, to complicate matters. Connected in parallel with the top resistor (R1) is a resistor that changes between one of two values, we'll call it Rx.

Rx is either 6,005 ohms or it's 3,551 ohms - always one of the two values.

R1 and R2 can each be of any value but once chosen must remain.

When Rx is 6,005 ohms Vout needs to be 1.2 volts
When Rx is 3,551 ohms Vout needs to be 2.6 volts

A set resistor of any value can be placed in series with Rx, ie if you add 1K to it then Rx becomes 7,005 ohms and 4,551 ohms respectably but that set resistor must remain the same value.

Maybe someone here can hit on it or maybe I'll feel better tomorrow and can think straightly again, right now everything's just a blur.

Can this be done and, if so, what would be the values of R1 and R2?

2. ### Ghar Active Member

Mar 8, 2010
655
73
The solution is approximately R1 = -8390 ohms and R2 = 6668 ohms, without a series resistance.

Adding a series resistance won't help, to give you a positive R1 that series resistance would need to be more negative than -2.5 kohm from my calculation...

So it doesn't look like it's possible.
The problem is that there just isn't enough difference between 3551 ohms and 6005 ohms. Adding series resistance just reduces the ratio between those values, that's why the condition for a series resistance ended up being negative; i.e. you need to increase the ratio between them.

The condition I have for R1 > 0 is that:

$\frac{\frac{5}{2.6} - 1}{Rx1} > \frac{\frac{5}{1.2} - 1}{Rx2}$

For you thats:
$\frac{\frac{5}{2.6} - 1}{3551} > \frac{\frac{5}{1.2} - 1}{6005}$

Last edited: Sep 25, 2010
3. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
202
Thanks, no wonder I was throwing fits over this, may have to add something active to amplify the output differential.

It really matters not how I do it so long as I end up with
When Rx is 6,005 ohms Vout needs to be 1.2 volts
When Rx is 3,551 ohms Vout needs to be 2.6 volts

The component Rx is a regular NTC B curve thermistor so I assume 1,201 and 710 ohms would present the same situation.

4. ### Ghar Active Member

Mar 8, 2010
655
73
You can easily do it with a comparator, I attached something that should work.

It's set to flip when the thermistor equals 4.3k, you can tweak that obviously by changing the reference divider R1& R2 or by changing R4 as the comparison resistance.

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5. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
202
Thanks, that I could do but I need the variable output.

I might try the equation with two or three thermistors in series for Rx today, the alternative gets too complex for what this needs to do.

That 1.2V - 2.6V output needs to be continuously variable as Rx changes.

6. ### Ghar Active Member

Mar 8, 2010
655
73
Oh I see, so it really does matter how you do it and it's not just the endpoints you're worried about
The transfer characteristic of a voltage divider like this would be very non-linear, not sure anymore if that's a problem for you.
Then again, to have those outputs you do need a non-linear transfer.

Putting more thermistors in series won't help because it's the ratio that's important, not the difference.
If you have 3 in series you just get 3*6005 and 3*3551 and you still have the same ratio between them.

With 3 identical thermistors in series you'd just increase that -8390 and 6668 ohms by 3, with 3 in parallel you'd divide those by 3.

Maybe if you could combine several varieties of thermistor... I can only see that helping though if the thermistor just naturally has a much larger variation for the same temperatures you're interested in.

Last edited: Sep 26, 2010
7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
1,188
You can try use this circuit:

The gain mus be equal to
Ku = (2.6V - 1.2V) / ( 2.848V - 2.195) = 2.14395
Then :
R1/R2 = 1.14395
And for Vin = 2.195V we wont Vout = 1.2V
So voltage divider voltage Vd must be larger then 2.195V by voltage drop on a R2.
For example
For R2 = 100K ---> R1 ≈ 114KΩ
Then Vd = 2.195 + (2.195 - 1.2V)/114K *100K = 3.067V

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8. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Wow, inspired! Well done Jony!

You can always count on an OpAmp...