Hi, guys.

I previously read a thread on voltage scaling and the quote below is from that thread.

The original question was for a similar input voltage of 3.2 to 4.2

I need to scale an input voltage of 3.2 to 4.5 volts down to 1 to 2v ( or 0 to 1v to drive a 1v panel meter)

Input voltage will be 5v.

If anyone can calculate the resistors for that would be great.

The input of pin 3 will have a 10k thermistor to positive and 10k resistor to negative to create a potential divider, so the voltage goes up with with increasing temperature.

Also, is there a transistor alternative rather than an op amp ?

So you need this circuit:

Resistors R3 and R4 need to be very accurate.

 

Close as I can come with 1% resistors. You will, off course, have to use a rail-to-rail output opamp. The 5V supply must be accurate to +-1% (5mV)

 

Transistors will not work as well as an op amp for this application.

 

Close as I can come with 1% resistors. You will, off course, have to use a rail-to-rail output opamp. The 5V supply must be accurate to +-1% (5mV)
View attachment 81346

Are you sure you don't want to do the write-up and submit the homework for the OP in addition to doing their work for them?

 

Thanks all for the reply to my first question as a new user to the site.
I totally forgot about simulation, I don't do as much hobby electronics as I used to.
But hopefully I can be of some use in helping others with my experiences, which is mainly RF electronics (I operate ham radio here in the uk).

 

Are you sure you don't want to do the write-up and submit the homework for the OP in addition to doing their work for them?

I'm sure you can tutor the OP in an algebraic method. As it is, I didn't notice it was posted in the Homework forum, and I view new posts, I sometimes don't notice which forum...

I did it based on experience, and trial and error. Let's see if the OP can come up with the equations...

 

Okay -- and I know that it can be easy to miss that a post is in Homework Help (I've certainly done it myself). Please try to watch a little closer to avoid just handing out solutions as opposed to trying to guide the OP into finding the solution for themselves.

 

Hi, I only came across this site a couple of days ago so i'm not used to it.
The project is simply for a practical hobby use, and is not part of any work for tuition.
I will look at where I am posting threads in the future.

 

Hi, I only came across this site a couple of days ago so i'm not used to it.
The project is simply for a practical hobby use, and is not part of any work for tuition.
I will look at where I am posting threads in the future.

Not to worry; you are just fine.

We have a general approach that is different for homework assistance compared to more general questions. The idea behind homework assistance is to give hints and suggestions and ask leading question but to try to get the poster to do as much of the work for themselves in the belief that this is, overall, the best way for them to learn effectively. As part of this, most of us that seriously haunt the Homework Help forum are pretty stringent about insisting that the poster show at least some effort, no matter how shaky, to solve their own homework. But, of course, it works better for some than others. Then there's always the issue of people posting homework in non-homework forums trying to trick folks into doing their homework for them. Sometimes it's pretty obvious and we move the thread, other times the person succeeds and gets a freebie here and there. There will always be hiccups along the way.

 

Hi, I only came across this site a couple of days ago so i'm not used to it.
The project is simply for a practical hobby use, and is not part of any work for tuition.
I will look at where I am posting threads in the future.

Good, that gets me out of hot water for giving away the store...

This problem is solved by fitting the two given points with a y=mx+b linear fit.

A little algebra that m = (2-1)/(4.5-3.2) = 0.769
and b = -1.462

The trick is to set the gain of the overall opamp gain(from the input via the non-inverting input) to be 0.769. I did that by making the gain = [R4/(R3+R4)]*[1 + R2/(R1//R5)]

R4/(R3+R4) = 3.83K/(3.83K + 6.04K) = 0.388
R1//R5 = R1*R5/(R1+R5) = 34K*14.3K/(34K + 14.3K) = 10.066K

So the actual overall gain is 0.388*[1+1oK/10.066K] = 0.388*1.993 = 0.773, pretty close to 0.769

Injecting a positive current into the inverting input from the 5V rail gives us the negative offset required.