# Dynamic shifting of resistance?

Discussion in 'The Projects Forum' started by psychoboy, Jan 9, 2014.

1. ### psychoboy Thread Starter New Member

Jan 9, 2014
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I googled Ohm Amplifier and got all the speaker links you might ever want.

so....I searched thru some threads on here and found some that got near what I was looking for, but I couldn't find a clear solution,

so..... I registered and thought I'd ask outright.

I have a level gauge that shows full at 2-5ohm and empty at 105-110ohm.

I have a level sender that gives 0ohm at full and 70ohm at empty.

I imagine I could piggy back a resistor on the sender and get the gauge to read correctly at one end, but it would come up half short or way long on the other.

Is there a way to amplify the resistance dynamically so I can be accurate on both ends?

if my math is correct, i'd need a flat 3ohm push at 0 (sender output), with a .5ohm per ohm linear rise.

full: 0+3+(0x.5) = 3
empty: 70+3+(70x.5) = 108

Thanks for any help!
Hope I'm not coming off as a complete newb.....

Last edited: Jan 9, 2014
2. ### MikeML AAC Fanatic!

Oct 2, 2009
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A current source applied to a variable resistor (your sender) makes a voltage proportional to the resistance value.

An opamp can be configured to provide independently adjustable gain and adjustable offset.

3. ### wayneh Expert

Sep 9, 2010
15,742
5,866
Your level gauge is likely not so much an ohmmeter as it is a current meter. That is, the needle deflection is proportional to the current passing through the meter. Just a guess. Or it's responding linearly to applied voltage.

4. ### crutschow Expert

Mar 14, 2008
21,125
6,011
Assuming the sender is a rheostat in series with the gauge you might try adding a resistor in parallel with the gauge to reduce its sensitivity. It's value will depend upon the gauge resistance. Can you measure that?

Last edited: Jan 9, 2014
5. ### psychoboy Thread Starter New Member

Jan 9, 2014
5
0
crutschow: I can measure it, but i don't have it in front of me at the moment.

wayneh: you are likely correct. in the gauge's original setting, it passes voltage to the sender which is grounded. I'm guessing the sender varies its resistance and the gauge reports the voltage thruput (as my non-EE brain sees it)

MikeML: if a pile of resistors doesn't solve it, that likely would. excellent idea.

I think i'm thinking about this from the sender end rather than the gauge end...and that's what's screwing me up (well, that and the fact that my electronics usage is usually limited to hooking up existing compatible components....not trying to build them)

6. ### MikeML AAC Fanatic!

Oct 2, 2009
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Some combination of supply voltage V2, R1 and R2 will let you fit it at the two end points. It is just trial and error...

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7. ### psychoboy Thread Starter New Member

Jan 9, 2014
5
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i assume, once i have the impedance of the gauge itself, there is a way to mathematically narrow down the ranges of R1 and R2?

V2 is ~13.5 volts.

R2 is going to be the shift (theoretically 3 ohm).

R1 is going to be balanced with the gauge to augment it and provide the additional/variable resistance?

or it is truly trial and error, putting a pair of pots in and narrowing the field manually?

8. ### MikeML AAC Fanatic!

Oct 2, 2009
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You could spend all day calculating it, or five minutes with two pots and a variable power supply

9. ### psychoboy Thread Starter New Member

Jan 9, 2014
5
0
look kids!, i'm not as dumb as i think i am!

thanks guys!

10. ### crutschow Expert

Mar 14, 2008
21,125
6,011
If you can measure the gauge resistance then it's not hard to calculate the required resistors values.