# Need 0.094v as op-amp input - Help, please

Discussion in 'The Projects Forum' started by Jack_K, Oct 5, 2009.

1. ### Jack_K Thread Starter Active Member

May 13, 2009
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I have an op-amp circuit that is based on the slope of a line (y=mx+b). The circuit needs to receive an input of 0.53v to 5.90v and put out 0.40v to 3.50v respectively.

That means the equation is Vin * 0.577 + 0.094v. The circuit, less the 0.094v input, is shown below.

How can I create a 0.094 volt or so source? Maybe the saturation of some semiconductor?

Any ideas?

Jack

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2. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
69
Use a diode to 0V, fed with a resistor from V+, then put a potential divider (or preset pot) across the diode to reduce it from around 0.7 to .094

As long as the feed resistor provides slightly more current than the pot takes, you should get a reasonably stable voltage.
eg. feed it with 4k7 and use a 1K pot across a 1N4148.

You could put another fixed resistor in series with the pot to reduce the adjustment range, ie. a 2k2 would put the pot max down to somewhere around 0.2V

3. ### Jack_K Thread Starter Active Member

May 13, 2009
115
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It seems like all that resistance would mess up the gain of the op-amp and make calculations pretty difficult (for me).

Jack

4. ### MikeML AAC Fanatic!

Oct 2, 2009
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I think your question is "where do I get a reference voltage" which is not effected by the cars battery voltage and temperature? Look at zeners, IC voltage reference chips, LM431, etc.

Another problem, the gas gauge sender may or may not be fed from a regulator somewhere in the car, in which case the voltage you are taking from across the sender could be ratiometric with the battery voltage, or not. You need to know which, otherwise your carefully derived equation is meaningless.

5. ### Jack_K Thread Starter Active Member

May 13, 2009
115
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No, I am unable to find a reference voltage source that will go that low (0.094v).

It's a gas gauge. I'm not concerned with accuracy, just the full and empty indications. Gas gauges are notoriously nonlinear, so it won't matter much if the reading moves a little with slight voltage changes.

Jack

6. ### Jack_K Thread Starter Active Member

May 13, 2009
115
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So, do we think that this will work then? Disregard the layout. It's from an earlier circuit.

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7. ### MikeML AAC Fanatic!

Oct 2, 2009
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Then just use a two resistor voltage divider with the top endconnected to the car's battery (13.5V?), and choose the resistors to produce 0.094V at the tap.

Vout = 0.094 = Vin* R2/(R1+R2).

8. ### Jack_K Thread Starter Active Member

May 13, 2009
115
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By using a resistive divider I would be changing the gain of both voltages. The resistors would be in the positive input of the op-amp to ground, wouldn't they?

How about a germanium diode with a Vf of 0.1 volt? I might be able to find one since I have a bunch of old germanium diodes.

Jack

9. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
I assume you are using a basic mA meter movement (or dc voltmeter) as the indicator. Look at this circuit. The input impedance is very high, so you do not load the existing sender. The gain is set with one resistor. The zero point is set with a trim pot, which is also unloaded. The two adjustments do not interact. The opamps could be almost anything, like LM358 or LM324 (but not a 741).

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10. ### Jack_K Thread Starter Active Member

May 13, 2009
115
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Mike,

The gas gauge is a real gas gauge. It's the kind that has a bi-metallic strip that gets heated and that moves the needle.

Thanks, though.

Jack

11. ### Jack_K Thread Starter Active Member

May 13, 2009
115
0
Thanks for everyone's help. The circuit works and I've designed and built a PCB and installed it. Here's what I ended up with.

Jack

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