Offset voltage on voltage follower

[rosborne]

I still think an adder will work.
1. voltage divide your dc source to 0.5 volts. If 5 VDC source 9k and 1k resistors.
2. voltage divide your 0-5 volts source down to 0-3 with a 2k and a 3k resistor.
3. add the two voltages for a 0.5 to 3.5 VDC output signal that track linearly with the 0-5 volt source.
4. apply both signals to the inverting input of the OPAMP through a 10k resistor. That is one 10k resistor for each source's output joined to the inverting input.
5. the non-inverting input goes to ground perhaps thru a 1k resistor.
6. the feedback resistor is 10k.
7. Oh yeah, and then invert the negative 0.5 to 3.5V signal.
8. Connect the output of the 1st OPAMP to another OPAMP's inverting input thru a 10k resistor. 10k feedback resistor. 1k from ground to the noninverting input(optional). The output will be 0.5 to 3.5 volts.<p>Nine (or 11) resistors and 2 OPAMPS and + and - 5 VDC sources one each for the OPAMPS. Yikes, there has to be a better way. I'm looking at what Dale and Al said again. This is spiralling out of control, but it will work. <p>can you make adders with the non inverting input???
-Rick

[rosborne]

Ok, got it down to 6 resistors and one opamp if you have the plus and minus 5VDC sources. The general idea is the same as before. . . voltage divide the signal down to 3/5. Divide the -5VDC source to -0.5 and configure OPAMP as a difference amplifier with all of the resistors equal to 10k. I can send you the simulation if you have ICAPS4 demo software.
-Rick
There are assumptions here, if you don't have a negative source for the OPAMP and the -0.5 VDC it won't work.

[rshayes]

Consider a 3 resistor network.
10K from input to output.
144K from +12 volts to output.
16.74K from output to ground.<p>If the input is 5 volts, the output will be 3.5 volts. If the input is ground, the output will be .5 volts.<p>If the source impedance is accurately known to be 10K, then this can replace the 10K resistor, and only two resistors are needed.<p>Otherwise, use an op amp as a unity gain buffer to drive the 10K resistor. The CA3160 can work with a single 12 volt supply and handle inputs and outputs down to its negative supply (ground). This avoids a negative supply.<p>The 100K load impedance can be compensated for by increasing the 16.74K resistor to 20.11K.

[rosborne]

Stephen,<p>I like it. Much more elegant solution than my flailing produced. The buffered output circuit looks better to me though, in case the instrument turbo describes is an active load. He has the OPAMP, it would be a shame not to use it and the test circuit with the 1k load would work, too.
-Rick

[Turbo46]

To Dale and all the other clever people!!<p>I dont have a have a part No, for the tranducer. It is a pressure transducer which is used in a marine application for measuring depth. At my part time summer job, i was given this spec/ problem and was told to solve it. I was told the CA3140 will do the job and i thought i would ask on this site for a bit of help.<p>Please forgive me i cant provide any more information that what was already given. I know its a wierd one, but i have been told it is realy relavant to the problem and my learing curve.<p>Thanks for all your input guys. I will try to simulate it on multisim later in the week. The thing which is different in real application, appose to calcultion is the input and output impedances dragging the circuit down.<p>Thanks again guys. Dont have to many sleepless nights!! lol :0)