OP-Amps and A to D

Sep 10, 2009 10 Replies

Let's say, for example, I have an Op-amp set up with a gain of 1, if I have



2.5V at the - input and 2.7V at the + input, would the output be 0.2V or
2.7V? If 2.7V, how do you set this up for an A-D converter?

My actual applications is wanting to take transducer signals, amplify them and read them with A/D converters. Many transducers already have a useable analog output but I'm also interested in load cells, thermocouples, and RTD's.



Thanks!



RogerN


Take it to sci.electronics.basics ;-)

...Jim Thompson

| James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.

What, you don't want to tell him where the gain set pin is on an op-amp?

www.wescottdesign.com

My education is over 25 years old now but IIRC on a 741, pin 2 was inverting, 3 non-inverting, 4 -V, 1&5 were offset, 6 output, 7 V+ and 8 n/c. Gain was set with rf/ri for inverting, rf/ri + 1 for non-inverting. I would guess the output to be 2.7V but perhaps some of the newer chips could cancel out common mode voltages. My real goal is to have a gain high enough to get a 0-5V signal with a 0 to 30mV signal from a bridge circuit in a load cell.

RogerN

H "I have an "if I have 2.5V at the - input and 2.7V at the + input, would the output be 0.2V or 2.7V?"

Think OpAmp fundamentals, then return for further tutoring ;-) ...Jim Thompson

| James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | An engineer is supposed to have an inquisitive mind and question unproven theories. Leftist weenies have neither attribute. Their behavior is of a religious nature. Thus, like all religious nut- cases, they should be culled from the fraternity and dispatched.

I'm sure I could easily get what I want in the lab with a + & - supply and a separate supply for load cell excitation and a bias pot on the NI input. I was hoping to come up with something a little more practical though, like a single supply rail to rail op-amp that could offset to output the differential voltage X gain.

But to answer my own question, I'm pretty sure with a 741 op amp I'd get

2.7V out. I'm not sure though if other op-amps and instrumentation amplifiers are the same or how CMRR affects this.

RogerN

Not relevant, really. Look again at the problem statement: the phrase "set up with a gain of 1" can mean inverting gain of 1 (which means (+) input is at GND), or noninverting gain of 1 (which means (-) input is resistor-connected to GND and the output), OR could be referring to a fully-differential instrumentation amplifier (not an op amp).

Instrument amplifier requires two or three op amps, let's rule out that possibility (since it could be a uA741 according to the later comment). Unbuffered fully-differential connection uses up both op amp terminals, only resistors are connected to the inputs, so that's out, too.

In neither remaining case can one arbitrarily connect signals to BOTH (+) and (-) op amp terminals. You used up one of those connections with the gain setting components.

Roger,

I guess it would be meaningful for us to know exactly what sort of circuit configuraiton you were referring when you said "I have an Op-amp set up with a gain of 1".

Is it a voltage follower circuit with the 741-type general op-amp, an amplifer with an input signal divider network that makes over-all gain to be unity, or one with output signal divider? Or is it some kind of instrumentation op-amp?

Atsunori

Please don\'t multi-post; crosspost instead. If you\'re using a difference amplifier with a gain of 1, your output voltage will be the reference plus the difference between the two voltages, 2.5V + 0.2 volts, 2.7V. For a single supply driving both the sensor and the opamp: (View in Courier:) +V>--+-----+------------+ | | | [R1] [R3] +---|--[R6]--+ | | | | | | +--[R5]--+--|-\\ | | | | >------+-->Vout +-----|-----------|+/ | | | [R2 [R4] | | | | GND>-+-----+------------+ R1R2 is the half-bridge sensor, and assuming that its output will be half of the supply when it\'s at rest, R3R4 will be used to set Vout to half of the supply and, for a gain of 1, R5 will be equal to R6. http://www.national.com/an/AN/AN-20.pdf http://www.national.com/an/AN/AN-31.pdf http://www.omega.com/LITERATURE/

If you have 2.5V _at the inverting input pin of the opamp_ and 2.7V _at the non-inverting input pin of the opamp_, then you have a broken opamp.

The rest is simply done with input and feedback resistors.

Hope This Helps! Rich

On Wed, 09 Sep 2009 22:36:18 -0500, RogerN wrote: ...

That's easy. I've used an LM324 to do exactly this.[1] Just set your input/ feedback resistors to give a gain of 5/0.03 . Admittedly, that's a lot of gain, but what's the highest freq. from your load cell? (i.e., do you have gain-bandwidth issues?)

Good Luck! Rich [1] Well, my input wasn't a load cell, it was a current shunt, but the principle is the same.

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