Op amp design help please!
- Thread starter hsilver
- Start date
What's so special about getting an output of 5 V? If you've designed it properly, then any pair of (reasonable) input voltages in which v2 is 1 V greater than V1 should give you an output of 5 V.
We can't help you unless you show at least an attempt to work the problem -- that gives us something to work with.
I believe the problem is to provide an output offset of -5v, which is not necessarily trivial with only one op amp in a differential configuration.
I agree with above.
Maybe a clue is to rewrite Vo=10(V2-V1)-5 into:
Vo+5=10(V2-V1)
then run +12V through voltage divider to provide you with +5V
then stack +5V and Vo
I agree that this is probably what the TS meant -- but the TS needs to learn to communicate what they want instead of being sloppy and expecting others to read their mind.
The problem is a bit tricky, but presumably they have seen both a single op-amp version of a summing amplifier and a difference amplifier, so they should be able to take a decent shot at the topology needed to take the difference between the sum of one pair of signals and the sum of another pair (though in this case one "pair" is just a single signal).
Analysis of this topology should yield an equation of the form:
Vout = A·V2 - B·V1 - C·Vref
Where the A, B, and C coefficients are functions of the resistances. They then have to decide what Vref is -- the intent is likely to use one of the supply voltages, though this isn't necessary the best choice, but unless the TS is free to introduce additional components such as zeners or bandgap references or something else, that is probably the best they can do. After that, it is a straight-forward matter of matching coefficients to the desired response to get needed ratios of values. In this case, those ratios are all pretty tame.
As you've already noted, the TS needs to show their best shot at a topology and their attempt to figure out the values.
Hi,
Why would getting a -5v output offset be difficult if we already have an op amp?
You mean because the non inverting input terminal voltage changes because this will be part of a difference amplifier?
Maybe we can find a away around that?
I'll wait for the next OP response though.
LATER:
Yes i found a version that looks like it works but it requires a voltage reference and a couple resistors and may only be applicable to certain circuits with one input source that can stand some offset current. So no second op amp but does require a good voltage reference diode or depending on accuracy, a low voltage zener. So there would be some limitations.
Why would getting a -5v output offset be difficult if we already have an op amp?
You mean because the non inverting input terminal voltage changes because this will be part of a difference amplifier?
Maybe we can find a away around that?
I'll wait for the next OP response though.
LATER:
Yes i found a version that looks like it works but it requires a voltage reference and a couple resistors and may only be applicable to certain circuits with one input source that can stand some offset current. So no second op amp but does require a good voltage reference diode or depending on accuracy, a low voltage zener. So there would be some limitations.
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Reading between the lines, the TS probably doesn't need to worry about accuracy or need a reference because they are still at a stage where everything is ideal, including the supply rails (i.e., they can use the power supply for their reference voltage).
Hi there,
Well for this circuit, it's a differential amplifier right?
That would mean that he cant use one of the power rails for a reference for the offset in the usual way, unless he adds another op amp, and using the reference diode would be an idea to help eliminate the need for another op amp. The difference is the reference diode can be used as a floating reference voltage (given proper biasing is satisfied).
I havent looked at every possible way though, there may be another way too where we can use the power rail. Maybe you can come up with a circuit.
What is this "in the usual way" that you refer to?
The circuit that deals with this is a trivial combination of a difference circuit and a summing circuit. Five resistors and one op amp.
Sorry, but I don't think it's as "trivial" as you think.
This extract from a pdf document may provide a clue.
MITMAS_836S11_read02_bias.pdf
MAS.836 Sensor Technologies for Interactive Environments
MIT OpenCourseWare http://ocw.mit.edu
MITMAS_836S11_read02_bias.pdf
MAS.836 Sensor Technologies for Interactive Environments
MIT OpenCourseWare http://ocw.mit.edu
Thus for example, if you used the Split Resistor differential circuit in RBR1317's post to meet the OP's requirements, you would need to solve simultaneous equations for the values of R3 and R4 so that the offset at Vo is +5V while their equivalent parallel impedance equals 10 times the value of R1.
Hello again,
Well i think that only works if we have a negative supply rail available to act as an actual negative offset, after all this seems to require a negative output offset. So the question becomes is there a negative rail available?
I had originally assumed that the only power rails available where the +Vcc and ground. Rereading the first post though i see that he stated that he has plus and minus 12v available so that answers that. If the -12v line is STABLE then it could work, but if it is not stable then it wont work very well.
You could always show your circuit too if you like
It's too early to show a solution. The OP isn't even 48 hours old yet -- the assignment may not be due and the TS may well be back. Let's give them a chance to show some work. I laid out an approach that is simple and doesn't even involve solving simultaneous equations, though that shouldn't be a huge factor even if it did.Hello again,
Well i think that only works if we have a negative supply rail available to act as an actual negative offset, after all this seems to require a negative output offset. So the question becomes is there a negative rail available?
I had originally assumed that the only power rails available where the +Vcc and ground. Rereading the first post though i see that he stated that he has plus and minus 12v available so that answers that. If the -12v line is STABLE then it could work, but if it is not stable then it wont work very well.
You could always show your circuit too if you like
You suggested it's a trivial combination of 5 resistors and an op amp, but is that different from the circuit in Post #12?
Yes.You suggested it's a trivial combination of 5 resistors and an op amp, but is that different from the circuit in Post #12?
Consider a traditional (inverting) summing amplifier:
Now consider a traditional one op-amp differential amplifier:
If you understand HOW a summing amplifier works, what is SO difficult about seeing how to add a fifth resistor to the diff amp such that it is taking the difference between V2 and the sum of two other signals?
Hi,
Sorry i dont see it. But i do see biasing the 'ground' to a negative value voltage (i wont say whether that really works for sure yet and leave that to the OP).
Yes I DO understand HOW a summing (and differential) amp works.
What makes is SO difficult is when you try to add a bias offset to the differential amp.
The differential amp depends upon the the ratio of Rg and R2 being identical to the the Ratio of Rf and R1.
Any added resistor must not disturb this ratio.
Thus for example, in the circuit in Post #12 the equivalent parallel resistance of R3 and R4 must equal the value of RL.
The Value of R3 and R4 must also be selected to give the desired 5V offset at the output.
I don't see how to get R3 and R4 to satisfy both criteria without solving two simultaneous equations for their values.
If you can, than I would like to see how.
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