I don't think you are understanding the term offset. In LTspice, it is a DC voltage you have 'added' to the signal. In other words the input signal sits on top of this voltage. If you wish to treat that offset as part of the signal then you need to design the amp to work correctly all the way down to DC.
So your 2.5 volt offset is correctly giving a DC voltage of 6.66 volts DC at the opamp output. If you make the offset minus 2.5 volts then the opamp output becomes minus 6.66 volts.
So your 2.5 volt offset is correctly giving a DC voltage of 6.66 volts DC at the opamp output. If you make the offset minus 2.5 volts then the opamp output becomes minus 6.66 volts.
I am getting that. My input signal is 2.5 dc offset with amp 1.2. and I want to implement a equation Y=mX-C
How to implement -C to circuitary? Is -C is another offset that needs to b added
How to implement -C to circuitary? Is -C is another offset that needs to b added
I assume Y is the output. What is m, X and -C. At the moment you have Vout=Vin*Gav where Gav is the gain.
Are you saying you want a minus term in the output ? For example +1 volt input would give -2.666 volts output.
Are you saying you want a minus term in the output ? For example +1 volt input would give -2.666 volts output.
Y=mX-C, this is equation I want to implement. m=gain=2.666 , Y=output,
X=input(1.2 amplitude with 2.5 dc offset), I want this -C to implemented.
I can do it by using a subtractor but I have only positive power supply.
What to do in such a situation?
or choosing an op amp having offset of -4.865 but getting this value is not possible?
X=input(1.2 amplitude with 2.5 dc offset), I want this -C to implemented.
I can do it by using a subtractor but I have only positive power supply.
What to do in such a situation?
or choosing an op amp having offset of -4.865 but getting this value is not possible?
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I still don't understand what -C is. Is it a negative gain factor you multiply the result by.
This shows your circuit and vout together with -vout. Is that what you want to do ? If so then C in this case is -1
This shows your circuit and vout together with -vout. Is that what you want to do ? If so then C in this case is -1
Sounds like OP-AMP 101, using an operational amplifier to solve an equation?
Unless the problem is that there is a DC offset on the input and you want zero offset on the output, something generally solved with either a capacitor or a servo?
Unless the problem is that there is a DC offset on the input and you want zero offset on the output, something generally solved with either a capacitor or a servo?
Mooly,
y = mx+c is the standard equation for a straight line function.
m = slope or gradient or "gain"
c = offset from origin.
Using an op amp, m is defined by gain ratio -Rf/Rin etc, and c can be an addition fixed input summed to the output.
AFAIR (a LONG time ago!) this will need two stages, the first setting the gain, the second adding the constant. Polarities will be tricky, even more so with only a V+ supply.
'sounds like H616148 has been set a college problem?
Once again this points out the need for the OP to fully state the problem at the beginning.
I need an inst. amp goes nowhere. I need to implement a y=mx+c function, and i only have a single supply would have got us straight to this point!
y = mx+c is the standard equation for a straight line function.
m = slope or gradient or "gain"
c = offset from origin.
Using an op amp, m is defined by gain ratio -Rf/Rin etc, and c can be an addition fixed input summed to the output.
AFAIR (a LONG time ago!) this will need two stages, the first setting the gain, the second adding the constant. Polarities will be tricky, even more so with only a V+ supply.
'sounds like H616148 has been set a college problem?
Once again this points out the need for the OP to fully state the problem at the beginning.
I need an inst. amp goes nowhere. I need to implement a y=mx+c function, and i only have a single supply would have got us straight to this point!
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Yes, tedious task it is. here value of C is coming out to be -4.865. How is it possible to subtract two signals without negative voltage? Even I cant change polarity of positive to get negative? I guess, I need to add 1 more op amp.
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A good 'ole analogue computer from way back when 😀
Thanks 🙂 Maths in pretty much any form has never been a strong point of mine. I'll be honest, seeing that equation didn't ring any bells either 😱
This is fundamental to any design... lay out your requirements, draw up a specification. A few posts back and it was mentioned that the input signal was AC and of limited bandwidth.
Not quite sure what you mean. If a circuit won't run LT throws up an error log, but its not always easy to decipher.
You have already asked this previously in another thread. Please don't keep opening new threads on the same topic in different forums.
^ Indeed 🙂
Threads have been merged and moved to Analogue Line Level.
H161648... I think there might be enough information here to help you. Search engines my friend, search engines ! Make use of them.
Threads have been merged and moved to Analogue Line Level.
H161648... I think there might be enough information here to help you. Search engines my friend, search engines ! Make use of them.
You have placed R1 in the wrong leg of U1.
R1 should go from R6 to the output of U1 and not to the input.
And R6 should go directly to the - input of U1.
If not using a minus supply, you should give both sine generators an offset like Sine(7.5 1.2 1000) and Sine (7.5 -1.2 1000)
Succes,
Hans
R1 should go from R6 to the output of U1 and not to the input.
And R6 should go directly to the - input of U1.
If not using a minus supply, you should give both sine generators an offset like Sine(7.5 1.2 1000) and Sine (7.5 -1.2 1000)
Succes,
Hans
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