Solution without inspiration: add an inverting amplifier after the summer.
Solution with inspiration: combine the concepts of the summing amplifier
and non-inverting amplifier. Textbooks usually show summing only into
the inverting input, but you can sum into the non-inverting input instead
(or even sum into both inputs at once). You can achieve C = 2A + 3B with
Good job! I'm glad you chose inspiration.
Time to use what we know about an ideal opamp:
1. The differential input voltage is zero.
2. Neither input draws or supplies any current.
Write an equation for V0 when V2 = 0 (shorted). Write another
equation for V0 when V1 = 0. Superposition is valid for this
linear circuit, so the total output V0 is the sum of the two
output voltages you just found. The rest is algebra.
For an ideal amplifier, the input offset current is zero so
resistor can be eliminated. In that case, the resistances
could be 1, 2, 3, and 4 to achieve C = 2A + B.
I am another student trying to obtain this Op Amp Circuit and saw your
post as I was posting on google. I am not getting the correct answer
when doing the algebra from your steps. Is there any way you can post
this or also further explain how those resistances achieve C* + B?
I'd greatly appreciate it. Thanks.
PMJI. The difference between summing at the + and - inputs is that
feedback keeps the - input reasonably still, which prevents the inputs
from talking to each other. (If you move the + input around, the +
signals will also talk to the - signals). Some sources don't mind that,
but others do--you can get all sorts of nasty intermodulation and so on.
If the circuit needs to be used with different sources, I'd spend
another dime and use a dual op amp. Sum the + signals in the first amp,
and send its output into the (-) input of the second amp, along with the
Hmm. As I understand it, one advantage of the inverting summer
configuration is that the summing node (the inverting input) is actively
held at ground voltage, so each of the individual inputs sees a
resistive load to ground (just the input resistor) which does not change
with input voltage on that input, and it also unaffected by the voltage
on any of the other inputs.
It seems that with the circuit layout you're suggesting, the summing
node voltage will vary in proportion to the output voltage (i.e. the sum
being calculated). Thus the current in each input resistor will depend
on the state of the other inputs, and the apparent input impedance will
vary between infinity (with all inputs at equal voltage) and a value
somewhat lower than the individual input resistors.
Now, if the inputs are actually coming from voltage sources (e.g. other
opamp outputs, the varying input impedance wouldn't matter, and you
would get the summed value that you wanted. But if the sources have
their own finite impedance, how does the unpredictable input impedance
of the summer affect the result? You could always avoid this by adding
a buffer between each input and the summer input resisters, but an extra
opamp for each input is now more complex than an inverting summer plus
one extra opamp to invert its output.
(not a student or an EE, but an electronics hobbyist)
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