Combining Inputs and Offsetting by Separate Grounds

May 06, 2009 2 Replies

Greetings,



I'm not really sure about the subject line if it's right or not, but it's basically what I'm looking to do.



I posted just yesterday a question about a circuit which was answered very well by one T. Wescott. It turns out I simplified the problem too much and a better representation is as follows.



Given inputs (P0, P, Q0, Q, R0, R)



P0 GND P > P0 Q0 GND Q > Q0 R0 GND R > R0



And constants (A, B, C)



A constant < 1 B constant < 1 C constant < 1



Desire these outputs:



P* = A(P - P0) + B(Q - Q0) + C(R - R0) + P0 Q* = A(P - P0) + B(Q - Q0) + C(R - R0) + Q0 R* = A(P - P0) + B(Q - Q0) + C(R - R0) + R0



Note that the operation is the same other than adding a different offset for each output.



Is there a simple solution for this? (e.g. just resistors)? Is there a systematic way to scale up to more and more inputs and constants? e.g. S0/S/D, T0/T/E, etc.



Any references on-line or otherwise are welcome. Any searches I do bring up operational amplifiers which might be overkill for this kind of application. And any searches I do regarding resistor circuits don't seem to go beyond elementary introductions.



Thank you for reading.



Regards,



YR


no. consider the case where A=B=C=0.99

if you put the put P0,Q0, and R0 near ground and the the same input signal on PQR you end up needing a gain of 2.97

and you can't get amplification from just resistors.

There are lots of solutions, and 'simple' isn't a good selection criterion. With least parts, you can sum signals with resistors onto two terminals of a transformer, and create your outputs by summing multiple secondary windings of the transformer with the inputs. That's one resistor for each term of the common section, one transformer to perform the subtraction.

Is that clear? The transformer has two input terminals, call one (+) and one (-); the secondary windings (you'll want three) will have the (-) terminals wired respectively to P0, Q0, R0, so those windings (+) terminals will produce the signals you describe.

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