# Summing Amplifier Question!

Discussion in 'Homework Help' started by aly34, Feb 16, 2012.

1. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
1. The problem statement, all variables and given/known data

2. Relevant equations

Absolutely no idea how to start this problem!

3. The attempt at a solution

A) I would think that there are 4 input combinations for this circuit.

B) 0110 = 6v?

C) No clue

D) 1011 = 11v?

E) No clue

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2. ### BSomer Member

Dec 28, 2011
433
106

A) it is a 4 bit combination

E) see (A)

As for B,C & D you will have to do some math to figure out what Vs is. It is not necessarily a binary - decimal conversion. The formula for Vs is pretty much given in the question B.

3. ### aly34 Thread Starter New Member

Jan 29, 2012
20
0
For E:

would it just be:
0000
0001
0010
...
1111
?

And B: Not sure what formula?

C: Would it just be sin(w*f*t) = sin(4000*f*t)

so, w*f*t = 4000*f*t

and so f = 2000hz?

4. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,222
This will be a job for the superposition theorem. Just Google superposition theorem. And you will get a lot of hits that will be helpful. Since the network is resistive. The vout will be Vs multiplied with a constant regardless of vs is DC or AC. Assuming ideal components

5. ### crutschow Expert

Mar 14, 2008
13,502
3,376
As the title of your thread states, it is a summing amplifier. So the output is the sum of all the inputs that are energized. And the gain for each input to output is just the feedback resistor divided by the input resistor. For those values of resistors give all the gains are less than one.

6. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
another simpler way to look at it: whatever input paths are connected to the source signal, puts those source resistors in parallel. Combine them to get an equivalent source resistor value and then the gain equation is simple.

7. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
I can tell you that is wrong. It's a lot more, but not so many you couldn't easily go through them one at a time by toggling each bit.