Summing analog signals

"Richard Holmes"

** Adding a signal source affects a resistor summing network - just adding a signal does not.

A "summing" op-amp is not affected by either.

Eg:

..... Phil

--
Yes, use an inverting opamp as a current to voltage converter: (View in
Courier)

                    V+
               +----|----+
               |    |   [R]
IN1>---[R1]--+-+---|-\   | 
             |     |  >--+--> 
IN2>---[R2]--+  +--|+/ 
             .  |   |
             . GND  V-
             .
             |
INn>---[Rn]--+

     
JF

S1--R1---+ S2--R2---+----Vo S3--R3---+

Where + is all connected is the basic idea. The problem is that you cannot draw any current from Vo without distorting the results. This is called loading the output. You can, for example, connect a scope probe to Vo and expect to see S1 + S2 + S3.

Now because we don't want to "load the output" we have to one of any number of methods to allow us to effectively load the output(i.e., still gives us Vo but let us draw a current from it without changing Vo).

To do this you must use some type of buffering/amplification technique. A simple op amp buffer will work but will not provide amplification. You'll essentially get Vo out of the buffer but be able to drive larger loads than the resistor network will.

The op amp buffer is at:

So you connect Vo to Vin of the buffer and then Vo of the buffer is your Vo but you can now drive larger loads(up to the op amps spec).

You can also implement buffers and amplifiers with bjt's and fets with effectively the same result(op amps are more precise).

Here is what you are trying to accomplish:

Make sure you get the resistor values right. If you have

S1--R1---+ S2--R2---+----Vo S3--R3---+

Then the sum is

(R1*S1 + R2*S2 + S3*R3)/(R1 + R2 + R3)

which is known as a weighted sum. Generally you'll want R1=R2=R3. Also, The R's should not be too large. Something in the 10k's should work. Nothing below 1k in general.

It should be noted that the output signal will be the sum of the INn (where 'n' is the number of the input) times the ratio of R/Rn. It's a very powerful circuit.

You must have done something wrong in the use of a summing op amp; you can add more inputs without any effect at the output o the other signals - ASSUMING the op-amp is in the linear range AND the frequencies aer well within the gain-bandwidth.

So, assuming two input signals of 1Vpp and 2Vpp, and a single supply op amp, the following circuit will work?

CA 3130 +12V +-----|----+ | | | | | [Trim pot] gain

. | | [R2] | |___| | GND Not shown:

1. R1 between +12V and non-invert IP.
2. 120pf cap pins 1 + 7

Sorry, I don't have the ASCII circuit program.

Many thanks,

Richard Holmes

"Richard Holmes"

** I see your pain.

A summing amp is inherently an * inverting stage* - ie positive inputs generate negative outputs.

If the input signals are DC coupled, ground referenced and swing positive or both ways - then you MUST use a suitable op-amp with dual supplies.

The schem you posted will only work with AC coupled input signals - ie there is a cap between the sources and each input and the ouput will have a +6 volt fixed bias.

.... Phil

One of the most often used is over at

Hasn't been updated in a while but it's quite useful.

--
Rich Webb     Norfolk, VA

Richard Holmes schrieb:

Hello,

using bipolar input signals with a single supply op amp is not a good idea, especially if you want bipolar output signals too.

Bye

How about DC offsetting the input signals?

Richard Holmes

"Richard Holmes"

** You will have to offset them in the negative direction.

Do read what has been posted.

..... Phil

Hmm, probably a silly idea with a mistake. But if you could reference the signals to some voltage half way between the supply voltages (call it common). You could then use the summing op-amp circuit. Then invert back about the common point and remove the DC offset at the end... sounds like a lot of trouble just to add two voltages with single supply opamp.

George H.