# Summer/Mixer Op amp

Discussion in 'Homework Help' started by tbinder3, Jul 2, 2013.

1. ### tbinder3 Thread Starter Member

Jun 30, 2013
30
1
This is my first post and hope to contribute to the forums as well! :

I took a test last Wed., and am rethinking what I did.

So I wish I had a schematic available, if someone can point me to a website, that would be great!

What I had was a 2 input summing amplifier (One input was DC, and one was AC) So after both inputs there was a resistor of course, and then the two inputs tied together at point "A" (Virtual common to the op amp, and also where the negative feedback is as well).

So what I'm trying to figure out is, if both the currents that were going through the resistors were positive (the currents weren't opposing each other) (And they joined at Virtual common), so when I add the input currents to the negative feedback, should I use the total rms of both the AC and the DC current? "total rms = sqrt(I^2ac+I^2dc)" to get the current going through the negative feedback?

This being my first post, don't hate on me too hard! haha
any questions just ask, i'll be checking back every couple hours as im visit parris island for my sister's marine graduating and got me thinking about that test i took, lol

2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Unless you are worried about power dissipation in the feedback resistor, you needn't be concerned with RMS.
Ifb=Iac+Idc

3. ### tbinder3 Thread Starter Member

Jun 30, 2013
30
1
See thats what I first thought it was, but then I remember average total rms sqrt(Idc ^ 2 + I ac ^ 2), so I figured since I have two different sources I would need to find the average total between AC and DC.

Last edited: Jul 2, 2013
4. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
RMS is a statistical measure of a (time) varying quantity. If you want the instantaneous current at any point in time, it is just the arithmetic sum of the two inputs.
In electronics and electricity, RMS is generally used when the voltage or current is converted to power by applying it to a load.

Shagas and tbinder3 like this.
5. ### WBahn Moderator

Mar 31, 2012
18,065
4,905
You need to spend time learning what these quantities mean. Presently, it sounds like they are just formulas to you and when you need a formula you rummage through your list of formulas and grab one that seems to fit.

So take a step back and take the time to truly understand what the whole concept of RMS values are all about. If you do that, you will not even begin to confuse when you need the instantaneous sum of two currents and when you need the RMS sum of two currents because you will KNOW which one in appropriate for a given circumstance.

6. ### LvW Active Member

Jun 13, 2013
674
100
Don`t forget the opamp output voltage that drives a current through this resistor.

7. ### tbinder3 Thread Starter Member

Jun 30, 2013
30
1
I"m talking like that with equations for your purposes so you can see what I'm thinking. As long as you know the basics you can make up your own equation to solve for almost anything.

@LvW - Of course, the schematic given was quite simple actually, so the current through the negative feedback was the output current.

I'll give an example: if we have a summer/mixer amp and 1 input is AC and 1 input is Dc. The AC is 10Vpeak and the DC is 10 Vrms (parallel). There is a 1k resistor after both power sources, and the two resistors connect at virtual common of the op amp, (since resistance is very high in the IC it acts like an open, so the current to and through the op amp is EXTREMELY low,) in the negative feedback there is a 1k resistor, what is the current through the negative feedback back to common?

Wouldn't I have to find the total rms of the AC & DC currents from the inputs? Im not tooo familiar with AC yet, but do know some.

(The total rms of an AC and DC voltage is definitely sqrt(AC^2 + DC^2)

Last edited: Jul 3, 2013
8. ### WBahn Moderator

Mar 31, 2012
18,065
4,905
But what is it that you are trying to find? The total current in the feedback resistor or the RMS current in the feedback resistor? The answer to that depends on why you are trying to find the current in the feedback resistor in the first place. If you understand what you need and you understand what each current tells you, then you can determine which one you want.