# summing voltages with an op-amp

Discussion in 'The Projects Forum' started by marshallf3, Oct 6, 2010.

1. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
201
While I'm familiar with the method used in an inverting situation is there a way to use a single op amp to add two voltages? I'd hate to add another device to the design since I only have one amp left in the quad. Single supply of +5.0V

I have Signal A, which varies from 0 - +1.0 volts

I have Signal B, which will be used as an adjustable positive offset, from 0 - +1.0V

In other words let's say Signal A varies from 0 - 1V but I'm wanting to be able to add a positive DC offset to it.

Input A = 0V, B = 0.5V, Output = 0.5V
Input A = 0.5V, B = 0.5V, Output = 1.0V
Input A = 0.5V, Input B = 0.75V, Output = 1.25V
Input A = 0.75V, B = 0.25V, Output = 1.0V

Easy enough to do in the inverting configuration then invert it again, but with one amp?

I'll swear this is simpler than I think, I'm just not 100% again tonight.

2. ### ifixit Distinguished Member

Nov 20, 2008
638
108
Would this work for you?

Have fun,
ifixit

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3. ### wayneh Expert

Sep 9, 2010
12,087
3,027
That's the textbook solution, so it ought to work, even with more inputs. My textbook pictures seem a bit neater than your's though!

4. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
201
Thanks, couldn't remember if I needed to have resistors or take the - back to the output. I assume the normal gain equation applies so I should keep tight tolerances on all the resistors?

Last edited: Oct 6, 2010
5. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
201
OK, I'm hoping thhis is going to turn out to be an exact gain of 1, I ordered a handful of 100K 1/4% resistors.

My circuit's input impedances are very low in comparison and the output impedance is high.

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
No one has stated what is probably obvious to those very familiar with the opamp circuit provided by ifixit. Interestingly, the two resistors in series with the two inputs to be summed form a resistive divider since the current into to the positive input to the opamp is nearly zero. This resistive divider attenuates each of the two inputs by 50%. Of course this assumes that the output impedance of each of the two input sources to be very low in order for the divider to work properly. The non-inverting opamp stage has a gain of two for signals applied to its positive input so that makes up for the 50% attenuation.

hgmjr

7. ### marshallf3 Thread Starter Well-Known Member

Jul 26, 2010
2,358
201
You're right, but it works. That threw me off at the first too, thought I was going to have to divide the output but then I got to looking.