# Voltage Summer Without Opamp

#### DickCappels

Joined Aug 21, 2008
7,863
Sure, just pick the right resistor values and voltages. Using a summing amplifier reduces the math and makes the output independent of load.

#### Ramussons

Joined May 3, 2013
1,086
Summing can be done with resistor networks IF the voltage is being monitored only by a DMM or an oscilloscope.
If the voltage is to be processed by another stage, the "loading" of that resistor divider network will have to be taken into account.
Use a 4 Flip Flop Shift register. This is for a 5 Volt Shift register. I have not taken the voltage drops of the Diodes in calculating the resistor values.
If you "load" the output, voltages will reduce.

• hoyyoth

#### hoyyoth

Joined Mar 21, 2020
136
h
Summing can be done with resistor networks IF the voltage is being monitored only by a DMM or an oscilloscope.
If the voltage is to be processed by another stage, the "loading" of that resistor divider network will have to be taken into account.
Use a 4 Flip Flop Shift register. This is for a 5 Volt Shift register.
View attachment 245355
I have not taken the voltage drops of the Diodes in calculating the resistor values.
If you "load" the output, voltages will reduce.
Hi Ramussons,

Thank you.
May I know how much voltage I need to give the I/P of D1,D2,D2. I have a 5V source in hand.

Regards
Hai

#### ericgibbs

Joined Jan 29, 2010
13,833
hi Hai,
Resistor values modified.
Note: The Shift register may not output exactly 5V per step, it could be 4.8V

What are you driving with the output of the Summer.??

E

#### Attachments

• hoyyoth

#### Ian0

Joined Aug 7, 2020
3,480
With a page of algebra and three simultaneous equations you can eliminate the diodes.
Replace all the resistances with conductances, and apply Kirchhoff’s current law.
1). -4G1 + G2 + G3 + G4 = 0
2). 2G1 - 3G2 + 2G3 + 2G4 = 0
3). ½G1 + ½G2 - 4½G3 + ½G4 = 0

Eliminating G1 from the first two equations gives 2G1 = G2
substituting back in gives G2 = 4G3
substituting in 3) gives ¾G2 = G4

Turning back to resistances
R1= 2R2
R3 = 4R2
R4 = 4/3 R2

They work out to be 15k, 7.5k, 30k and 10k (or decades thereof). Suspiciously all E24 values. Was this homework?

• hoyyoth

#### hoyyoth

Joined Mar 21, 2020
136
With a page of algebra and three simultaneous equations you can eliminate the diodes.
Replace all the resistances with conductances, and apply Kirchhoff’s current law.
1). -4G1 + G2 + G3 + G4 = 0
2). 2G1 - 3G2 + 2G3 + 2G4 = 0
3). ½G1 + ½G2 - 4½G3 + ½G4 = 0

Eliminating G1 from the first two equations gives 2G1 = G2
substituting back in gives G2 = 4G3
substituting in 3) gives ¾G2 = G4

Turning back to resistances
R1= 2R2
R3 = 4R2
R4 = 4/3 R2

They work out to be 15k, 7.5k, 30k and 10k. Suspiciously all E24 values. Was this homework?
Hi Ian,
Thank you, this was not homework, but an interview question was able to answer everything except this one If you don't mind could please explain your equations with a circuit diagram.

Regards
Hai

#### hoyyoth

Joined Mar 21, 2020
136
hi Hai,
Resistor values modified.
Note: The Shift register may not output exactly 5V per step, it could be 4.8V

What are you driving with the output of the Summer.??

E
Hi Eric,

Thank you very much.
May I know how to design the resistors.

Regards
Hai

#### Ian0

Joined Aug 7, 2020
3,480
Same circuit as @Ramussons in post #3, but without the diodes. The outputs are assumed to come from a CMOS 4017 or shift register, running off a 5V supply.
Replacing the resistances with conductances, R1 etc becomes G1 etc.
the current through each is the voltage across each multiplied by the conductance.
When Q1 is at 5V, Q2 and Q3 are at 0V, so the voltages across each resistance/conductance is -4V across R1, 1V across each of R2, R3 and R4. That gives equation 1).
Equations 2 and 3 follow the same way, except with Q2 and Q3 at 5V.

• atferrari and hoyyoth

#### ericgibbs

Joined Jan 29, 2010
13,833
hi Hai,
It would not be my choice of circuit design. As the circuit uses steering diodes, operating at different currents, their forward voltage drop would not be the same, so calculation would be messy, so I used an iterative approach for selection

E

#### Attachments

• hoyyoth

#### Ramussons

Joined May 3, 2013
1,086
With a page of algebra and three simultaneous equations you can eliminate the diodes.
Replace all the resistances with conductances, and apply Kirchhoff’s current law.
1). -4G1 + G2 + G3 + G4 = 0
2). 2G1 - 3G2 + 2G3 + 2G4 = 0
3). ½G1 + ½G2 - 4½G3 + ½G4 = 0

Eliminating G1 from the first two equations gives 2G1 = G2
substituting back in gives G2 = 4G3
substituting in 3) gives ¾G2 = G4

Turning back to resistances
R1= 2R2
R3 = 4R2
R4 = 4/3 R2

They work out to be 15k, 7.5k, 30k and 10k (or decades thereof). Suspiciously all E24 values. Was this homework?
Yes, I know.
Was just a little lazy to do all that algebra #### Ian0

Joined Aug 7, 2020
3,480
Yes, I know.
Was just a little lazy to do all that algebra There’s something really satisfying about a circuit which performs as predicted after a page or two of algebra. Much more so that one that works like its SPICE simulation.

#### ericgibbs

Joined Jan 29, 2010
13,833
hi Hai,
What is the load on the output of that circuit, what is it driving into.?

E

• hoyyoth

#### Ian0

Joined Aug 7, 2020
3,480
Curiously, the output impedance of the no-diode version is constant, regardless of output voltage.

#### crutschow

Joined Mar 14, 2008
27,908
There’s something really satisfying about a circuit which performs as predicted after a page or two of algebra.
And I can imagine few things more tedious than solving a page or two of algebra equations. • ericgibbs

#### hoyyoth

Joined Mar 21, 2020
136
hi Hai,
What is the load on the output of that circuit, what is it driving into.?

E
Hi Eric,

There is no load at the output.

Regards
Hari

• ericgibbs

#### Ian0

Joined Aug 7, 2020
3,480
And I can imagine few things more tedious than solving a page or two of algebra equations. I have to admit I did it at work - where I was watching a battery charge!

• atferrari

#### atferrari

Joined Jan 6, 2004
4,442
I have to admit I did it at work - where I was watching a battery charge!
Finally!! A genuine alternative to watch paint dry.

• Ramussons and Delta Prime