Voltage Summer Without Opamp

Thread Starter

hoyyoth

Joined Mar 21, 2020
136
Dear Team,

I need to generate a waveform as shown below.I was given a 5V supply .Will it be possible with resistor based voltage summer
may i know how to solve this.

1628662117670.png
Regards
Hai
 

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.
1628664242632.png
I have not taken the voltage drops of the Diodes in calculating the resistor values.
If you "load" the output, voltages will reduce.
 

Thread Starter

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
 

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?
 

Thread Starter

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
 

Thread Starter

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.
 

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

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 :D
 
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