[Solved] Why Doesn't This Resistor Mixer Sum the Input Voltages?

Thread Starter

johnyradio

Joined Oct 26, 2012
288
My understanding is that parallel resistors, tied together at their outputs, sum their input voltages
Vo= V1+ V2.. VN.



Therefor, why in this circuit is the output peak, 5V, the same as the input peaks?
These are 2 schmitt oscillators, btw.

 

Hymie

Joined Mar 30, 2018
1,100
If that was to work, I could configure 10 resistors each with an input of 5V and by some miracle there would be 50V at the commoned end.
 

ebp

Joined Feb 8, 2018
2,332
Back to basics

in the circuit at #1
use just two equal value resistors, one with the "input side" connected to 10 volts, the other with the input side connected to 0 volts. Assume no current flows into the "output"
Calculate the voltage at the output.

now use three resistors, two connected to 10 V and one to 0

now use three resistors, two connected to 10 V and one connected to 5 V

These are all easy to calculate.
 
Last edited:

ebp

Joined Feb 8, 2018
2,332
Something else to do:

Repeat all of the above, but with the output short-circuited to "ground" and instead of calculating voltages, calculate the currents.

Try another with different resistor values and three different voltages.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
288
Something else to do:

Repeat all of the above, but with the output short-circuited to "ground" and instead of calculating voltages, calculate the currents.

Try another with different resistor values and three different voltages.
To calculate, I only need the correct formula. Seems that website has the wrong formula.

Should be following, correct?

Vo= (V1+ V2.. VN) / N
 
Last edited:

ebp

Joined Feb 8, 2018
2,332
You should be able to solve the first two-resistor problem with Ohm's law (Kirchhoff's voltage law also applies, but this really should be a "by inspection" case). Don't just look for formulas to plug values into. This is absolutely fundamental stuff and you need to be able to work out at least the simpler problems easily. Sometimes redrawing the circuit can help by making things a bit more visually obvious.

Ohm's law and Kirchhoff's laws go a long way to solving simple circuits. You MUST understand these laws.

You also have access to a simulator. Try using it to find the answer and to show you details of voltages and currents. See if you can figure out how to arrive at the same answers.

Do you know how to find solutions of problems using simultaneous equations? (not required for the problems I suggested, but often very useful)
 

MrChips

Joined Oct 2, 2009
27,641
The formula is wrong. Why?
Because the correct formula is the sum of the currents, as in Kirchhoff's Current Law, not voltage.
For the sum of currents formula to be correct, you are missing something. What is missing?
You are missing a path for the current to flow. Supply a path for the sum of the currents to flow then the current formula will work.
Add an op-amp to the circuit to convert from current to voltage. Then the voltage formula will work.
 

Plamen

Joined Mar 29, 2015
101
My understanding is that parallel resistors, tied together at their outputs, sum their input voltages
Vo= V1+ V2.. VN.



Therefor, why in this circuit is the output peak, 5V, the same as the input peaks?
These are 2 schmitt oscillators, btw.

Petkan:
Your assumed formula Vout=V1+V2...etc is not correct. Clearly if you applay say 1V to all inputs, you will get 1V at the output.
Use superposition to calculate the transfer function. Signal sources are considered one at a time with the rest shorted. The total response is the sum of individual responses. Load at the output must be considered too as it affects the result.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
288
I def misread that article. My bad. More careful reading shows (I think) that Vo is the average of all Vin's.

Solved.
 

Hymie

Joined Mar 30, 2018
1,100
Thx for all replies.


In that case, what if each input is the supply voltage?
If all the inputs are at the supply voltage, then the op-amp output will be driven to the supply voltage (actually the negative supply voltage in this case, as it is an inverting summing configuration).

The same is true if the summed voltages equals or exceeds the supply voltage.
 
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