How Can a Circuit Compare Two Voltage Sources?

[iScience]

hi all, i need a favor; could someone design a simple circuit whereby

* it takes two constant potential sources (V1, V2) as inputs & then

* { outputs V1 , V1=V2 }
{ outputs 0 , V1≠V2 }

what i found tricky was that these were voltage sources, so that kinda eliminates being able to use any transistors

 

[phinds]

what i found tricky was that these were voltage sources, so that kinda eliminates being able to use any transistors

HUH ?

 

[davenn]

^^ +1
that really didn't make much sense

Please give us a FULL description of what you are trying to achieve

Dave

 

[Baluncore]

There is no such thing as equality or inequality of analogue voltages because of noise, so I guess you must be considering logic signals.

Truth table; { outputs V1 , V1=V2 } or { outputs 0 , V1≠V2 } Let V1= A, V2=B, Out = Z.
A B Z Z
0 0 A 0
0 1 0 0
1 0 0 0
1 1 A 1
So the only way to get Out=1 is when A=1 AND B=1. That simply requires an AND gate.

You can do that with one resistor and one diode.
Connect the resistor from A to Z. Connect the diode cathode to B, anode to Z.
Done.

 

[iScience]

Please give us a FULL description of what you are trying to achieve

problem: you have u two charged, conductive objects A and B. you attach a wire to each and these are your voltage sources. they are not big enough to serve as a current source they are not capacitors. there is no "ground," there are only the two charged objects.

i wanted a circuit with the function that outputted V1 if the two potentials were not equal ( ie |V1-V2|> some x%..), and did nothing if the |V1-V2|<x%.

i was looking for a circuit design that would be entirely field operated as opposed to current operated. ie i just wanted to just be able to use the two voltage sources that i had. I could indeed add a current source to my circuit but that's easy and i already know how to design a circuit with a current source. I'm wondering if there's a way to do this entirely with potential.

that really didn't make much sense

transistors require a current source which I'm trying to avoid. even FETs require a current source for 'source' and drain. hope this clears up some confusion

 

[Vanadium 50]

"Wanna"? That's not even slang. That's baby talk. Why would someone use baby talk to ask a serious question?

As Baluncore says, there is no such thing as equality of analog voltages because of noise, so the answer is that the output wire is disconnected.

 

[Baluncore]

( ie |V1-V2|> some x%..) percentage relative to what?
If there is no ground then how can you output a voltage, relative to what?

 

[Mike_In_Plano]

If you have two objects which are charged and their surroundings are all at the same voltage, than you might consider the surroundings as "ground potential." To ascertain the charge on one of the objects, you can contact it and drain the charge into the summing node of what's known as a charge amplifier. The charge amplifier resembles an op-amp based integrator in that it feeds back to it's inverting input through a capacitor.

When not using the charge amplifier, a reed relay can serve to short across the capacitor, thus resetting the amplifier.
Likewise, a reed relay can join your charged object to the inverting node of the op-amp giving you a voltage output that's
V=-Q/C

After a time, the reading tends to drift depending on the construction, cleanliness of the circuit (oh how I miss CCl4...), and input bias current of the op-amp. Anyway, if you built two charge amplifiers and get them fairly closely matched, you can measure the difference between the charges.

P.S. The circuit may drift quickly, so I recommend a data acquisition device to capture the values quickly. Also, glass capacitors are the best for this app though you may have to live with them not being matched. In that case, correct for gain in the math. Did I mention it needs to be really clean?

- Best Luck
Mike

 

[Baluncore]

We have no idea what voltages to expect. Are you considering 1 mV, 1V, 1kV or 1MV ?

Consider using a vibrating reed supporting the gate of a MOSFET. When the gate reed sensor is oscillating between say, the 40% and 60% points on a line between two differently charged objects, the MOSFET current will be modulated. The phase and amplitude of the MOSFET drain current will be proportional to the differential voltage. Synchronously detect the signal by multiplying the MOSFET current by the reed position, then low pass filter it. If the detected voltage is close to zero then the voltages are the same. A 100 Mohm resistor may be needed to bias the MOSFET gate into the active region.

Such a dynamic electric field sensor could be mounted on a reed oscillating in orthogonal directions, (driven by sine and cosine), that passes the two charged objects, ground, V0 and a third voltage, V3. Synchronously detecting the quadrants will identify V1 and V2. V3 can then be adjusted to match V1 when the difference between V1 and V2 is greater than an absolute threshold.