Two different voltage sources in parallel

[rattus]

Why not?

Why not?

Why don't we just say,

Vx = [(V1 - V2)/(R1 + R2)]*R1 + V1?

True that is until both batteries are ruined!

 

[gar]

100713-1205 EST

topgone:

For two DC voltage sources connected in parallel and assuming linear devices and a resistive load on the two sources, then the load voltage can be anything from 0 to somewhere between or equal to one of the values of the open circuit voltages of the two sources depending upon the load resistance. I do not have an answer for both voltage sources have zero internal resistance.

This means the output voltage can be below the smaller of the two source voltages, equal to the lower source voltage if its internal resistance is zero, or between the two source voltages if their internal resistances are non-zero, or equal to the larger voltage if its internal resistance is 0.

If you consider AC and circuits with L, R, and C, then the voltage could be higher than the sources.

.

 

[mxslick]

Theres an IEEE paper on a related theme thats quite intreresting; it looks at the scenario when a higher voltage transmission line connects with a lower voltage distribution line, ie a 33KV line drops onto a 11KV line.

The surprising (to me, anyway) conclusion was that most of the time the lower voltage "wins" so (using the example above) the 33KV line potential drops significantly, and the 11KV line rises slightly.

Of course, you hope the breakers open to sort the mess out before something really bad happens...

You mean as in something like this?


PART 1


PART 2


PART 3

 

[dbuckley]

You mean as in something like this?

Yep!

Man that's impressive...

 

[Pitt123]

I understand now that output voltage will be a function of source impedances and load impedance.

100713-0726 EST

rcwilson:

You can view a loadcell as some equivalent voltage source with some internal resistance when just viewing the two output terminals.

The approximate equation, but very close for small unbalance, for the source voltage would be vout = Vin*K*f where Vin is the excitation voltage, K a scaling constant, and f the applied force. The equation for internal resistance is two sets of two parallel resistors and then these in series.

If you want to create the equivalent circuit of the bridge but referenced to one of the excitation terminals, then you have two voltage sources with approximations of vout1 = Vin*0.5 + Vin*K1*f, and vout2 = Vin*0.5 - Vin*K2*f and the difference of vout1 and vout2 is your desired signal, or vout = Vin*K1*f - Vin*K2( -f ) = Vin*(K1+K2)*f. The source resistance for each side is the parallel resistance of the two bridge leg resistors on the associated side.

Ideally a bridge would have identical resistance in each leg except for the delta resistance from the applied force.

So a nominal 350 ohm output bridge also has a 350 ohm input and is made up of four 350 ohm resistors. Two 350 ohms in parallel = 175 ohms, and then two 175 ohms in series = 350 ohms.

Real straingage bridges are not quite built this way so input and output internal resistance measurements are not the same. This is because other resistances are associated with the bridge to provide standardization, zero balance, and temperature compensation. Measured with an ohmmeter the output resistance will be slightly less than the input resistance.

A transducer I pulled from the cabinet has 1,020 ohms input and 1000 ohms output resistance.

You can find a discussion of the application of Thevevin's Theorem to bridge circuits on p67 -71 of "Basic Electrical Measurements", by Melville B. Stout, Prentice-Hqll, 1950.

.

What if you have several of these load cells in parallel connected to a common input or amplifier? I have see an arangement where multiple load cells ( 5 wires each) connected in prallel all connected to a common input block. How do the mV outputs of each load cell in this case combine to produce a mV ouput signal?

 

[gar]

100714-2254 EST

Quite complex and probably substantial errors.

Loadcell bridges are not identical and internally have adjustment resistors. I do not want to try to set up some description of how bad this could be.

The way it could be done and get the best results would be to provide isolated excitation supplies for each bridge instead of a common excitation source. These would need ultra low leakage greater than 10 megohms.

Then each bridge is a floating low source voltage with a low internal resistance, like 10 MV max at maximum load and 0 at 0 load, and maybe 350 ohms resistance. The parallel connection would feed a low resistance, 0.35 ohms, shunt to measure current. This arrangement would sum the currents from the separate loadcells. If there were four loadcells at full load, then the voltage across the current shunt would be 40 microvolts. Very low with which to work.

.

 

[mivey]

Had this exact case while in the Army. This photog connected a 6V battery in parallel with the 4.5V in his flashgun. Could not understand why all the batteries were discharged in short order. Would not or could not believe me either.

Had a senior lab partner that bread-boarded a 12v and 5v supply to the same point and could not figure out why that would not work. Two semesters away from graduation-how is that possible? All I can say is a degree does not mean everything.

 

[rattus]

Had a senior lab partner that bread-boarded a 12v and 5v supply to the same point and could not figure out why that would not work. Two semesters away from graduation-how is that possible? All I can say is a degree does not mean everything.

I used to think I was smart until I got to college and met some brilliant kids. I like to think I am somewhere in the middle--able to recognize genius and stupidity. One guy with an MS figured the open circuit voltage of a voltage divider then surmised that this created a 3V forward voltage across a diode! Another dropped out in his 3rd year when it was discovered that he didn't understand Ohm's law.

But, the smartest guy I know was kicked out of school after one semester because he only attended classes to take the quizzes. No matter that he did well on the quizzes, the old fogeys insisted he sit through boring lectures about things he already knew!

 

[mivey]

the old fogeys insisted he sit through boring lectures about things he already knew!

Too full of themselves. Reminds me of the profs that continued talking after the bell and would call you out if you started to leave so you could make the class on time for the other arrogant prof who would call you out if you were late!