Two different voltage sources in parallel

[Pitt123]

What happens if you have 2 different sized voltaeg sources in parallel with a load. What voltage would the load see? For instance if i had a 5V and a 10V in parallel connected to a load in parallel what voltage would the load see?

I know there would be a circulating current between the two sources depending on the impedances of the sources?

Would you have to do a mesh analysis between the two circuit loops to determine the current and voltage output to the load?

 

[charlie b]

To answer your questions and observations in order:

• The higher of the two.
• 10
• Confirmed
• That is one way to find the currents. But the voltage would be the higher of the two. More precisely, the voltage would be that of the higher of the two sources, minus that source's internal resistance times the current flowing through the source.

 

[nakulak]

the answer above assumes that the lower voltage source can handle the increased current through it. When dealing with lead acid batteries, for instance, the lower voltage source might blow up, giving the same result, unless it melted into a short causing the other voltage source to blow up or short out also.

 

[Pitt123]

To answer your questions and observations in order:

• The higher of the two.
• 10
• Confirmed
• That is one way to find the currents. But the voltage would be the higher of the two. More precisely, the voltage would be that of the higher of the two sources, minus that source's internal resistance times the current flowing through the source.

Woul the two sources share the current then supplied to the load?

Would the same apply to load cells that were connected in parallel providing a mV feedback?

 

[charlie b]

Would the two sources share the current then supplied to the load?

I think not. The lower voltage source would become a second load on the higher voltage source. If you look at the mesh that involves the higher voltage source and the load, you would see current flowing out of the source, through the load, and back. If you look at the mesh that involves the two voltage sources, you would see current flowing out of the higher voltage source, through the lower voltage source, and back.

Would the same apply to load cells that were connected in parallel providing a mV feedback?

I do not understand this question. I don't know what a "load cell" is, nor what "mV feedback" means.

 

[gar]

100712-1153 EST

Pitt:

Do a little circuit analysis.

First, determine the source resistance (impedance) of each voltage source. Assume you have no problems as mentioned above.

Determine the equivalent circuit of the unloaded sources when tied in parallel. This will be an ideal source voltage equal to the open circuit voltage of the two in parallel. The series source resistance of the equivalent circuit will be the value of the two individual source resistances in parallel.

The load resistance on this equivalent circuit to prevent back-feeding into the low voltage source will occur when the voltage across the load resistance is equal to the lower voltage of the two original source voltages.

To your real question --- paralleling the outputs of two loadcell. This is a bad idea.

To sum or difference the two loadcells use two buffer amplifiers feeding a summing amplifier.

.

 

[Besoeker]

To answer your questions and observations in order:

• The higher of the two.

• 
  I'm rather astonished that a professional engineer would make such a comment.
  In actual fact, the voltages would be somewhere between the two.
  It depends on the internal impedances of both sources and that of the load.
  
  Assume, for simplicity, that the sources are DC impedances are resistive and linear.
  If both sources have equal internal resistance and very much lower than the load resistance. In such circumstances the load voltage would be roughly half way between the higher and lower voltages. For example, with the internal resistances at abut 1% of the load, the voltage is 7.46V - actually slightly closer to the lower voltage than the higher.
  
  Making the internal resistances and load resistance equal has an interesting outcome. Consider just the 10V side for a moment with the 5V side disconnected. The equal resistances would result in a load voltage of 5V. Connecting the 5V source would make no difference regardless of its internal resistance.
  In this specific instance, the load voltage would be the lower of the two, not the higher.

 

[rattus]

And:

And:

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.

 

[rcwilson]

A load cell or strain gauge is a resistive network that changes resistance in proportion to the force or weight applied to it. The signal our of a load cell is a milli-volt signal but it is not a voltage source. An excitation voltage is applied to the bridge circuit containing the load cell variable resistance and a voltage proportional to the force can be measured on the sensing terminals, very similar to measuring temperature with an RTD (Resistance Thermal Detector).

The voltage or mV/pound force signal depends on the excitation voltage and the design of the bridge circuit and measuring instrument. If one load cell puts out 10 mV and another 5 mV, the resultant when they are paralleled depends on the excitation voltage source and the sensing circuit design.

On large truck scales I worked on years ago, I recall paralleling the load cell outputs to add the signals to get total weight. I don't recall what type of sensing circuit the Toledo scale was using or how we actually wired the sensing lines.

If you are trying to parallel load cell signals, do a little more research or talk to a load cell supplier to find out the proper way to add the signals.

 

[charlie b]

I'm rather astonished that a professional engineer would make such a comment.

Perhaps you should take a closer look at the fourth bullet in that post.

Sometimes a simple answer to a simple question is better than a complex answer, even if it leaves out some relevant fact or truth. He was looking for a basic concept, and I gave him the basic answer. I inferred that he was wondering whether the answer would be the average of the two voltages, and I wanted to quickly steer him away from that notion. Later, I explained more fully.

 

[Besoeker]

Perhaps you should take a closer look at the fourth bullet in that post.

I did and had.
"That is one way to find the currents. But the voltage would be the higher of the two. More precisely, the voltage would be that of the higher of the two sources, minus that source's internal resistance times the current flowing through the source."
Once again, you repeat that point about it being the higher of the two.
That's just plain wrong.

Sometimes a simple answer to a simple question is better than a complex answer, even if it leaves out some relevant fact or truth.

Leaving out some facts is one thing but stating something that is simply not true, particularly if you know it not to be true, isn't acceptable in my book.
But you opted for a categoric statement, more than once, that it is the higher of the two voltages. Even when you included your final caveat, you couched it terms of the higher voltage.

He was looking for a basic concept, and I gave him the basic answer. I inferred that he was wondering whether the answer would be the average of the two voltages, and I wanted to quickly steer him away from that notion. Later, I explained more fully.

The simple and truthful answer would have been:
"Somewhere between the two."

 

[topgone]

""Somewhere between the two."

I agree with this one. If those were batteries, the lower voltage would be "charged" while the higher voltage would discharge to the lower voltage supply. The resulting voltage would be lower than the higher voltage supply but higher than the low voltage supply.

 

[LarryFine]

But, as for power supplies, there are more thyan one type. The typical rectifier-and-capacitor supply would probably not react to a higher voltage applied to its output, so the higher supply would prevail (and supply all current).

However, regulated and switching supplies are likely to attempt to force their outputs back to their settings, so I could see one or both give up the magic smoke if paralleled. I think the OP, Mr. Pitt, should experiment and report back.

We expect video.

 

[dbuckley]

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...

 

[charlie b]

More precisely, the voltage would be that of the higher of the two sources, minus that source's internal resistance times the current flowing through the source.

Are you telling me that this is not true?

 

[nakulak]

I'm no expert, but with the variety of power sources, and even with batteries, anything but a simple generalization of this type of (wrong) connection of power sources is pointless since the characteristics of the circuits are going to vary wildly.

 

[gar]

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.

.

 

[gar]

100713-0905 EST

charlie b:

Consider a DC circuit with two real world voltage sources. The assumption will be that we have stable linear devices. Each source can be described as an ideal voltage, V (constant voltage with zero internal resistance), and an internal resistance of R.

First voltage source is V1 and Ri1, and the second is V2 and Ri2. Connect the negative terminal of each source together and use this common point as the voltage reference point.

Next connect the two positive terminals together and call this point V3.

With no other load V3 = V1 + ( ( V2-V1 ) * Ri1 / (Ri1 + Ri2) ). By Thevenin's Theorem the original pair of voltage sources can be replaced with a new ideal voltage source of V3 with an internal resistance of Ri3 = Ri1*Ri2/(Ri1+Ri2).

For example:
V1 = 10 V, V2 = 8 V, Ri1 = 1 ohm, and Ri2 = 0.8 ohm.
The equivalent Thevenin's circuit is Ri = 1*0.8/(1+.8) = 0.444 ...... ohms, and V = 10 - 2*(1/1.8) = 10. - 2*0.555 ... = 8.888 ... V. If you load this with 2.0 amperes, then the load resistance is RL = 8/2 = 4.0 ohms and no current flows to or from V2.

Assuming I made no mistakes.

.

 

[topgone]

100713-0905 EST

charlie b:

Consider a DC circuit with two real world voltage sources. The assumption will be that we have stable linear devices. Each source can be described as an ideal voltage, V (constant voltage with zero internal resistance), and an internal resistance of R.

First voltage source is V1 and Ri1, and the second is V2 and Ri2. Connect the negative terminal of each source together and use this common point as the voltage reference point.

Next connect the two positive terminals together and call this point V3.

With no other load V3 = V1 + ( ( V2-V1 ) * Ri1 / (Ri1 + Ri2) ). By Thevenin's Theorem the original pair of voltage sources can be replaced with a new ideal voltage source of V3 with an internal resistance of Ri3 = Ri1*Ri2/(Ri1+Ri2).

For example:
V1 = 10 V, V2 = 8 V, Ri1 = 1 ohm, and Ri2 = 0.8 ohm.
The equivalent Thevenin's circuit is Ri = 1*0.8/(1+.8) = 0.444 ...... ohms, and V = 10 - 2*(1/1.8) = 10. - 2*0.555 ... = 8.888 ... V. If you load this with 2.0 amperes, then the load resistance is RL = 8/2 = 4.0 ohms and no current flows to or from V2.

Assuming I made no mistakes.

.

Yep, you're right! At a load of 2 amps (on your example), V2 cannot feed the load as V1 voltage drops to just within 8 volts (10-(1ohm*2 amps) =8 volts) and is hugging the load requirement! FWIW, it it still correct to say the voltage at the interconnection point will be lower than V1 but should be higher that V2(or thereabouts)!

 

[Besoeker]

Are you telling me that this is not true?

Let's look again at the four bullet points from your post #2.
• The higher of the two.
• 10
• Confirmed
• That is one way to find the currents. But the voltage would be the higher of the two. More precisely, the voltage would be that of the higher of the two sources, minus that source's internal resistance times the current flowing through the source.

The first two points are just plain wrong, of course.
The third is correct provided you exclude the particular circumstances I noted in in the last paragraph of my post #7. I suppose I should add that if there are circulating currents the lower voltage supply would have to be capable of accepting as well as delivering current.
In point 4 you say "But the voltage would be the higher of the two again." Then you add the caveat. I have no disagreement with that caveat. But it is added as a caveat and still couched in terms of the higher of the two voltages.

Now, anyone without knowledge of circuit calculations might be forgiven for taking from your points that you were indicating that the higher of the two voltages fundamentally determines what the load voltage will be - after all, you stated this categorically four times. And yet no mention at all of the lower voltage.

Your point in post #10 indicated that you were trying to give a simple answer. With respect, I think it is both confusing and incorrect.