Parallel circuit of current sources

[BAJ]

Assume I have a parallel circuit of current sources with 5 current sources generating 1mA each. Therefore from both the terminals I will get 5mA total. Now if I add a battery to one of the terminals where current(i) will flow into my circuit. If i = 50mA, will I get 55mA current from another terminal of my circuit??

 

[Baluncore]

Welcome to PF.
Fixed current sources regulate a current flowing from a more positive node of the circuit to a more negative terminal. They do not fix the voltage, that is decided by the way external components respond to the sourced current.

If there are five 1mA sources in parallel, you will get the equivalent of a 5mA current source.

Now if I add a battery to one of the terminals where current(i) will flow into my circuit.

How do you “add a battery” to “one terminal”. You need to draw a couple of circuit diagrams showing the sources with and without the battery. Are you going to be charging that battery or is it an arbitrary fixed voltage source.

If i = 50mA, will I get 55mA current from another terminal of my circuit??

I can see no way that your statement could be true. So my answer is probably No.

But produce your circuit diagrams, (drag and drop them onto the edt window), so we can work out exactly how you modify the circuit when you add the battery.

 

[BAJ]

Actually by adding a battery I meant connecting the battery in a way that it is able to deliver the current to one terminal. However here is my proposed circuit.

 

[FOIWATER]

Replace the five current sources by a single 5 mA current source in series with a battery and resistor.

What would be the value of 'i' then?

 

[BAJ]

55mA. Isn't it?

 

[FOIWATER]

If you have a single loop and you know the value of current through anyone component, isn't that the value of the current through out the circuit?

In this circuit the current source dictates the current. The voltage source would dictate the voltage level.

The ideal current source allows any voltage to be developed across it. The ideal battery allows any amount of current to flow through it.

That's the way I would look at this, some one else may have more insight.

 

[BAJ]

But won't the resultant current add up although the current source is adding current to the circuit?

 

[FOIWATER]

No the current source's value will determine total circuit current is my understanding of this. The voltage developed across the resistor will then be determined by ohm's law, and the voltage developed across the current source will be determined by the value of the battery voltage.

The amount of current through the battery is dictated by the current source is what I am saying.

The issue lies when we say "the battery supplies 50mA and the current sources supply 5mA.." That wouldn't happen.

 

[BAJ]

"No the current source's value will determine total circuit current is my understanding of this." ----- didn't understand. can u explain a bit??

 

[BAJ]

And moreover I don't want to consider it as equivalent of 5mA in series with the battery. Because if you consider only the parallel circuit and the current source, then both pts A and B will have 5mA current. Now when battery delivers 50mA current in pt A, then definitely backflow is going to happen for which the pt B will have a current reading of 55mA (theoretically). It looks correct, no?

 

[FOIWATER]

Sure.

Have you ever read about what constitutes a current source? like you already know batteries and generators are examples of voltage sources right? what is an example of a current source?

have you studied transistors yet? This is something that comes up in very basic transistor circuits. Transistors are examples of current sources. The current through a transistor is a function of its amplification factor, and the base current. Not the voltage tied to the collector of the transistor.

Take a basic transistor circuit:

This is quite similar to your circuit. You have a battery, Vcc, you have a resistor, and you have a current source. The current through the battery, resistor, and current source is affected only by the base current of the transistor. The voltage supply is a constant, and the current it delivers is a function of the current source. The current source is drawing current from the battery. They cannot be different..

If you have a circuit with no battery, you assume the current source is able to draw its current from an in-explicit source of energy that has been absorbed into the current source symbol.. but this is how I understand it..

Someone may come along and set us both straight, though.

 

[FOIWATER]

both pts A and B will have 5mA current. Now when battery delivers 50mA current in pt A...

The battery can't deliver 50mA in pt A. It can't deliver more than the current sources are drawing.

 

[BAJ]

Thanks for your quick explanation, cleared the matter more. However consider this, the current sources are not drawing current from the battery, they are simply graphite placed parallely like the current sources drawing current from an electrolytic solution. And I have experimented earlier (without attaching the battery) that orienting 5 graphite blocks in this way can fetch you ~1mA of current. Now if I subject this parallel circuit to an external power supply, like the diagram I have provided, what will be the value of 'i' in this scenario.

I am a Biotech engineer, only have basic knowledge on electrical, sorry for my ignorance.

 

[FOIWATER]

No apologies required ! I love the discussion..

The current sources would drop sufficient voltage to limit the current supplying capabilities of the voltage source to levels dictated by the current sources themselves. The current sources don't 'pick-up' the current injected by the battery and add it to their own current like that. They deliver what they deliver, and they drop voltage to limit the current through the battery to their own levels.

That may be wrong but it is my understanding.

I can say with certainty that the current in a series circuit has to be a constant. Different sources will not contribute different current levels that 'add up' that way.

 

[FOIWATER]

You see, the current injected by the battery is determined itself by the current source. Current sources behave that way. That is, they drop voltage supplied externally (in your case the battery) and they just supply current.

If this makes sense and you are looking for a way to remember it from here on out, remember:

Current sources determine current levels.
Voltage sources determine voltage levels.

Current sources will drop any amount of voltage to ensure they continue to supply their current (ideally)
Voltage sources will sink any amount of current to ensure they supply their voltage (ideally)

 

[BAJ]

But as I explained the components I am calling current sources are not conventional current sources, they draw current from an electrolysis of a solution, not from the battery. Or are you saying, if I attach the parallel circuit with battery, these graphites will draw current from the battery?

I agree to that in series, current has to be constant. But "The current sources would drop sufficient voltage to limit the current supplying capabilities of the voltage source to levels dictated by the current sources themselves." ... can u explain a bit more?

 

[FOIWATER]

Sure.

Let me answer your question with a question. In your series circuit, what is the voltage drop across the current source?

Would you agree that, if the current source drops voltage, any voltage, that it limits the amount of current that the battery injects?

My understanding is that the current source will drop whatever voltage is necessary to ensure the current through the whole loop is determined by it alone. The batteries ability to inject current is commandeered by the current source.

Consider this: Consider replacing the current source with a resistor. For sure the battery would not be able to supply as much current then. Same is true for the current source, right? it has to drop some voltage?

That means the current source DOES in fact influence the batteries ability to source a current. Question then becomes, how much does it influence it? answer is, it influences it infinitely. The current source is a supplier of current and a dropper of voltage. It will drop whatever is required to supply its current.

It's the boss when it comes to current !

 

[BAJ]

The voltage drop across the current source must be determined by the circuit it is connected to. In my case, it may be ~50mV or something, I forgot actually. And yes I agree to your 2nd sentence too if the resistance in the circuit is considered constant. If the resistance is variable then it will be different matter.

 

[FOIWATER]

Right. So my advice would be... make the circuit ! Make the circuit and see if the current is limited to the value of the current supply.

Ensure in reality that the current source is rated for at least the voltage of the battery, though ! and vice versa for current ! The discussion here has been 100% theoretical using 'ideal' components.

In reality, there are levels to the voltages and currents components can accept.. as you know

 

[BAJ]

Ok sure. Thanks for all of your explanations. It helped to get a clear understanding. "It's the boss when it comes to current !" ------ I will definitely remember this :)

 

[Baluncore]

But as I explained the components I am calling current sources are not conventional current sources, they draw current from an electrolysis of a solution, not from the battery. Or are you saying, if I attach the parallel circuit with battery, these graphites will draw current from the battery?

So the current sources are not regulated sources, but electrochemical cells. That probably makes them voltage sources rather than current sources. They will have a high series resistance, with open circuit voltage dependent on ionic species and current dependent on the concentration in the electrolyte and the load resistance.

By explaining what the “sources” really are we might be able to come up with a model circuit that is applicable to your situation. If you add a battery you will either strip ions of the graphite into the electrolyte or plate them onto an electrode.
Why are you even considering a circuit? What are you trying to do here? What is the experiment testing?

 

[BAJ]

Ok. Explaining what I have done so far. The components I am calling current sources are commutator carbon brushes, attached to a wire which is attached to the circuit, parallely (as shown in my diagram ... the red part). They are dipped in sewage water where, through microbial respiration all the electrons that are released, are being captured and transferred to the circuit.
Now here a thing has to be considered since different amount of bacteria may adhere to each graphites on which the release of electrons are dependent. I assumed that the graphites are distributing all the electrons to both pts A and B since there is no P.D in this case. This is why I want to attach an external power supply (upto 2.5V) so that I can force this electron emission in only 1 point.

Hope u can understand the scenario. If further explanation is required, do tell.

 

[Baluncore]

The components I am calling current sources are commutator carbon brushes, attached to a wire which is attached to the circuit, parallely (as shown in my diagram ... the red part).

If all four single brushes hang in the water, and all brushes are in parallel, then where is the second electrode? Is it the water container, or maybe another set of brushes somewhere else?

They are dipped in sewage water where, through microbial respiration all the electrons that are released, are being captured and transferred to the circuit.

That is a bold assumption. If a few electrons collect on an electrode then they will polarise that electrode negative and so prevent more electrons from moving that way.

I assumed that the graphites are distributing all the electrons to both pts A and B since there is no P.D in this case.

If electrons are moving they must have somewhere to go.
Your circuit diagram shows a short circuit at the end of the line of parallel “sources”.

This is why I want to attach an external power supply (upto 2.5V) so that I can force this electron emission in only 1 point.

It seems like you recognise that you need to make a closed circuit.
You now have two independent electrodes in the solution. If you apply a small DC voltage between the electrodes you will get a current flow that is proportional to the reaction rate at the electrode surfaces. If you apply a voltage greater than 1.23 volt you will start to electrolyse the water and force other species out of solution and onto the electrodes.
https://en.wikipedia.org/wiki/Electrochemical_cell#Half-cells

I expect that you would carry out an experiment by very slowly ramping up the voltage while measuring the current. That current graph will identify the species present by their electrochemical series cell voltage, the integral will be the total charge that flows which will identify concentration of each species. Once you have plated the electrodes with various species, you can reverse the circuit and strip those species from the electrode.
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

You have not said why you are doing this. Do you want to measure the rate of reaction to assess something like water quality, or do you want to control the rate of reaction by providing the energy needed?
Is the electrolyte flowing in a channel or is it a static pool?
Do you know anything about the bacterial reaction that releases the electrons?
Do you know the energy or electrochemical voltage of the bacterial respiration chemical reaction?

 

[BAJ]

If all four single brushes hang in the water, and all brushes are in parallel, then where is the second electrode? Is it the water container, or maybe another set of brushes somewhere else?

Yes that electrode is placed above the circuit. It is also made of graphite powder and separated from the circuit using a cloth, mainly a polyester-cloth like J-cloth. It is also called air-cathode as it is in contact with the air. I also attached a copper wire to it for taking readings. Because when I attach a multimeter to take reading, the circuit gets completed as the cathode reaction would be 4(H+) + O2 + 4e- = 2H2O.
Actually during this microbial respiration or metabolism they breakdown the glucose content of the sewage as C12H22O11 + 13H2O = 12CO2 + 48H(+) + 48e- inside the anode compartment. This e- are captured by the graphite and the H+ travels towards the cathode.

That is a bold assumption. If a few electrons collect on an electrode then they will polarise that electrode negative and so prevent more electrons from moving that way.

Agree but why would they prevent? As I am taking readings periodically, the electrons that were accumulated on the graphite surface would be moving along the circuit towards the cathode for completing the reaction.

If electrons are moving they must have somewhere to go.
Your circuit diagram shows a short circuit at the end of the line of parallel “sources”.

Yes that was kept intentional. You see building up of electrons in those graphite surface takes time (minimum an hour). Each of the brushes generated current in microAmpere range. To get a good reading I had to ensure the generated electrons don't go to the cathode and forms water through the reaction.

It seems like you recognise that you need to make a closed circuit.
You now have two independent electrodes in the solution. If you apply a small DC voltage between the electrodes you will get a current flow that is proportional to the reaction rate at the electrode surfaces. If you apply a voltage greater than 1.23 volt you will start to electrolyse the water and force other species out of solution and onto the electrodes.
https://en.wikipedia.org/wiki/Electrochemical_cell#Half-cells

That means if I apply around 1.20 volt of power supply it won't electrolyse the water and might serve my purpose too, no? But I didnt understand this "...force other species out of solution and onto the electrodes." If you are referring to the bacterial species, they are already attached to the electrode (i mean the carbon brushes), its their nature to accumulate on a substratum, when provided.

I expect that you would carry out an experiment by very slowly ramping up the voltage while measuring the current. That current graph will identify the species present by their electrochemical series cell voltage, the integral will be the total charge that flows which will identify concentration of each species. Once you have plated the electrodes with various species, you can reverse the circuit and strip those species from the electrode.
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

The thing is I conducted this experiment in my college with a very very low funding. Around 1000/- (about 20 dollars). But now I am working in a different field and that's why want to continue my experiment with good materials. And by various species you meant coating with Pt/Ti/Zn/Mn?

You have not said why you are doing this. Do you want to measure the rate of reaction to assess something like water quality, or do you want to control the rate of reaction by providing the energy needed?
Is the electrolyte flowing in a channel or is it a static pool?
Do you know anything about the bacterial reaction that releases the electrons?
Do you know the energy or electrochemical voltage of the bacterial respiration chemical reaction?

The thing I worked upon is to upgrade 'Microbial fuel cell (or MFC)' with electrode configurations. The theory is about a century old and research is going on all over the world to scale it up. The main purpose of MFC is to generate sustainable energy while treating wastewater. One of the primary challenges in scaling up is to reduce the internal resistance of the wastewater (which comes in range of 2K to 3K range). So formulated the idea to increase the current output so that it can reduce it (if we apply the standard Ohm's law). In parallel circuit for current sources, current will add up but voltage will remain constant. Thats why implemented this way.

The electrolyte can be static or flowing, both. However researches proved flowing electrolyte with a retention time of about 10 hrs can increase the voltage output.

The way bacterial metabolism (sorry not respiration) is generating the electrons is a complex but fully enzymatic reaction. It is called ETC or electron transport chain that occurs within the mitochondrion of a cell. You can check it up in google.

The metabolism doesn't yield energy in its raw form but in form of ATP required to drive cellular functions. It is about 2ATP that is about 20Kcals/mole of glucose.

 

[Baluncore]

Thanks for your realistic description.

Now that I think about it, if you are trying to generate power there will be little advantage in biasing the circuit with a voltage source such as a battery. The energy cost of bias will be greater than the return. You might try to make a line of chemical cells that recycled water can flow through so the voltages will add up in series.

If you do not want to generate power now, but instead want to monitor the chemical cell current for research purposes, then a simple op-amp set up as a current to voltage converter would give you an estimate of metabolism in less than a couple of seconds. You could set the cell bias voltage and monitor the current at the same time. That would make a simple and useful research instrument in your field.

 

[BAJ]

Yes I want to generate power. In fact I want to increase the output. Making a line of chemical cells to add up voltages is not feasible. Thought of that first but how many will I add up to get to light a LED bulb? Thats why wanted to increase by looping the output and keeping the load in between. However any way you can give an idea to decrease the internal resistance more by attaching something to the circuit? By this setup I got it to a minimum of 220 ohms. Want to decrease it further in a range of 10-50 ohms.

 

[Baluncore]

What is the open circuit voltage of your reaction cell when you are not drawing any current?

Voltage without current is not power. Current without voltage is not power. You need both current and voltage. There is no way I can see to multiply voltage in one pool of electrolyte. But you can increase current in a single pool by going for much greater electrode areas with less separation. Look at the construction of a car battery to see how the cells are isolated and the electrodes have a great area.

A “battery” of cells would increase the voltage by adding many cell voltages. That battery could be a cascade of isolated pools in a closed and electrically insulated tube. Each pool would be one cell in the higher voltage battery.

If you can produce a reactive gas such as oxygen in a biochemical reactor then you could remove that for use in a separate fuel cell. Why must the chemical and electrical conversion be done in one place when the growth of bacterial films on the electrodes may contaminate the surfaces and reduce the current?

The “species” I referred to are ionic chemicals, such as Na and Cl that migrate within the electrochemical cell.

 

[BAJ]

Sorry to reply late. Was very busy since Tuesday. When I connected the multimeter to take readings (1 end to cathode and other end to anode), it showed around 400mV max and 2mA max. I didnt measure by connecting both anode ends. And to prevent contamination of electrode, I used graphite. And current will only be reduced if the layer of the bacteria on those graphites are thick. else bacterial layer is needed to get the electrons. However, in your 2nd last reply you said "The energy cost of bias will be greater than the return" ... can you explain it bit more?

 

[Baluncore]

You are making an electrochemical cell from which you will extract power, W = V·I. Power is the rate of flow of energy. You need to get electrical energy from the cell so you must load the cell in such a way as to get the maximum power, W = V · I.

If you place another identical electrochemical cell in series with the first you will get twice the voltage, 2V, but the same current, I, (determined by the reaction rate), so the power will be W = 2V·I, twice that of a single cell.
If you place another identical electrochemical cell in parallel with the first you will get twice the current, 2I, but the same voltage, V, so the power will be W = V · 2I, twice that of a single cell.

The maximum voltage is set by the chemical reaction half-cell potential of the cell. That voltage will only be available when no current is being drawn by a very high resistance load. W = V · 0 = 0.
The maximum current is limited by the reaction rate and so the area of the plates. Maximum current will flow when the load resistance is zero, but then the voltage will also be zero, so W = 0 · I = 0.

There is a compromise, as you first begin to draw current the cell voltage will begin to fall slightly due to the internal resistance of the cell. Then when the current reaches that limited by the reaction rate, the voltage will fall rapidly without a significant increase in current. So you need to load the cell carefully to maximise the power extracted. That is called the maximum power point, MPP. For general power supplies that occurs when the output resistance is equal to the internal resistance of the source. For solar and chemical cells it occurs on the tightest part of the bend in the VI curve.

"The energy cost of bias will be greater than the return"

If you force more current by adding a series voltage source then you will need to provide energy from that extra source. The energy from the added voltage source will always be greater than the extra energy extracted from the cell. That is because the cell current, I, will flow through the voltage source, V, and that power W = V · I, will have to come from somewhere else. You would do better to add more series or parallel electrochemical cells to your battery of cells.