Why is the voltage drop created by a current source added?

Question

A few days ago a similar question was asked for the simple circuit below consisting of three elements in series (voltage source, resistor and current source). Despite professional explanations, the OP could not understand why in the lower diagram the voltage drop across the resistor was added to the source voltage and not subtracted.

I was watching the discussion with interest because I have encountered this circuit in many interesting electronic circuits. I made a brief comment on the nature of voltage and current sources... and at this point the OP impulsively asked me to explain why VR1 was added to V1 to obtain Vo. How can you not respond to such a touching request? I started thinking about what the OP's problem with understanding was and how to solve it.

Unexpectedly for me, however, the OP removed his/her question. But I still decided to finish my answer and publish it under a more precise question...

Answer 1

Such circuits, in which voltage drops are added/subtracted according to KVL, can be visualized in an attractive manner by voltage bars (in red) with a proportional height. If we ground the circuit, we can observe the four combinations below between the source directions. Let's consider them.

1. Positive voltage, positive current. This is the usual case when a voltage source with positive voltage is discharged by a load. The interesting thing here is that the role of the load is performed by a current source (more precisely, sink); so the voltage source is discharged with constant current.

Fig. 1. Current source (sink) discharging a positive voltage source

As in the classic circuit with a passive load (e.g., a resistor), the voltage drop VR1 across the resistor R1 is subtracted from the voltage V1 and the resulting voltage Vo across the current source is zero (V1 -VR1 = V1 - I.R1 = 10 - 10 = 0 V). It is interesting that VR1 is constant… and if V1 varies, VR1 will not vary… so Vo will follow V1 variations. You can think of R1 as another "battery" with voltage VR1 connected in series with the main battery V1.

This effect can be observed in a common-emitter amplifier stage where if the supply voltage varies, the collector voltage follows it. Also, it is used in some op-amps to "shift down" the voltage variations.

At the OP's conditions (V1 = 10 V, R1 = 5 ohm and I1 = 2 A), the voltage drop VR1 is equal to the voltage V1; so the output voltage Vo across the current source is zero (like a virtual ground). I have considered this situation in more detail in Fig. 6 below. It would be interesting to increase the current and to see what Vo will be.

2. Positive voltage, negative current. Let's reverse the current source (the OP's problem). Now the voltage source becomes a "load" which is charged by the current source.

Fig. 2. Current source charging a positive voltage source

The voltage drop VR1 across the resistor R1 is added to the voltage V1 and the resulting voltage Vo across the current source is two times higher (V1 + VR1 = V1 + I.R1 = 10 + 10 = 20 V). Interesting… is it a voltage doubler?

Since VR1 is constant when V1 varies, Vo will follow V1 variations. Again, you can think of R1 as a floating "battery" with voltage VR1 connected in series and in the same direction with the main battery V1. So there is nothing special in this case either. See also Fig. 5 where a conceptual internal circuit of the current source is shown.

The most typical application is charging a rechargeable battery with internal resistance R1. Also, the weird negative impedance converter (INIC) resembles this circuit.

3. Negative voltage, positive current. This is the same arrangement as in Fig. 1; only the battery is grounded with its positive terminal.

Fig. 3. Current source discharging a negative voltage source

As in Fig. 1, the voltage drop VR1 is equal to the voltage V1 and the output voltage Vo across the current source is zero. And here it would be interesting to increase the current and to see what Vo will be.

4. Negative voltage, negative current. And this arrangement is equivalent to Fig. 2.

Fig. 4. Current source charging a negative voltage source

5. Inside the negative current source. I think the main problem of understanding this arrangement was that OP (of the original question) did not know what was inside this circle with an arrow. That is why, in the conceptual picture beloe, I have shown a possible implementation of a constant current source. It is connected according to Fig. 2.

Fig. 5. The negative current source - a possible implementation with "dynamic voltage source"

As you can see, this is a real but "dynamic" voltage source with internal resistance RI and "self-varying" voltage VI. The idea is simple but clever - if V1 varies, VI follows it ("shifted" with a constant value). As a result, the voltage drop VR1 and, accordingly the current I1, stay constant. I have explained this current-creating technique in my answer to the question, How do we create current sources?

6. Inside the positive current source (sink). Let's now see the same implementation of the constant current source (sink) by a "dynamic" voltage source but for the case shown in Fig. 1 (positive current). I have redrawn Fig. 5 in a more appropriate form so it has become more beatiful, symmetric and tidy - Fig. 6. Note that the elements with positive voltage across them (V1 and R1) and belonging to them voltage bars are drawn above the zero voltage level (ground); the elements with negative voltage across them (VI and RI) and their voltage bars are drawn below the ground. Now we can try to explain it.

Fig. 6. The positive current source (sink) - a possible implementation with "dynamic voltage source"

It is interesting to compare the classic Ohm's circuit (e.g. the left circuit in Bruce Abbott's answer) with this circuit. In the first, the lower end of R1 is grounded so it has zero voltage... while here it is "virtually grounded" and also has zero voltage. The short circuit in the first is a "piece of wire" while here it is a network of a resistor RI and voltage source VI in series. In the first circuit R1 is "pulled down" to ground by the very ground while here it is "pulled" to ground by the negative voltage source VI through RI.

Op-amp inverting amplifier (Fig. 7) is a well-known application of this conceptual circuit. Here the op-amp output serves as the dynamic voltage source VI and the resistor R3 as RI. Both they constitute the current sink IIN (I1). Also, R1 is R1 and VIN is V1.

Fig. 7. The op-amp inverting amplifier is a typical application of the arrangement in Fig. 6

Basically, this is the same arrangement as above (Vo = 0) but, in addition, a negative feedback is introduced. The op-amp current sink (OA + R2) adjusts its current drawn from VIN through R1 so that the voltage drop VR1 is always equal to VIN. It does it by "observing" the virtual ground.

VIN and R1 act as an input current source. So, we can consider the whole arrangement of four elements as a current source and current sink in series.

Answer 2

First let me make an example. The flow of water only happens from a higher potential to a lower one, for example from a mountain to a valley. Now if you want to send the water from valley to mountain you have to use a motor or something to put energy behind the water.

As for the electricity the same thing happens. When we say current we mean the amount of electrical charge moving in a interval of time as we say:

I = dq/dt

In other words current is just the flowing of electrons in time. Now as with our example the electrons only go from a point with a high voltage to low voltage just like the water goes from a high potential point to a lower one.

About the circuit. If we notice we have one loop , so the current going in the ground is the same current that as passed the resistor and since we have said that electrons move from a high voltage to a lower one , so we figure that the left side of the resistor has definitely a higher voltage than its right hand side, otherwise the current should have been negative (in other direction), so that's why we said voltage drop has been added because the left hand side of the resistor is positive , since the current is in that direction.

Now when we say 'convention' its because of a simple fact that we said the flow of electrons (with respect to time) is defined as the current , we could have said the flow of holes(loosely speaking:the position of electrons when it leaves) with respect to time is the current , in that case everything would be reversed in the other direction but still the overall answer would be completely the same.

Answer 3

It's actually very simple. I1 is generating a current that flows through R1 no matter what else is in the series circuit - and the voltage across it has the same polarity that it would have if a voltage produced that current.

Therefore in the upper circuit the resistor must have positive on the left because the current is going through it from left to right, and in the lower circuit it must have positive on the right because the current is going from right to left.

^{simulate this circuit – Schematic created using CircuitLab}

In the case of voltage applied across a resistor, the resistor takes on the polarity applied and draws a current according to Ohm's law (I = V/R). In the case of current forced through a resistor it behaves exactly the same, with V = I*R.

Answer 4

not only i want to answer to this question completely but i want to make an application with this simple circuit so we would know what was the purpose of creating these simple circuits.

let's say in our circuit we have 2 constant elements which are the current source and the resistor and we have a battery which we could remove it and make the voltage source as ground. so everything i have is this circuit and i want to write mu name and others names with this single circuit at the vo or output.

analysis of the circuit : part 1:

in the above we know how this circuit works. in summary because the current is fixed the vo would be equal to 0 so that we have a current = 2A passing through the resistor and going to ground.(v0=0). now i see the question arises here that people are asking what would happen now to the current source when both sides are 0 volt.

if you pay attention the current source is equal to 2A and more importantly its an independent current source. by definition an independent current source with the value of A ampere has always A ampere current regardless of what the load is whether the load is open circuit or short circuit. that's why we say in real world there is nothing like an independent current source, because we can't make such a thing and that's why the above circuitry created by @Circuitfantasist (in the above response in part 5) is wrong even though its just a simple simulation. with no circuit you could make an independent current source. but let's say why use it? well now the vo = 0 even if remove the current source which currently does not have any effect but i don't want to do that. what we want to do here is to remove the battery or in other words connect that node to the ground like this:

what i mean is that instead of a constant voltage of 10v , you imagine that you have a voltage source which oscillates between 10 and 0. now in this case we would have:

v0 = 5(-2) = -10v

so as you see we only have two states for the output voltage of vo. its whether 0 when we have 10v at input or its -10v when we have 0 at input.

now lets create our application:

from now on instead of -10,0 i want to say 0 and 1. so whenever we have -10 let it be 0 and when we have 0 volt at the output think of it as 1.

before we move on i tell you to assign a number to each english alphabet respectively. so a would be 0, b is 1, and so on until we get to 25 which is z. now i tell you also that in each interval of 8 seconds i send some data consisting of 8 zero or one and call it one byte. then i wait for 2 seconds and move to next data and consider each voltage level 1 bit.

now i do this:

second 1: put the voltage 0 as the input ----> we have 0 at output

second 2: input = 0 ---> output=-10v

second 3: input = 0 ----> output = -10v

second 4: input = 0 ----> output = -10v

second 5: input = 10 ----> output = 0v

second 6: input = 10 ------> output = 0v

second 7: input = 0 -----> output = -10v

second 8: input = 0 -----> output = -10v

so its like we keep the input at 0 for four seconds , then for two seconds we put 10v as the input and again for 2 seconds we make it zero again and we write the vo at each seconds.

considering the face that we said -10 = 0and 0 = 1 so we would have this values for the above:

-10 -10 -10 -10 0 0 -10 -10

0 0 0 0 1 1 0 0

00001100 in binary is equal to 12 in decimal and we just said that we encode numbers as letter based on this(just convention for all parts):

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

a b c d e f g h i j k l m n o p q r s t u v w x y z

and so 12 would be equal to m .

then i wait for 5 seconds and repeat everything some how that i could make these two numbers:

00000110 -->6 ----> g

00000111 --->7----> h

and therefore the person that is writing values of vo at the input miles away from the input!!! that i am changing it could know that its "mgh" written at the output.

the reason that i made this application was the same reason that an independent current source has been created for these circuits. something that could hold the output node while we are changing the input. then this idea has been moved to transistors. in transistor the same thing happens only then transistor is like a dependent current source which somehow @Circuitfantasist has shown how we could implement it using these elements in the above example since that's just a dependent current source.

Answer 5

first lets say we have a circuit like this :

since we have i = V/R so the smaller resistor gets more current and we would have:

current passing 4k resistor:

i = 5/4 mA (pardon the Kilo and therefore mA , old habits for electronics)

current passing 2K resistor:

i = 5/2 mA

what it means is that for two resistors in parallel with R1 and its current as i1 and R2 and its current R2 we would have this relation :

i1/i2 = R2/R1

now in our example the current passing through 4k would be 1/2 the current of the 2k resistor.

now lets say we increase the current of second resistor and we have:

now we have:

i1 = (R2/R1)i2 => i1 = 0.5/4i1 => i1 = 0.125 i2

so current of i1 is almost 0.1 current passing through second resistor, in other words the second resistor or the smaller one takes 90% of the current i (which might be from the rest of circuit which we haven't wrote) . now lets say we again decrease it more, the more we decrease the smaller resistor the more current of i it takes and the smaller the current of i1 would become. now lets say we make the resistor2 so small , as small as 0 which means that we only have a short circuit. in this case all the current would pass the second resistor and therefore the current passing through the resistor one (4k) would be equal to zero and it would act as an open circuit , in other words we would have:

now we get to our circuit and the reason that i say we can't imagine a circuit for our independent current source :

and now we make a resistor parallel to it and decreasing the value of the resistor until its a short circuit (both sides equal to zero ) exactly simulating the problem that has been discussed above:

now one thing that has been asked is we make an approximation for the current source , so we could say:

one thing that one may expect is that like before the resistor parallel to it (0) would take all the current coming from other branch and its like we have one short circuit alongside the new resistor and new voltage source. but the problem here is that our current now is indeed dependent on the voltage of its node and its become a dependent current source. that's why i'm saying we can't create an independent current source.