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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.

Original question

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...

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  • \$\begingroup\$ @user255378, A few days ago, you asked me to explain "why VR1 was added to V1 to obtain Vo". I started thinking about how to solve your problem. But unexpectedly you removed your question. I still decided to finish my answer and publish it under a more precise question. You probably have done your homework. But remember that life is not just a series of homeworks. It requires knowledge and understanding circuits if you have decided to dedicate yourself to this field. So I recommend that you read my answer and take a stand... \$\endgroup\$ Jun 21, 2020 at 21:03
  • \$\begingroup\$ I would have said that the arrow of current points from the more positive end of a resistor to the more negative end. (Conventional current.) If you don't know the direction, you just pick one and stick with it when working out the KVL for a loop, for example. \$\endgroup\$
    – jonk
    Jun 21, 2020 at 22:12
  • \$\begingroup\$ @jonk, For the purposes of intuitive understanding, I always use the actual current directions and voltage polarities. These colorful pictures are "snapshots" at certain points of the circuit operation. There is no way to imagine the circuit operation if they are arbitrarily chosen. The understanding is a qualitative activity that needs qualitative tools... while the calculation requires quantitative means... \$\endgroup\$ Jun 22, 2020 at 5:51
  • \$\begingroup\$ The conventional current direction is more intuitive since we know from our experience that (electric, water, air...) fluids start where there is a lot (+) and go where there is not (-). And also, when they encounter an impeding element (R), pressure (+) occurs at the input of the element. This may seem too primitive but this is the way one imagines the invisible electrical quantities... \$\endgroup\$ Jun 22, 2020 at 6:11
  • \$\begingroup\$ Why would you assume that the OP has a "cognitive problem"? Perhaps you are not a native English speaker but this would often be considered a rude statement, particularly from a teacher. With just a touch of humility you might wonder if the feedback provided to the OP was the source of any continuing confusion, rather than some intellectual deficit on the part of the OP. \$\endgroup\$ Jun 22, 2020 at 15:43

5 Answers 5

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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.

Positive voltage, positive 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.

Positive voltage, negative current

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.

Negative voltage, positive current

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.

Negative voltage, negative current

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.

Inside negative current source

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.

Inside positive current source

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.

Inverting amplifier

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.

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  • \$\begingroup\$ I don't understand you. Step 2 answers exactly the OP's question (from the removed question) about "why VR1 is added to V1 to obtain Vo". In this case, VR1 adds to V1, so the voltage across the current source must be higher than V1. I have shown this case also in Fig. 5 where you can see the internal voltage VI of the current source (V1 < Vo < VI). \$\endgroup\$ Jun 22, 2020 at 3:49
  • \$\begingroup\$ You can not assume that the current source is the load in your first figure, based solely on the information in that schematic. It is quite possible that the current source is providing power to the circuit and that the voltage source is absorbing power. \$\endgroup\$ Jun 22, 2020 at 15:02
  • \$\begingroup\$ Definitely... V1 can be a Zener diode... and this configuration (current source supplying a Zener diode) is widely used in power supplies. The opposite (voltage source supplying a transistor) is also common. And finally, both they can be sources (current source charges a rechargeable battery). Only the fourth combination is missing (both are not sources:) \$\endgroup\$ Jun 22, 2020 at 15:27
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    \$\begingroup\$ I think you are using the terms incorrectly. In your original schematic you have a voltage source and a current source. They are always sources by definition. The question is whether either or both is supplying power to the circuit. \$\endgroup\$ Jun 22, 2020 at 15:34
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    \$\begingroup\$ And that is why your explanations bother me. You choose to redefine terms to have a different meaning than every other author I have encountered. This does not help students, it confuses them. \$\endgroup\$ Jun 22, 2020 at 15:50
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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.

everything depends on the current passing and also how we define the current(flowing of the electrons or holes)

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.

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  • \$\begingroup\$ An excellent explanation... a result of an imaginative thinking... Only to say that, in my opinion, in this case the voltage drop VR1 is subtracted from V1. Am I right? \$\endgroup\$ Jun 22, 2020 at 9:20
  • \$\begingroup\$ @Sparky256, It has been "V/R=I" since 1826... As it is so simple, would you explain how it is possible that Vo = 0? The current source should be just a "piece of wire" (short circuit). Can you see this situation in some electronic circuit today? Or is it just a homework assignment? \$\endgroup\$ Jun 23, 2020 at 6:22
  • \$\begingroup\$ @ Mgh Gh, I want to ask you something... but my request is that you answer me honestly. What do you imagine there is inside the circle with an arrow (the current source) in your drawing, when V0 = 0? Can you draw it? \$\endgroup\$ Jun 23, 2020 at 8:13
  • \$\begingroup\$ @Mgh Gh, I'm reading it right now... Just to ask you... Do you have any idea about basic electronic circuits implemented with transistors and op-amps? \$\endgroup\$ Jun 23, 2020 at 10:19
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    \$\begingroup\$ @Circuitfantasist. We can be aggressive interms of a better answer, but not benevolent. I gave you a +1 once I understood your use of virtual points, something new to us who like to keep it simple. A 1D or 2D matrix is allowed as long as you make it clear to us. \$\endgroup\$
    – user105652
    Jun 24, 2020 at 10:47
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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.

schematic

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.

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  • \$\begingroup\$ It is interesting to compare your left circuit where the lower end of R1 is grounded so it has zero voltage... with Fig. 6 in my answer where the R1 end is "virtually grounded" and also has zero voltage. The short circuit in your case is a "piece of wire" while in my circuit it is a network of a resistor RI and voltage source VI in series. In your circuit R1 is "pulled down" to ground by the very ground while in my circuit it is "pulled" to ground by the virtual ground (by the negative voltage source VI through RI). \$\endgroup\$ Jun 23, 2020 at 20:16
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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:

enter image description here

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. enter image description here

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: enter image description here

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

enter image description here

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.

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  • \$\begingroup\$ This is the most incredible explanation (application) of a simple electrical circuit through digital concepts I have ever come across. As far as I understand, you are trying to explain the transmission of digital signals in serial form (serial interface) through current (current interface). If so, I will note that in a current interface, the input signal (0 or 1) controls a dependent current source. The idea is that the current as a carrier does not depend on interferences along the line. Here we do not need a voltage source. We need only a resistor to convert the current to voltage. \$\endgroup\$ Jun 23, 2020 at 10:54
  • \$\begingroup\$ The interesting thing is what you explain in the first part of the text - how the changes 0 -> 10 V are translated ("moved", "shifted") into changes -10 -> 0 V. You can observe this phenomenon of "voltage shifting" in various electronic circuits. I would like to repeat my question, "Do you imagine what is inside the symbol of the current source in the case when Vo = 0 V?" I would be really happy if you can see in it a famous op-amp circuit... \$\endgroup\$ Jun 23, 2020 at 11:10
  • \$\begingroup\$ Interesting... That is why I stay here instead in Wikipedia, Wikibooks or ResearchGate... In some inexplicable way for me, the questions constantly asked by thinking young people like you, excite my imagination and make me rethink circuit ideas in a way I did not suspect existed... \$\endgroup\$ Jun 23, 2020 at 11:29
  • \$\begingroup\$ That's why I want to know what you imagine is inside the symbol. I am preparing another figure 6, which will correspond to Fig. 1 (Fig. 5 corresponds to Fig. 2). My question is addressed to all who follow (I guess with interest) our discussion. BTW we will soon be invited to continue chatting. \$\endgroup\$ Jun 23, 2020 at 11:36
  • \$\begingroup\$ Electrical symbols of elements are something abstract. You know the symbol and the definition of its behavior. But at one point this becomes insufficient for a deep understanding of the circuit... as here, in you circuit... and you begin to wonder what is inside this "black box" (not concretely, but conceptually). What is it in the specific case that the current is constant... independent on V1 and R1? \$\endgroup\$ Jun 23, 2020 at 12:35
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first lets say we have a circuit like this :

enter image description here

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 :

enter image description here

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:

enter image description here

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:

enter image description here

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

enter image description here

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:

enter image description here

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

enter image description here

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.

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  • \$\begingroup\$ As far as I understand, you are discussing a parallel circuit for regulating the current by diverting part of it. But it is more suitable for voltage regulation (for example in a Zener diode voltage stabilizer). Usually, current sources are made with series regulation - the internal voltage or resistance of the current source is changed to keep the desired current. For example, in Fig. 5 and Fig. 6 (that I have just added) the voltage is varying so that to keep a constant voltage drop across RI. Have you understood this ingenious idea that makes the 5 ohm resistor RI behave as "infinite"? \$\endgroup\$ Jun 23, 2020 at 17:21

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