parallel resistance Rs in current source

Hi

I have always thought of a current source as a very, very high voltage supply with a very, very large resistor, Rs, in series (I mean this large resistor would make series connection with the load resistor, RL). The larger the series resistor the less effect the change in load resistor has on the load current. The Rs should be a variable resistor which is adjusted to get the wanted current through the RL. In other words, a current source is also a special type of voltage source.

In a regular voltage source the Rs is very small which can be adjusted to vary the voltage across the RL. If you want more volts to appear across the RL then you need to decrease the resistance of Rs.

Please correct the stuff above if you found something wrong. Thanks

Now I'm coming to the main question(s).

Please have a look on the linked diagram:
https://img843.imageshack.us/img843/3518/imgan.jpg

A current source is represented as shown in Fig. 3 in the diagram with the Rs in parallel configuration. I don't get it. Why? As I say above, in my view, a current source is a special kind of voltage source. They say, Rs, is an internal resistance and makes a parallel connection (as shown in Fig. 3). I don't get it. Do you get where I'm having problem? Please help me. Thanks

No, a current source is just that--a source that provides a certain fixed current. Of course, it also delivers voltage, but its purpose is to provide a certain current, and no more. (It can't guarantee against less current, of course, since you can't force current through a circuit.) The way this is implemented within the actual power supply may involve variable voltages, but that's only a technical detail, not the principle of the current source.

I think Rs is just the internal source resistance of the current source.

PG1995 said:

Hi

I have always thought of a current source as a very, very high voltage supply with a very, very large resistor, Rs, in series (I mean this large resistor would make series connection with the load resistor, RL). The larger the series resistor the less effect the change in load resistor has on the load current. The Rs should be a variable resistor which is adjusted to get the wanted current through the RL. In other words, a current source is also a special type of voltage source.

In a regular voltage source the Rs is very small which can be adjusted to vary the voltage across the RL. If you want more volts to appear across the RL then you need to decrease the resistance of Rs.

Please correct the stuff above if you found something wrong. Thanks

Now I'm coming to the main question(s).

Please have a look on the linked diagram:
https://img843.imageshack.us/img843/3518/imgan.jpg

A current source is represented as shown in Fig. 3 in the diagram with the Rs in parallel configuration. I don't get it. Why? As I say above, in my view, a current source is a special kind of voltage source. They say, Rs, is an internal resistance and makes a parallel connection (as shown in Fig. 3). I don't get it. Do you get where I'm having problem? Please help me. Thanks

Click to expand...

Hi,

Fig 1 is a non ideal current source. Fig 2 is an ideal current source. Fig 3 is an equivalent non ideal current source, where the current is Vs/Rs from the Fig 1....it's the Norton equivalent circuit and works the same as Fig 1.

If you want to understand this better, evaluate Fig 1 with a couple resistor values for the 'load', then using Is=Vs/Rs for Fig 3 (get Vs and Rs from Fig 1), evaluate that circuit for those same resistor values. Compare results. The results will show you first hand how this works.

It's Thevinin and Norton's theorems at it again. It basically states that an ideal voltage source can be replaced by an ideal current source (Rs = infinity) in parallel with a resistor and al current source can be replaced by an IDEAL voltage source in series with a resistor.

The CATCH here is ideal and real. The effective output Z of a current source should be as high as possible/required and the effective Z of a voltage source should be as low as possible/required.

Thank you, carbonzit, MrAl.

MrAl said:

Hi,

Fig 1 is a non ideal current source. Fig 2 is an ideal current source. Fig 3 is an equivalent non ideal current source, where the current is Vs/Rs from the Fig 1....it's the Norton equivalent circuit and works the same as Fig 1.

If you want to understand this better, evaluate Fig 1 with a couple resistor values for the 'load', then using Is=Vs/Rs for Fig 3 (get Vs and Rs from Fig 1), evaluate that circuit for those same resistor values. Compare results. The results will show you first hand how this works.

Click to expand...

Hi MrAl:

Part 1:
You have used three terms for a current source. I think Fig. 1 somewhat tells how a 'real' current source works. It's a 'real' current source - not some kind of mathematical model which is used to make analysis easy. If you ask me I would say in Fig. 2, I have only used a single symbol for the current source circuit shown in Fig. 1. So, in my view (which I'm almost sure is wrong! ) that is also a non ideal source.

Perhaps, the current source shown in Fig. 3 is only some kind of mathematical abstraction.

Part 2:
For Fig. 1: Let's say Vs is 10V and Rs is set at 1 ohm, and RL is 200 ohm. The current delivered would be: 10/(1+200) = 0.05A, and the voltage across RL would be: 9.95V.

For Fig. 3: Using values from Fig. 1 above. Is = Vs/Rs => Is = 10A. Using current divider rule: I_RL = (Rt/RL)I_total => I_RL = 0.05A.

Part 3:
Perhaps, somewhere in mind I'm trying to understand the working of a current source from 'actual' point of view. How it really works. Please don't go into too much technical details because I won't be able to understand them and appreciate your effort. Thanks.

Part 3:
Mathematics and the real world are different, For instance, in the real world zero doesn't always exist. In the real world 1E16 might be effectively the same number as 2 E16. You have to remember that. A 2 x 4 is not 2" x 4". Get the idea?

Ideal has specific definitions that help analysis of circuits. No more, no less. Sometimes, you can use the ideal model and sometimes you cannot.

You don't know yet, that feedback in an amplifer can lower the effective output resistance. You don;t know yet that the output Z of a finite current source is delta I/delta V. You can make some changes at the input of a circuit and refer it to the output. That's really how the numbers come about.

Here is a datasheet on an older current source: **broken link removed** It's output R > 1e14 ohms on the 1 nA scale. Yep, it's a big number, but it's not infinite. It could be considered infinate compared to 1 ohm, but not 1e13 ohms if those resistances were in parallel.

PG1995 said:

---Part 3:
Perhaps, somewhere in mind I'm trying to understand the working of a current source from 'actual' point of view. How it really works. Please don't go into too much technical details because I won't be able to understand them and appreciate your effort. Thanks.

Click to expand...

Let me take a stab at that.

A typical actual current source uses feedback to approximate (and it can be a very good approximation within limits) of the ideal current source (current independent of the voltage).

In the constant current circuit the output current is monitored (typically the voltage across a small resistor in series with the output) and this is used to control the current.

This current may be provided by a voltage source or a high impedance circuit (such as the collector of a bipolar transistor or drain of a MOSFET).

The negative feedback signal from the current monitor is then simply used to control the voltage or high impedance circuit to keep the output current at the programmed value. For a voltage source is will adjust the value of the voltage. For the transistor source it will adjust the base current or gate voltage. It either case the output current will stay constant. The limit is when the voltage has to to higher than the available supply voltage of the circuit to maintain the current.

How close this comes to the ideal is determined mainly by the gain of the feedback loop. With typical gains available in modern op amp circuits it can maintain the current within a very small fraction of a percent.

PG1995 said:

Thank you, carbonzit, MrAl.



Hi MrAl:

Part 1:
You have used three terms for a current source. I think Fig. 1 somewhat tells how a 'real' current source works. It's a 'real' current source - not some kind of mathematical model which is used to make analysis easy. If you ask me I would say in Fig. 2, I have only used a single symbol for the current source circuit shown in Fig. 1. So, in my view (which I'm almost sure is wrong! ) that is also a non ideal source.

Perhaps, the current source shown in Fig. 3 is only some kind of mathematical abstraction.

Part 2:
For Fig. 1: Let's say Vs is 10V and Rs is set at 1 ohm, and RL is 200 ohm. The current delivered would be: 10/(1+200) = 0.05A, and the voltage across RL would be: 9.95V.

For Fig. 3: Using values from Fig. 1 above. Is = Vs/Rs => Is = 10A. Using current divider rule: I_RL = (Rt/RL)I_total => I_RL = 0.05A.

Part 3:
Perhaps, somewhere in mind I'm trying to understand the working of a current source from 'actual' point of view. How it really works. Please don't go into too much technical details because I won't be able to understand them and appreciate your effort. Thanks.

Click to expand...

Hi again,


If you look back at my first post you'll see that i suggested that you analyze the circuits in Fig 1 and Fig 3 and compare results. If you did that you probably wouldnt have to ask more questions I dont think i made the procedure clear enough however so it wasnt really your fault.

The current source in Fig 3 is called the "Norton equivalent circuit". It's used to help simplify the analysis of some circuits.

Now if you analyzed Fig 1 and Fig 3 for three different resistor values you'd get the following results...

Lets say Vs=100v, and Rs=1Megohm in Fig 1. If we then calculate the current through the load resistor RL making it first 10 ohms, then 100 ohms, then 1000 ohms, we get the following current levels for each resistance respectively:
I1=9.99990000099999e-005
I2=9.99900009999e-005
I3=9.99000999000999e-005

and from inspection we can see that the voltage source and large resistance is acting very closely to a true current source as expected.

Now we go to Fig 3, and as i mentioned before we make Is=Vs/Rs from Fig 1, and since Vs=100 and Rs=1000000 we get Is=0.0001 amps.
Since once we connect RL to the circuit in Fig 3 we will have Rs and RL in parallel now (instead of in series) we have to calculate the current through RL using current division. The current through RL thus works out to:
i=Is*Rs/(Rs+RL)
where Is=0.0001 so we have:
i=0.0001*Rs/(Rs+RL)
Computing this current for the same three values of resistance for RL, we get the following results:
I1=9.99990000099999e-005
I2=9.99900009999e-005
I3=9.99000999000999e-005

Now we compare these last three results with the previous three results, and we find that they are exactly the same.

Dont feel bad if it takes you a while to start to understand current sources. It often takes a little thinking and doing some circuits before it starts to take root in your head. This happens with many people when they first learn about current sources because most of us are so conditioned to thinking in terms of voltage sources. It's just one of those things, and it takes a little while to get the picture of what is going on.

The best way to overcome this is first to simply accept that the true current source is just that, and forget about the non ideal source for now. Using the ideal current source, analyze a bunch of circuits with only a current source in it. Compare your results with a circuit simulator to check that you are getting the right answers and go from there. Keep analyzing new circuits until it sinks in. They dont have to be complex circuits either. You'll eventually find that some circuits with current sources are actually much easier to evaluate than some circuits with voltage sources.

Also, looking at Thevenin and Norton equivalents taken together as source transformations they just state two main ideas:
1. A voltage source with a series resistance is equivalent to a current source in parallel with a resistance (of the right values of course).
2. A current source in parallel with a resistance is equivalent to a voltage source in series with a resistance (of the right values again).
These are often just referred to as "Source Transformations".
In the above calculations we did we used #1 to convert from a voltage source to a current source. The new current is equal to the voltage divided by the series resistance, and the new resistance is just the old resistance. So the new current sources current Is is just the voltage sources voltage Vs divided by the voltage sources series resistance Rs, and the new current sources parallel resistance is just the voltage sources resistance Rs. So we go from having Vs in series with Rs to having Is in parallel with Rs.
You might want to call this an 'abstraction' if you like, but then we might be going the other way around, from the current source in parallel with Rs to a voltage source in series with Rs, so which one is the real abstraction? The answer is that they are BOTH abstractions of what we find in the real world, but they are mathematical tools that can help us simplify some problems significantly and thus come in very handy sometimes.

Many, many thanks everyone, KeepitSimpleStupid, crutschow, MrAl. I have to say this. You guys are really nice.

MrAl, I really appreciate your advice(s). Thanks.

A 2 x 4 is not 2" x 4". Get the idea?

Click to expand...

No, I don't get the idea? What is 2" and 4"?

____________________________________________________________________________________________________

I was reading that when you intend to turn off a voltage source you replace it with a short circuit, and to turn off a current source you replace it with an open circuit. Why is so? I understand that I have been asking several questions about power supply and answers to which might be obvious. In that case please excuse my ignorance and guide me. Thanks.

A 2 x 4 is a US standard for wall construction in a home and businesses, but it's not 2 inches x 4 inches in cross section.

MrAl, you are so nice. Thank you very much. You and many other good people are making every effort to make it simple for me. But I'm still stuck. You can blame my stupidity for that; what can I do about that. No one likes to be stupid!

It's quite funny when you yourself don't really know what is bugging you and why you are stuck at one point (well, one thing one could be sure of is his stupidity!. Moreover, understanding of the concepts becomes hard especially when you don't have an opportunity to see those concepts in action such as laboratory. I'm sorry to ask many of the questions again but I'm still having difficulty understanding it. I hope you won't mind my asking again.

I will use the linked circuit diagrams to make the gaps in my understanding clear so that you can help me effectively.

1: What does it really mean when your short a voltage source in 'real' world? Obviously you wouldn't connect +ve and -ve terminals with a wire!

2: What does it really mean when your open a current source in 'real' world? Would you simply disconnect its +ve and -ve terminals from the circuit?

Please have a look on Figure 4.12 in the link #1.

3: To build up that circuit I need two voltage supplies and one current supply. There is no parallel resistance for the current source in the circuit. Why? Does the 'actual' lab power supply have a parallel resistance in it?

4: What does it mean when Rth is zero when Thevenin equivalent circuit?

1: https://img710.imageshack.us/img710/5271/powersuppy1.jpg
2: https://img200.imageshack.us/img200/2150/powersupply2t.jpg

hi,
Look thru these links

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/visource.html

http://en.wikipedia.org/wiki/Current_source

http://en.wikipedia.org/wiki/Voltage_source

Shorting / Opening is an analysis technique.

1. The fuse pops. Something else pops. The unit goes into whatever current compliance technique that's been designed in.
2. Open a Voltage source, means it put's out the most voltage it can or is allowed to.
3. A lab supply has an "effective" parallel resistance and it may or may not influence the analysis. Reality states that you CANNOT build that circuit, but you can analyize it.
4. Probably means the voltage source is ideal.

PG1995 said:

MrAl, you are so nice. Thank you very much. You and many other good people are making every effort to make it simple for me. But I'm still stuck. You can blame my stupidity for that; what can I do about that. No one likes to be stupid!

It's quite funny when you yourself don't really know what is bugging you and why you are stuck at one point (well, one thing one could be sure of is his stupidity!. Moreover, understanding of the concepts becomes hard especially when you don't have an opportunity to see those concepts in action such as laboratory. I'm sorry to ask many of the questions again but I'm still having difficulty understanding it. I hope you won't mind my asking again.

I will use the linked circuit diagrams to make the gaps in my understanding clear so that you can help me effectively.

1: What does it really mean when your short a voltage source in 'real' world? Obviously you wouldn't connect +ve and -ve terminals with a wire!

2: What does it really mean when your open a current source in 'real' world? Would you simply disconnect its +ve and -ve terminals from the circuit?

Please have a look on Figure 4.12 in the link #1.

3: To build up that circuit I need two voltage supplies and one current supply. There is no parallel resistance for the current source in the circuit. Why? Does the 'actual' lab power supply have a parallel resistance in it?

4: What does it mean when Rth is zero when Thevenin equivalent circuit?

1: https://img710.imageshack.us/img710/5271/powersuppy1.jpg
2: https://img200.imageshack.us/img200/2150/powersupply2t.jpg

Click to expand...

Hi again,

What circuit are you trying to build?

Although you dont have a real life lab you do know there are free circuit simulators out there where you can do experiments in software and learn quite effectively a lot about these subjects. The LT simulator is free and is quite good so i would suggest downloading it and learning to use it. It doesnt take too long before you'll be building circuits on your computer screen and doing experiments with all this stuff, including of course voltage and current sources both ideal and non ideal.

You should realize that some things that we do to more easily analyze a circuit on paper we would never do in real life because it could ruin something or even cause fire or explosion. Shorting a voltage source is one of those things. We dont short voltage sources, but we could open circuit them first and short the two terminals they 'used' to connect to and that would be the same thing as shorting the supply but we would not harm the power supply or anything else.
We also dont open circuit a current source really, what we do is first short it out and then disconnect it from the circuit, leaving the two terminals on the current source shorted while leaving the two terminals in the circuit open. Thus we opened the two terminals where the current source 'was', but we shorted the actual current source itself only.

So you see to think of shorting a voltage source first think about disconnecting it from the circuit and only shorting the two circuit terminals not the source itself, and to think of open circuiting a current source first think about shorting it out and then disconnecting it from the circuit leaving the circuit terminals open.

Real life sources depend greatly on their design no matter what kind they are. You have to consult the manufacturers specs to see what you can get away with. You can also build your own.