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
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.
Let me take a stab at that.---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.
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.
A 2 x 4 is not 2" x 4". Get the idea?
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
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?