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: http://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 Cheers
With a variable voltage source, the voltage is variable. The Thevenin resistance is usually far, far too small to have significant effect on the load voltage if it is adjusted. Same with constant current sources. You adjust the current setting (or the voltage setting in your mathematically equivalent voltage source), not the resistance. That stays constant. In this case, changing the load resistance will slightly change the total amount of current; it's not ideal. That's the entire point of including source resistance in the diagrams: to model non-ideal sources (which EVERY source is, to one degree or another). A current source can be modeled as a high voltage source with high series resistance; that is know as "source transformation", and is a very basic part of circuit analysis. Likewise, a voltage source can be modeled as a high current source with a small parallel resistance. The series resistance of a non-ideal voltage source forms a voltage divider with the load. As more current is drawn from the source (lower load resistance), more voltage is lost in the source resistance, so the load voltage goes down. Likewise, the parallel resistance of a non-ideal current source forms a current divider with the load. As more voltage is created by the source (higher load resistance), more of the current is lost through the parallel source resistance, and the load current goes down.