Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

series and parallel resistance

Status
Not open for further replies.

electronics_kumar

Advanced Member level 2
Joined
Nov 29, 2004
Messages
657
Helped
34
Reputation
68
Reaction score
9
Trophy points
1,298
Location
Tamilnadu
Activity points
5,552
what we are achieving by putting series resistance in constant Voltage source and a parallel resistance across current source
 

This "resistance" is used to model the internal resistance of a practical voltage/current source.
 

Hi.

The series and parallel resistances is used to represent the Thevenin and Norton equivanence circuits of a voltage and current sources, respectively.
 

when we are trying to model a none-ideal voltage source or current source, we may need to put some resistors, I guess that is what you are asking.
for voltage source, we want it to have 0 output impedence so no matter how big the loading is, its output voltage remains the same, however realistic voltage source can not do it, to put a small resistor in serial with ideal voltage source can model such a case.
same thing happens for current source, we expect an ideal current source has infinit output resistance, but to model a practical one, we can put a big resistor in parrelle with an ideal current source to represent it.

does that sounds helpful?
 

then what should be the value of that series and parallel resistance for V and I sources respectively
 

I think you r little bit confused about the model of the actual voltage and current source.

The resistance that is in series with a voltage source represents the internal resistance of a voltage source. For an example think about a battry(a AA size ... ;-) ). I think you have had experiance that the increasing of the current drawing from the battry decreases the treminal voltage. Think about using V = iR.

We can get the maximum output power by connecting a resistive load which is equal to the value of the internal resistance.

For a current source this could be explained in the same way.

I think I did not confuse you further more ;-)

cya

Randika K
 

I think it is to minimize the voltage divider effect of the internal resistance of the source
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top