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parallel resistance Rs in current source

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PG1995

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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
 

kak111

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There is two ways to examine common circuits



Thévenin's theorem

Replace voltage sources with short circuits, and current sources with open circuits.
Calculate the resistance between terminals A and B. This is RTh.

Norton's theorem

Replace independent voltage sources with short circuits and independent current sources with open circuits.
The total resistance across the output port is the Norton impedance RNo.
or
Use a given Thevenin resistance: as the two are equal.


A Norton equivalent circuit is related to the Thévenin equivalent by the following equations:

R_{Th} = R_{No}
V_{Th} = I_{No} R_{No}
V_{Th} / R_{Th} = I_{No}


Thévenin's Theorem

Any linear voltage network which may be viewed from two terminals can be replaced
by a voltage-source equivalent circuit comprising a single voltage source E and a single
series impedance Z. The voltage E is the open-circuit voltage between the two terminals
and the impedance Z is the impedance of the network viewed from the terminals with all
voltage sources replaced by their internal impedances.

Norton's Theorem

Any linear current network which may be viewed from two terminals can be replaced by a
current-source equivalent circuit comprising a single current source I and a single shunt
admittance Y. The current I is the short-circuit current between the two terminals and
the admittance Y is the admittance of the network viewed from the terminals with all current
sources replaced by their internal admittances.

Thévenin and Norton Equivalence

The open circuit, short circuit and load conditions of the Thévenin model are:

Voc = E
Isc = E / Z
Vload = E - IloadZ
Iload = E / (Z + Zload)

The open circuit, short circuit and load conditions of the Norton model are:

Voc = I / Y
Isc = I
Vload = I / (Y + Yload)
Iload = I - VloadY

Thévenin model from Norton model
Voltage = Current / Admittance
Impedance = 1 / Admittance
E = I / Y
Z = Y -1

Norton model from Thévenin model
Current = Voltage / Impedance
Admittance = 1 / Impedance
I = E / Z
Y = Z -1

When performing network reduction for a Thévenin or Norton model, note that:
- nodes with zero voltage difference may be short-circuited with no effect on the network current distribution,
- branches carrying zero current may be open-circuited with no effect on the network voltage distribution.



Read more...............................

Thévenin's theorem - Wikipedia, the free encyclopedia

Norton's theorem - Wikipedia, the free encyclopedia

Electronics/Thevenin/Norton Equivalents - Wikibooks, open books for an open world

Regards KAK
 
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PG1995

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Many, many thanks, KAK. I genuinely appreciate your help.

A new question.

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.
 

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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?
Presuming you have understood the nature of a voltage respectively current source from the previous discussion, you should be able to answer this questions yourself:
- What's the equivalent of an ideal voltage source with Vemf(or Vth) equal to zero?
- What's the equivalent of an ideal current source with zero current?
 

kak111

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You can think of it in terms of changes (dV and dI)

Z = dv / di


For a ideal voltage source, the output voltage is constant as the output current changes, so

Z = dv / di = 0 / di = 0


But for a ideal current source, the output current is constant as output voltages changes so

Z = dv / di = dv / 0 = infinity


Voltage source can provide infinite amount of current.
And a current source can provide an infinite amount of voltage.

Voltage and Current Sources

Regards
KAK
 

PG1995

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Presuming you have understood the nature of a voltage respectively current source from the previous discussion
Hi FvM: Unfortunately I'm still struggling to understand the concepts.

Thank you, kak, for your reply.

Hi

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

KerimF

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I will use the linked circuit diagrams to make the gaps in my understanding clear so that you can help me effectively.
Ok, so let us start from the beginning.
What is the purpose of this problem (Example 4.5)?
It is given not only to calculate the current i (in 3 Ω) but to do it by taking advantage of what is known as "the superposition theorem".
As you know, the circuit conssits of 3 loops. So one can writes the 3 loop equations and solve their 3 unknow currents. Then the current i will be equal to the difference of the currents of the lower two loops. I mean by this that we can find out the i value without applying the superposition theorem. OK?

So why we have to learn and apply a new theorem! Actually it is supposed to make some circuits easier to analyze. In real life, no one will force you to use any specific theorem or method to solve the circuit your work on. But it is better to know as much theorems as possible so we can have later a longer list to choose from... right?

The main idea of our theorem here is that the requested unknown current i could be seen as the sum of many currents. The number of these currents is equal to the number of the circuit sources (voltage or current). Each of these currents is generated by one source only while all others are disabled (as sources).

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!
So let us see how we can disable a voltage source in our calculation.
In a real circuit, a voltage source could be a battery. And when this battery loses all its charges, what will we get? We get a 0V battery ;-) while it keeps us pass in it a current as if it is just a wire. But let us talk more professionally. A battery has also an internal resistance. So it is not uncommon that we model a battery (or any real voltage source) by two elements; an ideal voltage source in series with a resistor. So when the voltage of the battery becomes zero, its internal resistor will not vanish so its effect in the circuit won't change. So now when we say we like to disable a voltage source we let its voltage equal zero so it can be replaced by just a wire (though we keep its series internal resistor), exactly as if it is an empty battery. OK?

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?
I am afraid it may be a hard for me to give you a real current source that you are familiar with because good current sources are possible using active devices only as transistors for example (in electronics). Very briefly, you will learn that a transistor has 3 terminals; input (called base), output (called collector) and common (called emitter). Its collector can sink a current if we inject a current into its base. So if we let the base current be zero, the collector will float as if it is cut from the rest of the circuit (open circuit). I mean if we like to disable a current source we let its current be zero so it looks like disconnected. Here also there is an internal resistor but in parallel with the source. In our example of the transistor, though practically we may say that its collector current is zero (when its base current is zero), there is a very small leakage current which could be modelled by a large resistance in parallel which always exists even if we disable the current source (making it zero).

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?
You are right. If you have a current source in the lab, it certainly has an internal parallel resistance. And I won't be surprised if its value is about 1,000,000 Ω. Even it is as low as 1,000 Ω only its effect on the result of your given problem will be very small. Of course, your remark will be important if this internal resistance is comparable with the values of the circuit of interest. In this case it shouldn't be omitted.

4: What does it mean when Rth is zero when Thevenin equivalent circuit?
It means that the two terminals can deliver to an outside load (or circuit) any high current we can imagine of and their voltage will be unchanged and remains equal to their open circuit one (in other words, the voltage drop is zero since I_load * Rth = I_load * 0 = 0 V always).
 
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Regarding 1+2: In the real world, you have neither ideal sources nor ideal switches. And you most likely won't short a voltage source or open the path of a current source. It's just a model. You have been previously talking about replacing a voltage source with a short rather than shorting a voltage source with a wire. That's better, because shorting an ideal voltage source with an ideal wire would cause an infinite current. And opening the current source terminals infinite voltage.

Regarding 3: The exercise problem is obviously based on ideal sources. But why you are missing a parallel resistor associated with the current source?. The current source don't need it. The purpose of the parallel resistor is to model the finite internal resistance of a real current source. But you rarely won't find it in a real circuit.

A lab power supply usually hasn't a current source as it's building block, so it also don't need a parallel resistor.
 
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