# parallel resistance Rs in current source

Discussion in 'General Electronics Chat' started by PG1995, Jun 3, 2011.

1. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
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

2. ### steveb Senior Member

Jul 3, 2008
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469
The Ebook here is going into this (see this link). You are asking about the well-known Thevenin-Norton equivalent circuits.

http://www.allaboutcircuits.com/vol_1/chpt_10/10.html

Your concept of a current source (and maybe even the voltage source) is not really rigorously correct.

Last edited: Jun 3, 2011
3. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
300
The parallel configuration is correct, to turn a perfect current source into an imperfect one with some dependency between current and voltage. Connecting a resistor in series with a perfect current source would not have any effect on the current delivered, as the source is, well, perfect!

In fact these networks can be described as duals of each other. If both circuits contain equal resistances, and the current source value is set to equal the voltage source value divided by the resistance, they are actually indistinguishable as measured from the external connections.

Such networks are commonly used as simplified equivalent circuits of bigger circuits, where the voltage source in series with resistance is callled a Thévenin equivalent circuit. http://en.wikipedia.org/wiki/Thévenin's_theorem

The current source in parallel with resistance is called a Norton equivalent circuit.http://en.wikipedia.org/wiki/Norton_theorem

4. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thanks a lot, steveb, Adjuster.

I have one new question. Please help me with it.

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.

5. ### #12 Expert

Nov 30, 2010
16,330
6,818
I am reminded of vacuum tube amplifiers...voltage driven. Open their output and the voltage from the output transformer starts punching holes in the insulation. A transistor amp must be left open because it will dump all the amps it can find into a short.

This might not be what you are after today, but it's another aspect of the same thing. Store it in your memory. You will meet it someday.

6. ### Adjuster Well-Known Member

Dec 26, 2010
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If you wish to calculate an equivalent resistance for a circuit, then these are the correct procedures. An ideal voltage source has zero output resistance, while a current source has an infinite one.

That said, in a practical context you normally would not short-circuit a voltage source to shut it down. Doing that with (say) your house current supply or a battery could lead to a dangerous accident. We turn off the supply from such things with an open-circuit switch.

Similarly, some (rather rarer) current sources are hazardous if open-circuited. Apart from the approximately constant current output from some electronic valve (tube) amplifiers that someone else has just mentioned, you might encounter the secondary of a current transformer, used for sensing current in a wire: the output from this also approximates to a constant current.

The output voltage from either of these can rise to dangerous levels into an open circuit. The current transformer output can be shorted safely, as can that from some simple tube amplifiers. I would not recommend shorting any tube amplifier output at random however, as in some cases it is possible for damage to occur, particularly in larger amplifiers using negative feedback.

7. ### Adjuster Well-Known Member

Dec 26, 2010
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I hope you won't take this amiss, but in my opinion what you have said is not quite right. It is indeed unsafe to open-circuit the output from a tube amplifier, but this is not because they have a "voltage" output. (They are voltage-driven as far as the input grid circuit is concerned, but that is another matter.)

In fact, the pentode and beam tetrode tubes which are most commonly used as output devices have essentially constant current outputs. It is for precisely this reason (combined with transformer coupling) that their output voltages can rise so dangerously. Triodes have much lower output impedances, and may not present so much of a risk in case of an open output, but it's not an experiment I would care to try.

Here is a sample datasheet for a beam power output tube, giving characteristics for normal configuration and a configured as a virtual triode. Note the "flat" output characteristic in normal operation. http://www.drtube.com/datasheets/kt88-sed1999.pdf

8. ### steveb Senior Member

Jul 3, 2008
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As Adjuster said, this is the correct procedure for ideal sources. One can put it into even simpler terms.

Think about the voltage across a short. It's always zero for ideal wires, right? What better way to make an ideal voltage source that has zero Volts?

Think about the current through an open. It's always zero for perfectly insulating media, right? What better way to make an ideal current source that has zero Amps?

Keep in mind that these are mathematical rules used for abstract analysis at the circuit level. They only apply to the ideal sources, so if you make a more realistic source model that includes a resistance, then use that resistance instead of the short or open.

These rules are particularly useful when using another mathematical trick, - namely the superposition principle. For linear circuits and systems, the total effect of many sources can be computed by adding the contributions from each source taken alone. The way to remove the other sources is to use the rules you just quoted.

Last edited: Jun 4, 2011
9. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

I think quoted post by Adjuster somewhat has helped me to understand where I'm having problem inside my mind (it's quite funny when you yourself don't really know what is bugging you). 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

Sep 20, 2005
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11. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,960
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In 'real' world we don't short the voltage source on purpose, sometimes we short the voltage source by accident, 'real' voltage sources don't like work in "short circuit".
As for the current source. The "current sources" does not even exist in real world. The current source is a theoretical device. Only we can do is to make the device which will be behave (in limited range) just like a ideal current source.
And if we have build the circuit that you post, we don't need to short or left open any of the sources in the circuit. We simply can use voltage mete and current meter to measure all voltage/currents in the circuit.
We short the voltage source and we left open the current source only for analysis purpose. In real word we don't use this methods.
We use this methods only to simplify the analysis of the circuit.

12. ### steveb Senior Member

Jul 3, 2008
2,433
469
It's not clear to me either exactly what part is giving you trouble, but I can try to answer these questions you posed.

First of all, the rule to short the voltage source in circuit analysis is better thought of as replacing the ideal voltage source with a wire (or short). If you think of it this way, then you may have less issue with the idea of shorting the source. From a mathematical point of view, shorting an ideal source is problematic and creates an indeterminant mathematical condition. It's sort of like asking, "What happens when an infinite force pushes against an immovable object?". The ideal source will give infinite current into a zero Ohm source resistance. So, what is the voltage drop on the source resistor? You can't really say, hence you don't know what the output voltage is. However, if you just replace the ideal source with a wire, there is no issue.

Now, for your question. In the real world, shorting the source does mean to connect +ve to -ve, but normally this is not a good idea unless the system has been designed for this. Shorting a car battery, for example, is a very BAD idea, and will likely damage the battery and anything nearby. However, a lab power supply may be designed to invoke a current limit, thus turning the voltage source into a current source. In fact, many power supplies are really dual current/voltage sources because of the current protection mechanism, which is really a feedback system operating around a voltage source. Ironically, the voltage source itself may be a current source (i.e. transistors) with a feedback system operating around them to make them look like a voltage source. Bottom line is that the real world is much more complex than this issue of shorting the sources for the purpose of using the superposition principle.

Yes, exactly. Again, as Adjuster pointed out, this is often a bad idea. However, some systems are designed to accept the open output condition. However, this is a protection mechanism and the circuit is no longer operating as a current source when the terminals are open.

An actual lab power supply is fairly complex and so there is much more than just a parallel resistance. However, we can make a mathematical model of an equivalent ideal current source and a parallel resistance. It's not a perfect model, but is often good enough.

What kubeek said.

Last edited: Jun 5, 2011