Confusing text about ideal and non ideal voltage source, bad writing or am I stupid?

Thread Starter

xan137

Joined Feb 5, 2021
5
Hello,
I am looking for a website where I can ask questions related to circuits, I hope I can ask them here.
I am reading All-in-One Electronics Guide by Cammen Chan and there is a confusing text about non-/ideal voltage and current sources.
Should I understand everything and only then continue reading or can I skip few thing? I don't know and that is why I am here.
What I hope for, is a person with time and knowledge, that can explain he basic idea of this paragraph. I am sorry to the wall of text...
He writes so often non-idea, ideal, not-so-pefect world, finite intenal resistance, input, ouput...
If you can only answer one question, please tell me what input is in a voltage source.
"If node A is the output voltage, input would be the same as the output."
V goes into node A, and 5 volt leaves node A. Input is negative terminal?

Page 44, title DC Circuit:
A non-ideal voltage source can be viewed as a voltage divider. Figure 1.22 .
If it were an ideal voltage source, internal resistance would be zero Ω. The
voltage at node A will be exactly the same as voltage originating from the voltage source. If node A is the output voltage, input would be the same as the output.

In a not-so-perfect world, voltage source would have finite internal resistance. This finite resistance originating from the voltage source makes the circuit look just like a voltage divider.
Voltage at node A is no longer the same as the original voltage source. In a non-ideal
world, when you connect a voltage source to a resistor, the output will not be exactly the same as the input. High quality power supplies offer extremely low internal resistance (still non-zero), and your output is “almost” the same as the input. It’s for this reason voltage divider is seldom used as a constant voltage source. Using Figure 1.22 as an example, if the original voltage source on the left is 10 V, the intended voltage output is 5 V at node A. By design, we set both resistors to have the same values (voltage divided by half) so that 5 V at node A can be obtained. In reality, the voltage at node A won’t be constant at 5 V. Firstly, any changes from the original input source will change the voltage at node A (again by the voltage divider action). Secondly, any change in the resistances (e.g., caused by temperature variations) will also change the voltage at node A. To achieve a more stable voltage output, low drop-out and switching regulators are used, which will be discussed later in this book.

To show you that I tried to understand the text by myself, here are my notes on the paragraph which I don't quite understand, after that I post whole Paragraph.

Ideal voltage source:
Zero internal resistance
Non-ideal voltage source:
Non-zero (finite) internal resistance
non-ideal voltage source (VS) = voltage divider
If ideal Vs = internal resistance 0 = V node A same as VS.
"If node A is the output voltage, input would be the same as the output" (what does input does in a VS? The black lead of a laboratory supply? Negativ terminal? The input can't be the same, the output Voltage is "reduced" by component, last node should have 0V.)
Non ideal VS is voltage divider, because of iternal resistance.
Sorry I can't write down my notes, because I have to many questions.

non ideal v source.png
Looking forward for your answers.
 

jpanhalt

Joined Jan 18, 2008
11,088
Welcome to AAC, xan.

My interpretation of, "If node A is the output voltage, input would be the same as the output," is that with a perfect voltage source (Vs), input (to node A) = Vs . Hence, in that situation, input to node A = output from node A = Vs.

If you have more questions, let's address them one by one. It would be a futile exercise to try to re-write a book most of us may not have seen.
 

Thread Starter

xan137

Joined Feb 5, 2021
5
Thank you jpanhalt. I hoped after writing my post I would understand the text better. I guess I just skip it.
 

BobTPH

Joined Jun 5, 2013
3,306
I think the writer has a talent for making something simple complicated.

An ideal source will always have the same output voltage. A non-ideal source will not, it will vary with the load.

Saying it has an internal resistance is, however, an oversimplification. A regulated supply will typically drop it’s voltage only slightly with increasing load until a limit is reached, then it will fail quickly.

Bob
 

MrChips

Joined Oct 2, 2009
23,505
An ideal voltage source outputs the same voltage independent of the load.
The internal resistance of an ideal voltage source is zero.

A real-life "non-ideal" voltage source has internal resistance. The output voltage falls as the output current increases.
You have to take into account this internal resistance and place it in series with the load. Hence you now have a voltage divider. The output voltage has been reduced because of the current flowing through the internal resistance, (Vloss = I x RS, where RS represents the source internal resistance).

An ideal current source outputs the same current independent of the load.
The internal resistance of an ideal current source must therefore be modeled as being infinite.
The output voltage of any current source has to change according to the load in order to meet the requirements of Ohm's Law (VLOAD = I x RL, where RL represents the load resistance).
 

crutschow

Joined Mar 14, 2008
27,171
A good regulated power supply can be close to an ideal voltage source since it's internal resistance can be basically just the resistance of the output wire until it reaches its current limit.

Batteries all have some internal resistance, although it can be small, so are not ideal.
 

MrChips

Joined Oct 2, 2009
23,505
A well designed power supply or opamp circuit (even one with non-zero internal resistance) can be modeled as an ideal voltage source because it uses voltage feed-back to control the output voltage. When the voltage feed-back comes from sense connections at the load, the control will compensate for the resistance of the output wires to the load.
 
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