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enter image description here

The realistic(practical) voltage source is modeled as an ideal voltage source plus a series resistor R in the above figure. In an ideal voltage source whatever the current load is the voltage would remain the same.

Apparently in a practical voltage source when the load current increases the imaginary series resistor will cause a higher voltage drop.

I think the wire resistance between the source and the load is not the issue here, since we are talking about the voltage change accross the power-supply terminals.

My question is then: What is this series resistor in a power-supply? What causes it to exist? Is that the change in Thevenin equivalent resistor?

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Real power supplies come in all sorts of varieties, so the details of the source resistance vary. Common to all power supplies is the fact that they have internal conductors. These conductors (internal wiring) will act to drop voltage with current. This wiring will create a "real" series resistance.

Batteries have another effect. Current is produced by electrochemical reactions, and these reactions only occur so fast. In a lead-acid battery, for instance, sulfuric acid reacts with the lead/lead oxide terminals, and once the sulfuric acid has reacted new acid must take its place, and this takes time. As a result, the maximum current is limited by the geometry of the battery plates and this shows up as drop in voltage as current increases. This is modeled as a series resistor, although there is no physical unit which causes the voltage drop.

A good power supply these days will sense the output voltage, and adjust the voltage of the internal source to compensate for internal voltage drops, and for slowly changing loads will approximate an ideal voltage source quite closely.

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  • \$\begingroup\$ +1. @jjuserjr, the battery acid reaction time mentioned here is very real, and scales such that it gets worse with battery age, as the compacted byproducts make the plates less accessible and harder to charge. This reaction and battery age creates an internal resistance that has a definite presence and can be experimentally quantified. \$\endgroup\$ – Sean Boddy Dec 12 '15 at 17:11
  • \$\begingroup\$ this question is about linear regulated power supplies or smps, not batteries. i forgot to mention it. \$\endgroup\$ – user16307 Dec 12 '15 at 17:17
  • \$\begingroup\$ is that the change in Thevenin equivalent? \$\endgroup\$ – user16307 Dec 12 '15 at 17:19
  • \$\begingroup\$ @jjuserjr - I'm not sure I understand. What is "that" when you ask "is that the change"? \$\endgroup\$ – WhatRoughBeast Dec 12 '15 at 19:32
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    \$\begingroup\$ @jjuserjr - When modeling a power source as a voltage and a series resistance, then pretty much by definition the resistance is the Thevenin resistance. And no, it is not necessarily a constant. For a battery, for instance, it varies with current levels, and for "regular" power supplies it varies with frequency (the circuits can't handle high frequencies and fast transients perfectly) and also changes with current when the current limit of the supply is reached. In effect, the value of R increases greatly for some limiting value of current and the output voltage drops precipitously. \$\endgroup\$ – WhatRoughBeast Dec 12 '15 at 19:42
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In a battery, it has to do with the chemical reactions, something which I know very little about so I will not comment further on this.

In a stand-alone power supply, it is mainly the loss in the cables, and the amount of loss depends on the cable resistance. A good bench power supply will have sense wires that you can connect at the point where you want your voltage to be for example 5.0 volts, and it will then make sure that the voltage at the connector will be slightly higher than 5.0 volts, so it will be exactly 5.0 volts after the loss in the cables.

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All "power supplies" have some internal series resistance. It is just a fact of nature. Because of this some "power supplies", e.g., signal generators, set a series resistance, 50 ohms being typical, so that it is a known quantity which may be considered by the user. When using such a signal generator it is important to consider this 50 ohm internal resistance to understand why, e.g., that the voltage measured at a 50 ohm load is half the signal generator's voltage setting.

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    \$\begingroup\$ The 50 ohm output impedance in a signal generator has nothing to do with having a "known quantity". It is because it can be used with a matched cable to remove reflections on the signal. en.wikipedia.org/wiki/Characteristic_impedance \$\endgroup\$ – pipe Dec 12 '15 at 17:47
  • \$\begingroup\$ Since there are many different cables with different characteristic impedances one had to be chosen. 50 ohms is very common but there are many others. Besides, signal generators are used for other applications such as driving logic circuits where the distances are "small" compared to the wavelength of the exciting signal. Knowing the series impedance of the signal generator is necessary for using it in such situations. \$\endgroup\$ – user34299 Dec 12 '15 at 23:23
  • \$\begingroup\$ And it has nothing at all to do with the original question. What you are talking about is deliberately added, for a completely unrelated reason, and this answer just confuses the issue and someone who is trying to understand the root cause. \$\endgroup\$ – pipe Dec 12 '15 at 23:55

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