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In any application, during initial start-up/load step, there is an high current requirement. Due to this high current requirement, there is a dip in that connected voltage rail.

I researched on why there is a dip in the voltage rail during start-up/load step condition. Upon researching, I found that the voltage dip is caused due to the internal resistance of the voltage source. During high currents, some voltage is dropped across the internal resistance of the voltage source and the remaining voltage is only is available at the voltage source terminals.

But the internal resistance of the voltage source is very minimal,right? If it is so minimal, then why do we have a voltage dip that is observable at the load side?

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2 Answers 2

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But the internal resistance of the voltage source is very minimal,right? If it is so minimal, then why do we have a voltage dip that is observable at the load side?

It's only 'minimal' compared to the normal load current. Many loads draw much higher current when power is first applied, and then the internal resistance of the power supply is not minimal.

Some examples:-

  1. Incandescent light bulb. Filament resistance is 10 times lower when cold.

  2. DC electric motor. Stall current is typically around 3-5 times higher than rated operating current.

  3. Power input filter capacitors (used in almost all electronic devices). internal resistance may be 0.01Ω or less.

If the power supply is regulated then its internal resistance may vary depending on load current. In a regulated supply the output voltage is usually held constant by applying negative feedback. This produces a much lower effective internal resistance, but only when the voltage regulation circuit is working. A sharp increase in load current will cause a momentary voltage drop until the feedback has time to take effect. If the unregulated input voltage drops below the 'dropout' voltage of the regulator then resistance will increase because there is no 'headroom' left to regulate from.

Most regulated supplies also have a current limit, which greatly increases resistance to stop current from going above the limit. Some supplies have 'fold-back' current limiting which reduces current to a lower value until the 'short' is removed. Some switching power supplies shut off completely, and (may) then restart a short time later. .

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  • \$\begingroup\$ Thank you. But could you just give a simple example like scenario with some numerical , easy to grasp. And finally, you are saying, the bottom-line is Internal Resistance, right? \$\endgroup\$
    – user220456
    Commented Sep 7, 2019 at 15:28
  • \$\begingroup\$ Example 1: 12V 1A unregulated supply with Ri = 1 Ohm. Connect 12V 6W lamp which normally draws 0.5A. Cold resistance is 2.4 Ohms so initial current is 12V/(1+2.4) = 3.5A and voltage is 12-(3.5*1) = 8.5V. Upon heating up the bulb resistance increases to 24 Ohms and voltage is 12-(12/24*1) = 11.5V. Example 2: 12V 1A supply with regulator, current limited to 1.2A. Initial voltage is 1.2A*2.4 Ohms = 2.9V. Bulb heats up (slower, due to lower current) to 24 Ohms and voltage rises to 11.95V (regulated Ri is 0.1 Ohm). regulated supply has lower Ri, but voltage sags more when overloaded. \$\endgroup\$
    – Bruce Abbott
    Commented Sep 7, 2019 at 21:42
  • \$\begingroup\$ I think the equation for hot bulb misses supply resistance, should be 12/25. Nevertheless, both examples show that supply resistance technically does not cause the drop, as OP question implied, only the current does. Reduce source resistance to zero and there still will be a drop due to wiring (OP is asking about "dip that is observable at the load side"). And with regulated supplies this whole question of source resistance becomes moot entirely. \$\endgroup\$
    – Maple
    Commented Sep 8, 2019 at 18:07
  • \$\begingroup\$ For fair comparison against the regulated supply I am assuming that both put out 12V with no load. The difference is 'minimal' anyway, ie. close enough to be ignored in this rough analysis. Wiring resistance is another matter, but would be considered a part of the 'power source' from the device's perspective. The OP said 'During high currents, some voltage is dropped across the internal resistance of the voltage source' so I think (s)he is aware that both resistance and current are required to cause voltage drop. \$\endgroup\$
    – Bruce Abbott
    Commented Sep 8, 2019 at 20:29
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I don't know about "any application", but most of them do have various amounts of capacitors to smooth out the supply voltage. When you power up the circuit there is an inrush current to charge those caps, which can be severe enough to cause voltage dip.

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  • \$\begingroup\$ So, Internal resistance does not a play a role in the voltage dip? Only the reservoir capacitors cause voltage to drop? \$\endgroup\$
    – user220456
    Commented Sep 7, 2019 at 5:08
  • \$\begingroup\$ Per Kirchhoff's voltage law the sum of voltages in closed circuit is 0. Oversimplified this means that the higher voltage drop the less voltage you have available. And the voltage drop is proportional to current, so I'd say inrush current defines the drop. But of course by the same law you should look at the entire circuit, including wire and source resistance too. Also keep in mind that some sources simply have limited power capacity. They cannot deliver more V*I than they can, so with higher I the V inevitably drops. \$\endgroup\$
    – Maple
    Commented Sep 7, 2019 at 5:22
  • \$\begingroup\$ @Newbie: Internal resistance (or better, impedance) is indeed the cause of voltage drops. You seem to want your notion of what's going on confirmed - the answer Maple gave could be understood to say what you wanted to hear. This answer is not complete, however. It leaves out virtually everything and only mentions the filter capacitors in AC/DC powersupplies. What about batteries? Why do the lights in your house dim when the airconditioner compressor kicks in? Why do the head lights on your car dim when you crank the engine? None of that is addressed by this answer. \$\endgroup\$
    – JRE
    Commented Sep 7, 2019 at 6:58
  • \$\begingroup\$ You should account for voltage drop across your connecting wires as well. These are often overlooked, and if you're measuring at the load end, then this definitely comes into play. \$\endgroup\$
    – Soldersmoke
    Commented Sep 8, 2019 at 13:34

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