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I am watching a course on Udemy about hardware design. The lecturer has provided a tutorial on how to calculate the current consumption.

I got quite confused when he started to talk about the input current of the voltage regulator.

Schematic description of used symbols.

The relation used in the video for the input current of the voltage regulator reads \$I_\mathrm{IN} = \frac{V_\mathrm{OUT}\cdot I_\mathrm{OUT}}{V_\mathrm{IN}}\$

There is some equality between input power and output power, but why should it be like this? Looking back at the theory on voltage regulators and going through some answers at this site, it makes sense that the input current should be quite similar to the output current (plus the small current drawn by diodes.)

Furthermore, the LDOs are known to not be as efficient as switched regulators. Since a lot of power is dissipated in heat, it is quite weird that input power and output power would be the same.

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  • \$\begingroup\$ LDO is basically a resistor, so aside from some tiny Iq, current in equals current out. \$\endgroup\$
    – user1850479
    Commented Oct 31, 2023 at 14:59

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\$I_\mathrm{IN} = \frac{V_\mathrm{OUT}\cdot I_\mathrm{OUT}}{V_\mathrm{IN}}\$ applies to a switching regulator, not a linear regulator.

When you change voltage using a switching regulator, the input current and output current are related through the consumed power.

For a linear regulator, \$I_{IN} = I_{OUT}+I_{Regulator}\$ - that is, the input current is the output current plus the little bit that the regulator itself consumes.

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The equation listed is for a lossless DC-DC.

For an LDO, input current is Iq+output current.

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  • \$\begingroup\$ Note that there's a difference between quiescent current and ground current; the ground current (i.e. current out the ground pin, i.e. additional current needed to bias the internal circuitry) is what you actually need to consider, not Iq. Ignd can vary significantly with output current, or it can hardly vary at all, depending on the regulator design. When efficiency is a concern (but you need to use a linear regulator, for noise or whatever other reason), you care about low Ignd, not low Iq. \$\endgroup\$
    – Hearth
    Commented Oct 31, 2023 at 15:20
  • \$\begingroup\$ @Hearth Interesting. I’ve never seen them separately listed. \$\endgroup\$
    – winny
    Commented Oct 31, 2023 at 19:13
  • \$\begingroup\$ The NCP718 lists both specifications, for an example. \$\endgroup\$
    – Hearth
    Commented Oct 31, 2023 at 19:46
  • \$\begingroup\$ @Hearth Whoa! You learn something every day. Thanks. \$\endgroup\$
    – winny
    Commented Oct 31, 2023 at 19:57
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You are correct, the formula can't be right for an LDO.

Suppose \$V_{IN}\$ = 10 V, \$V_{OUT}\$ = 5 V and \$I_{OUT}\$ = 1 A. That would make \$I_{IN}\$ = 0.5 A

In an LDO the input current must be greater than the output current, it's basically the current through the output pin plus the current through the ground pin (the current used by the regulator circuitry). Current goes in one pin, comes out two pins, those currents have to algebraically sum to zero.

In a switching regulator the output current can be higher than the input current, and for an ideal switching regulator that formula would be correct, output power and input power would be the same, so \$V_{OUT} < V_{IN}\$ would give \$I_{OUT} > I_{IN}\$.

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