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I have 5 V coming in from a USB power bank to an LDO voltage regulator that drops it down to 3.3 V. On the 3.3 V line I have several ICs and IR sensors. One of the IR sensor consumes quite a bit of current in short bursts (I have a 10 µF cap across it).

Whenever that power hungry IR sensor turns on, it causes some other parts of my circuit to behave oddly for a split second. I figured adding a large capacitor to the 3.3 V rail would help eliminate that, which it did. But I also noticed that I could instead add a significantly smaller capacitor on the 5 V side, and that also solved the problem.

Why is it that the capacitor is more effective on the input side of the regulator rather than the output? I figured the charge would be "more readily available" to the system if it was on the output/3.3 V side, where the sensor is.

(I just tinker with electronics and have no formal knowledge beyond basic physics E&M.)

*Edit: Before the problem/experimentation I already had on either side of the regulator a 0.1uF cap, a 1uF cap, and two 10uF caps (totaling 21.1uF on either side). I began to add extra caps after the problem.

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    \$\begingroup\$ Prior to your experimentation, did you have any caps on the input and output of the LDO? \$\endgroup\$ – Dan Laks Aug 1 '16 at 7:14
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    \$\begingroup\$ Most (if not 99.9% of regulators) need both so, when you didn't fit one or both you were more than likely not following the recommendations in the data sheet. When you don't fit both you are asking for trouble. \$\endgroup\$ – Andy aka Aug 1 '16 at 10:40
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The voltage drop during a transient at the utilization point is roughly composed of the following:

  1. inductance of the wire and the source before the regulator. In case of a typical system that uses long and thin power supply cable this is usually significant because the inductance of the cable is high.

  2. inductance of the wire/PCB track after the regulator. This is usually short if the utilization is near the regulator but may be significant if the system uses a big PCB or perhaps more interconnected PCBs.

  3. response time of the regulator. There are two major events that the regulator should respond to: input voltage variations, output load variations. These parameters can be found in its datasheet.

During a transient at the output of the regulator, the following happens:

  1. the voltage in the output capacitor drops
  2. the control loop of the regulator senses the voltage deviation and tries to conduct more. This takes time (the load regulation response time in the datasheet), and during this, the voltage falls more.
  3. the regulator conducts more and pulls more current from the input capacitor.
  4. the voltage difference between the cap and the supply voltage before the cable causes the current to begin flowing through the cable filling back the input capacitor. This takes time because (roughly speaking) the inductance limits how fast the current can start flowing.

If the input capacitor can not hold enough charge until it is filled back by the source the voltage drops below the regulator's minimum allowed input voltage. The regulator can not do anything: the output voltage remains below the nominal level until the input reaches the minimal level.

Forcing the regulator out of its designed operating region may have other serious drawbacks. If the originally closed loop control opens, the pass device may saturate. It is also possible that the input voltage is not enough to reliably power the internal circuitry and the device may shut down due to undervoltage lockout functionality or just not work properly. The recovery time from these situations may be much longer than the typical load response when there is enough input voltage. You should avoid this happening.

This can occur even if the output capacitor is large. The voltage across it will drop, and the regulator senses and tries to keep the output voltage and fill it back. If the cap is too large, the regulator will pull high current from the input side. The first problem is that it comes from the input capacitor so even if you have a large cap at the output the above situation can occur. The second problem is that it is possible that the current may be high enough to trigger the overcurrent protection which in itself slows down the response plus the recovery from overcurrent may be slower than the load regulation time. You should keep the regulator in normal operating conditions to achieve the best performance.

The output capacitor should be as small as possible, just enough to bridge the time when the regulator responds and compensates for the increased load. Roughly speaking, if you increase the output cap you just hardening the work of the regulator.

The best real-world approach is to start with a sufficiently large cap on the input side and a small one on the output side. Read the datasheet for recommendations. Check the transient on the output side with an oscilloscope. If it is not satisfactory, try increasing the output cap or replacing it with one that has a lower series inductance. Then examine the transient at the input and try to reduce the input cap. Keep some safety margin at both sides.

EDIT:

The impedance of the wire/PCB track after the regulator...

...has the same effect mentioned before: during transients or also in case of continuous but high frequency loading, at the utilization point there will be voltage notch (or continuous drop). If you compare the signal with an oscilloscope at the output of the regulator and at the utilization point, you will see that at the regulator there will be much smaller noise.

The inductance of the wire/track combined with the capacitor at the output of the regulator is an LC low-pass filter, effectively dampening the HF components.

This is good, because the noisy load does not distort the voltage of the regulator (too much). You can supply the MCU or other (analog) circuits all independently from the regulator in a star topology. This will effectively reduce the interference. If the track's inductance is not high enough, you can deliberately include inductors in the line. This can be seen often in equipments similar to yours: high power transient loads combined with sensitive analog/digital control.

High supply impedance is also bad, because you want smooth supply on every load, but this can be fixed with adding (low-ESR) capacitors to every utilization point. If you examine a PC motherboard for example, you will see hundreds of ceramic caps everywhere for that very reason.

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    \$\begingroup\$ +1: excellent explanation of the mechanics of voltage regulation! \$\endgroup\$ – Lorenzo Donati Aug 1 '16 at 14:10
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With a capacitor on the output, if the input voltage drops below what is required to achieve output regulation, there will be a dropout in the supply, and the output capacitor will droop.

With a capacitor on the input the regulator will always have a voltage reserve, and if it holds above the minimum input voltage the output regulation can be maintained even with no capacitor (with somewhat compromised higher frequency impedance).

With rectified AC this effect would be very evident. With your 5 V supply it seems to point towards rather less current capability than your sensors need.

Try and get a look at the supply ripple waveforms with a scope. Consider having dedicated regulators if the budget and specifications can justify it. This will prevent a sensor from affecting the other parts.

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Because dQ = C*dV.

Unless you're running the regulator right on its limits, you can tolerate a larger dV on the input capacitor, allowing a smaller C.

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  • \$\begingroup\$ This does not explain a significantly smaller, since the voltage only goes from 5 to 3.3 volts. Of course, no one knows what a significantly smaller capacitor is. \$\endgroup\$ – pipe Aug 1 '16 at 9:40
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    \$\begingroup\$ @Pipe : it does. Input 5V to 3.3V : dV = 1.7V. Output 3.3V to probably 3.0V (or whatever the datasheet says Vddmin is) : dV = 0.3V. Some variability depending on the LDO's dropout voltage, but about 5:1 change is a significant difference. \$\endgroup\$ – Brian Drummond Aug 1 '16 at 9:57
  • \$\begingroup\$ It would probably be worth expanding this answer to explain the term line regulation. \$\endgroup\$ – The Photon Aug 1 '16 at 14:58
  • \$\begingroup\$ You sir, really understand what is going on. "Just add more cap" "Sure buy why?" I ask my colleagues in hope they will start to think about where any why energy are stored and used. \$\endgroup\$ – winny Aug 1 '16 at 22:07
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The basic premise of the question is invalid and not universally applicable. Certainly regulators (of whatever variety) need to have reasonably smooth (filtered) raw power to work with. Few if any will operate on the pulsed-DC comping out of a typical AC source and rectifier stage. This is where we typically see the large "bulk" filter capacitors.

HOWEVER, there are some cases where large capacitance is required to hold up the power supply bus in the presence of large, intermittent loads such as the one given as an example in the question.

It is not a question of "more effective before or after". Those are two separate and independent cases and cannot be logically combined as in the question as asked.

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    \$\begingroup\$ But those loads that require large capacitors are typically not tied directly to a regulator and especially not to the MCU power rail. Almost always a separate, adequate supply have to be used or at least an inductive coupling. Rarely makes sense to tie a large cap directly to the output of a regulator. \$\endgroup\$ – Gábor Móczik Aug 2 '16 at 17:43
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A capacitor on the output side of a regulator won't even start trying to do anything useful unless or until the output voltage changes. A capacitor on the input side will start supplying current when the input voltage drops. A typical regulator will try to minimize the extent to which changes on the input voltage affect the output, so the input-voltage drop necessary to make the input-side capacitor start supplying energy will typically not cause any significant output-voltage change.

In some cases, a regulator might not be able to react instantly to a sudden current demand, and in such cases an output capacitor may be helpful (if not required) to supply some current to the output during the time it takes the regulator to react to an increased load. The output cap won't be able to feed current very effectively without the output voltage dropping noticeably, but it may be able to feed enough to give the regulator time to react to the increased demand.

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