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is 100uF/16V capacitor across this wall wart safe?

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mik3ca

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Currently I have a circuit where the input and output of an LM2940-CT5 voltage regulator is decoupled with 100uF capacitors rated for 16V. I use batteries to test the circuit and it works.

Next, I want to use a wall wart. The crazy thing with my unregulated wart is that it shows 6V 900ma on the box. I measured the connector with a voltmeter on DC and the reading was 10.2V. I measured its voltage while it was powering a circuit before and the reading was about 9V.

I thought the math is use a capacitor with voltage rating of at least 2x the voltage of the circuit. The overall circuit itself is 5V max but I'm worried about the capacitor across the power supply (16V/100uF). If the regulator shoots out 10V then 10 x 2 = 20 which is over 16 and cap would malfunction? or am I clueless?

So is it safe enough to use a capacitor rated for 16V here or will it likely explode?
 

dknguyen

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You're fine...unless you actually have good reason to expect the regulator to suddenly overshoot it's rated output voltage by 4x.
 

MikeMl

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Do you have a 'scope? If so measure the peak voltage at the output of the wall-wart with no load on except the 10meg scope probe. Get a capacitor that has a DC working Voltage rating about 50% higher than the max voltage you see on the scope.
 

audioguru

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An unregulated wall-wart means that its output voltage is higher without a load than its rated output voltage with a load.
Why would the 5V regulator shoot out a voltage higher than 5.25V (5V plus 5%)?
 

MikeMl

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I worry that 100uF at the input of the voltage regulator is too small to act as a proper filter capacitor. The capacitor has to store enough charge so that it doesn't discharge below the dropout voltage of the regulator before the next current pulse arrives from the full-wave rectifier inside the wall-wart. If your actual load current is ~100mA, then the filter would have to bigger than 200uF. The maximum load current on your regulator that 100uF will support without dropping out is about 50mA.
 
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dknguyen

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I worry that 100uF is too small to act as a proper filter capacitor. The capacitor has to store enough charge so that it doesn't discharge below the dropout voltage of the regulator before the next current pulse arrives from the full-wave rectifier inside the wall-wart. If your actual load current is ~100mA, then the filter would have to bigger than 200uF. The maximum load current on your regulator that 100uF will support without dropping out is about 50mA.
Perhaps, but that sounds like a different concern than the capacitor voltage rating that was originally asked about. I'm pretty sure they get out of dropout fast enough to beat aa slowly rising 60Hz waveform and stop it from getting out of hand.
 

mik3ca

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An unregulated wall-wart means that its output voltage is higher without a load than its rated output voltage with a load.
Why would the 5V regulator shoot out a voltage higher than 5.25V (5V plus 5%)?
Just to clarify, My whole circuit has a regulator circuit with it (LM2940-CT5) in which the input (wall wart connection section) is connected to ground via a 100uF/16V capacitor. The output (5V section) is connected to ground via a 100uF/16V capacitor. The capacitor I'm most worried about is the input capacitor that's connected to the wall wart. Do I need to include a picture?
 

dr pepper

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Ac is usually specified as the Rms voltage, when you rectify Ac you end up with a voltage close to the peak not Rms.
The peak is about 1.414 x the Rms, so 6v x 1.414 = 8v, pretty close to your 9v.
When the wall wart is loaded to its mA capacity the voltage will drop, probably close to 6v, thats because its unregulated.

10v on a 16v capacitor is fine, massively underating a capacitor isnt a good thing, applying a voltage higher than the capacitor is rated for though is a lot worserer.
 
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dknguyen

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if you massively underrate a capacitor like 5v on a 450v capacitor its lifetime will not be as long, 10v on a 16v cap is fine.

I've never heard of this. Can you direct to any references? I routinely use 50V ceramic caps at 3.3V all the time because that's just what's easiest to stock.
 

dr pepper

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Apologies my original post didnt make sense so I edited it, however that sentence still stands.

Heres a page that touches on this, this lot say 80% though I find that varies a bit.
https://www.eetimes.com/author.asp?doc_id=1279791
I dont really know how underrating a cap affects it internally, some weird chemical affect on the electrolyte.
The biggest effect is temperature, if you run a cap hot it'll fail quicker, voltage rating has less of an effect.

Edit: heres a cap life calculator, have a fiddle with some values and see:
http://www.illinoiscapacitor.com/tech-center/life-calculators.aspx
 
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MikeMl

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mik3ca,

The dropout voltage for your regulator is 0.5V, so the filter capacitor V(filt) must never dip below 5.5V. Here is my best WAG at what is in your wall-wart. This shows that the maximum current that can be supplied by a 100uF is ~75mA @ 60Hz power line frequency. Situation is worse at 50Hz.

74.png

Green = 25mA load, Red = 50mA load, ... Violet = 100mA load.

btw: 100uF across the output of the regulator is totally wasted. You would be much better off if you moved it to the input, and used only the recommended 22uF low-voltage one on the output.
 
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dknguyen

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Apologies my original post didnt make sense so I edited it, however that sentence still stands.

Heres a page that touches on this, this lot say 80% though I find that varies a bit.
https://www.eetimes.com/author.asp?doc_id=1279791
I dont really know how underrating a cap affects it internally, some weird chemical affect on the electrolyte.
The biggest effect is temperature, if you run a cap hot it'll fail quicker, voltage rating has less of an effect.

Edit: heres a cap life calculator, have a fiddle with some values and see:
http://www.illinoiscapacitor.com/tech-center/life-calculators.aspx
Oh, what that article is saying is that there are two things that can affect cap lifetime:

1. Voltage
2. Heat

Combined with one other fact:
Higher voltage rating caps inherently have higher ESR, which increases heating from ripple currents.

So you end up with two opposing design parameters and if you go too far into underrating the voltage on your cap, you end up with diminishing returns. The reduction in lifetime due to increased heating from increased ESR will start to outweigh the increased lifetime from underrating the cap voltage.

Makes sense. You had originally kind of made it sound like there was an inherent chemical reaction issue from underrating a cap that reduces it's lifetime, similar to parasitic loads on a battery.
 
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audioguru

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Why worry if the additional capacitor is "only" 100uF? It is an AC-DC wall-wart that already has an output capacitor. Since it is rated to produce 900mA then it probably has 1000μF. The additional capacitor is not added at the wall-wart, instead it is added at the pins of the voltage regulator to prevent it from oscillating. The datasheet of the LM2940 regulator shows only 0.47μF, nothing more and nothing less. They do not say what type of capacitor but show a non-polarized one (not electrolytic). The output capacitor has a critical minimum value of 22μF and a critical narrow range of ESR (0.1 ohm to 1 ohm) because this is a low dropout regulator that has a PNP pass transistor with voltage gain, not the no-gain NPN emitter-follower used in ordinary regulators.

22μF is the minimum stated value for the output capacitor that probably has a tolerance of -20% so I would use a 33μF output capacitor.
 

MikeMl

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Why worry if the additional capacitor is "only" 100uF?
In his other thread, he implied that he didn't have space for a larger filter capacitor. Here, he says that he has 100uF on both input to the reg and on its output. I was simply suggesting that he would do better with 200 uF on the input side rather than 1oouF on both in and out...
It is an AC-DC wall-wart that already has an output capacitor.
Not necessarily. Out of about 30 AC-to-DC wall-warts I have here in a box, a third of them do not have any filter capacitors inside the plastic enclosure. As measured with a scope, they put out full-wave rectified non-filtered pulsating DC. I would guess that these were originally used to charge NiCd or Lead Acid batteries, so no capacitor was needed. These are marked as DC output.
Since it is rated to produce 900mA then it probably has 1000μF. The additional capacitor is not added at the wall-wart, instead it is added at the pins of the voltage regulator to prevent it from oscillating.
As covered in the OP's previous thread, he had insufficient ripple filtering (i.e. most likely a wall-wart with no built-in capacitor).
The datasheet of the LM2940 regulator shows only 0.47μF, nothing more and nothing less. They do not say what type of capacitor but show a non-polarized one (not electrolytic). The output capacitor has a critical minimum value of 22μF and a critical narrow range of ESR (0.1 ohm to 1 ohm) because this is a low dropout regulator that has a PNP pass transistor with voltage gain, not the no-gain NPN emitter-follower used in ordinary regulators.

22μF is the minimum stated value for the output capacitor that probably has a tolerance of -20% so I would use a 33μF output capacitor.
So use the minimum required value on the output, and sufficient filtering (not dictated by stability) on the input. I showed that 100uF is marginal for all but minimal output current.
 

audioguru

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In the other thread he fixed the problem by simply adding a missing capacitor required by the crystal.
Now he is worried about the voltage rating of the input capacitor of the voltage regulator, but wrongly said the regulator's output capacitor.
 
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