Power supply cap considerations??

Discussion in 'General Electronics Chat' started by Mike33, Oct 28, 2011.

  1. Mike33

    Thread Starter AAC Fanatic!

    Feb 4, 2005
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    Hi,
    I'm working with a solenoid (part of yet another model train setup!) which will be tripped by a relay...the switching part of the circuit is fine; now it's time to deal w/another issue.

    The train owner (my father in law, lol) is trying to come up with a scheme to avoid the very momentary loss of voltage when the solenoid energizes (inrush). My thought is to build a bank of capacitors across the PS back where the transformer is...they would charge up on power-on and provide a 'reserve' for the infrequent times he switches tracks with this solenoid. The amount of time involved is less than 10ths of a second.

    So what I am looking at is a reservoir...I'm not familiar with how this is done, however....would there be required resistances? Drawbacks to doing this (I'd put in a bleeder network, of course). He's not set on whether to run the solenoids from AC or DC....seems to me, DC would be better if we're adding caps....
    Can this be done, or should he maybe consider power JUST those solenoids (3, I think) from another power supply??
    Thanks!
     
  2. tom66

    Senior Member

    May 9, 2009
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    A diode might be helpful here. You could run the low current stuff off a diode and a small capacitor. When the solenoid engages, the cap(s) won't discharge due to the diode and will keep the voltage stable for a while.

    One thing to be aware of though is that a good diode will drop around 0.5V - 1V so a 12V supply will look closer to 11V - 11.5V; depending on the circuit this may or may not matter.
     
  3. Mike33

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    Feb 4, 2005
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    I see....sort of like, instead of trying to provide a 'battery' using caps for the high-draw stuff, protect the low=current stuff using a diode and cap...hmmm, pretty interesting idea, Tom - thanks! Comes at it from the other point of view...I like it.

    What does the diode do, and would the caps be in parallel with the diode before or after them? I think he's running his lights on AC (I'll clarify that pretty quickly. We do this via phone, and he has a hard time understanding "ac, dc, rippling dc, measure the resistance, measure current", etc). I need to get over there and see what he's been up to! (300 miles)
     
  4. AlexR

    Well-Known Member

    Jan 16, 2008
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    A solenoid should not produce any "inrush" current when it is turned on. Inductors resist change in current and do not produce inrush current so what you may be seeing could be a voltage spike that occurs when the solenoid is de-energised. A diode across the solenoid would fix that. Connect the diode directly across the solenoid coil so that the supply voltage reverse biases the diode (when the solenoid is on the diode does not conduct but when the solenoid gets de-energised the diode will absorb the back EMF spike). On the other hand the problem could be a loss of regulation in the power supply as the solenoid adds an extra load in which case adding capacitors to the power supply might help.
     
  5. djsfantasi

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    Apr 11, 2010
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    Search for "Model Train Capacitor Discharge Unit". There are many solutions out there, from using an electrolytic capacitor with resistors and diodes to protect the solenoid switch machine and capacitor, to circuits built with an SCR for fast recovery time. This link about Capacitor Discharge Units has several examples of various circuits of which I am familiar, but note I have not used any of these specifically.
     
  6. Mike33

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    Feb 4, 2005
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    Thanks for the replies, guys. I'll read up on the Cap Discharge stuff....

    Alex - No inrush current, when you energize a coil of wire????????? Such as turning on an electric motor?????? What builds the magnetic field that would move the rod, then?? The 'solenoid' is actually an Atlas relay; it's 'working part' is a low-ohm coil of wire. Any time it is energized, the little streetlights etc. that he has wire up blink off for the slightest time as that coil sucks juice.

    I thought an inductor only opposed changes in current AFTER the field is built (?)
    And that capacitors only oppose changes in voltage AFTER they have charged up.....but both, in their 1st,2nd, 3rd TC, should be pulling voltage and current..... :confused:

    I'm going to correct the lack of a snubbing diode, for sure, anyway - when the field collapses, yup - he is susceptible to back EMF causing damage elsewhere....
     
  7. Adjuster

    Well-Known Member

    Dec 26, 2010
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    No: the opposition to current change in an ordinary coil is present at all times, barring saturation. The operation of a number of devices such as switched-mode power supplies depends on this principle, so it is worth trying to understand it.

    A simple inductor will draw a current which at least initially builds up progressively from zero. If a fixed voltage V is applied to an the inductor of L henries, the current initially builds with a constant slope (of V/L amps/second). If this continues long enough, the rate of rise of current may increase if the coil core material becomes magnetically saturated. At any event, for any ordinary coil the current will finally be restrained from further increase either by the coil's own resistance or by the capabilities of its power source.

    Motors can show more complicated behaviour, and in particular some motors have severe start-up current surges due to the absence of back-emf when the armature is stationary. An electromagnetic actuator contains moving magnetic parts so that its effective inductance varies, perhaps quite abruptly, as it operates. The current waveform may not be simple.

    Capacitors also oppose voltage build-up even before they have any charge. You may discover this the hard way if you are too generous with the capacitors used to discourage brownouts, particularly if you connect them to the supply with no intervening resistance. At switch-on, said capacitors form a virtual short-circuit, and load the supply heavily until they have charged. This can blow fuses, or even rectifiers etc. if it gets out of hand.

    You may be tempted to object that many power supplies contain reservoir capacitors, and should be affected by this issue. The simple answer is that they all are, but typically at least for small supplies the numbers come out to acceptable values. Managing inrush current can become problematic if a large value reservoir capacitor is used.
     
  8. Mike33

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    Feb 4, 2005
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    That was what I was saying...that, to a power supply, an inductor of very low resistance appears - like a very low resistance, and draws an appreciable current while building its magnetic field.
    And yes, to a PS, a discharge capacitor appears as a dead short (I build tube amps, believe it or not). Thus, a small resistance, even 100R, is traditionally placed before it/them in an R/C network to control the charging process.

    "A simple inductor will draw a current which at least initially builds up progressively from zero" >> yes, and this is where, in my problem, current is screaming into the solenoid, maxing the power supply, and making any other items connected to the PS brown out. After which time, the inductor will oppose changes in current by alternately building and collapsing its field. When it has 'finished', I assume the field is collapsing and back EMF is present, which should be handled with an anti-snubbing diode...

    If it's not called 'inrush current', well, ok. I'm a seat-of-the-pants electronics guy, and my engineer friend explained that any coil you apply power to has inrush, but if it's not a motor perhaps the term is not being used correctly (by me). Perhaps it is just the low resistance drawing current, not specifically an "inrush"? At any rate, thanks - I can see what's up here....the accessory side of this is a better place to try to mitigate the effect, rather than on the 'power products' end!! :)
     
  9. AlexR

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    Jan 16, 2008
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    You need to read up on how inductances work.
    When you apply voltage to an inductor the current starts off at zero and builds up over time to a maximum which is limited by the coil resistance. The back EMF of the coil (caused by the build-up of the magnetic field) acts to oppose the increasing current hence at turn-on there is no "current screaming into the solenoid" but rather a steady and relatively slow build-up of current which will continue till the current is limited by the solenoid resistance. The current flowing through the solenoid will AT NO TIME be greater than the supply voltage divided by the DC resistance of the coil.
     
  10. Adjuster

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    Dec 26, 2010
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    It would be as well to understand how inductors really behave, as much or their applications to valve (tube) amplifiers as elsewhere.

    The idea of the current building first then the opposing back-emf building up later really only applies to some things such as motors, where mechanical acceleration is involved. Actually your "solenoid" may show some such behaviour on a transient basis as its armature moves, but unless the coil is switched off at the end of its travel the current will not remain lowered.

    The magnetic field will not collapse of its own accord either, unless the supply is removed, by a limit switch or otherwise. Of course, it is going to be turned off at some point, so snubbing is still advisable.

    None of these arguments rule out the possibility that connecting a high-current inductor to a power supply will produce a brownout, especially if the coil current is close to the supply ratings or actually exceeds them.
     
  11. Mike33

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    Feb 4, 2005
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    Thanks again, guys - I do appreciate the clarification of these principles, even if I am stubborn with respect to 'my view'....many ppl, including myself, operate on very basic assumptions, and 'get away with it'. We don't use proper terms online, so we don't look 'professional', I guess. Nothing I build or work on requires me to read a text on induction or to learn all of those cumbersome equations associated with magnetics....I have found that my very simple understanding works pretty well for most situations, and if I were to ever really go inside very deeply, I would probably have to undertake a more advanced study! Although many older designs employ them,I don't use chokes in amps, BTW. :) I understand that current flow will slow in an inductor once it has charged and will oppose changes in current, and basically why it does so.

    What I am explaining, in my own way, as "current screaming in" is the concept that in an otherwise small and stable current setting, I have a few VERY low resistance components...the solenoids. They come in and "mess things up", lol. I would be surprised if they measured out at 20 to 50 ohms. So, at 16V (we'll just assume DC here), that is about .3 to .8A, which are demanded instantly. That may NOT be inrush current - again, my use of a term, probably in the wrong way. If using a transformer rated for 2A or so, and it already has various items drawing from it, it's not unlikely that the extra draw is asking too much of it, albeit for a very brief period. These solenoids "snap" - they turn on and then very quickly drop out.

    I believe I just need to have him set up his little 'street lights' and stuff (accessories) to be located on a separate feed directly from the trafo with an R/C network before them, to be determined by incrementally increasing the capacitance.

    If that is no-go, then a separate power supply may be called for - either for the power stuff or the accessories. In fact, if it were MY train, I'd just do that....go to RS and get a 25V transformer and make them their own power supply - end of problem!
     
  12. studiot

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    Nov 9, 2007
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    Mike, I am sorry that this reply is so scruffy but I tried to dash it off whilst you were still online.

    Anyway Adjuster and Alex are correct in telling you that there is no 'inrush' current when a direct voltage is suddenly applied to an inductor.

    As stated, the current starts at zero and builds up to a maximum. You cannot suddenly 'apply' a current to an inductor.

    The attachment shows graphically what happens and gives the mathematical equations which describe the build up of current and also its decay if you suddenly remove the voltage.

    Remember this is for Direct voltage - For Alternating voltage the situation is more complicated.

    go well
     
  13. tom66

    Senior Member

    May 9, 2009
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    In theory what has been said about inductor (or solenoid) inrush current is true. However, all wires are inductors, as is the transformer used to power the circuit. And as has been said, inductors resist changes in current. So although the solenoid may not pull inrush current itself, it does pull some current and this will briefly cause a voltage sag due to the inductance of the wires and transformer, and potentially the main bulk capacitor may sag briefly too before the next cycle of the mains charges it.
     
  14. Mike33

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    Feb 4, 2005
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    Coming back to this JUST a tad late (busy at work!), but thanks a bunch for the replies and well-thought-out information, guys! And for the graphic, studiot - helpful, all!

    For a 'macro' circuit like this little train exercise, the inductance of the wiring is probably negligible. What is going on, when I used the term "inrush current", is simply the sudden demand for MORE current that a solenoid will cause when energized. At this level, subtle (and important if dealing with more sensitive circuits!) effects are swamped by these large, in-your-face problems. A circuit pulling 200mA suddenly calling for 2A, for example.....it is dramatic, lol.

    The solution, in the end, seems to be to use my CB power supply (13.8VDC, 3A) to power the accessories for the train (which is VERY impressive, and fills a whole room with scenery, a truck stop, street lights, crossing signals, etc...). Keeping the "power products" on their own supply should really help with any light blinking, and also would allow some buffering, if necessary, to be applied to the 'low current side', which is only running those LEDs. Most of my work is done with audio circuits - so this approach reminds me of 'star grounding', sort of. Keep the noisy stuff away from the sensitive stuff!
    It is working well so far :D
     
  15. SgtWookie

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    Jul 17, 2007
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    Gee, sorry I missed this one before.

    I just drew up a couple of CDUs (Capacitive Discharge Unit) for another fellow on the forum. You use AC for the input, but DC can be used as well interchangeably.

    Have a look at the attached; there are a couple of versions. The 1st one is a bit better. If you want to build the 2nd one, I suggest adding D3 like in the 1st schematic.
     
  16. jimkeith

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    Oct 26, 2011
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    Hey Mike,
    You are getting AC confused with DC.

    The AC solenoid has very low inductance initially and switches to high inductance when picked up due to the completion of the low reluctance magnetic circuit--hence the inrush. A transformer can also have inrush, but it is different--it depends upon the polarity of the residual magnetic field (set) in the core that can cause saturation on the first half-cycle.

    With DC, there is no inrush as there is no way the current can ever exceed E/R. In addition, the current builds slowly due to the effect of the inductance.
     
  17. studiot

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    Nov 9, 2007
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    Take care here.

    The inductance is a constant property of the inductor, whether current is flowing or not.
    It is fixed by the geometric configuration of the coils and the core material permeability.

    There are variable reluctance devices, but they rely on mechanical intervention to change the inductance.

    I said it in post # and I will repeat it more stongly.

    It is impossible to suddenly change the current through an inductor.

    This applies whether the change is from zero to some actual value or from one value to another.
     
  18. SgtWookie

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    Jul 17, 2007
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    Studiot,
    A solenoid has a movable core, so it's inductance actually starts out rather low (almost an air core), and increases very significantly as the core gets pulled into the body of the solenoid.

    Since the inductance is low when power is first applied and most of the core is extended from the body, current can indeed build quite quickly.
     
  19. studiot

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    Hi Sgt Wookie,

    That's exactly what I said!

    I would say, though, that a solenoid is simply a cylindrical helically wound coil.
    It may have a moveable core or a fixed core or even no core at all.

    See this article from th University of Surrey.

    http://info.ee.surrey.ac.uk/Workshop/advice/coils/air_coils.html

    see also

    http://en.wikipedia.org/wiki/Solenoid

    You are right in observing that the mechanical intervention may be self reinforcing, with an appropriate mechanical arrangement.

    :)
     
  20. colinb

    Active Member

    Jun 15, 2011
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    But wouldn't the dc resistance of the coil remain constant and limit the maximum current no matter what? When voltage is first applied, coil current would still follow the same type of curve from zero current to I=E/R, even if that happens in a shorter duration of time due to low inductance of the air-core inductor when the solenoid is extended.
     
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