Author Topic: Quick and simple question about de-energizing Capacitors.. and BM235 questions..  (Read 3262 times)

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Offline PcmakerTopic starter

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Capacitor questions

1, Are capacitors the only things that hold electrical charges on a circuit board?

2, I know you have to short both pins to de-energize a capacitor, but how long do you have to hold both pins for the voltage it holds to be considered safe or to go down to 0? Is it instant?

3, Sometimes, I'm having a hard time touching both pins with my needlenose or screwdriver. Can I de-energize a capacitor with double ended alligator clips where I put a clip on each pin? Do I have to worry about gauge of the aligator clip wire?



4, Is the charge held in a capacitor always DC and will a multimeter be able to accurately measure the amount of charge it's holding?

Brymen 235 questions. My box didn't come with a manual, so here's some basic questions

5. I have a Brymen 235 multimeter I just bought and I was wondering does the "Auto V LoZ" feature of the multimeter automatically detect current in AC or DC? I thought it did, but when I put it on this setting and I tried testing an outlet, it tripped the GFCI.

6. What is the LoZ feature?

7. When I checked for ohms on a resistor, the resistor is 0.47 ohms, but the meter only showed 0.4. Is the Brymen 235 incapable of showing that second digit? It has K and M and I'm assuming K is thousands and M is Millions?
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Offline katzohki

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1. No, sometimes batteries do, but when a given device is unpowered that's usually where you see some voltage hanging around.
2. Depends on the resistance of the "short". If it's a simple wire near 0 ohms, then it should be almost instantaneous (time constant would be driven by the ESR of the capacitor). Note that this is not necessarily a good thing and can cause sparks. It could be dangerous with very large caps.
3. Yes. I recommend caution, but if you're just talking about small surface mount 0603s it's probably fine.
4. Always DC. (Assuming any connected circuit is "off")
5. I don't know this meter off hand, sorry
6. Probably low-Z; meaning low impedance
7. It might be, that's a low value. Correct on k and M
 
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Offline PcmakerTopic starter

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1. No, sometimes batteries do, but when a given device is unpowered that's usually where you see some voltage hanging around.
2. Depends on the resistance of the "short". If it's a simple wire near 0 ohms, then it should be almost instantaneous (time constant would be driven by the ESR of the capacitor). Note that this is not necessarily a good thing and can cause sparks. It could be dangerous with very large caps.
3. Yes. I recommend caution, but if you're just talking about small surface mount 0603s it's probably fine.
4. Always DC. (Assuming any connected circuit is "off")
5. I don't know this meter off hand, sorry
6. Probably low-Z; meaning low impedance
7. It might be, that's a low value. Correct on k and M

Thanks!

The Brymen 235 is what Dave Jones from EEVblog rebranded and is currently selling. It's a great meter for the price. I've seen multiple video reviews of teardowns and accuracy testings and it passed with flying colors.
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Offline MeerkatAirforce

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I'll try to answer some of the capacitor questions:
1. Strictly speaking, capacitors are the only passive element that holds a charge (batteries or generators or the like would be active elements). However, you find "parasitic" (unintended and unwanted) capacitance in many other elements (a basic capacitor is formed by two close conductive plates, so you essentially have unavoidable mini-capacitors in your resistors and transistors and other elements). This parasitic capacitance is often negligibly small, but if you're doing anything precise, it can mess you up if you don't account for it.

Note: inductors don't hold charge. They create magnetic fields that push charges (electrons) through themselves, creating a current.

2. Capacitors do, in fact, take some small time to charge and discharge. The charge time and discharge time are described by a couple of exponential equations you can look up.
Voltage vs Time Graphs during Charge and Discharge: http://www.electronics-tutorials.ws/rc/rc2.gif

These are characterized by a "time constant" Tau (the Greek letter Tau) as you can see on the time axis in the graph. This time constant is Tau = R*C (Resistance times Capacitance) (note that the units for capacitance, Farads, are equal to seconds/ohm, so multiplying R*C gives you a value in seconds). In this case, the resistance is whatever you're using to "short" the capacitor (the screwdriver or wire still has some tiny amount of resistance).

Generally, we say that a capacitor is fully charged or discharged after about 5Tau, or 5 times the length of the "time constant." Since your capacitors are usually going to be absolutely tiny, it happens almost instantly, but for larger capacitors, it can take a little while.

3. You have to be careful if you're using just a thin wire like on those alligator clips to discharge a capacitor. Remember, you're running all the charge from capacitor through the resistance in the wire, and the energy of that charge will be dissipated as heat. If it's a large capacitor discharging through a very thin wire, it could melt the coating (though you probably won't face that with small capacitors). A wire is definitely fine for tiny caps, but it's usually best practice to use something that can definitely take more energy passing through it. If you can't get the tools in there to short it, you can use a regular small resistor to "short" the ends of the capacitor. The little wire legs on those boys can usually get into the small places you need (and you can hold it with the pliers if you need more reach).

4. Short answer: Basically yes. Long answer: Basically sometimes, but kinda no.
Short: If you put your multimeter probes on either pin of a capacitor while it's in an RC circuit (only resistance and capacitance) where the input is DC (flowing one way), it will basically give you the DC voltage across it. If you take the capacitor out of your circuit and try to measure it, the capacitor will basically discharge through the resistance inside your multimeter, and so you'll just read a continually dropping voltage.

Long: (Sorry in advance for the pedantry about charge and current)
  • Situation 1: you have a capacitor in your circuit somewhere, but the input voltage is coming from somewhere else.
    A capacitor stores up charge, which in itself can't be DC or AC. DC means that the current (the flow of those charges) is going one way. AC means that the flow of those charges is going back and forth (though some capacitors can only be used for DC). If you have an AC input or a switching input, you can't expect to read a constant voltage with your meter (You would need an oscilloscope to look at what's going on).
  • Situation 2: you charged up a capacitor using some voltage source, then you disconnect the voltage source and use the charge on the capacitor to run a circuit.
    In a circuit with just elements that use up charge like resistors (like a light bulb for example), the capacitor will just act like a battery, basically, and keep pushing charge in one direction to be used up by the resistors/light bulbs/whatever-you-have until it's fully discharged. It will be DC.

    Where things get funky is when you have a capacitor and an inductor in the same circuit (see LC circuits). It basically bounces the charge backward and forward around the circuit, creating an alternating current
    See this image: https://upload.wikimedia.org/wikipedia/commons/8/80/Tuned_circuit_animation_3_300ms.gif
    Notice how the charge builds up on one plate of the capacitor, then flows through the circuit to build up on the other plate, then flows back to the first plate.
    If you toss a resistor in there, it basically does the same thing, except that the resistance dissipates some of the energy each time the current switches until all of the energy is gone from the circuit, and charges eventually stop going back and forth.


I hope all of this was correct, and any of it was helpful.
<Edited for formatting>
« Last Edit: October 19, 2017, 12:39:05 am by MeerkatAirforce »
 
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Offline PcmakerTopic starter

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How exactly does the charge dissipate when you short the 2 leads on a capacitor? Where does it go?
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Offline MeerkatAirforce

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Short answer: The charges themselves don't dissipate. The potential energy we stored in the capacitor from the difference in charge is converted to other types of energy, some of which is heat, as the electrons (negative charges) move through the circuit.

If you "short" the two leads of the capacitor, you touch the metal leads together. Those metal leads have a very small resistance. It's so small that we say that is effectively a "short," even though it technically is not.  When you put a voltage across a resistance to push a current through it, some of that potential energy (the voltage) eventually gets converted to heat energy.

Electric space heaters generate heat this way; they pump current through what is essentially a big old resistor for the purpose of turning it into heat. Light bulbs are similar, but they dissipate the energy as both heat and light. Though in our case, it is not purposeful, because there is a small amount of resistance in just those metal wires, it acts like a tiny version of the space heater, and some of our energy is dissipated as heat.

Note: When I say "dissipated as heat," really the energy is just being changed from one form to another. We say that it's being "dissipated" because it's being converted to a form we don't want it to be, and so it's "wasted" (e.g. if we had a light that didn't generate any heat, much more energy would be being converted to light, so our batteries would last much longer).


Long answer:
A capacitor is effectively two parallel metal plates. It stores a surplus of negative charge (electrons) on one side, and we say that it stores a positive charge (a relative lack of electrons) on the other plate. This difference in charge results in a voltage, which is a difference in potential energy between the two plates. When there's a difference in potential energy, it can be turned into mechanical energy. In our case, the potential energy is this voltage across the capacitor, and the mechanical energy will be the physical movement of the electrons through the circuit (i.e. an electrical current).

(Other examples: Elastic potential energy: use a mousetrap spring to power a mousetrap car; Chemical potential energy: burn gas to break the bonds between the atoms to release the energy and power a real car; Nuclear potential energy: break apart the atom itself to release the energy stored within it to power a real city)

The way we build up this surplus of electrons is by using an outside voltage source like a power supply or a battery to push a whole bunch of electrons to one side of the capacitor.
                     ___
                       |
                       | <---wire connected to positive voltage source (electrons got pushed out this way)
                       |
                       |
+++++++++++++++++++++
+++++++++++++++++++++   <---positive charge (lacking electrons) Note: not entirely lacking electrons. Just some got pushed out.
___________________________   
                                                    <---metal plates (not touching)
___________________________
--------------------------------------   <---negative charge (surplus electrons)
--------------------------------------
                      |
                      |
                      |<---wire connected to ground (electrons got pulled in this way)
                      |
                   ____
                     __
                      _

Once we have this buildup of different charges, resulting in a difference in the potential energy, we can put it into a circuit with a resistor and power the circuit from there.

       _____________
       |+                   |
CAPACITOR        RESISTOR
       |-___________|
          <--current--

Once we put the capacitor in the circuit with the resistor, the natural reaction is for the negatively charged particles to push away from each other and move toward the positively charged side. Since the two metal plates of the capacitor are separated, the electrons will move from the negative side to the positive side, through the resistor, until there is no difference in the build-up of charges on either side of the capacitor.

These moving electrons have kinetic energy associated with them, as they are particles that have both charge and mass (like a moving boat has energy associated with it). A resistor, as the name implies, resists the movement of these electrons as they are getting pushed through the resistor. If that boat from before was an ice-breaker in the arctic, plowing through some ice, it clearly has enough energy to keep moving through that ice, but the ice is slowing it down a bit, and it's imparting some of that energy to the ice by hitting it, breaking the ice.

This is an example of an object being resisted, imparting kinetic energy to what's resisting its movement. Electrons moving through resistors is an electrical version of that. The electron is running into a whole lot of tiny electrical fields from the particles in the atoms in the resistor, and it's getting slowed down and pushed around. Just like the ice-breaker pushes back against the resisting ice, the electrons push back against the resisting fields ("Something something Forces, something something Equal and Opposite Reaction" - Isaac "Big Daddy" Newton).

Since energy can't be created or destroyed, the energy has to be transferred completely into some other form (or forms). We're releasing the potential energy (the voltage) into kinetic energy (the motion of the electrons), and some of the energy from the moving electrons gets transferred to the material of the resistor as some other form of kinetic energy. This is most often heat, but can be light (e.g. light-bulbs or something that's "red hot") or motion (e.g. motors) or sound (e.g. speakers), for example.

As we said before, even when you "short" the leads of the capacitor together, there is a tiny bit of resistance in the metal leads, so "shorting" the capacitor still gives you the situation described above, where the energy from the voltage is eventually converted into heat energy.

Side Note: As more electrons move from one side of the capacitor to the other, the difference in charge drops, so the difference in energy levels across the capacitor drops, so the voltage drops. That's why we see the voltage going down in that voltage vs. time graph in my previous post.

<EDITED to show where surplus electrons go/come from while charging a capacitor in Fig. 1>
« Last Edit: October 19, 2017, 03:33:20 am by MeerkatAirforce »
 
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Offline 2N3055

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Brymen 235 questions. My box didn't come with a manual, so here's some basic questions


No manual? No problem... Dave has it for download..... There you are..
http://www.eevblog.com/files/BM235-Manual-EEVblog.pdf
 
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Offline PcmakerTopic starter

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Thanks, guys!
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Offline IanMacdonald

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Strictly speaking, inductors also hold charge in the form of circulating current flow. However, due to the relatively lossy nature of typical inductors it will rarely persist for more than a fraction of a second.

One exception is with superconducting magnets, where the circulating current can persist for years. Look up MRI scanner magnets on YouTube for more info.

BTW, never short large caps to discharge them. Use a suitable resistor. Shorting large caps develops a pulse current of several kiloamps which can damage adjacent components.  For work on audio amplifiers or the like it's handy to have a dedicated tool with two insulated prods and a suitable wirewound (say 100R 10W) for this purpose as you use it often.
 

Offline MeerkatAirforce

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I was under the impression that inductors don't technically store charge (aside from with parasitic capacitance), but rather that the magnetic field induced in them creates an impetus for continued current flow.

Can you describe how an inductor stores charge, or point me to a resource that talks about inductors technically storing charge? I would like to read more about that.
 


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