Caps and Diode setting on meters

Capacitors have impedance, a frequency dependent relationship between the voltage across them and the current through them. Impedance is two dimensional, and can be expressed as magnitude (of ohms) and phase shift (polar form), versus frequency, or in artesian form as real and imaginary parts of the total.

Capacitance, itself, is measured in farads. A farad of capacitance experiences a 1 volt change across its terminals for each coulomb of charge that passes through it. Since a coulomb per second is called an ampere, we can say that an ampere passing through a Farad of capacitance changes the voltage across it by a volt, every second.

As a formula, this is I=C*(dv/dt), where I is current in amperes, C is capacitance in farads, and dv/dt is the rate of change of voltage in volts per second.

Sort of, except that capacitance does not consume energy, but stores it and gives it back, depending on whether the voltage is rising or falling in magnitude. The energy stored in a cap is energy=1/2 * C * V^2, where energy is in watt seconds or joules, C is capacitance in farads, and V is the volts across the terminals.

It holds more electrons shifted from one plate to the other with less applied voltage. Think of the capacitor as a rigid tank with a rubber membrane stretched across the middle, and a pipe into the vessel on each side of the membrane. In this analogy, the pressure difference (voltage) between the two pipes (terminals) is proportional to the amount of fluid removed from the volume on one side of the membrane and forced into the volume on the other side of the membrane. The deformation of the membrane is what stores the energy. In a capacitor, it is the electric field between the plates that stores energy as the dielectric is stressed with voltage.

The charge is localized to the surface of the plates, but the energy is stored in the electric field between the plates these surface charges produce.

If the voltage applied between their terminals is similar, as it would be if they were connected in parallel, the 1uF cap would be passing a million times more current than the 1 pF cap. In other words, the impedance of the 1 uF cap is a million times lower than the impedance of the 1 pF capacitor. The energy stored at any voltage is also a million times higher.

But if you connect these two capacitors in series, and pass the same current through each, this is like connecting a very low and a very high ohm resistor in series. Most of the total voltage drop occurs across the higher value resistor, and across the very high impedance capacitor, which is the smaller one. The magnitude of capacitive impedance is ohms = 1/(2*pi*f*C), where f is frequency in hertz, and C is capacitance in farads. Ohms is a word that means volts per ampere.

If they are forced to discharge their voltage at the same rate (volts per second) the 1 uF cap will deliver a current a million times larger than the 1 pF cap does during this discharge. If they are drained at the same current, the 1 uF cap will have a discharge time a million times longer than the 1 pF cap does.

The main thing that complicates capacitors that is missing from the understanding of resistors is time. You cannot think clearly about what capacitors do in a circuit or in isolation, if you don't include time. The only thing you can say without referring to time is the instantaneous energy stored in the cap, that is purely a function of capacitance and voltage.

(snip)

It is delivering a regulated current and displaying the millivolts of drop that current produces as it passes through the leads and the component they are connected to. The reading usually goes off scale if the leads are open circuited. You can use a separate volt meter to measure how high this voltage goes, open circuit. Something around 2 volts is common (2000 on the meter). Silicon diodes may drop, between .3 and .6 (300 to 600 millivolts) when passing this small (and also measurable with a second milliamp meter) current.

If you test a large capacitor, you should see something near zero (uncharged) that rises smoothly to the meter overload (cutoff) as the current ramps up the capacitor voltage, as charge is stored in the cap. Not a good idea to test the charge capacitor in reverse, because the meter is not expecting its voltage to go negative. Touch the capacitor leads together to dump the stored charge, to repeat the test.

Once you calibrate this capacitor test (zero to cut off time for a given capacitance) you can roughly estimate other capacitances by their charge time, relative to this one. For instance, if it takes 2 seconds for a 100 uF cap to charge to meter cut off, it should take about 20 seconds for a 1000 uF cap to do the same thing.

Darn, John you wrote a mini-textbook here ... nice answer!

Thank you.

I just remember how long I was frustrated with not finding explanations of simple electronic principles when I was trying to teach myself electronics. So many sources were either so simplified that they didn't really explain anything, or they dove directly into differential equations and left me behind.

I tried to keep the differential equations to a minimum.

Capacitance is the ability to store electric charge, much like a balloon stores air.

Capacitance is measured in Farads. Impedance is measured in ohms. Which unit you use depends on what you're measuring. Impedance can be calculated from capacitance and frequency, if the signal is a sine wave. For other shapes, you have to consider each frequency component (harmonics etc) of the wave separately; each sees a different impedance.

It can be, but not always. When dealing with AC signals, where the capacitor is part of a filter, it makes sense to think of it that way, because each component (frequency) of the signal will see a different impedance. For DC or DC-like signals (long-duration square waves, for example, like the 555, or non-periodic signals), it's better to think of them as storage devices.

Both, sort of. Capacitance depends on the surface area of the plates and the distance between them.

Vin ------| |------* Vout Ic ---> | >

| Gnd ---------------* Gnd

Let's say Vin is increasing. The voltage across the capacitor won't change immediately, so the voltage across the load does. This voltage drop causes Ic. Ic causes the cap to charge, which increases the difference between Vin and Vout - effectively reducing Vout.

If the capacitance is smaller, the charge reacts faster (takes less time to charge), thus Vout stays closer to zero. A larger capacitance takes longer to charge, letting Vout go further from zero.

The neat part about sine waves is that the integrals and derivatives (math formulas that relate charge to current) are also sine waves (although of various amplitudes and phases). If you put waves of other shapes through the above circuit, you won't get the same shape out. So when Vin is a sine wave, Ic and Vout are also sine waves, although at different amplitudes and phases. When looking at the sine wave amplitudes, one can think of the capacitor and resistor above like a voltage divider, hence one can calculate the "impedance" (or effective resistance at that frequency) of the capacitor.

If the applied voltage is the same, yes. However, it takes more or less total current to do so, depending on the capacitance.

Assuming all else is the same, yes. That is, if two capacitors have the same charge-induced-voltage and load resistance, they'll both start out with the same discharge current. However, the larger capacitor will be able to sustain that current longer before it's discharged as much as the smaller one. Example: two caps at 10v with

1k load resistors. One is a higher capacitance than the other. The larger one will take longer to discharge to 5v than the smaller one.

I think it's just putting a small fixed current through the diode, and measuring the voltage drop across it. When forward biased, the diode passes the current with a drop of Vf. When reverse biased, the diode does not pass the current, and the whole power supply's voltage is across it.

So, if you test the diode forward-like, the meter tells you the voltage drop across it. If the number makes sense, the diode is good.

The meter puts a fixed current (1mA is common) through the diode and the display is actually showing the voltage drop across that diode. So yes, the display is in VDC.

Dave.

Well, well !!

;-)

--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
                                             (Stephen Leacock)

Damn spell checker!

--
Deep! ;)

Thanks John, DJ and David for taking the time to answer my question. I will let it sink in. Uriah