# Caps and Diode setting on meters

Discussion in 'Electronic Basics' started by Uriah, Jun 13, 2007.

1. ### UriahGuest

Believe it or not, I have been into electronics for a long time but
for some reason some things don't sink in. Like what exactly is
capacitance? I know it is measured in ohms. I know it is like a
frequency dependent variable resistor. But when the capacitance goes
up in uF what is going on with the electrons? Does it hold more
Electrons? Or is the field between the plates holding a higher
charge? If you have a 1uF cap and a 1pF cap and you send a sign wave
through it of a particular frequency of 1k. What is happening that is
different between the two? They both charge to the same potential? Do
the discharge at a different rate? I can't see the forest for the
trees, I think, what ever that means.

Also,

When you are testing something with your DVM and you have it set on
the diode setting. You usually get a .400 number and when reversing
the leads you get an open. What is the meter doing? Is it putting out
1 volt? And then is it just measuring the difference? Is the.400 in
VDC? Can someone explain what a meter is doing on this setting.

Thanks for all of the help
Uriah

2. ### John PopelishGuest

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 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.
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.

3. ### CharlesGuest

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

4. ### John PopelishGuest

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.

5. ### DJ DelorieGuest

Without going into too many details...

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.

6. ### David L. JonesGuest

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.

7. ### Fred AbseGuest

Well, well !!

;-)

8. ### John PopelishGuest

Damn spell checker!

10. ### UriahGuest

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