Capacitors in DC Circuits?

CaVeDoG · Sep 16, 2003

I'm trying to understand how a simple flashing LED circuit works
(http://cavedog.ath.cx/fledsch.gif), but don't get how the capacitors work
with DC (the only thing I don't get is how they discharge). In all of the
many articles I've read on capacitors, they say that the electrons build up
on the positive plate until it's completely charged, at which point the
current stops...and that's where they start on about discharing in an AC
circuit. How and when does a capacitor discharge in a DC circuit?

John Popelish · Sep 17, 2003

CaVeDoG wrote:

I'm trying to understand how a simple flashing LED circuit works
(http://cavedog.ath.cx/fledsch.gif), but don't get how the capacitors work
with DC (the only thing I don't get is how they discharge). In all of the
many articles I've read on capacitors, they say that the electrons build up
on the positive plate until it's completely charged, at which point the
current stops...and that's where they start on about discharing in an AC
circuit. How and when does a capacitor discharge in a DC circuit?

Flashers require some switching circuit to dump the charged capacitor
through the LED, and then switch off, allowing the cap to recharge.
In low voltage designs, the cap charges to almost the battery voltage
and then is reconnected in series with the battery, to effectively
double the voltage to supply the LED. Other designs use an oscillator
and a step up transformer to generate higher voltage to a cap that
gets dumped to an LED after it reaches some voltage. Others use an
inductor much like an ignition coil (connecting the coil to the
battery till a given current builds up, and then opening that circuit,
allowing the inductance to generate a larger, reverse voltage that is
delivered to a cap.

--
John Popelish

Jim Large · Sep 17, 2003

CaVeDoG wrote:

I'm trying to understand how a simple flashing LED circuit
works (http://cavedog.ath.cx/fledsch.gif), but don't get
how the capacitors work ...

It's only just a few weeks since I was scratching my head
trying to understand that same circuit.

Let's start with Q2 turned on, and Q1 is off. The voltage
on the collector of Q1 (and the left hand terminal of C1) is
pretty close to Vcc, and the voltage on the base of Q2
(right hand terminal of C1) is about one diode drop above
zero (somewhere around 0.7 volts).

NOTE: You haven't labelled Vcc or ground in your
diagram, but it's easier for me to think in those
terms. I'm calling the "+" terminal of the battery
Vcc, and I'm calling the "-" terminal ground.

Now, Q1 suddenly turns on. Don't worry about why, we'll get
to that soon. The voltage on the collector of Q1 is pulled
to ground, which means that the voltage on the left hand
side of C1 is pulled down too.

What does the capacitor do? Well one equation says that the
voltage across the terminals of the capacitor is equal to
the charge divided by the capacitance. The capacitance
hasn't changed, and since we're only looking at just one
instant in time (i.e., before any current gets to flow) the
charge hasn't changed either. So that means that the
voltage between the two terminals must be the same as it was
a moment ago.

Well the voltage on the left just dropped by almost the full
supply voltage, so if the difference between the two
terminals hasn't changed, that means that the voltage on the
right MUST have dropped by the same amount. The voltage on
the right is now NEGATIVE (i.e., less than ground).

Since the voltage on the right terminal is also the voltage
on the base of Q2, that means that Q2 will switch off.

O.K., the voltage on the base of Q2 can't stay negative
forever because R2 is trying to pull it up to Vcc. This is
where we might say "the capacitor charges." When the
voltage on the base of Q2 eventually comes back up to that
magic 0.7 volts (a few tens of milliseconds given the
component values in your circuit), Q2 will switch on, and
ITS collector will be pulled low, and the base of Q1 will be
driven negative, and Q1 will turn off.

After a while, C2 "charges", allowing Q1 to suddenly turn
on, and there we are! One full cycle.

-- Jim Large

P.S. Here's your circuit in SwitcherCAD. Cut everything
below the dotted line, and paste into a file named FLED.ASC.
SwitcherCAD is an awesome toy. Get it from this page if you
don't already have it. (http://www.linear.com/software/).
.............................................................
Version 4
SHEET 1 880 680
WIRE -112 400 -112 304
WIRE 336 400 336 304
WIRE -464 400 -464 128
WIRE -464 48 -464 -288
WIRE -112 -288 -112 -240
WIRE 48 -288 48 -112
WIRE 176 -288 176 -112
WIRE 176 -32 176 64
WIRE 176 64 224 64
WIRE 48 -32 48 64
WIRE 48 64 0 64
WIRE -64 64 -112 64
WIRE -112 64 -112 -32
WIRE -112 208 -112 64
WIRE 336 208 336 64
WIRE 336 -112 336 -288
WIRE 288 64 336 64
WIRE 336 64 336 -32
WIRE 176 64 176 144
WIRE 176 144 64 256
WIRE 64 256 -48 256
WIRE 48 64 48 144
WIRE 48 144 160 256
WIRE 160 256 272 256
WIRE -112 -176 -112 -112
FLAG -112 400 0
FLAG 336 400 0
FLAG -112 -288 Vcc
FLAG 48 -288 Vcc
FLAG 176 -288 Vcc
FLAG 336 -288 Vcc
FLAG -464 -288 Vcc
FLAG -464 400 0
SYMBOL npn 272 208 R0
SYMATTR InstName Q2
SYMATTR Value 2N2222
SYMBOL npn -48 208 M0
SYMATTR InstName Q1
SYMATTR Value 2N2222
SYMBOL res -128 -128 R0
SYMATTR InstName R1
SYMATTR Value 330
SYMBOL res 32 -128 R0
SYMATTR InstName R2
SYMATTR Value 47k
SYMBOL res 160 -128 R0
SYMATTR InstName R3
SYMATTR Value 47k
SYMBOL res 320 -128 R0
SYMATTR InstName R4
SYMATTR Value 560
SYMBOL cap 288 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C2
SYMATTR Value 10ľ
SYMBOL cap 0 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 2.2ľ
SYMBOL voltage -464 32 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 12
SYMBOL LED -128 -240 R0
SYMATTR InstName D1
SYMATTR Value QTLP690C
TEXT -498 506 Left 0 !.tran 1.5

CaVeDoG · Sep 17, 2003

Ahh, thank you very much, it makes a lot more sense now. My only question
now is how is one transistor on when you start?

"Jim Large" <moc.liam@egral.mij> wrote in message
news:4TO9b.231$Cs1.90@nwrdny02.gnilink.net...

CaVeDoG wrote:

I'm trying to understand how a simple flashing LED circuit
works (http://cavedog.ath.cx/fledsch.gif), but don't get
how the capacitors work ...

It's only just a few weeks since I was scratching my head
trying to understand that same circuit.

Let's start with Q2 turned on, and Q1 is off. The voltage
on the collector of Q1 (and the left hand terminal of C1) is
pretty close to Vcc, and the voltage on the base of Q2
(right hand terminal of C1) is about one diode drop above
zero (somewhere around 0.7 volts).

NOTE: You haven't labelled Vcc or ground in your
diagram, but it's easier for me to think in those
terms. I'm calling the "+" terminal of the battery
Vcc, and I'm calling the "-" terminal ground.

Now, Q1 suddenly turns on. Don't worry about why, we'll get
to that soon. The voltage on the collector of Q1 is pulled
to ground, which means that the voltage on the left hand
side of C1 is pulled down too.

What does the capacitor do? Well one equation says that the
voltage across the terminals of the capacitor is equal to
the charge divided by the capacitance. The capacitance
hasn't changed, and since we're only looking at just one
instant in time (i.e., before any current gets to flow) the
charge hasn't changed either. So that means that the
voltage between the two terminals must be the same as it was
a moment ago.

Well the voltage on the left just dropped by almost the full
supply voltage, so if the difference between the two
terminals hasn't changed, that means that the voltage on the
right MUST have dropped by the same amount. The voltage on
the right is now NEGATIVE (i.e., less than ground).

Since the voltage on the right terminal is also the voltage
on the base of Q2, that means that Q2 will switch off.

O.K., the voltage on the base of Q2 can't stay negative
forever because R2 is trying to pull it up to Vcc. This is
where we might say "the capacitor charges." When the
voltage on the base of Q2 eventually comes back up to that
magic 0.7 volts (a few tens of milliseconds given the
component values in your circuit), Q2 will switch on, and
ITS collector will be pulled low, and the base of Q1 will be
driven negative, and Q1 will turn off.

After a while, C2 "charges", allowing Q1 to suddenly turn
on, and there we are! One full cycle.

-- Jim Large

P.S. Here's your circuit in SwitcherCAD. Cut everything
below the dotted line, and paste into a file named FLED.ASC.
SwitcherCAD is an awesome toy. Get it from this page if you
don't already have it. (http://www.linear.com/software/).
............................................................
Version 4
SHEET 1 880 680
WIRE -112 400 -112 304
WIRE 336 400 336 304
WIRE -464 400 -464 128
WIRE -464 48 -464 -288
WIRE -112 -288 -112 -240
WIRE 48 -288 48 -112
WIRE 176 -288 176 -112
WIRE 176 -32 176 64
WIRE 176 64 224 64
WIRE 48 -32 48 64
WIRE 48 64 0 64
WIRE -64 64 -112 64
WIRE -112 64 -112 -32
WIRE -112 208 -112 64
WIRE 336 208 336 64
WIRE 336 -112 336 -288
WIRE 288 64 336 64
WIRE 336 64 336 -32
WIRE 176 64 176 144
WIRE 176 144 64 256
WIRE 64 256 -48 256
WIRE 48 64 48 144
WIRE 48 144 160 256
WIRE 160 256 272 256
WIRE -112 -176 -112 -112
FLAG -112 400 0
FLAG 336 400 0
FLAG -112 -288 Vcc
FLAG 48 -288 Vcc
FLAG 176 -288 Vcc
FLAG 336 -288 Vcc
FLAG -464 -288 Vcc
FLAG -464 400 0
SYMBOL npn 272 208 R0
SYMATTR InstName Q2
SYMATTR Value 2N2222
SYMBOL npn -48 208 M0
SYMATTR InstName Q1
SYMATTR Value 2N2222
SYMBOL res -128 -128 R0
SYMATTR InstName R1
SYMATTR Value 330
SYMBOL res 32 -128 R0
SYMATTR InstName R2
SYMATTR Value 47k
SYMBOL res 160 -128 R0
SYMATTR InstName R3
SYMATTR Value 47k
SYMBOL res 320 -128 R0
SYMATTR InstName R4
SYMATTR Value 560
SYMBOL cap 288 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C2
SYMATTR Value 10ľ
SYMBOL cap 0 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 2.2ľ
SYMBOL voltage -464 32 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 12
SYMBOL LED -128 -240 R0
SYMATTR InstName D1
SYMATTR Value QTLP690C
TEXT -498 506 Left 0 !.tran 1.5

R. Steve Walz · Sep 17, 2003

CaVeDoG wrote:

Ahh, thank you very much, it makes a lot more sense now. My only question
now is how is one transistor on when you start?
---------------

The two transistors and their RC circuits are never equal, and one
goes HI or LO first, and it starts itself. If they were too equal,
they wouldn't start as easily. It's wise to use unequal components
for symmetrical oscillators so they will start easily.

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public

"Jim Large" <moc.liam@egral.mij> wrote in message
news:4TO9b.231$Cs1.90@nwrdny02.gnilink.net...
CaVeDoG wrote:

I'm trying to understand how a simple flashing LED circuit
works (http://cavedog.ath.cx/fledsch.gif), but don't get
how the capacitors work ...

It's only just a few weeks since I was scratching my head
trying to understand that same circuit.

Let's start with Q2 turned on, and Q1 is off. The voltage
on the collector of Q1 (and the left hand terminal of C1) is
pretty close to Vcc, and the voltage on the base of Q2
(right hand terminal of C1) is about one diode drop above
zero (somewhere around 0.7 volts).

NOTE: You haven't labelled Vcc or ground in your
diagram, but it's easier for me to think in those
terms. I'm calling the "+" terminal of the battery
Vcc, and I'm calling the "-" terminal ground.

Now, Q1 suddenly turns on. Don't worry about why, we'll get
to that soon. The voltage on the collector of Q1 is pulled
to ground, which means that the voltage on the left hand
side of C1 is pulled down too.

What does the capacitor do? Well one equation says that the
voltage across the terminals of the capacitor is equal to
the charge divided by the capacitance. The capacitance
hasn't changed, and since we're only looking at just one
instant in time (i.e., before any current gets to flow) the
charge hasn't changed either. So that means that the
voltage between the two terminals must be the same as it was
a moment ago.

Well the voltage on the left just dropped by almost the full
supply voltage, so if the difference between the two
terminals hasn't changed, that means that the voltage on the
right MUST have dropped by the same amount. The voltage on
the right is now NEGATIVE (i.e., less than ground).

Since the voltage on the right terminal is also the voltage
on the base of Q2, that means that Q2 will switch off.

O.K., the voltage on the base of Q2 can't stay negative
forever because R2 is trying to pull it up to Vcc. This is
where we might say "the capacitor charges." When the
voltage on the base of Q2 eventually comes back up to that
magic 0.7 volts (a few tens of milliseconds given the
component values in your circuit), Q2 will switch on, and
ITS collector will be pulled low, and the base of Q1 will be
driven negative, and Q1 will turn off.

After a while, C2 "charges", allowing Q1 to suddenly turn
on, and there we are! One full cycle.

-- Jim Large

P.S. Here's your circuit in SwitcherCAD. Cut everything
below the dotted line, and paste into a file named FLED.ASC.
SwitcherCAD is an awesome toy. Get it from this page if you
don't already have it. (http://www.linear.com/software/).
............................................................
Version 4
SHEET 1 880 680
WIRE -112 400 -112 304
WIRE 336 400 336 304
WIRE -464 400 -464 128
WIRE -464 48 -464 -288
WIRE -112 -288 -112 -240
WIRE 48 -288 48 -112
WIRE 176 -288 176 -112
WIRE 176 -32 176 64
WIRE 176 64 224 64
WIRE 48 -32 48 64
WIRE 48 64 0 64
WIRE -64 64 -112 64
WIRE -112 64 -112 -32
WIRE -112 208 -112 64
WIRE 336 208 336 64
WIRE 336 -112 336 -288
WIRE 288 64 336 64
WIRE 336 64 336 -32
WIRE 176 64 176 144
WIRE 176 144 64 256
WIRE 64 256 -48 256
WIRE 48 64 48 144
WIRE 48 144 160 256
WIRE 160 256 272 256
WIRE -112 -176 -112 -112
FLAG -112 400 0
FLAG 336 400 0
FLAG -112 -288 Vcc
FLAG 48 -288 Vcc
FLAG 176 -288 Vcc
FLAG 336 -288 Vcc
FLAG -464 -288 Vcc
FLAG -464 400 0
SYMBOL npn 272 208 R0
SYMATTR InstName Q2
SYMATTR Value 2N2222
SYMBOL npn -48 208 M0
SYMATTR InstName Q1
SYMATTR Value 2N2222
SYMBOL res -128 -128 R0
SYMATTR InstName R1
SYMATTR Value 330
SYMBOL res 32 -128 R0
SYMATTR InstName R2
SYMATTR Value 47k
SYMBOL res 160 -128 R0
SYMATTR InstName R3
SYMATTR Value 47k
SYMBOL res 320 -128 R0
SYMATTR InstName R4
SYMATTR Value 560
SYMBOL cap 288 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C2
SYMATTR Value 10ľ
SYMBOL cap 0 48 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 2.2ľ
SYMBOL voltage -464 32 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 12
SYMBOL LED -128 -240 R0
SYMATTR InstName D1
SYMATTR Value QTLP690C
TEXT -498 506 Left 0 !.tran 1.5

Animesh Maurya · Sep 17, 2003

"CaVeDoG" <asdf@asdf.asdf> wrote in message news:<6MJ9b.154$JP6.35048@news.uswest.net>...

I'm trying to understand how a simple flashing LED circuit works
(http://cavedog.ath.cx/fledsch.gif), but don't get how the capacitors work
with DC (the only thing I don't get is how they discharge). In all of the
many articles I've read on capacitors, they say that the electrons build up
on the positive plate until it's completely charged, at which point the
current stops...and that's where they start on about discharing in an AC
circuit. How and when does a capacitor discharge in a DC circuit?

Circuit which you mentioned is a Astable Multivibrator, and is
commonly used to generate square waveform which glows that LED
alternately.

Frequency at which LED glows depends upon charging and discharging of
capacitors through resistor R2 and R3.

Frequency of asymmetrical Astable Multivibrator (as in your case) is
given by
1 / [0.69(R2*C1+R3*C2)].

Animesh Maurya

Rich Mechaber · Sep 18, 2003

Some past help I and others have received on understanding
multivibrators:
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&threadm=ultel0kf5fatd1%40corp.supernews.com&rnum=1&prev=/groups%3Fhl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26selm%3Dultel0kf5fatd1%2540corp.supernews.com

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&safe=off&frame=right&th=f205a13466864195&seekm=397ACEFC.755466E5%40ttlc.net#link1

Rich Mechaber

CaVeDoG wrote:

I'm trying to understand how a simple flashing LED circuit works
(http://cavedog.ath.cx/fledsch.gif), but don't get how the capacitors work
with DC (the only thing I don't get is how they discharge). In all of the
many articles I've read on capacitors, they say that the electrons build up
on the positive plate until it's completely charged, at which point the
current stops...and that's where they start on about discharing in an AC
circuit. How and when does a capacitor discharge in a DC circuit?

Capacitors in DC Circuits?

CaVeDoG

Guest

John Popelish

Guest

Jim Large

Guest

CaVeDoG

Guest

R. Steve Walz

Guest

Animesh Maurya

Guest

Rich Mechaber

Guest

Welcome to EDABoard.com

Sponsor

Online statistics

Forum statistics

Capacitors in DC Circuits?

CaVeDoG

Guest

John Popelish

Guest

Jim Large

Guest

CaVeDoG

Guest

R. Steve Walz

Guest

Animesh Maurya

Guest

Rich Mechaber

Guest

Log in

Welcome to EDABoard.com

Sponsor