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decoupling capacitors

G

Grassy Knollington

Jan 1, 1970
0
when using 9v battery as power source is decoupling capacitor still
necessary

Can anyone explain to newby purpose of decoupling and capacitor size
choice (I read it was to get 0 input impedance, but dont fully understand
that...)

Or pointers to a good web site...

ta.
 
i guess you are talkin abt bypass cap.
bypass cap provide a low impedence path for noise to ground so that you
hv a clean Vcc.
the impedence of a cap is calculated as Xc = 1/(2*PI*f*C) which f for
frequency and C for capacitance.
for DC, whose f = 0Hz, cap is basically a open circuit. for high
frequency noise, it's a short circuit.

if you are talkin abt decoupling cap in RF applications. the reason is
pretty simple.
to prevent RF energy goes from one stage to another.

hope that helps.
 
G

Grassy Knollington

Jan 1, 1970
0
i guess you are talkin abt bypass cap.
bypass cap provide a low impedence path for noise to ground so that you
hv a clean Vcc.
the impedence of a cap is calculated as Xc = 1/(2*PI*f*C) which f for
frequency and C for capacitance.
for DC, whose f = 0Hz, cap is basically a open circuit. for high
frequency noise, it's a short circuit.

if you are talkin abt decoupling cap in RF applications. the reason is
pretty simple.
to prevent RF energy goes from one stage to another.

hope that helps.

Yeah ta, its possibly the RF side included so you can use circuit as a
building block, but also as its DC pulse generator maybe its for clean Vcc?

I have seen .0047uf used and 47uf in another was wondering why that was?

In a timer app using 555 where you are generating DC pulses which is sort
of AC what size is used? What does it depend on? power source is battery,
so that would be clean anyway wouldnt it?, not like a noisy DC adaptor?
 
J

John Popelish

Jan 1, 1970
0
Grassy said:
when using 9v battery as power source is decoupling capacitor still
necessary

Can anyone explain to newby purpose of decoupling and capacitor size
choice (I read it was to get 0 input impedance, but dont fully understand
that...)

Or pointers to a good web site...

ta.
Many circuits will still benefit from having small bypass capacitors
connected across the 9 volt supply lines. The point of these
capacitors is to have some charge storage very close to any load that
can change its current draw in a very short amount of time. Without
the cap right up against such a load, the current changes pass through
all the wiring all the way back to the battery, bouncing the supply
rail voltage around with the voltage that wiring creates do to its
inductance. Remember the formula that relates inductor voltage to the
rate of change of current, V=L*(di/dt). This bounce couples all the
sub circuits together and can cause all sorts of trouble. So you
decouple this path by putting enough supply bypass capacitance at each
changing load. If a load is really a bad actor, you may want to add
additional inductance in series with its supply line (with even more
capacitance after the inductor), to let one side bounce without
dragging the rest of the supply rail with it.

Much logic is capable of drawing significantly different current in
sub nanosecond times, and many opamps can, too. And when the average
currents get larger (motors, relay coils, switching power supplies)
the rate of change can be significant, even when the transition time
is fairly long.
 
G

Grassy Knollington

Jan 1, 1970
0
Many circuits will still benefit from having small bypass capacitors
connected across the 9 volt supply lines. The point of these
capacitors is to have some charge storage very close to any load that
can change its current draw in a very short amount of time.

that makes sense!
without
the cap right up against such a load, the current changes pass through
all the wiring all the way back to the battery, bouncing the supply
rail voltage around with the voltage that wiring creates do to its
inductance. Remember the formula that relates inductor voltage to the
rate of change of current, V=L*(di/dt). This bounce couples all the
sub circuits together and can cause all sorts of trouble. So you
decouple this path by putting enough supply bypass capacitance at each
changing load. If a load is really a bad actor, you may want to add
additional inductance in series with its supply line (with even more
capacitance after the inductor), to let one side bounce without
dragging the rest of the supply rail with it.

Much logic is capable of drawing significantly different current in
sub nanosecond times, and many opamps can, too. And when the average
currents get larger (motors, relay coils, switching power supplies)
the rate of change can be significant, even when the transition time
is fairly long.

Thanks!, thats explained a lot, leaves/creates a few questions

so how do you calculate capacitor size, is it possible to put too
large a capacitor as the decoupler.

What size of capacitor do you need before you can feel a discharge?.

Building a Hulda Clarke "zapper" and that says .0047uF for decouple, If
you use a larger one, is that entire charge delivered for each pulse

I guess I need to read more about capacitor charge times and discharge
rates.
 
G

Grassy Knollington

Jan 1, 1970
0
that makes sense!


Thanks!, thats explained a lot, leaves/creates a few questions

so how do you calculate capacitor size, is it possible to put too
large a capacitor as the decoupler.

What size of capacitor do you need before you can feel a discharge?.

Building a Hulda Clarke "zapper" and that says .0047uF for decouple, If
you use a larger one, is that entire charge delivered for each pulse

I guess I need to read more about capacitor charge times and discharge
rates.

looking at another circuit seems using 47 times the circuits
other capacitance (ratio maintained across 2 similar but not same circuits)
 
J

John Popelish

Jan 1, 1970
0
Grassy said:
Thanks!, thats explained a lot, leaves/creates a few questions

Excellent! If I somehow explained everything in one paragraph, there
would be no reason to go on living. ;-)
so how do you calculate capacitor size, is it possible to put too
large a capacitor as the decoupler.

The formula that relates current to the rate of change of capacitor
voltage is a lot like the one that relates inductor voltage to the
rate of change of current through it. I=C*(dv/dt)
I is in amperes, C in farads, and dv/dt in volts per second. This
tells you that you can pass an ampere through a 1 farad capacitor and
the voltage will change 1 volt per second. The trick is to apply this
to short and variable spikes of current and calculate how the voltage
will change. Lots of people skip the calculus and find out the needed
capacitance by experiment (look at the supply voltage with a scope and
keep increasing the capacitance till the bounce gets small enough).
What size of capacitor do you need before you can feel a discharge?.

That is more dependent on the voltage stored across the capacitor,
than the size. As long as the voltage is less than about 50 volts,
dry skin will limit the current to less than a painful amount. Tough
your tongue across one, and the voltage has to be a lot lower to make
the experience not "shocking". The size of the capacitor just varies
how long the shock will go on, before the voltage has run down too low
to feel.
Building a Hulda Clarke "zapper" and that says .0047uF for decouple, If
you use a larger one, is that entire charge delivered for each pulse

Decoupling caps do not normally discharge during circuit operation.
They just sit there across the supply lines, holding a nearly constant
voltage. I would have to see a schematic to be sure what they are
describing is really a decoupling cap, and not some other circuit
function.
I guess I need to read more about capacitor charge times and discharge
rates.

They are very versatile components with lots of applications.
 
J

John Popelish

Jan 1, 1970
0
Grassy said:
looking at another circuit seems using 47 times the circuits
other capacitance (ratio maintained across 2 similar but not same circuits)

You went right over my head with that.
 
G

Grassy Knollington

Jan 1, 1970
0
Decoupling caps do not normally discharge during circuit operation.
They just sit there across the supply lines, holding a nearly constant
voltage. I would have to see a schematic to be sure what they are
describing is really a decoupling cap, and not some other circuit
function.

http://www.zapperlab.com/ZapSchematic.gif

This is schematic I am looking at, description of circuit at

http://www.zapperlab.com/

Also gives 1/4v positive offset to pulses

I just built a 555 timer test circuit from
http://www.doctronics.co.uk/555.htm

So I am now more familiar with the formula to control the frequency.
They are very versatile components with lots of applications.

Yeah, Cool! Thanks for your help!
 
J

John Popelish

Jan 1, 1970
0
Grassy said:
(snip)

The only capacitor performing a decoupling function in this schematic
is C1. And it is not bypassing the battery lines, but 2/3rds of that
voltage divided down by an internal resistor divider in the 555. The
CV label on the pin refers to the use of this fraction of the supply
as a control voltage (that is compared to the voltage on pin 6, to
decide when to reset the output flip flop). See:
http://www.uoguelph.ca/~antoon/gadgets/555/555.html
Here, the decoupling action is greatly improved by the resistors
between the CV pin and the supply rails.
 
G

Grassy Knollington

Jan 1, 1970
0
(snip)

The only capacitor performing a decoupling function in this schematic
is C1. And it is not bypassing the battery lines, but 2/3rds of that
voltage divided down by an internal resistor divider in the 555. The
CV label on the pin refers to the use of this fraction of the supply
as a control voltage (that is compared to the voltage on pin 6, to
decide when to reset the output flip flop). See:
http://www.uoguelph.ca/~antoon/gadgets/555/555.html
Here, the decoupling action is greatly improved by the resistors
between the CV pin and the supply rails.

Thats a useful reference ta, do you mean the test circuit on this
page where 2 resistors and 2 leds cross and join the output pulse line

On the zapschematic.gif, if I put a 4.7 uf capacitor (or what size
would you recommend) across the power terminals would that help the
circuit at all?

WOW, wish I had continued to study electronics 20 years ago when I
started... got into software design/prog instead... Never too late I guess!
 
J

John Popelish

Jan 1, 1970
0
Grassy said:
Thats a useful reference ta, do you mean the test circuit on this
page where 2 resistors and 2 leds cross and join the output pulse line

No. I am talking about the 3 equal resistors inside the 555 that
divide the supply voltage by 1/3 and 2/3, to act as reference voltages
for the trigger and threshold comparators.
On the zapschematic.gif, if I put a 4.7 uf capacitor (or what size
would you recommend) across the power terminals would that help the
circuit at all?

The 555 draws a very brief spike of current, each time the output
changes state. Any load on the 555 output has to be considered, also.
Your 4.7uF sounds like a fine guess as a bypass capacitor, unless
you find that the circuit is showing mysterious malfunctions.
WOW, wish I had continued to study electronics 20 years ago when I
started... got into software design/prog instead... Never too late I guess!

Not till you are dead.
 
G

Grassy Knollington

Jan 1, 1970
0
No. I am talking about the 3 equal resistors inside the 555 that
divide the supply voltage by 1/3 and 2/3, to act as reference voltages
for the trigger and threshold comparators.

? I've still a lot to learn!
The 555 draws a very brief spike of current, each time the output
changes state. Any load on the 555 output has to be considered, also.
Your 4.7uF sounds like a fine guess as a bypass capacitor, unless
you find that the circuit is showing mysterious malfunctions.

the load is going to be resistance of my body holding two copper tubes
positive to earth... difficult to calculate but could use multimeter?
Not till you are dead.

Long time yet I hope...
 
J

John Popelish

Jan 1, 1970
0
Grassy said:
? I've still a lot to learn!

The 555 tutorial I pointed you toward shows what is inside a 555. It
is essentially a flip flop with a high current output, a reset input,
a second output that can only pull down to the negative supply rail or
turn off (to discharge the timing capacitor, in some configurations,
and two comparators. One sets the flip flop, and one resets it.

A comparator is just a high gain amplifier that switches its output
when the voltage on one of its inputs passes that in the other. In
other words, it compares two voltages and its output tells you which
is more positive. A very common quad comparator is the LM339. Good
for lots of projects.

One comparator has one input connected to the internal divider that
sits at 1/3rd of the supply and the other has one input connected to
2/3rds of the supply. When the voltage on the other input of the
first comparator goes lower than 1/3rd of the supply, the flip flop is
set (output goes high). When the input to the second comparator goes
higher than 2/3rds of the supply voltage the flip flop is reset
(output goes low).
 
i think that capacitor is to reduce ripple on the voltage. Other reason
that i know is to provide some energy if there are energy demand that
come very fast
 
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