li_gangyi wrote:
Hey hey~! I'm not a Mr. yet~! at least in a "lawful" manner coz I'm 14...and
yeah...I have messed with some filters...but I guess it's my component
values that need changing...coz they dun cut at the frequency that I
need...and I dun use transistors...only passive...meaning I've got no
boost...right?? Can you point me in the right direction to build a
simple...filter...

Well, the simplest (first order) filter is one resistor and one
capacitor. The resistor is in series, the capacitor is shunted to
ground. Like this (you'll need to have a fixed-width font to see this
properly):

o--------/\/\/\/\-------+------o
R |
|
input ----- output
-----
C |
|
o-----------------------+------o

The value of R should be at least an order of magnitude lower than the
impedance at the output (in other words, in parallel with C). For most
tube amps this will be quite high, 100k to 1M or so, usually the volume
control pot. So, for example, if it's 500k, you'd want to choose R to be
around 50k. Let's choose 47k.

Then choose your corner frequency. Let's say you want the rolloff to
start at 50 Hz. Multiply the frequency F by the resistance R, and divide
that into 159,000 for the value of C in microfarads. In our example,
47,000 * 50 = 2,350,000. Divide that into 159,000 and we get about 0.068 uF.

The formula is derived from the formula for capacitive reactance. The 3
dB corner will be that frequency at which the reactance equals the
resistance R. It's given by the equation:

F=1/(2*Pi*R*C) with R in ohms, C in farads, F in Hertz, Pi ~ 3.1416.

Cheers,
Fred
--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects, Vacuum Tubes & other stuff: |
| http://www.dogstar.dantimax.dk |
+--------------------------------------------+

Fred Nachbaur wrote:

li_gangyi wrote:
Hey hey~! I'm not a Mr. yet~! at least in a "lawful" manner coz I'm 14...and
yeah...I have messed with some filters...but I guess it's my component
values that need changing...coz they dun cut at the frequency that I
need...and I dun use transistors...only passive...meaning I've got no
boost...right?? Can you point me in the right direction to build a
simple...filter...

Well, the simplest (first order) filter is one resistor and one
capacitor. The resistor is in series, the capacitor is shunted to
ground. Like this (you'll need to have a fixed-width font to see this
properly):

o--------/\/\/\/\-------+------o
R |
|
input ----- output
-----
C |
|
o-----------------------+------o

The value of R should be at least an order of magnitude lower than the
impedance at the output (in other words, in parallel with C). For most
tube amps this will be quite high, 100k to 1M or so, usually the volume
control pot. So, for example, if it's 500k, you'd want to choose R to be
around 50k. Let's choose 47k.

Then choose your corner frequency. Let's say you want the rolloff to
start at 50 Hz. Multiply the frequency F by the resistance R, and divide
that into 159,000 for the value of C in microfarads. In our example,
47,000 * 50 = 2,350,000. Divide that into 159,000 and we get about 0.068 uF.

The formula is derived from the formula for capacitive reactance. The 3
dB corner will be that frequency at which the reactance equals the
resistance R. It's given by the equation:

F=1/(2*Pi*R*C) with R in ohms, C in farads, F in Hertz, Pi ~ 3.1416.

Cheers,
Fred

Thanks Fred, you have laid out a splendid path for our lad to tread,
I hope he steps carefully, and asks many questions on the way;
if not, he isn't learning much.
A few old electronics text books might help him along some more, and the answers
to his questions are in them, which might make our lives easier.

Patrick Turner.

--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects, Vacuum Tubes & other stuff: |
| http://www.dogstar.dantimax.dk |
+--------------------------------------------+

Patrick Turner wrote:

Fred Nachbaur wrote:


li_gangyi wrote:

Hey hey~! I'm not a Mr. yet~! at least in a "lawful" manner coz I'm 14...and
yeah...I have messed with some filters...but I guess it's my component
values that need changing...coz they dun cut at the frequency that I
need...and I dun use transistors...only passive...meaning I've got no
boost...right?? Can you point me in the right direction to build a
simple...filter...

Well, the simplest (first order) filter is one resistor and one
capacitor. The resistor is in series, the capacitor is shunted to
ground. Like this (you'll need to have a fixed-width font to see this
properly):

o--------/\/\/\/\-------+------o
R |
|
input ----- output
-----
C |
|
o-----------------------+------o

The value of R should be at least an order of magnitude lower than the
impedance at the output (in other words, in parallel with C). For most
tube amps this will be quite high, 100k to 1M or so, usually the volume
control pot. So, for example, if it's 500k, you'd want to choose R to be
around 50k. Let's choose 47k.

Then choose your corner frequency. Let's say you want the rolloff to
start at 50 Hz. Multiply the frequency F by the resistance R, and divide
that into 159,000 for the value of C in microfarads. In our example,
47,000 * 50 = 2,350,000. Divide that into 159,000 and we get about 0.068 uF.

The formula is derived from the formula for capacitive reactance. The 3
dB corner will be that frequency at which the reactance equals the
resistance R. It's given by the equation:

F=1/(2*Pi*R*C) with R in ohms, C in farads, F in Hertz, Pi ~ 3.1416.

Cheers,
Fred


Thanks Fred, you have laid out a splendid path for our lad to tread,
I hope he steps carefully, and asks many questions on the way;
if not, he isn't learning much.
A few old electronics text books might help him along some more, and the answers
to his questions are in them, which might make our lives easier.

Patrick Turner.


Thanks, Patrick. Yes, there's nothing better than learning the basics
well, and thinking things out oneself. It's nice to have a starting
point though.

You gave some good tips in your reply also, and in reading it I realized
that I should have mentioned the point about having to have a low input
impedance into the filter. Furthermore, the input impedance may have to
be accounted for in choosing the value for "R".

For example, if the input to the filter is the output of a typical CD
player, the impedance is usually on the order of a thousand ohms or so,
and therefore can be safely ignored. However, if it's the output of a
tube preamp, it might be considerably higher; say, on the order of
10,000 ohms or maybe even more. In this case, this value has to be
subtracted from the calculated value of "R" in order for the filter to
work as expected.

Cheers to all RATs, young, old, and inbetween,
Fred
--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects, Vacuum Tubes & other stuff: |
| http://www.dogstar.dantimax.dk |
+--------------------------------------------+

Fred Nachbaur wrote:

Patrick Turner wrote:

Fred Nachbaur wrote:


li_gangyi wrote:

Hey hey~! I'm not a Mr. yet~! at least in a "lawful" manner coz I'm 14...and
yeah...I have messed with some filters...but I guess it's my component
values that need changing...coz they dun cut at the frequency that I
need...and I dun use transistors...only passive...meaning I've got no
boost...right?? Can you point me in the right direction to build a
simple...filter...

Well, the simplest (first order) filter is one resistor and one
capacitor. The resistor is in series, the capacitor is shunted to
ground. Like this (you'll need to have a fixed-width font to see this
properly):

o--------/\/\/\/\-------+------o
R |
|
input ----- output
-----
C |
|
o-----------------------+------o

The value of R should be at least an order of magnitude lower than the
impedance at the output (in other words, in parallel with C). For most
tube amps this will be quite high, 100k to 1M or so, usually the volume
control pot. So, for example, if it's 500k, you'd want to choose R to be
around 50k. Let's choose 47k.

Then choose your corner frequency. Let's say you want the rolloff to
start at 50 Hz. Multiply the frequency F by the resistance R, and divide
that into 159,000 for the value of C in microfarads. In our example,
47,000 * 50 = 2,350,000. Divide that into 159,000 and we get about 0.068 uF.

The formula is derived from the formula for capacitive reactance. The 3
dB corner will be that frequency at which the reactance equals the
resistance R. It's given by the equation:

F=1/(2*Pi*R*C) with R in ohms, C in farads, F in Hertz, Pi ~ 3.1416.

Cheers,
Fred


Thanks Fred, you have laid out a splendid path for our lad to tread,
I hope he steps carefully, and asks many questions on the way;
if not, he isn't learning much.
A few old electronics text books might help him along some more, and the answers
to his questions are in them, which might make our lives easier.

Patrick Turner.


Thanks, Patrick. Yes, there's nothing better than learning the basics
well, and thinking things out oneself. It's nice to have a starting
point though.

You gave some good tips in your reply also, and in reading it I realized
that I should have mentioned the point about having to have a low input
impedance into the filter. Furthermore, the input impedance may have to
be accounted for in choosing the value for "R".

Indeed, but when he starts using equations, and working out what
a filter response should be, and he doesn't mesure that, then he should ask, why
is this not working out? where is the extra resistance?
I asked questions like this, and went looking, and then more questions,
and eventually, I got to bed at 5 in the morning.



For example, if the input to the filter is the output of a typical CD
player, the impedance is usually on the order of a thousand ohms or so,
and therefore can be safely ignored.

Indeed.
Usually if a signal source impedance is 1/10 of the minimum filter input
impedance, then the results measured about tally with results calculated.

However, if it's the output of a
tube preamp, it might be considerably higher; say, on the order of
10,000 ohms or maybe even more. In this case, this value has to be
subtracted from the calculated value of "R" in order for the filter to
work as expected.

Indeed.
You can't tell the youngans everything, you gotta let them
find some things out, their empty minds are easily filled,
but not so common is a burning passion to learn.



Cheers to all RATs, young, old, and inbetween,
Fred

Indeed,

Patrick Turner.


--
+--------------------------------------------+
| Music: http://www3.telus.net/dogstarmusic/ |
| Projects, Vacuum Tubes & other stuff: |
| http://www.dogstar.dantimax.dk |
+--------------------------------------------+