R
ryan wiehle
Guest
denidoank@gmail.com wrote:
http://www.rocktechnology.com/Sale/BlueTooth.htm
here is some more info:
http://www.rocktechnology.com/Sale/BlueTooth.htm
N
Nog
Guest
"davidd31415" <davidd31415@yahoo.com> wrote in message
news:1119043272.326881.115190@o13g2000cwo.googlegroups.com...
It's crap. I like my butane soldering iron. Heats up quick and I can control
the temp.
news:1119043272.326881.115190@o13g2000cwo.googlegroups.com...
It's crap. I like my butane soldering iron. Heats up quick and I can control
the temp.
J
John G
Guest
<thomson.eric@gmail.com> wrote in message
news:1119060802.483102.260530@z14g2000cwz.googlegroups.com...
Why do you want a constant current supply?
Most of the electronics experiments you will encounter will require
constant Voltage and these supplies are every where because everyone
uses them for most everyting.
--
John G
Wot's Your Real Problem?
news:1119060802.483102.260530@z14g2000cwz.googlegroups.com...
Nobody as asked:-
Why do you want a constant current supply?
Most of the electronics experiments you will encounter will require
constant Voltage and these supplies are every where because everyone
uses them for most everyting.
--
John G
Wot's Your Real Problem?
J
John Popelish
Guest
thomson.eric@gmail.com wrote:
http://www.national.com/ds/LM/LM117.pdf
All you need for a variable current regulator is an isolated voltage
source (battery or wall wart) and a variable, low value resistor, or a
selector switch that selects various low value resistors.
The current setpoint is 1.2 volts/resistance. The LM317 may well need
a heat sink for currents above 100 mA. At 1 amp, the resistor will
also produce some significant heat ((1.2 volts^2)/resistance).
regulator LM317:
http://www.national.com/ds/LM/LM117.pdf
All you need for a variable current regulator is an isolated voltage
source (battery or wall wart) and a variable, low value resistor, or a
selector switch that selects various low value resistors.
The current setpoint is 1.2 volts/resistance. The LM317 may well need
a heat sink for currents above 100 mA. At 1 amp, the resistor will
also produce some significant heat ((1.2 volts^2)/resistance).
J
John Popelish
Guest
John G wrote:
of hypothetical circuits that contain voltage sources and current
sources, to demonstrate the analytical principles. If you are an
experimentalist, you need both sources to verify that they are not
lying to you.
Have you ever gone through basic electronics text book? They are full
of hypothetical circuits that contain voltage sources and current
sources, to demonstrate the analytical principles. If you are an
experimentalist, you need both sources to verify that they are not
lying to you.
J
John Fields
Guest
On 18 Jun 2005 03:57:31 -0700, alienonearth@gmail.com wrote:
+-IN>---[1N4148>]-------+--A
| Y--+------A
+--B | EXOR Y-->OUT
| [R] +--B
| | |
[1K] +---+
| |
| [C]
| |
GND>--------------------+--------+
Any gate will work at the junction of the diode and resistor since
its purpose is to provide a totem-pole output to source and sink
current to charge and discharge the cap.
--
John Fields
Professional Circuit Designer
---
+-IN>---[1N4148>]-------+--A
| Y--+------A
+--B | EXOR Y-->OUT
| [R] +--B
| | |
[1K] +---+
| |
| [C]
| |
GND>--------------------+--------+
Any gate will work at the junction of the diode and resistor since
its purpose is to provide a totem-pole output to source and sink
current to charge and discharge the cap.
--
John Fields
Professional Circuit Designer
B
Ban
Guest
alienonearth@gmail.com wrote:
differentiate both outputs, so the pulse length is short compared to the
original frequency. this signal you can feed into two comparators and
or-them by two diodes or another gate.
--
ciao Ban
Bordighera, Italy
You invert the digital pulse train and then with a small capacitor you can
differentiate both outputs, so the pulse length is short compared to the
original frequency. this signal you can feed into two comparators and
or-them by two diodes or another gate.
--
ciao Ban
Bordighera, Italy
A
Andrew Holme
Guest
redhat wrote:
capacitance of varactor D1.
First, you need to know the capacitance of D1. This varies with the DC bias
voltage. In your case, the bias voltage is Vcc. There should be a graph of
capacitance vs voltage on the varactor data sheet.
You can then combine the varactor capacitance with that of C5 using the
equation for capacitors in series:
C = (C1*C2) / (C1+C2)
The resonant frequency of the LC circuit can then be found from :
2*pi*f = 1 / sqrt(LC)
You have a series LC circuit with inductor L, capacitor C5, and the
capacitance of varactor D1.
First, you need to know the capacitance of D1. This varies with the DC bias
voltage. In your case, the bias voltage is Vcc. There should be a graph of
capacitance vs voltage on the varactor data sheet.
You can then combine the varactor capacitance with that of C5 using the
equation for capacitors in series:
C = (C1*C2) / (C1+C2)
The resonant frequency of the LC circuit can then be found from :
2*pi*f = 1 / sqrt(LC)
J
John Popelish
Guest
redhat wrote:
resonance worth mentioning. if the output is a short to ground, the
resonant frequency is the series equivalent capacitance of the
varactor and the variable capacitor with the inductor w=1/(sqrt(L*C)).
If the output has other impedance, there is a different solution.
indeterminate. I(f the output is open circuit, the circuit has no
resonance worth mentioning. if the output is a short to ground, the
resonant frequency is the series equivalent capacitance of the
varactor and the variable capacitor with the inductor w=1/(sqrt(L*C)).
If the output has other impedance, there is a different solution.
A
Andrew Holme
Guest
redhat wrote:
Please post the complete circuit.
J
John Popelish
Guest
redhat wrote:
The Vcc bias alters the capacitance of the varactor. It also makes
sure that the varactor stays reverse biased during the signal cycle,
so that the device always looks like a capacitor. If the varactor
ever becomes forward biased, it acts like a low value resistor in
parallel with any junction capacitance.
The Vcc bias alters the capacitance of the varactor. It also makes
sure that the varactor stays reverse biased during the signal cycle,
so that the device always looks like a capacitor. If the varactor
ever becomes forward biased, it acts like a low value resistor in
parallel with any junction capacitance.
D
Don Bowey
Guest
On 6/19/05 8:48 AM, in article
1119196136.998452.306600@g43g2000cwa.googlegroups.com, "redhat"
<aezzat1@gmail.com> wrote:
the part number you can obtain that information. Since it is the FM
"modulator," off-hand I think it's capacitance value will not affect the at
rest frequency more than a couple hundred kHz.
Don
1119196136.998452.306600@g43g2000cwa.googlegroups.com, "redhat"
<aezzat1@gmail.com> wrote:
The vericap is spec'd with a plot of capacitance:voltage. If you can find
the part number you can obtain that information. Since it is the FM
"modulator," off-hand I think it's capacitance value will not affect the at
rest frequency more than a couple hundred kHz.
Don
L
Lord Garth
Guest
"Tom MacIntyre" <tom__macintyre@hotmail.com> wrote in message
news:cq9bb1hvmn87jgnu49lec1jnittc3dumfh@4ax.com...
<snip>
The equivalent is misandrist (a person who hates persons of the male sex), a
rare word but seemingly much sought-after. The corresponding noun for the
attitude is misandry.
They also spell the above word 'misogynist'.
news:cq9bb1hvmn87jgnu49lec1jnittc3dumfh@4ax.com...
<snip>
Askoxford says this:
The equivalent is misandrist (a person who hates persons of the male sex), a
rare word but seemingly much sought-after. The corresponding noun for the
attitude is misandry.
They also spell the above word 'misogynist'.
L
Larry Brasfield
Guest
<kamizu83@hotmail.com> wrote in message
news:1119199598.178755.153170@g14g2000cwa.googlegroups.com...
from its current Y form to the equivalent delta form. See
http://www.answers.com/topic/y-delta-transform
Once that is done, the topology can be step-wise converted
using the rules you should know.
R1 R3 R4 +
R2 R3 R4 +
R1 R2 R5 +
R1 R3 R5 +
R2 R3 R5 +
R1 R4 R5 +
R2 R4 R5)
/
(R1 R2 +
R1 R3 +
R2 R3 +
R2 R4 +
R3 R4 +
R1 R5 +
R3 R5 +
R4 R5)
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
news:1119199598.178755.153170@g14g2000cwa.googlegroups.com...
One way to solve it would be to convert the R1,R2,R3 set
from its current Y form to the equivalent delta form. See
http://www.answers.com/topic/y-delta-transform
Once that is done, the topology can be step-wise converted
using the rules you should know.
(R1 R2 R4 +
R1 R3 R4 +
R2 R3 R4 +
R1 R2 R5 +
R1 R3 R5 +
R2 R3 R5 +
R1 R4 R5 +
R2 R4 R5)
/
(R1 R2 +
R1 R3 +
R2 R3 +
R2 R4 +
R3 R4 +
R1 R5 +
R3 R5 +
R4 R5)
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
T
Tom MacIntyre
Guest
On Sun, 19 Jun 2005 17:41:09 GMT, "Lord Garth" <LGarth@Tantalus.net>
wrote:
Tom
wrote:
But of course...
Tom
A
Andrew Holme
Guest
redhat wrote:
I'm forced to assume the modulating input is coupled via a large resistor or
a radio-frequency choke.
I'm also assuming it's AC-coupled.
Correct?
Then the average DC voltage across D1 will be Vcc - get the capacitance from
the datasheet for this voltage. It's probably in the region of a few pF.
The series resonant frequency of D1, L1, C5 can then be found.
I*R3 is negligible.
BTW Is that supposed to be an inductor in Q1 collector?
You could have included a bit more to the left!
I'm forced to assume the modulating input is coupled via a large resistor or
a radio-frequency choke.
I'm also assuming it's AC-coupled.
Correct?
Then the average DC voltage across D1 will be Vcc - get the capacitance from
the datasheet for this voltage. It's probably in the region of a few pF.
The series resonant frequency of D1, L1, C5 can then be found.
I*R3 is negligible.
BTW Is that supposed to be an inductor in Q1 collector?
T
Tom Biasi
Guest
"davidd31415" <davidd31415@yahoo.com> wrote in message
news:1119055855.222801.231650@g47g2000cwa.googlegroups.com...
would be working with sensitive components wouldn't use the thing in the
first place.
Tom
news:1119055855.222801.231650@g47g2000cwa.googlegroups.com...
Maybe you can't find anyone who has damaged components because anyone who
would be working with sensitive components wouldn't use the thing in the
first place.
Tom
J
Jamie
Guest
redhat wrote:
J
Jonathan Kirwan
Guest
On 19 Jun 2005 09:46:38 -0700, kamizu83@hotmail.com wrote:
no current flows through R3 and you can eliminate it from your
analysis.
But don't be too hard on yourself if they aren't equal, as this is the
ever-popular unbalanced Wheatstone bridge problem and it takes an
amount of algebra to solve it via simultaneous equations and a branch
current analysis. (Or you can use mesh analysis, which is perhaps a
little easier, algebraically speaking.)
Here is how a spice program might solve the problem. First, spice
uses conductances for the resistors, instead of resistance values.
(Which is a reason why spice hates resistances of 0 ohms.)
So it would do something like:
Treat the resistances as conductances,
G1 = 1/R1
G2 = 1/R2
G3 = 1/R3
G4 = 1/R4
G5 = 1/R5
Set known voltages,
V(1) = V
V(4) = 0 (Spice needs a ground ref, so it might as well be here.)
Set up an equation for the rest (KCL/KVL):
V(2)*(G1+G2+G3)-V(3)*G3-V(1)*G1-V(4)*G2 = 0
V(3)*(G3+G4+G5)-V(2)*G3-V(1)*G4-V(4)*G5 = 0
One way of viewing the above setup for these two "= 0" equations (and
it's only one of the ways, but I think it may help here) is to look at
each node like this: Current _spills_ away from a node based on the
voltage at the node and each of the conductances away from it and
current _spills_ back into a node based on the voltages at the nearby
nodes and the conductances back in.
So when I wrote:
V(2)*(G1+G2+G3)-V(3)*G3-V(1)*G1-V(4)*G2 = 0
I was thinking,
"Hmm. V(2) will spill current away based on the sum of the
conductances leaving the node. This is G1, G2, and G3, so the current
spilling away is simply V(2)*(G1+G2+G3). But, current is spilling
into the node based on the three different voltages at the nearby
nodes coming back through these same conductances, so this amount will
be of opposite sign and will be -V(1)*G1, -(V3)*G3, and -V(4)*G2.
Since the total current arriving into a node must be the same as the
total current leaving it, the sum of these two must be zero -- or else
we are in for big trouble!"
Similar thinking gets you the results for V(3), too. Later on, I'll
apply this thinking to the node at V(1) [V(4) would work, as well] in
order to compute the total current.
Okay, back to reality. Since V(4)=0 and V(1)=V, the above reduces
slightly to:
V(2)*(G1+G2+G3)-V(3)*G3-V*G1 = 0
V(3)*(G3+G4+G5)-V(2)*G3-V*G4 = 0
In the above pair, you have two unknowns. These are V(2) and V(3).
You also have two equations, luckily. So let's rewrite them into
slightly different form which calls out the constants a little more
clearly:
V(2)*[G1+G2+G3] + V(3)*[-G3] = [V*G1]
V(2)*[-G3] + V(3)*[G3+G4+G5] = [V*G4]
which can be rewritten as:
a*x + b*y = c
d*x + e*y = f
where x is V(2) and y is V(3) and with the obvious substitutions for
a, b, c, d, e, and f.
The solution to such a pair of equations can be done through simple
algebraic manipulation, but it is often shown through setting up a
basic matrix form because it is easy to visualize without getting too
caught up in the detailed manipulations:
[ a b ] [x] [c]
[ ] * [ ] = [ ]
[ d e ] [y] [f]
The solution is, of course:
[ a b ]-1 [c] [x]
[ ] * [ ] = [ ]
[ d e ] [f] [y]
So you just need to compute:
c*e - f*d (G1*(G3+G4+G5) + G3*G4)
V(2) = x = --------- = V * ----------------------------
a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2
and,
f*a - c*d (G4*(G1+G2+G3) + G3*G1)
V(3) = y = --------- = V * ----------------------------
a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2
To solve for the voltages at the two nodes. After you have the
voltages, you should be able to find all the currents, of course.
You can look up matrix solutions in most any pre-calculus (preparation
for linear systems) or decent algebra math book (solving intersections
of two different lines is a very common task in algebra 2, I think.)
Or you can work through the solution of a two-line intersection
problem, keeping the terms abstract, and see it just as clearly.
(You might also want to visualize the actual "lines" involved and how
they relate back to the diagram, but that's for another time.)
Anyway, what about the total current through the system? Well, you
can apply the same reasoning I was applying before, to think like
this:
"The total current flowing into node V(1) from the voltage source
would give me what I need to know. So let's do up an equation for
that node. The total current spilling into the node from the voltage
source must be equal to the net of what happens when I only consider
the other connections. So, current spilling out through R1 and R4 are
simply V(1)*(G1+G4) and spilling back in through those resistances
must be -V(2)*G1 and -V(3)*G4. What arrives through the only other
connection, which is from the voltage source, must be this amount.
So: I(total) = V(1)*(G1+G4) - V(2)*G1 - V(3)*G4"
Since you already now have V(2) and V(3), computing this becomes easy.
Let's select some exact values for the resistors.
R1 = 1200
R2 = 2700
R3 = 330
R4 = 1800
R5 = 680
and,
V = V(1) = 12V.
If you do your equations as I did (and if I didn't get all of this
dead wrong), it should result in:
V(2) = 5.7881 V
V(3) = 4.7872 V
And the total voltage source current is 9.184 mA. Net resistance of
the bridge is about 1306.66 ohms.
Jon
Assume:
If the ratios of R1/R2 and R4/R5 are equal, it becomes trivial because
no current flows through R3 and you can eliminate it from your
analysis.
But don't be too hard on yourself if they aren't equal, as this is the
ever-popular unbalanced Wheatstone bridge problem and it takes an
amount of algebra to solve it via simultaneous equations and a branch
current analysis. (Or you can use mesh analysis, which is perhaps a
little easier, algebraically speaking.)
Here is how a spice program might solve the problem. First, spice
uses conductances for the resistors, instead of resistance values.
(Which is a reason why spice hates resistances of 0 ohms.)
So it would do something like:
Treat the resistances as conductances,
G1 = 1/R1
G2 = 1/R2
G3 = 1/R3
G4 = 1/R4
G5 = 1/R5
Set known voltages,
V(1) = V
V(4) = 0 (Spice needs a ground ref, so it might as well be here.)
Set up an equation for the rest (KCL/KVL):
V(2)*(G1+G2+G3)-V(3)*G3-V(1)*G1-V(4)*G2 = 0
V(3)*(G3+G4+G5)-V(2)*G3-V(1)*G4-V(4)*G5 = 0
One way of viewing the above setup for these two "= 0" equations (and
it's only one of the ways, but I think it may help here) is to look at
each node like this: Current _spills_ away from a node based on the
voltage at the node and each of the conductances away from it and
current _spills_ back into a node based on the voltages at the nearby
nodes and the conductances back in.
So when I wrote:
V(2)*(G1+G2+G3)-V(3)*G3-V(1)*G1-V(4)*G2 = 0
I was thinking,
"Hmm. V(2) will spill current away based on the sum of the
conductances leaving the node. This is G1, G2, and G3, so the current
spilling away is simply V(2)*(G1+G2+G3). But, current is spilling
into the node based on the three different voltages at the nearby
nodes coming back through these same conductances, so this amount will
be of opposite sign and will be -V(1)*G1, -(V3)*G3, and -V(4)*G2.
Since the total current arriving into a node must be the same as the
total current leaving it, the sum of these two must be zero -- or else
we are in for big trouble!"
Similar thinking gets you the results for V(3), too. Later on, I'll
apply this thinking to the node at V(1) [V(4) would work, as well] in
order to compute the total current.
Okay, back to reality. Since V(4)=0 and V(1)=V, the above reduces
slightly to:
V(2)*(G1+G2+G3)-V(3)*G3-V*G1 = 0
V(3)*(G3+G4+G5)-V(2)*G3-V*G4 = 0
In the above pair, you have two unknowns. These are V(2) and V(3).
You also have two equations, luckily. So let's rewrite them into
slightly different form which calls out the constants a little more
clearly:
V(2)*[G1+G2+G3] + V(3)*[-G3] = [V*G1]
V(2)*[-G3] + V(3)*[G3+G4+G5] = [V*G4]
which can be rewritten as:
a*x + b*y = c
d*x + e*y = f
where x is V(2) and y is V(3) and with the obvious substitutions for
a, b, c, d, e, and f.
The solution to such a pair of equations can be done through simple
algebraic manipulation, but it is often shown through setting up a
basic matrix form because it is easy to visualize without getting too
caught up in the detailed manipulations:
[ a b ] [x] [c]
[ ] * [ ] = [ ]
[ d e ] [y] [f]
The solution is, of course:
[ a b ]-1 [c] [x]
[ ] * [ ] = [ ]
[ d e ] [f] [y]
So you just need to compute:
c*e - f*d (G1*(G3+G4+G5) + G3*G4)
V(2) = x = --------- = V * ----------------------------
a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2
and,
f*a - c*d (G4*(G1+G2+G3) + G3*G1)
V(3) = y = --------- = V * ----------------------------
a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2
To solve for the voltages at the two nodes. After you have the
voltages, you should be able to find all the currents, of course.
You can look up matrix solutions in most any pre-calculus (preparation
for linear systems) or decent algebra math book (solving intersections
of two different lines is a very common task in algebra 2, I think.)
Or you can work through the solution of a two-line intersection
problem, keeping the terms abstract, and see it just as clearly.
(You might also want to visualize the actual "lines" involved and how
they relate back to the diagram, but that's for another time.)
Anyway, what about the total current through the system? Well, you
can apply the same reasoning I was applying before, to think like
this:
"The total current flowing into node V(1) from the voltage source
would give me what I need to know. So let's do up an equation for
that node. The total current spilling into the node from the voltage
source must be equal to the net of what happens when I only consider
the other connections. So, current spilling out through R1 and R4 are
simply V(1)*(G1+G4) and spilling back in through those resistances
must be -V(2)*G1 and -V(3)*G4. What arrives through the only other
connection, which is from the voltage source, must be this amount.
So: I(total) = V(1)*(G1+G4) - V(2)*G1 - V(3)*G4"
Since you already now have V(2) and V(3), computing this becomes easy.
Let's select some exact values for the resistors.
R1 = 1200
R2 = 2700
R3 = 330
R4 = 1800
R5 = 680
and,
V = V(1) = 12V.
If you do your equations as I did (and if I didn't get all of this
dead wrong), it should result in:
V(2) = 5.7881 V
V(3) = 4.7872 V
And the total voltage source current is 9.184 mA. Net resistance of
the bridge is about 1306.66 ohms.
Jon
J
Jonathan Kirwan
Guest
On Sun, 19 Jun 2005 19:53:44 GMT, Jonathan Kirwan
<jkirwan@easystreet.com> wrote:
c*e - f*b (G1*(G3+G4+G5) + G3*G4)
V(2) = x = --------- = V * ----------------------------
a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2
Typo. In this case, it makes no difference in the calculation as 'b'
and 'd' are the same variable and sign.
Jon
<jkirwan@easystreet.com> wrote:
Correction. This is:
c*e - f*b (G1*(G3+G4+G5) + G3*G4)
V(2) = x = --------- = V * ----------------------------
a*e - b*d (G1+G2+G3)*(G3+G4+G5) - G3^2
Typo. In this case, it makes no difference in the calculation as 'b'
and 'd' are the same variable and sign.
Jon