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Sound
Joined: Jun 06, 2006
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 Posted: Tue Feb 16, 2010 5:06 am Post subject:
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| I don't get it. Ohms law only says I=V/R. Says that 100Ω resistor will drive 150mA at 15V. independently of the consume of your circuit. |
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fonik
Joined: Jun 07, 2006
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| Posted: Tue Feb 16, 2010 6:15 am Post subject:
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when I is given (0.02A) and resistance is given (100R) you can solve to V (2V).
i thought that's why we want heavy gauge for our distro (low resistance). when current draw changes the voltage won't change that much then. and that's why we use regulated PSUs, which regulate the output, even with changing current draw (different PSUs have different specs for that, some better, some worse)?
maybe i am missing something very important? again, i am acutally interested in the wattage the ferrite/resistor has to dissipate...
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rosch
Joined: Oct 03, 2009
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Location: germany
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| Posted: Tue Feb 16, 2010 6:55 am Post subject:
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hi! i'd like to buy some ferrite beads but can't find them. Reichelt doesn't seem to have them. can someone pleas give me a tip where to look?
thanks |
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Dego
Joined: Apr 22, 2008
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| Posted: Tue Feb 16, 2010 8:51 am Post subject:
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| mouser |
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frijitz
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| Posted: Tue Feb 16, 2010 9:38 am Post subject:
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| fonik wrote: | | ...or am i completely wrong about the ferrite's resistance? |
Ferrite beads don't have any resistance. Where do you get that? They are a piece of heavy wire surrounded by ferrite material.
Ian |
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Sound
Joined: Jun 06, 2006
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| Posted: Tue Feb 16, 2010 10:21 am Post subject:
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Great Ian,  I also thought that they have little resistance, but looking at technical documentation it specifies impedance - in some examples more than 100Ω- but at 100Mhz!!!
http://www.murata.com/products/catalog/pdf/o63e.pdf
Well I expended a little time with this and some experiments - I burnt 3 resistors an one finger- and my conclusion is I should study resistance, impedance, a little more |
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Sound
Joined: Jun 06, 2006
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| Posted: Tue Feb 16, 2010 10:30 am Post subject:
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Well after see that the ferrite bead have resitence at very high frequencies... completely out of the audio range...
Against what exactly they protect? |
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numbertalk
Joined: May 05, 2008
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EdisonRex
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| Posted: Tue Feb 16, 2010 12:05 pm Post subject:
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| Sound wrote: | Well after see that the ferrite bead have resitence at very high frequencies... completely out of the audio range...
Against what exactly they protect? |
Resistance is not exactly impedance. Without going into AC theory in depth, it's easier to think of AC components which ride on DC signals as adding energy to the "signal". What the inductors do is to help to suppress those transients and help to keep the DC stable as seen by the circuit. Where the resistor causes a voltage drop (V=IR) depending on the current drawn by the circuit, and corresponding P=IV such that it acts like a cheap fuse, the inductor will only affect transients. The inductor therefore will not protect your circuit from being plugged into the supply backwards, and will only affect the supply if transients are present on the supply. This is also protection, just not brute force protection.
Remember many transients can happen on the supply. Spikes or oscillations in the supply can and do have an effect on the performance of a circuit (as Ian and others have noted). The inductors can help mitigate that effect. They will not solve all problems, but they can be effective for what they do.
Best protection against miswiring is probably not to miswire... doh!
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fonik
Joined: Jun 07, 2006
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| Posted: Tue Feb 16, 2010 12:18 pm Post subject:
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so i've been correct with voltage drop according ohm's law but what i entirely missed: ferrites don't have any noticeable resistance for DC!
that's great, guys. thank you.
BTW: was my calculation of the wattage correct?
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matthias
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Sound
Joined: Jun 06, 2006
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| Posted: Tue Feb 16, 2010 12:48 pm Post subject:
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Thanks for the link Numbertalk seems interesting Ill check it.
| EdisonRex wrote: |
What the inductors do is to help to suppress those transients and help to keep the DC stable as seen by the circuit. Where the resistor causes a voltage drop (V=IR) depending on the current drawn by the circuit, and corresponding P=IV such that it acts like a cheap fuse, the inductor will only affect transients. The inductor therefore will not protect your circuit from being plugged into the supply backwards, and will only affect the supply if transients are present on the supply. This is also protection, just not brute force protection.
Remember many transients can happen on the supply. Spikes or oscillations in the supply can and do have an effect on the performance of a circuit (as Ian and others have noted). The inductors can help mitigate that effect. They will not solve all problems, but they can be effective for what they do. |
OK, two things:
Referring resistors , I don't see how apply the Ohms law to the voltage drop... How it works? Say you have a module with 40mA consume at the positive rail... and you put a 22Ω resistor at the input... 0.04*22=0.88V you will have 0.88 V at the positive rail? seems more real 0.88V referring voltage drop... but why is the voltage drop? I don't see the sense.
Referring ferrite beads, I understand now, it has sense , they can protect against spikes and fast transients.
But! as I see they will not protect against audio signal in the voltage rail! isn't it? if a oscillator is in the positive or negative rails , the ferrite bead will not attenuate it! |
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frijitz
Joined: May 04, 2007
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| Posted: Tue Feb 16, 2010 1:14 pm Post subject:
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| Sound wrote: | | Well after see that the ferrite bead have resitence at very high frequencies... completely out of the audio range... |
Again, they have reactance (also measured in Ohms) not resistance. They work -- as do the 22R resistors, in conjunction with the caps to ground that follow them. These L-C or R-C networks form a low-pass filter whose purpose is to kill RF spikes on the PS lines.
You really cannot rely on these to remove low frequency signals, because you can not (practically) put in large enough valued caps. The way to avoid low-frequency cross-talk is to avoid ground loops by using proper power supply wiring.
Ian |
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slacker
Joined: Nov 18, 2007
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| Posted: Tue Feb 16, 2010 2:17 pm Post subject:
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| fonik wrote: | s
BTW: was my calculation of the wattage correct? |
Yes it's correct 
| Sound wrote: |
OK, two things:
Referring resistors , I don't see how apply the Ohms law to the voltage drop... How it works? Say you have a module with 40mA consume at the positive rail... and you put a 22Ω resistor at the input... 0.04*22=0.88V you will have 0.88 V at the positive rail? seems more real 0.88V referring voltage drop... but why is the voltage drop? I don't see the sense.
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The resistor is in series with the module, so like you said earlier it's like a voltage divider the resistor is the top of the divider and the module is the bottom. According to Ohms law if the module is drawing 40mA then there must also be 40mA going through the resistor so that gives you 0.88 volts across the resistor.
That means after the resistor you will have 15 - 0.88 = 14.12 volts going to the module. So the resistor is said to drop 0.88 volts. |
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Sound
Joined: Jun 06, 2006
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| Posted: Tue Feb 16, 2010 2:49 pm Post subject:
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| frijitz wrote: | | Sound wrote: | | Well after see that the ferrite bead have resitence at very high frequencies... completely out of the audio range... |
Again, they have reactance (also measured in Ohms) not resistance. They work -- as do the 22R resistors, in conjunction with the caps to ground that follow them. These L-C or R-C networks form a low-pass filter whose purpose is to kill RF spikes on the PS lines.
You really cannot rely on these to remove low frequency signals, because you can not (practically) put in large enough valued caps. The way to avoid low-frequency cross-talk is to avoid ground loops by using proper power supply wiring.
Ian |
Ah Ok Ian, this has sense.
One thing... so the ferrite lead work together with the capacitor... how much the combination ferrite lead/capacitor improves the capacitor alone.
its to big the difference. Could it be achieved increasing the 10μf capacitors?
Do you had any experience on that adding the ferrite lead was a palpable improvement? What it was?
Thaaanks! |
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Sound
Joined: Jun 06, 2006
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| Posted: Tue Feb 16, 2010 3:55 pm Post subject:
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| slacker wrote: |
The resistor is in series with the module, so like you said earlier it's like a voltage divider the resistor is the top of the divider and the module is the bottom. According to Ohms law if the module is drawing 40mA then there must also be 40mA going through the resistor so that gives you 0.88 volts across the resistor.
That means after the resistor you will have 15 - 0.88 = 14.12 volts going to the module. So the resistor is said to drop 0.88 volts. |
mmm,
Slacker, I did some tests,
I have a circuit that consumes 34.1mA in both, positive and negative rails.
I have done some measurements over the positive rail. with different resistor value. Table shows: Resistor value, mA (circuit + resistor), Voltage after the resistor and voltage drop related to the power supply +15V.
A) R=0Ω----34.1mA----15V-----Vdrop=0
B) R=22Ω---33.6mA----14.3V--Vdrop=0.7
C) R=100Ω--32.2 mA---11.8V--Vdrop=3.2
D) R=500Ω--25.6mA----2.1V---Vdrop=12.9
Now I see! so if IR=V for
A) 0*0.0341=0V
B)22*0.0336=0.7392V
C)100*0.0322=3.22V
D)500*0.0256=12.8V
So all coincides! but mA change with the resistor values, so we need to know more data in addition to the mA that consumes the circuit. Is it right?
How could we measure the impedance of this circuit for example? |
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blue hell
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| Posted: Tue Feb 16, 2010 4:18 pm Post subject:
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| Sound wrote: | | How could we measure the impedance of this circuit for example? |
Actually you just did that 
R = V / I and you measured the voltage over the circuit and the current through it ... so just do some math now for all the values.
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Jan
also .. could someone please turn down the thermostat a bit.
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BananaPlug
Joined: Jul 04, 2007
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Location: Philly
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| Posted: Tue Feb 16, 2010 8:59 pm Post subject:
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| rosch wrote: | hi! i'd like to buy some ferrite beads but can't find them. Reichelt doesn't seem to have them. can someone pleas give me a tip where to look?
thanks |
Mouser Part #: 623-2743001112LF |
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scriptstyle
Joined: Jan 22, 2008
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| Posted: Tue Feb 16, 2010 10:37 pm Post subject:
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| while semi on topic, what application do you guys use 0ohm resistors? |
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numbertalk
Joined: May 05, 2008
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Location: Austin, TX
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| Posted: Wed Feb 17, 2010 1:41 am Post subject:
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| scriptstyle wrote: | | while semi on topic, what application do you guys use 0ohm resistors? |
As jumpers. |
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Sound
Joined: Jun 06, 2006
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| Posted: Wed Feb 17, 2010 9:42 am Post subject:
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| Blue Hell wrote: | | Sound wrote: | | How could we measure the impedance of this circuit for example? |
Actually you just did that
R = V / I and you measured the voltage over the circuit and the current through it ... so just do some math now for all the values. |
So, that circuit has an impedance of =~440Ω, so my idea of calculate it as the R2 of a voltage divider does not match the test results... anything is missed.
Very interesting the link of the other thread above.
To me, and after read all this, seems that the better solution, if there is not interference issues is just wire jumpers. Isnt it?
In case of interference issues should be studied the circuit and apply a solution for that particular case. |
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kkissinger
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| Posted: Wed Feb 17, 2010 10:17 am Post subject:
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| scriptstyle wrote: | | while semi on topic, what application do you guys use 0ohm resistors? |
I've created some by accident. 
I've created some 0.0 uf caps by accident, too.
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peng
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| Posted: Wed Feb 17, 2010 1:36 pm Post subject:
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| kkissinger wrote: |
I've created some 0.0 uf caps by accident, too. |
Same here.
Quite dangerous, too!
p. |
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blue hell
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| Posted: Wed Feb 17, 2010 2:14 pm Post subject:
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| Sound wrote: | So, that circuit has an impedance of =~440Ω, so my idea of calculate it as the R2 of a voltage divider does not match the test results... anything is missed.
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Did you calculate it for all cases A, B, C and D?
_________________
Jan
also .. could someone please turn down the thermostat a bit.
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Sound
Joined: Jun 06, 2006
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| Posted: Thu Feb 18, 2010 8:02 am Post subject:
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| Blue Hell wrote: | | Sound wrote: | So, that circuit has an impedance of =~440Ω, so my idea of calculate it as the R2 of a voltage divider does not match the test results... anything is missed.
|
Did you calculate it for all cases A, B, C and D? |
Yes,
If R=V/I, and the circuit V=15, Amp=0.0341 seems that the impedance* of the circuit is 439.88Ω
So if voltage divider is Vout=Vin(R2/R2+R1) the resistor input will be R1 and circuit impedance R2.
B) Vout=15(439.88/439.88+22 )=~14.28V; Difference to test=0.02V
C) Vout=15(439.88/539.88 )=~12.22; Difference to test=0.42V
D) Vout=15(439.88/939.88 )=~7.02V; Difference to test=4.92V
So it doesn't match the test... it was an idea... I have not clear how all this work...impedance, resistance...
But seems that the circuit resistance doesn't go to ground... because maybe is a bipolar circuit?
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blue hell
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| Posted: Thu Feb 18, 2010 10:41 am Post subject:
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what i meant was to calculate the circuit's impedance for all four cases ... you'll get four different values ... meaning that the impedance of the circuit depends on the voltage over it ... and this explains the mismatch you get in your calculation ... in general the impedance of a complex circuit will not be a constant value.
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Jan
also .. could someone please turn down the thermostat a bit.
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