John Curl's Blowtorch preamplifier part III

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Max, by pity, it does not make any sense.
Listen. I am one of the most awful subjectivists. The sworn enemy of the objectivists of this forum who dream to send me in the fires of hell.
Imagine that, from time to time, I even chose circuits that measure less well because I prefer the way they sound.

But if there is one thing that I do not forget, it is to measure first, to be wary of my perceptions, and, above all, never to draw hasty conclusions, acrobatic generalisation, or to make far-fetched assumptions about things that I do not understand.
Wouldn't that make you an objectivist? :scratch:
 
It's known as mechanically breaking through a previously oxidized layer. Do you actually believe it's something else? Like working a bad switch back and forth, I have an old DVM that does this. I thought you were taking a break from picking at scabs. Maybe we should revisit your battery that creates charge out of one terminal or some of your other fantasies. Battery, switch, transmission line, yes when you close the switch there is instantaneously current in both terminals of the battery. You can measure it, it's easy.

Scott,

In the past you have maintained that it required more energy than that to clear a reluctant contact. Or do you think the mechanical energy travels the length of the microphone cord to where the bad contact may reside? Having fixed more than a few of these it is often the wall jack connection that gets replaced.

Now to properly misinterpret you, are you really claiming current In a closed circuit is faster than light?
 
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Now to properly misinterpret you, are you really claiming current In a closed circuit is faster than light?

Are you serious, jn and I have already gone over this with you months (years?) ago? Electrons (current) do not travel at c (or 2/3c) in a wire. This problem is in every entry level EM text book. Do you need some pictures?

Ignoring minute issues with losses a 100m of RG58 terminated in 50 Ohms is indistinguishable from a 50 Ohm resistor you connect a 1V source and you have 20mA flowing at t = 0+. Why don't you borrow a fast current probe and see for yourself?
 
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What does full wave rectification have to do with it? Are you trying to use a bog standard radio shack meter to measure wire?
Try an AP. I suspect they didn't use a full wave bridge at their inputs.
I mean to measure the noise of each polarity half wave separately, not combined/averaged as in most metering instruments, capice ?.

Dan.

OK, next. The above makes no physical sense at all.
As per above post, look for different current noise according to current direction.

Dan.
 
Are you serious, jn and I have already gone over this with you months (years?) ago? Electrons (current) do not travel at c (or 2/3c) in a wire. This problem is in every entry level EM text book. Do you need some pictures?

Ignoring minute issues with losses a 100m of RG58 terminated in 50 Ohms is indistinguishable from a 50 Ohm resistor you connect a 1V source and you have 20mA flowing at t = 0+. Why don't you borrow a fast current probe and see for yourself?

Current flow is not the same as electron motion (drift), the energy propagates as a wave, energy is voltage and current, why do you keep confusing this?

Have you ever used a Time Domain Reflectometer?
 
I am saying that it seems that there is difference in current noise spectrum according to current direction.

Okay, suppose you take a wire, fold it in half, and cut at the fold. Now there are 2 wires each pointing in the, shall we say, opposite noise direction. Solder them together like that in parallel. Does the directionality of the parallel structure then cancel out?
 
Current flow is not the same as electron motion (drift), the energy propagates as a wave, energy is voltage and current, why do you keep confusing this?

Have you ever used a Time Domain Reflectometer?

Yes, why do you keep this up? Why don't you answer a simple question, battery, switch, and transmission line terminated in its characteristic impedance? Close the switch the current flows out one terminal of the battery and in the other immediately anything else violates conservation of charge. Why do you keep saying this has anything to do with super-luminal propagation?

BTW it is perfectly normal to refer to the DC current in a circuit as flowing, and as we showed ages ago both ways of looking at these problems (lumped or T-line) converge on the same answer because they have to since the both conserve energy and charge.

BTW you're the confused one, I'm too lazy to go back to when you admitted some ancient thesis you read was wrong all along. You forget easily that you claimed closing the switch had charge going out one terminal and there was a round trip delay before it started going into the other terminal.
 
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Ah no.

So let us start at improving our communications.

I use voltage to mean potential or Electromotive Force (EMF). It can be direct current or alternating current, but I will skip the AC bits for now.

Current is measured in amperes (amps) and is related to the voltage and the load resistance. In an ohmic load the current is proportional to the load resistance. There are loads that are not ohmic or linear and often someone may refer to a load that draws one amp at one volt as a one ohm load even if at 2 volts it draws 10 amps. Precision would call this a nonohmic load that behaves as....

Even gizmos sold as resistors and operated with their ratings limits are not trully ohmic. The error may be as high as parts per thousand or as low as parts per billion. But nothing is poifect.

Power measured in watts is the current draw times the voltage potential.

So I refer to current not electron flow or drift and voltage as a level not a flow.

Now as to a TDR measuring a coaxial cable;

If I take a length of 50 ohm coaxial cable and leave the far end open I will see a reflection pulse at some time we can call T. On my system the pulse will be a positive voltage.

If I short the far end I will at time T see a negative pulse.

If I then place a perfect variable resistor of maximum 100 ohms at the end of the cable, when it is set for 100 ohms I will see a positive spike of lesser voltage at time T.

As I reduce the load the spike will decrease until there is a minimum at 50 ohms. As I go below 50 ohms the spike will go negative and increase in level.

The time required will always be T.

Now Scott are you telling me at 50 ohms the time for the power of the pulse to reach the resistor is zero?
 
Scott and Simon, I suggest that you enter the war hatchet.
It's sad, though, that two high-value partners are torn apart by virtual fantasies on the other's image.

I believe you meant to say "bury the hatchet"...as in peace.

Jn

I am saying that it seems that there is difference in current noise spectrum according to current direction.
I do not know of any suitable instruments that are able to discriminate according to signal polarity.


Dan.
Without any test equipment capable of this measurement at your disposal, how did you decide that this was happening?

Jn

Have you ever used a Time Domain Reflectometer?
I use one all the time, it's when you put a mirror behind a clock movement to see the clock guts you can't look at from the front of the case.

Now, give me a hard question, that was a lob directly over the plate..just too easy.

Jn
 
Ah no.

So let us start at improving our communications.


Now Scott are you telling me at 50 ohms the time for the power of the pulse to reach the resistor is zero?

What did I ever say to give you that idea? Lets start over. You take a battery, switch, ideal transmission line (length l or time T) terminated in its characteristic impedance.

You close the switch and launch a TEM wave but at t = 0+ the line also looks to the battery like a plain old resistor so immediately the poor slow drifty electrons flow through the battery just like there was a resistor there. So the battery immediately supplies energy to the system. As the wave travels down the line V and I always satisfy the characteristic impedance equation so as the wave passes it kicks those electrons to join. In this case when the wave gets to the end the story is over and all the power is now delivered to the load.

But what happened to the power that the battery delivered for time T? It's stored in the line because the bulk C is now charged to V and the L has current I. I think that you will find that if you take the parameters of any ordinary coax that the numbers work out.

Again what did I say that made you think the power was delivered to the load instantaneously? I simply said when you close the switch the battery sees current in both leads and delivers energy to the system immediately. This has to be so.
 
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Scott

When you throw a pebble into a pond, the water was there before and after you threw it. The ripples of the wave still take time to reach the shore. Apparently we agree on that.

Now why are you mixing RLC in?

With a TDR when the line is shorted why does the pulse go negative? Does the time required reflect the length of the cable slowed by the propagation velocity or twice that?

Don't get excited this is just a response to your repeating issues you took your own interpretation on before.

The issue we started with was tapping a microphone to start it working. You then got off to current flow. It is now the wave model of signal propagation and mirroring.

JN you must like taking big risks. When you place the mirror behind the clock, if you are not careful you might see sonething terrifying!
 
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