BTW, what are you using to log the data and make the graphs?
It would be cool to graph out the temperature characteristics vs. power consumption of the huge peltier-cooled cooler I'm building.. lol
It would be cool to graph out the temperature characteristics vs. power consumption of the huge peltier-cooled cooler I'm building.. lol
On graph #3:
Now it looks like the resistance is more like a linear constant again.
Except after it is exposed to higher voltages it has a higher resistance by the time it returns to 300. (I'm assuming the path goes from top to bottom instead of bottom to top.)
It seems there is confusion in the term "negative resistance" here. We're really talking about a change in the value of the resistance. In ideal resistors, it should be constant and unchanging, but we are seeing the resistance change at high voltages.... This latest graph seems to confirm that the changes are being caused by heat and changes in the physical structure/properties of the graphite.
Now it looks like the resistance is more like a linear constant again.
Except after it is exposed to higher voltages it has a higher resistance by the time it returns to 300. (I'm assuming the path goes from top to bottom instead of bottom to top.)
It seems there is confusion in the term "negative resistance" here. We're really talking about a change in the value of the resistance. In ideal resistors, it should be constant and unchanging, but we are seeing the resistance change at high voltages.... This latest graph seems to confirm that the changes are being caused by heat and changes in the physical structure/properties of the graphite.
- Dec 9, 2000
- 16,528
- 4
- 0
They gave me a program called Test Point. It's like Visual Basic for digital sourcemeters. I program the machine through a serial cable, tell it what I want, it takes the readings and sends them back to me and I auto export them to Excel. It's neat, but the sourcemeter costs $5,000, so I can't have one at home.
Ouch.Originally posted by: JohnCU
They gave me a program called Test Point. It's like Visual Basic for digital sourcemeters. I program the machine through a serial cable, tell it what I want, it takes the readings and sends them back to me and I auto export them to Excel. It's neat, but the sourcemeter costs $5,000, so I can't have one at home.
I need to win the fscking lottery. lol
Ah, just skimming through that link about negative resistance.
There are two types of resistance:
Static resistance... It's the R in V=IR or R=V/I
Dynamic resistance... r= dV/dI = (change in V)/(change in I)
The dynamic resistance or change in resistance is what they're talking about when they say negative resistance. It's really negative "dynamic" resistance or negative change in resistance.
The static resistance is always positive...
There are two types of resistance:
Static resistance... It's the R in V=IR or R=V/I
Dynamic resistance... r= dV/dI = (change in V)/(change in I)
The dynamic resistance or change in resistance is what they're talking about when they say negative resistance. It's really negative "dynamic" resistance or negative change in resistance.
The static resistance is always positive...
Why didn't you just say that the change in resistance was negative in the first place? Did you fail calculus??
- Dec 9, 2000
- 16,528
- 4
- 0
The change in resistance is not negative. The slope (derivative) of the line is the resistance, and the slope is negative, so the resistance appears to be negative, but if you take the Voltage at any point on that line, and the current at that point, and divide them, the resistance is positive.
If the graph looks like /, the slope is constant and the graph of the derivative looks like -------. Likewise, \ derivative graph looks like negative ------------ The resistance isn't changing just because the slope is negative. It's constant and appears negative.
Considering i as a function of v.
i(v) = v/R
derivative of i with respect to v is 1/R, or C.
If the graph looks like /, the slope is constant and the graph of the derivative looks like -------. Likewise, \ derivative graph looks like negative ------------ The resistance isn't changing just because the slope is negative. It's constant and appears negative.
Considering i as a function of v.
i(v) = v/R
derivative of i with respect to v is 1/R, or C.
I don't know how you calculate resistance, etc, but on the graph the line is labelled as "detector resistance' which never becomes positive anyway. If that is correct, isn't slope of the line just change in resistance, meaning that the resistance of the graphite/clay goes down at about 300v and then stays at some low point at 400v?
- Dec 9, 2000
- 16,528
- 4
- 0
No, that line is wrong and I forgot to delete it, it's for something else.
The slope of the line IS resistance, or conductance in this case, the reciprocal of resistance.. y = mx + b, where m is the slope and b is the y-intercept. i=v/R, where the slope is 1/R, or C.Originally posted by: JohnCU
Nope, I passed Calculus 1, 2, and 3 with A's and a B in Calc 4. Maybe you failed it, since you think dV/dI is the change in resistance.
I must have failed it, because I thought dV/dI is the change in resistance too.
- Dec 9, 2000
- 16,528
- 4
- 0
Originally posted by: tikwanleap
Originally posted by: JohnCU
Nope, I passed Calculus 1, 2, and 3 with A's and a B in Calc 4. Maybe you failed it, since you think dV/dI is the change in resistance.
I must have failed it, because I thought dV/dI is the change in resistance too.
No, dV/dI is the slope of the line, which is resistance. The change in resistance is the second derivative.
Originally posted by: tikwanleap
Originally posted by: JohnCU
Nope, I passed Calculus 1, 2, and 3 with A's and a B in Calc 4. Maybe you failed it, since you think dV/dI is the change in resistance.
I must have failed it, because I thought dV/dI is the change in resistance too.
Oh ok, I was confusing myself....
For a V = IR graph... R is the slope if I is on the x axis.... ok so dV/dI is the instantaneous slope or instantaneous resistance at that point in the graph.... ok my brain hurts now...doh
Originally posted by: JohnCU
Originally posted by: tikwanleap
Originally posted by: JohnCU
Nope, I passed Calculus 1, 2, and 3 with A's and a B in Calc 4. Maybe you failed it, since you think dV/dI is the change in resistance.
I must have failed it, because I thought dV/dI is the change in resistance too.
No, dV/dI is the slope of the line, which is resistance. The change in resistance is the second derivative.
Yep, you're right.
TRENDING THREADS
-
Discussion Zen 5 Speculation (EPYC Turin and Strix Point/Granite Ridge - Ryzen 9000)- Started by DisEnchantment
- Replies: 25K
-
Discussion Intel Meteor, Arrow, Lunar & Panther Lakes + WCL Discussion Threads
- Started by Tigerick
- Replies: 24K
-
Discussion Intel current and future Lakes & Rapids thread- Started by TheF34RChannel
- Replies: 23K
-
-
Facebook
Twitter