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I have question about R L Mathematics
I have beads* on a coax and want to know the R and the L.
I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. |
I have question about R L Mathematics
On 1/28/2014 1:03 PM, amdx wrote:
I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. I had a thought, I measured the R by dividing Voltage by Current. So that means, my current was limited my the L also. The Total impedance is 3,350 ohms, this includes R and L. Mikek |
I have question about R L Mathematics
In rec.radio.amateur.antenna amdx wrote:
I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. The real resistance should not change with frequency so just measure it with an ohmmeter. Total impedance is the square root of the sum of the squares of resistance and reactance. The phase angle will tell you if the reactance is inductive or capacitive. -- Jim Pennino |
I have question about R L Mathematics
Hi, Mike -
On 1/28/2014 1:14 PM, amdx wrote: On 1/28/2014 1:03 PM, amdx wrote: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I will assume that Z is 3350 ohms at 3.85MHz. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. The impedance of the reactance (alone) IS the reactance (itself). Can anyone solve this for me? I will try. I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? We do (based on your numbers)... Z = 3350 @ 23.5 degrees. R = Z * COS(23.5) and X = Z * SIN(23.5) Therefore, R = 3072 ohms and X = 1336 ohms As a sanity check, Z = sqrt(R^2 + X^2) = 3350 Good! HTH, John S Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. I had a thought, I measured the R by dividing Voltage by Current. So that means, my current was limited my the L also. The Total impedance is 3,350 ohms, this includes R and L. Mikek |
I have question about R L Mathematics
El 28-01-14 20:03, amdx escribió:
I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. Did you actually measured R (say Re(Z) ) , or |Z|? Can you provide us some info on your setup? -- Wim PA3DJS www.tetech.nl Please remove abc first in case of PM |
I have question about R L Mathematics
On Tue, 28 Jan 2014 13:03:35 -0600, amdx wrote:
I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. Measure it again. That's an awfully high resistance for a piece of coax cable of any length. Knowing the type of coax and the length would be handy. Hopefully, you're not measureing the resistance of teh broken pot cores. That won't work. * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. The inductance of gapped and non-gapped ferrites are quite different. Check to see if the inductance moves when you move the coax. Also, RG-59/u is not the best coax on the planet. Try to find some RG-6/u instead. You might want to read through these papers on ferrites (especially the first): http://www.audiosystemsgroup.com/K9YC/K9YC.htm -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On 1/28/2014 2:25 PM, Wimpie wrote:
El 28-01-14 20:03, amdx escribió: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. Did you actually measured R (say Re(Z) ) , or |Z|? I measured the voltage, the current and the phase relationship. Can you provide us some info on your setup? I can, it might turn into a word war, but have used it very successfully 100s of times at 600kHz, I'm not sure of the accuracy at 10 MHz. But with time I will refine it as needed. Here's a diagram of the setup. http://s395.photobucket.com/user/Qma...awing.jpg.html Ignore the green lines and print for now. Here's the board with scope probes attached. http://s395.photobucket.com/user/Qma...notes.jpg.html (Note the curly Qs holding the probes, very useful for measurements were the leads cause ringing.) You connect a frequency generator and "Device to be Measured", adjust the frequency, and set the output level. I like to set the voltage at 1vpp, but it doesn't matter. On the scope you will see the voltage and the current as measured across the sense resistor. Say for your current you have 5 units pp on .02v scale, across a 100 ohm sense resistor. 5 x .02 / 100 = 0.001 amps pp. Then 1vpp / 0.001amps pp = 1000 ohms. Your 100 sense resistor is in series, so must be subtracted out, 1000 - 100 = 900 ohms. You can use the scope to see the phase. You can drop the pp, the numbers all come out the same. Now the fun part. I original used a device like this when I was measuring the R and C of bonded piezos in water. That brings me to the green lines in the first drawing, where it shows the capacitor, I had a variable inductor. I would adjust the inductor to tune out the capacitance of the piezo, (set the inductor so the scope shows zero phase difference.) I would then use the scope readings to calculate the R of the piezo. Then I would short the piezo connection and use the scope readings to calculate the impedance of the inductor, which is the same as the capacitance of the piezo with reverse sign. Then the higher functioning brains would calculate transformer and inductors for the amplifier. Any questions? Learned this from Henry. Mikek |
I have question about R L Mathematics
On 1/28/2014 2:12 PM, John S wrote:
Hi, Mike - On 1/28/2014 1:14 PM, amdx wrote: On 1/28/2014 1:03 PM, amdx wrote: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I will assume that Z is 3350 ohms at 3.85MHz. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. The impedance of the reactance (alone) IS the reactance (itself). Can anyone solve this for me? I will try. I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? We do (based on your numbers)... Z = 3350 @ 23.5 degrees. R = Z * COS(23.5) and X = Z * SIN(23.5) Therefore, R = 3072 ohms and X = 1336 ohms As a sanity check, Z = sqrt(R^2 + X^2) = 3350 Good! HTH, John S Thank you John. Mikek |
I have question about R L Mathematics
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I have question about R L Mathematics
On 1/28/2014 3:18 PM, amdx wrote:
On 1/28/2014 2:12 PM, John S wrote: Hi, Mike - On 1/28/2014 1:14 PM, amdx wrote: On 1/28/2014 1:03 PM, amdx wrote: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I will assume that Z is 3350 ohms at 3.85MHz. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. The impedance of the reactance (alone) IS the reactance (itself). Can anyone solve this for me? I will try. I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? We do (based on your numbers)... Z = 3350 @ 23.5 degrees. R = Z * COS(23.5) and X = Z * SIN(23.5) Therefore, R = 3072 ohms and X = 1336 ohms As a sanity check, Z = sqrt(R^2 + X^2) = 3350 Good! HTH, John S Thank you John. Mikek It makes me feel good that I could assist. So, I thank you as well. Cheers, John S |
I have question about R L Mathematics
On 1/28/2014 2:26 PM, Jeff Liebermann wrote:
On Tue, 28 Jan 2014 13:03:35 -0600, amdx wrote: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. Measure it again. That's an awfully high resistance for a piece of coax cable of any length. Knowing the type of coax and the length would be handy. Hopefully, you're not measuring the resistance of teh broken pot cores. That won't work. It is a lot. But when I put it together 15 years ago, I seem to remember about 4000 ohms. No, I'm measuring shield end to end. Here's a picture, maybe that will give you a different opinion. http://s395.photobucket.com/user/Qma...40099.jpg.html * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. The inductance of gapped and non-gapped ferrites are quite different. For sure, but I didn't care, I just wanted to see how it would act when I put RG-58u (said 59 before, wrong) thru a bunch of cores. Check to see if the inductance moves when you move the coax. Also, RG-59/u is not the best coax on the planet. Try to find some RG-6/u instead. You will see in the picture I stabilized it and tapped the whole thing. The center hole is not big enough for RG-6/u. You might want to read through these papers on ferrites (especially the first): http://www.audiosystemsgroup.com/K9YC/K9YC.htm Will do. When I did this I was thinking about choke baluns hams use on coax driving antennas. The whole think is just a curiosity. Thanks, Mikek |
I have question about R L Mathematics
On Tuesday, January 28, 2014 2:12:53 PM UTC-6, John S wrote:
Therefore, R = 3072 ohms and X = 1336 ohms Looks like it might be #77 ferrite material. |
I have question about R L Mathematics
On 1/28/2014 4:02 PM, wrote:
In rec.radio.amateur.antenna amdx wrote: On 1/28/2014 1:32 PM, wrote: In rec.radio.amateur.antenna amdx wrote: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I want to calculate the impedance of the reactance. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Thanks, Mikek * it is actually a bit more than beads. Years ago, we were sent a box of ferrite potcores, the cores arrived broken. I slide 42 broke halves onto a piece of RG59, and now I'm measuring it. The real resistance should not change with frequency so just measure it with an ohmmeter. I'm not sure, it might change with frequency, this is a ferrite around a wire, so the ohm meter won't work. It's loss in the ferrites. Correct, but if there is any significant loss in the ferrites at the frequency of interest you probably shouldn't be using that ferrite. Something like an AIM 4170 is very handy for measuring this and a whole bunch of other things though a bit pricy. Ok. |
I have question about R L Mathematics
On Tuesday, January 28, 2014 4:02:39 PM UTC-6, wrote:
Correct, but if there is any significant loss in the ferrites at the frequency of interest you probably shouldn't be using that ferrite. OTOH, if one doesn't dissipate the common-mode energy, the reactance of the ferrite choke may make common-mode problems worse by bringing the common-mode circuit impedance to series resonance, i.e. low impedance. I personally prefer to dissipate common-mode energy rather than risk making the mischief worse. -- 73, Cecil, w5dxp.com |
I have question about R L Mathematics
On 1/28/2014 3:50 PM, W5DXP wrote:
On Tuesday, January 28, 2014 2:12:53 PM UTC-6, John S wrote: Therefore, R = 3072 ohms and X = 1336 ohms Looks like it might be #77 ferrite material. It's a Ferroxcube 3B7 material. Here's some info. http://www.ferroxcube.com/Ferroxcube...asheet/3b7.pdf Don't know how it compares to #77. Mikek |
I have question about R L Mathematics
W5DXP wrote:
On Tuesday, January 28, 2014 4:02:39 PM UTC-6, wrote: Correct, but if there is any significant loss in the ferrites at the frequency of interest you probably shouldn't be using that ferrite. OTOH, if one doesn't dissipate the common-mode energy, the reactance of the ferrite choke may make common-mode problems worse by bringing the common-mode circuit impedance to series resonance, i.e. low impedance. I personally prefer to dissipate common-mode energy rather than risk making the mischief worse. -- 73, Cecil, w5dxp.com Yeah, I was thinking of other stuff... Sometimes you want it lossy, sometimes you don't; for a choke you do. -- Jim Pennino |
I have question about R L Mathematics
On Tuesday, January 28, 2014 5:16:36 PM UTC-6, amdx wrote:
It's a Ferroxcube 3B7 material. Overall, #77 is much better for HF with the resistance/reactance curves crossing at about 18 MHz. 3B7 is better for MF with the resistance/reactance curves crossing at about 1.5 MHz. The curves indicate that it should be more reactive than resistive at 3.8 MHz. Makes me wonder if the phase angle is actually 90-23.5=66.5 degrees? -- 73, Cecil, w5dxp.com |
I have question about R L Mathematics
On Tue, 28 Jan 2014 15:44:38 -0600, amdx wrote:
It is a lot. But when I put it together 15 years ago, I seem to remember about 4000 ohms. No, I'm measuring shield end to end. Here's a picture, maybe that will give you a different opinion. http://s395.photobucket.com/user/Qma...40099.jpg.html The resistance from shield to shield should fairly close to zero. Same with center pin to center pin. If you're not getting near zero, you have an open somewhere in the circuit. My guess(tm) would be the crappy old PL-259 connectors and UG-175 reducers. Check your continuity. You will see in the picture I stabilized it and tapped the whole thing. Perhaps you meant "taped" as in wrapped with duct tape? The center hole is not big enough for RG-6/u. Peel off the RG-6/u outer jacket and it will probably fit. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On 1/28/2014 8:11 PM, Jeff Liebermann wrote:
On Tue, 28 Jan 2014 15:44:38 -0600, amdx wrote: It is a lot. But when I put it together 15 years ago, I seem to remember about 4000 ohms. No, I'm measuring shield end to end. Here's a picture, maybe that will give you a different opinion. http://s395.photobucket.com/user/Qma...40099.jpg.html The resistance from shield to shield should fairly close to zero. My Kelvin resistance device says the the shield is 0.047 ohms end to end. Same with center pin to center pin. The center conductor is 0.105 ohms end to end. If you're not getting near zero, you have an open somewhere in the circuit. Looks good here. My guess(tm) would be the crappy old PL-259 connectors and UG-175 reducers. Check your continuity. I just did. You will see in the picture I stabilized it and tapped the whole thing. Perhaps you meant "taped" as in wrapped with duct tape? That is what I meant. The center hole is not big enough for RG-6/u. Peel off the RG-6/u outer jacket and it will probably fit. No need for that, This is just a curiosity. Not sure where we have disagreement, do you disbelieve the 3,350ohms at 3.58MHz? If I short one end shield to center pin, and measure the other end would that double impedance? I'm going to find out. Mikek |
I have question about R L Mathematics
On 1/28/2014 8:34 PM, amdx wrote:
On 1/28/2014 8:11 PM, Jeff Liebermann wrote: No need for that, This is just a curiosity. Not sure where we have disagreement, do you disbelieve the 3,350ohms at 3.58MHz? If I short one end shield to center pin, and measure the other end would that double impedance? I'm going to find out. Mikek Nope, almost zero ohms, shorting one end and measuring center to shield on the other end at 3.58MHz. To reiterate, measuring this 12ft of RG58/u with 42 half cores of 3B7 (material) (size 3019) slide over the coax, shows an impedance of about 3,350 ohms with an inductive phase angle of 22.5*. The whole system is touchy, putting your hand on the coax changes the current shown on the scope, also reorienting the coax will change the current. What are your thoughts, Jeff specifically and anyone else. Thanks, Mikek |
I have question about R L Mathematics
On Tue, 28 Jan 2014 20:34:21 -0600, amdx wrote:
Not sure where we have disagreement, do you disbelieve the 3,350ohms at 3.58MHz? Yep. I am a heretic. You stated: "I have measured the R at 3.85MHz, It is 3,350 ohms" From where to where did you measure 3.35K, presumably with an ordinary ohms-guesser? If I short one end shield to center pin, and measure the other end would that double impedance? I'm going to find out. It's like winding two turns on a toroid in opposite directions. The inductances cancel and you get zero inductance. Replace the coax cable with a single loop length of wire and see if it makes more sense. Only turns that go AROUND the toroid in the same direction provide useful inductances. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On Tuesday, January 28, 2014 8:11:33 PM UTC-6, Jeff Liebermann wrote:
The resistance from shield to shield should fairly close to zero. Jeff, he is measuring the choking impedance of a w2du choke-balun. It is like running one wire through a number of ferrite beads/toroids. -- 73, Cecil, w5dxp.com |
I have question about R L Mathematics
On Tue, 28 Jan 2014 21:05:07 -0600, amdx wrote:
Nope, almost zero ohms, shorting one end and measuring center to shield on the other end at 3.58MHz. To reiterate, measuring this 12ft of RG58/u with 42 half cores of 3B7 (material) (size 3019) slide over the coax, shows an impedance of about 3,350 ohms with an inductive phase angle of 22.5*. You previously said "I have measured the R at 3.85MHz, It is 3,350 ohms." Now, you change it to impedance. Please make up your mind. However, I screwed up in my previous message. I forgot that you're working with coax cable, where the center wire is shielded from the effects of the inductors by the outside shield. So, you'll see the equivalent of what you would get with a single wire going through all the cores. Got an Al value for a single core? I can't find a data sheet on the new and useless Ferroxcube web pile. Here's roughly how I would do it if I had the Al of your cores. L(mH) = Al * N^2 * n / 10^6 N = number of turns, which in this case is 1. n = number of cores, which in this case is 42. Xl = 2 * Pi * f * L Xl = 6.28 * 3.85*10^6 * L(mH)/10^3 The DC resistance is so small that it can be neglected. You obtained a 22.5 degree phase angle which might be the capacitance of the coax cable. I don't really know where it came from. That angle would normally come from a resistance in the loop, but the coax is nearly zero ohms. If there were any resistance, the phase angle would be: phase-angle = arctan(Xl/R) If it is the cazapitance: For RG-59/u 16.2 pf per foot for 12 ft would be 203 pf. Xc = 1 / (2 * Pi * f * C) Xc = 1 / (6.28 * 3.85*10^6 * 200*10-12) Xc = 207 ohms The vector sum of the reactances should give you the impedance. Now, all you have to do is either supply a measured inductance or find the Al of your cores. Are you sure it's 3.85MHz and not the more common 3.58MHz? I smell a transposition of numbers here. The whole system is touchy, putting your hand on the coax changes the current shown on the scope, also reorienting the coax will change the current. That's mostly because of the broken cores causing Al to change as they move. More duct tape. What are your thoughts, Jeff specifically and anyone else. My thoughts are that I'm going to throw up. However, it's not your questions or academic exercises. It's the junk food I excavated from the back of the office fridge. Time to recycle everything. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On Tue, 28 Jan 2014 20:19:40 -0800 (PST), W5DXP
wrote: On Tuesday, January 28, 2014 8:11:33 PM UTC-6, Jeff Liebermann wrote: The resistance from shield to shield should fairly close to zero. Jeff, he is measuring the choking impedance of a w2du choke-balun. It is like running one wire through a number of ferrite beads/toroids. Thanks. Enlightenment arrived a few milliseconds after I clicked send. I forgot that it was a coax cable and that the center wire was shielded from the effects of the inductors by the coax shield. Sorry for the muddle. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On 1/28/2014 9:51 PM, Jeff Liebermann wrote:
On Tue, 28 Jan 2014 20:34:21 -0600, amdx wrote: Not sure where we have disagreement, do you disbelieve the 3,350ohms at 3.58MHz? Yep. I am a heretic. You stated: "I have measured the R at 3.85MHz, It is 3,350 ohms" From where to where did you measure 3.35K, presumably with an ordinary ohms-guesser? The cable is 12 ft long with a PL259 on each end. I plugged each PL259 into an SO239 panel mount connectors. I then drove 2ma of current through the shield from panel mount connector to panel mount connector in series with a 100 ohm resistor. I measured the voltage across the resistor and calculated the current. This was done at 3.58MHz. The input voltage was divided by the current to get the impedance. btw, at one point I took a measurement and got around 800 ohms, took my a second to figure out I had lowered my frequency to 350 kHz. Reread, (or read) my response (3:17pm) to Wimpie about my setup and look at the pictures, I think I laid it out pretty clearly. I was impressed when I first started using the "setup". Simple, obvious idea, but the strays need to be worked to a minimum. The "setup" is cheap and simple and I'd like to refine for those that have a scope and frequency generator but no other way to measure complex impedances. If I short one end shield to center pin, and measure the other end would that double impedance? I'm going to find out. It's like winding two turns on a toroid in opposite directions. The inductances cancel and you get zero inductance. Replace the coax cable with a single loop length of wire and see if it makes more sense. Only turns that go AROUND the toroid in the same direction provide useful inductances. You're correct, I measured very low ohms and no phase difference. Curious to hear your input on the "setup" and what else I can do to give you more confidence in my measurement, OR fix my "setup". I'll measure some known parts tomorrow evening. Thanks Jeff. |
I have question about R L Mathematics
On 1/28/2014 10:29 PM, Jeff Liebermann wrote:
On Tue, 28 Jan 2014 21:05:07 -0600, amdx wrote: Nope, almost zero ohms, shorting one end and measuring center to shield on the other end at 3.58MHz. To reiterate, measuring this 12ft of RG58/u with 42 half cores of 3B7 (material) (size 3019) slide over the coax, shows an impedance of about 3,350 ohms with an inductive phase angle of 22.5*. You previously said "I have measured the R at 3.85MHz, It is 3,350 ohms." Now, you change it to impedance. Please make up your mind. I did say that, I wrongly thought it was 3,350 ohms with 22.58 phase shift, implying some as yet unknown amount of inductance. If you follow the thread, I immediately followed my own thread with, "The Total impedance is 3,350 ohms, this includes R and L." I apologize for my screwups making it difficult for anyone reading to follow along. However, if I new what I was doing, I wouldn't be posting a question. However, I screwed up in my previous message. I forgot that you're working with coax cable, where the center wire is shielded from the effects of the inductors by the outside shield. So, you'll see the equivalent of what you would get with a single wire going through all the cores. Yes, a loss resistance and inductance of one turn. Got an Al value for a single core? I believe it is 9660 +/- 25%. This is for two core sandwiched. However, does that have any use for calculations? I'm using the core in a manner that is not normal. I don't remember how I placed the core halves, face to face, back to back, mixed? I can't find a data sheet on the new Ya, I found an exchange where I was asking for that info back in back in 2007. I got it privately from a friend with an old catalog. and useless Ferroxcube web pile. I thought the pile was ok, it just didn't have the 3B7, must be old and obsolete. Here's roughly how I would do it if I had the Al of your cores. L(mH) = Al * N^2 * n / 10^6 N = number of turns, which in this case is 1. n = number of cores, which in this case is 42. Xl = 2 * Pi * f * L Xl = 6.28 * 3.85*10^6 * L(mH)/10^3 The DC resistance is so small that it can be neglected. You obtained a 22.5 degree phase angle which might be the capacitance of the coax cable. The phase is inductive, E leads I. I don't really know where it came from. That angle would normally come from a resistance in the loop, but the coax is nearly zero ohms. If there were any resistance, the phase angle would be: phase-angle = arctan(Xl/R) If it is the cazapitance: For RG-59/u 16.2 pf per foot for 12 ft would be 203 pf. Xc = 1 / (2 * Pi * f * C) Xc = 1 / (6.28 * 3.85*10^6 * 200*10-12) Xc = 207 ohms The vector sum of the reactances should give you the impedance. Now, all you have to do is either supply a measured inductance or find the Al of your cores. I thought I had, but apparently it was in my secret code. John S calculated "Therefore, R = 3072 ohms and X = 1336 ohms As a sanity check, Z = sqrt(R^2 + X^2) = 3,350" So if X = 1336 ohms at 3.85 MHz inductance is 55uH. I'm a little surprised, I would have thought the reactance would have been higher than the resistance. Are you sure it's 3.85MHz and not the more common 3.58MHz? I smell a transposition of numbers here. I picked 3.85MHz to be in the 80 meter ham band. The whole system is touchy, putting your hand on the coax changes the current shown on the scope, also reorienting the coax will change the current. That's mostly because of the broken cores causing Al to change as they move. More duct tape. Yes and more, just putting your hand by the coax causes the current to change. What are your thoughts, Jeff specifically and anyone else. My thoughts are that I'm going to throw up. However, it's not your questions or academic exercises. It's the junk food I excavated from the back of the office fridge. Time to recycle everything. I had to stop eating cashews last night, I was moving in that green feeling direction. Time to get ready for work. Thanks, Mikek |
I have question about R L Mathematics
On Wednesday, January 29, 2014 7:08:25 AM UTC-6, amdx wrote:
I'm a little surprised, I would have thought the reactance would have been higher than the resistance. According to the graphs it should be. That's why I guessed it was #77 material. |
I have question about R L Mathematics
On Wed, 29 Jan 2014 07:08:25 -0600, amdx wrote:
However, if I new what I was doing, I wouldn't be posting a question. If I knew what you were trying to do, I wouldn't be asking you questions. Got an Al value for a single core? I believe it is 9660 +/- 25%. This is for two core sandwiched. However, does that have any use for calculations? I'm using the core in a manner that is not normal. I don't remember how I placed the core halves, face to face, back to back, mixed? Argh. It's a pot core that's designed to have all its turns inside the two halves of the pot core on a bobbin. What you've apparently done is used it in a non-standard and unpredictable manner by shoving a wired through the hole normally reserved for either a ferrite adjustment slug, or a nylon mounting screw. The spec sheet Al value is worthless. However, all is not lost. You can take 1/2 of a core, shove a wire through the hole, and measure the inductance. That's inductance as in uH not the calculated reactance or guessed impedance. I presume that there are 42 half-cores so just multiplying the measured value by 42 will give a tolerable approximation of the total inductance. You could also measure the total inductance of all 42 cores to see if that really works. Note that I'm suggesting that you measure the inductance with a single wire going through the cores, and not with a loop produced by shorting one end of the coax cable. That takes some of the mystery out of the measurements. You can see what shorting one end does later. Ya, I found an exchange where I was asking for that info back in back in 2007. I got it privately from a friend with an old catalog. I have several old Ferroxcube catalogs from the 1970's and 1980's. You can't have them. The vector sum of the reactances should give you the impedance. Now, all you have to do is either supply a measured inductance or find the Al of your cores. I thought I had, but apparently it was in my secret code. John S calculated "Therefore, R = 3072 ohms and X = 1336 ohms As a sanity check, Z = sqrt(R^2 + X^2) = 3,350" So if X = 1336 ohms at 3.85 MHz inductance is 55uH. I'm a little surprised, I would have thought the reactance would have been higher than the resistance. The only thing I can be sure of here is that the resistive component is *NOT* 3000 ohms because it can be directly measured with an ohms-guesser. The number is (obviously) wrong because nowhere in your circuit is anything resembling a resistor of that high a value. If you work backwards and assume a DC resistance of zero, then Z = Xl. I'm still not sure what's causing the 22 degree phase angle. My best guess(tm) is that it's the capacitance of the coax cable, but that should have disappeared when you shorted one end of the coax. Dunno. Yes and more, just putting your hand by the coax causes the current to change. My hand is firmly attached to my arm and is nowhere near your setup. Perhaps you meant your hand? I had to stop eating cashews last night, I was moving in that green feeling direction. Time to get ready for work. I compounded my culinary error by eating about 1/3 a salted dark chocolate bar before going to bed. It's now 7AM and I've had about 3 hours of erratic sleep due to the caffeine overdose. I've done this before and should have known better, but it was sooooooo good. At least I'm now caught up on paying my bills and reading various reports. My next challenge will be to see if I can drive to the office without falling asleep. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On 1/29/2014 9:04 AM, Jeff Liebermann wrote:
On Wed, 29 Jan 2014 07:08:25 -0600, amdx wrote: However, if I new what I was doing, I wouldn't be posting a question. If I knew what you were trying to do, I wouldn't be asking you questions. Got an Al value for a single core? I believe it is 9660 +/- 25%. This is for two core sandwiched. However, does that have any use for calculations? I'm using the core in a manner that is not normal. I don't remember how I placed the core halves, face to face, back to back, mixed? Argh. It's a pot core that's designed to have all its turns inside the two halves of the pot core on a bobbin. What you've apparently done is used it in a non-standard and unpredictable manner by shoving a wired through the hole normally reserved for either a ferrite adjustment slug, or a nylon mounting screw. The spec sheet Al value is worthless. I have a feeling you didn't look at any of my pretty pictures, I think at least one layer of fog should have dissipated if you did. They're really pretty. However, all is not lost. You can take 1/2 of a core, shove a wire through the hole, and measure the inductance. That's inductance as in uH not the calculated reactance or guessed impedance. I presume that there are 42 half-cores so just multiplying the measured value by 42 will give a tolerable approximation of the total inductance. You could also measure the total inductance of all 42 cores to see if that really works. Note that I'm suggesting that you measure the inductance with a single wire going through the cores, I'm doing that now, only with 42 halves. I can try it with one, but I question with my "setup" will resolve that inductance. You can see what shorting one end does later. I already did that, zero inductance or resistance. Ya, I found an exchange where I was asking for that info back in back in 2007. I got it privately from a friend with an old catalog. I have several old Ferroxcube catalogs from the 1970's and 1980's. You can't have them. Do you need any potcores? The vector sum of the reactances should give you the impedance. Now, all you have to do is either supply a measured inductance or find the Al of your cores. I thought I had, but apparently it was in my secret code. John S calculated "Therefore, R = 3072 ohms and X = 1336 ohms As a sanity check, Z = sqrt(R^2 + X^2) = 3,350" So if X = 1336 ohms at 3.85 MHz inductance is 55uH. I'm a little surprised, I would have thought the reactance would have been higher than the resistance. The only thing I can be sure of here is that the resistive component is *NOT* 3000 ohms because it can be directly measured with an ohms-guesser. Since you have used "ohms-guesser" I guess I need to ask, What is that? and does it use ac or dc for the measurement? The number is (obviously) wrong because nowhere in your circuit is anything resembling a resistor of that high a value. I'm sure you understand that ferrite beads on a transistor lead, show up as a resistive and inductive. Why is this different? Here's a pdf with a graph page 4 right side showing R, X, and Z. Hand picked to show what I want it to! Although I should have secretly altered the frequency range. http://www.vishay.com/docs/ilb_ilbb_enote.pdf If you work backwards and assume a DC resistance of zero, then Z = Xl. I'm still not sure what's causing the 22 degree phase angle. My best guess(tm) is that it's the capacitance of the coax cable, but that should have disappeared when you shorted one end of the coax. Dunno. It's inductive. Yes and more, just putting your hand by the coax causes the current to change. My hand is firmly attached to my arm and is nowhere near your setup. Perhaps you meant your hand? LOL, you're right, it was my hand. I had to stop eating cashews last night, I was moving in that green feeling direction. Time to get ready for work. I compounded my culinary error by eating about 1/3 a salted dark chocolate bar before going to bed. It's now 7AM and I've had about 3 hours of erratic sleep due to the caffeine overdose. I've done this before and should have known better, but it was sooooooo good. At least I'm now caught up on paying my bills and reading various reports. My next challenge will be to see if I can drive to the office without falling asleep. Good luck. I'm in Fl. where people are acting like the end is near. We are slightly below freezing and businesses are closing, schools out. The weather men are have fits of frenzy. If I was back in Michigan, we would be happy it got warmer, because we were getting tired of starting the car at 7* in 8 inches of snow. Mikek |
I have question about R L Mathematics
On 1/29/2014 4:12 PM, Fred Abse wrote:
On Tue, 28 Jan 2014 13:03:35 -0600, amdx wrote: I have beads* on a coax and want to know the R and the L. I have measured the R at 3.85MHz, It is 3,350 ohms. I have also measured the phase shift, voltage leading by 17ns. The period of 3.85Mhz is 260ns. I'm not sure I understand what you're trying to measure. I assume that it's the effect of the "beads" on the inductance of the outer conductor, That is correct. in which case, 3350 ohms looks way OTT. No pot core material I know of is *that* lossy. That's what everyone seems to think. I'm the odd man out. If you haven't looked at the picture, here it is. The ferrite is 18 inches long. That is a quarter in front. http://s395.photobucket.com/user/Qma...40099.jpg.html How long is the coax sample? 17ns is the delay of approx. 5 meters of wire at 3.85MHz, without any inductive loading. It is 11 ft long. You did connect to the outer conductor at both ends, didn't you. Yes, outer conductor connected at both ends. To different inputs of the measuring device. Preferably short inner to outer, at both ends. That I have not done. I'll try that. ........ ....... OK, I'm back. Shorting the center pin to the shield, (at both ends) made absolutely no difference in the magnitude or phase of the measurement. I want to calculate the impedance of the reactance. LOL, I had to go back and see if I said that. Now I'll say what I meant. I want to know the magnitude in ohms of the reactance. John S already solved that for me, thanks. He knows how to solve for what I want, not what I ask for. :-) There's no such thing. Reactance is merely the imaginary part of a (complex) impedance. Already restated. Can anyone solve this for me? I would like to see the math, because I want to measure again at 7.5MHz. My first step was to find the phase angle, 23.5*. Do we agree there? Lets look at what you have: You have voltage/current=3350 ohms. That is the *magnitude* of the impedance, at an angle of 23.5 degrees, current lagging. That's 3350 angle 23.5 ohms. We now do a polar to rectangular conversion on that, giving: 3072 +j1335 ohms. Good, agreement with John S. 3072 seems way too high for the loss component, 1335 ohms is 55 microhenries,at 3.85MHz. Everyone agrees the loss component is to high. Oh, except me. :-) I think I have stated, I thought the L would be higher than the R. That's not what I'm measuring. I have no experience in ferrite losses, and no education regarding losses in ferrite. But I think my measurement are in the ballpark. One thing I suggest is that you do the whole thing again, without, and then with, the "beads". That way, you can eliminate propagation delays. I'll try another piece of RG-58/U, I can't get the ferrite of the cable without cutting off a PL259. This evening I'll wind a 55uH coil and find a 3,072 resistor. I'll put these in series and see how it measures compared to my lossy ferrites beads on a cable. I already know this measures about 6% high, probably because of the sense resistor. Thanks, Mikek |
I have question about R L Mathematics
On 1/29/2014 6:20 PM, amdx wrote:
On 1/29/2014 4:12 PM, Fred Abse wrote: Lets look at what you have: You have voltage/current=3350 ohms. That is the *magnitude* of the impedance, at an angle of 23.5 degrees, current lagging. That's 3350 angle 23.5 ohms. We now do a polar to rectangular conversion on that, giving: 3072 +j1335 ohms. Good, agreement with John S. 3072 seems way too high for the loss component, 1335 ohms is 55 microhenries,at 3.85MHz. Everyone agrees the loss component is to high. Oh, except me. :-) I think I have stated, I thought the L would be higher than the R. That's not what I'm measuring. I have no experience in ferrite losses, and no education regarding losses in ferrite. But I think my measurement are in the ballpark. One thing I suggest is that you do the whole thing again, without, and then with, the "beads". That way, you can eliminate propagation delays. I'll try another piece of RG-58/U, I can't get the ferrite of the cable without cutting off a PL259. This evening I'll wind a 55uH coil and find a 3,072 resistor. I'll put these in series and see how it measures compared to my lossy ferrites beads on a cable. I already know this measures about 6% high, probably because of the sense resistor. Thanks, Mikek Well, I found the 6% is actually the difference between my scope probes. That's only my first problem. I measured a 55uH inductor and 3090 ohm resistor in series, 3.85MHz and got 2778 ohms 19.9* phase difference. The calculated numbers are Z = 3,364 and I don't know how to calculate the phase angle. Later I'll check this at 100kHz and so if strays are causing errors. Mikek PS, I should have some new probes tomorrow. |
I have question about R L Mathematics
On Wed, 29 Jan 2014 09:45:51 -0600, amdx wrote:
On 1/29/2014 9:04 AM, Jeff Liebermann wrote: I have a feeling you didn't look at any of my pretty pictures, I think at least one layer of fog should have dissipated if you did. They're really pretty. http://s395.photobucket.com/user/Qmavam/media/TheBoardwithnotes.jpg.html http://s395.photobucket.com/user/Qmavam/media/TheBoardDrawing.jpg.html I clicked, I looked, I saw, I failed to understand, I blundered onward. What did I miss? You really want a vector impedance meter: http://www.ebay.com/itm/181276724052 Yeah, I know it's an antique. There are probably much better models available today, but not at the price. I had one at a previous employer and used it for measuring almost everything. 0.5 to 110 Mhz. I've been looking for a broken one (because I blew up the one I was using often enough to be familiar with the repair procedures) but can't seem to find one in my price range (i.e. free). Make sure it comes with the probe and power cord as both are difficult to find. Note that I'm suggesting that you measure the inductance with a single wire going through the cores, I'm doing that now, only with 42 halves. I can try it with one, but I question with my "setup" will resolve that inductance. Never mind. You won't see it. At the time, I was trying to correlate the measured value with the theoretical Al. However, since the cores are not being used in the normal manner, that's not going to work or yield any useful results. Might as well measure the whole string at once. Do you need any potcores? No thanks. I gave smoking the stuff in college. You might be able to sell them in Colorado. Since you have used "ohms-guesser" I guess I need to ask, What is that? and does it use ac or dc for the measurement? An ohms-guesser(tm) is a Harbor Freight DVM or equivalent. Usually sells for $5 or less. Capable of producing wrong values to at least 3 1/2 digit accuracy. I usually have several available for the inevitable visitors that roll into my parking lot wanting to borrow a meter to fix their vehicle electrical system. Ohms-guessers use DC current to measure resistance. There is also the volts-guesser and amps-guesser which offer similar features and lack of accuracy. The number is (obviously) wrong because nowhere in your circuit is anything resembling a resistor of that high a value. I'm sure you understand that ferrite beads on a transistor lead, show up as a resistive and inductive. Why is this different? Here's a pdf with a graph page 4 right side showing R, X, and Z. Hand picked to show what I want it to! Although I should have secretly altered the frequency range. http://www.vishay.com/docs/ilb_ilbb_enote.pdf ARGH! I goofed. I forgot how ferrite really work and was assuming that it represented a pure inductance. Please ignore everything I wrote about the resistive component. I tried to find a reactance vs frequency graph for the 3B7 material and failed. Maybe later tonite. I'm buried in broken machines and an office in desperate need of untrashing. Thanks (grumble). Incidentally, consider yourself off the hook for the MFJ-1800 yagi fiasco if you promise not to tell anyone how badly I messed up here. I'm in Fl. where people are acting like the end is near. Huh? Are they expecting Florida to sink into the ocean under the added weight of the snow? We are slightly below freezing and businesses are closing, schools out. The weather men are have fits of frenzy. If I was back in Michigan, we would be happy it got warmer, because we were getting tired of starting the car at 7* in 8 inches of snow. Mikek In the People's Republic of Santa Cruz, it has been mostly 60-70F highs and 35-45F lows for most of the alleged winter. Quite pleasant and comfortable. The problem is that we've had less than an inch of rain this season, where normal would be about 15-20 inches by this date. The forest looks awful with dead trees and shrubbery everywhere. Looks like we're going to have a severe drought here. Even worse is the lack of snow for skiing. Right now, I could use some snow or rain. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
I have question about R L Mathematics
On Wednesday, January 29, 2014 6:20:16 PM UTC-6, amdx wrote:
This evening I'll wind a 55uH coil and find a 3,072 resistor. I'll put these in series and see how it measures compared to my lossy ferrites beads on a cable. A coil of that large size will exhibit transmission line characteristics that will render the lumped-circuit model inaccurate because the current into the coil will not be equal to the current out of the coil. Better to use the distributed network model which is closer to Maxwell's equations. Here is an inductance calculator that will yield more accurate results for coils that are greater than ~15 degrees long electrically. Using the axial propagation factor (Beta) of the the coil, one can calculate how many degrees it occupies. Then using the characteristic impedance (Z0) of the coil, one can analyze the circuit like a transmission line using the distributed network model. If you will describe the characteristics of the coil in detail (total turns, length, wire size, etc.) I will do the analysis for you. -- 73, Cecil, w5dxp.com |
I have question about R L Mathematics
On Thursday, January 30, 2014 6:41:53 AM UTC-6, W5DXP wrote:
Here is an inductance calculator that will yield more accurate results for coils that are greater than ~15 degrees long electrically. Sorry, I forgot to post the URL of the inductance calculator: http://hamwaves.com/antennas/inductance.html |
I have question about R L Mathematics
On 1/29/2014 10:26 PM, Jeff Liebermann wrote:
On Wed, 29 Jan 2014 09:45:51 -0600, amdx wrote: On 1/29/2014 9:04 AM, Jeff Liebermann wrote: I have a feeling you didn't look at any of my pretty pictures, I think at least one layer of fog should have dissipated if you did. They're really pretty. http://s395.photobucket.com/user/Qmavam/media/TheBoardwithnotes.jpg.html http://s395.photobucket.com/user/Qmavam/media/TheBoardDrawing.jpg.html I clicked, I looked, I saw, I failed to understand, I blundered onward. What did I miss? Probably not much. :-( Very simple concept, Measuring current by use of a sense resistor. Then comparing voltage to current to get impedance. You really want a vector impedance meter: http://www.ebay.com/itm/181276724052 I have recently acquired an HP 3570A Network analyzer and a 3330B synthesizer that are GPIB connected to a computer with custom software. It has not been powered for at least 10 years. The only problem I know about is the computer cmos battery most likely has failed. I've had it a couple months now and have not got it setup or attempted any use. I need to figure out what to do about the computer first. I don't have the original software disc, so I need to copy the program first thing. Yeah, I know it's an antique. There are probably much better models available today, but not at the price. I had one at a previous employer and used it for measuring almost everything. 0.5 to 110 Mhz. I've been looking for a broken one (because I blew up the one I was using often enough to be familiar with the repair procedures) but can't seem to find one in my price range (i.e. free). Make sure it comes with the probe and power cord as both are difficult to find. Note that I'm suggesting that you measure the inductance with a single wire going through the cores, I'm doing that now, only with 42 halves. I can try it with one, but I question with my "setup" will resolve that inductance. Never mind. You won't see it. At the time, I was trying to correlate the measured value with the theoretical Al. However, since the cores are not being used in the normal manner, that's not going to work or yield any useful results. Might as well measure the whole string at once. Do you need any potcores? No thanks. I gave smoking the stuff in college. You might be able to sell them in Colorado. Since you have used "ohms-guesser" I guess I need to ask, What is that? and does it use ac or dc for the measurement? An ohms-guesser(tm) is a Harbor Freight DVM or equivalent. Usually sells for $5 or less. Capable of producing wrong values to at least 3 1/2 digit accuracy. I usually have several available for the inevitable visitors that roll into my parking lot wanting to borrow a meter to fix their vehicle electrical system. Ohms-guessers use DC current to measure resistance. There is also the volts-guesser and amps-guesser which offer similar features and lack of accuracy. You may have figured out, I'm not using this type of ohms guesser. The number is (obviously) wrong because nowhere in your circuit is anything resembling a resistor of that high a value. I'm sure you understand that ferrite beads on a transistor lead, show up as a resistive and inductive. Why is this different? Here's a pdf with a graph page 4 right side showing R, X, and Z. Hand picked to show what I want it to! Although I should have secretly altered the frequency range. http://www.vishay.com/docs/ilb_ilbb_enote.pdf ARGH! I goofed. I forgot how ferrite really work and was assuming that it represented a pure inductance. Please ignore everything I wrote about the resistive component. I tried to find a reactance vs frequency graph for the 3B7 material and failed. Maybe later tonite. I'm buried in broken machines and an office in desperate need of untrashing. Thanks (grumble). Incidentally, consider yourself off the hook for the MFJ-1800 yagi fiasco if you promise not to tell anyone how badly I messed up here. My deed here has been accomplished, I now officially close this thread. yippee, yahoo, hallelujah, praise be to any omnipotent powers. I'm in Fl. where people are acting like the end is near. Huh? Are they expecting Florida to sink into the ocean under the added weight of the snow? We are slightly below freezing and businesses are closing, schools out. The weather men are have fits of frenzy. If I was back in Michigan, we would be happy it got warmer, because we were getting tired of starting the car at 7* in 8 inches of snow. Mikek In the People's Republic of Santa Cruz, it has been mostly 60-70F highs and 35-45F lows for most of the alleged winter. Quite pleasant and comfortable. The problem is that we've had less than an inch of rain this season, where normal would be about 15-20 inches by this date. The forest looks awful with dead trees and shrubbery everywhere. Looks like we're going to have a severe drought here. Even worse is the lack of snow for skiing. Right now, I could use some snow or rain. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 I have found through some tests and exchanges here that the mechanics of my "setup" work fine below 1MHz or a little higher. But I think, strays are causing some problems when I go higher in frequency. I should get my new probes today and I'll be working at tightening up the circuit and maybe adding shielding. I built this "setup" as as duplicate of the one I used, I tried to use it at higher frequency and I should not have. When I was working with ultrasonics, our standard frequency was around 660 kHz. We used this daily to characterize piezo transducers to calculate matching to an amplifier. I sure wish the company made some money, I really enjoyed that job. btw, Need any 2" disc piezo ceramic discs 1/16" thick PZT-8 material? I have a few other sizes also. Do you know any fun things I could do with them, besides give people shocks. Thanks, Mikek |
I have question about R L Mathematics
Jeff Liebermann scribbled thus:
In the People's Republic of Santa Cruz, it has been mostly 60-70F highs and 35-45F lows for most of the alleged winter. Quite pleasant and comfortable. The problem is that we've had less than an inch of rain this season, where normal would be about 15-20 inches by this date. The forest looks awful with dead trees and shrubbery everywhere. Looks like we're going to have a severe drought here. Even worse is the lack of snow for skiing. Right now, I could use some snow or rain. All that rain is over here... Snow threatened :-( -- Best Regards: Baron. |
I have question about R L Mathematics
On 1/30/2014 9:37 AM, Baron wrote:
Jeff Liebermann scribbled thus: In the People's Republic of Santa Cruz, it has been mostly 60-70F highs and 35-45F lows for most of the alleged winter. Quite pleasant and comfortable. The problem is that we've had less than an inch of rain this season, where normal would be about 15-20 inches by this date. The forest looks awful with dead trees and shrubbery everywhere. Looks like we're going to have a severe drought here. Even worse is the lack of snow for skiing. Right now, I could use some snow or rain. All that rain is over here... Snow threatened :-( We're up to 34*, the icicles are dripping. Supposed to be 61* tomorrow. Glad January is about over, maybe February will be better. The global warming has been brutally cold this year. Mikek |
I have question about R L Mathematics
On 1/30/2014 9:33 AM, amdx wrote:
(snip) btw, Need any 2" disc piezo ceramic discs 1/16" thick PZT-8 material? I have a few other sizes also. Do you know any fun things I could do with them, besides give people shocks. Thanks, Mikek Hi, Mike - I would like to buy a couple to experiment with and, possibly, learn something. How many $ do you want and how do we connect? Thanks, John S |
I have question about R L Mathematics
On 1/30/2014 1:17 PM, Fred Abse wrote:
On Wed, 29 Jan 2014 20:26:16 -0800, Jeff Liebermann wrote: You really want a vector impedance meter: He's actually doing what a vector impedance meter does - the hard way. A vector voltmeter might be of more use in Mike's case. More versatile. Where is one available and what would it cost? |
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