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On Thu, 2 Dec 2004 23:20:40 +0000, "Ian White, G3SEK"
wrote: What does a typical "test jig" consist of Ian? That's one thing that's always puzzled me about these kind of measurements; particularly at UHF++. -- "What is now proved was once only imagin'd." - William Blake, 1793. |
Paul Burridge wrote:
On Thu, 2 Dec 2004 23:20:40 +0000, "Ian White, G3SEK" wrote: What does a typical "test jig" consist of Ian? That's one thing that's always puzzled me about these kind of measurements; particularly at UHF++. One that minimizes unwanted or uncontrolled lead lengths. In general, one that is based on solid lumps of metal and large, broad, low-inductance conducting surfaces. I had simply bent the resistor wires so that one end pushed into the centre of the VNA's N socket, and the other wire was literally tied onto the body of the socket. However, the measurement showed that most of the small inductance could be accounted for by those two wires - which you'd never leave as long as that in a practical layout. To home in on the inductance of the resistor body itself, I'd have to build a jig that allows the wire lengths to be reduced almost to zero. Harold W4ZCB sent a picture of something he uses, which is just a brass plate soldered to the back of an SMA connector. The Device Under Test is then soldered directly between the centre pin and somewhere on the plate. But I'm afraid my only visit to the workshop last weekend was to dump yet another cardboard box on the floor. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
On Tue, 7 Dec 2004 07:31:08 +0000, "Ian White, G3SEK"
wrote: To home in on the inductance of the resistor body itself, I'd have to build a jig that allows the wire lengths to be reduced almost to zero. Harold W4ZCB sent a picture of something he uses, which is just a brass plate soldered to the back of an SMA connector. The Device Under Test is then soldered directly between the centre pin and somewhere on the plate. How about cutting the resistor's leads off completely and just clamping it between say two 1" cube copper blocks? But then I guess you would still have the problem of connecting it to the VNA. :-( -- "What is now proved was once only imagin'd." - William Blake, 1793. |
How about cutting the resistor's leads off completely and just clamping it between say two 1" cube copper blocks? But then I guess you would still have the problem of connecting it to the VNA. :-( -- Would possibly work OK if your standards were fabricated in the same way. When it makes a measureable difference whether you're measuring to the front face of the SMA center pin and the back of that same pin, you can start to develop an appreciation for the accuracy possible with the VNA. I haven't measured the capacity between two copper faces one inch square and separated by the length of a quarter watt resistor, but it would be appreciable. (and measureable if I were really interested.) W4ZCB |
In article , Paul Burridge
writes: On Tue, 7 Dec 2004 07:31:08 +0000, "Ian White, G3SEK" wrote: To home in on the inductance of the resistor body itself, I'd have to build a jig that allows the wire lengths to be reduced almost to zero. Harold W4ZCB sent a picture of something he uses, which is just a brass plate soldered to the back of an SMA connector. The Device Under Test is then soldered directly between the centre pin and somewhere on the plate. How about cutting the resistor's leads off completely and just clamping it between say two 1" cube copper blocks? But then I guess you would still have the problem of connecting it to the VNA. :-( That is more trouble to set up than is needed. In order to measure any inductance component, the only requirement is to find the DIFFERENTIAL between a direct short across the bridge/RLC-meter connections and the device itself (in this case a resistor). Neil Hecht's excellent little LC Meter II does this automatically by the zeroing button that subtracts the shorting inductance from the device measurement, done arithmetically in the internal microcontroller's registers. The inductance of a standard gauge round wire is fairly well known and has been around for years in texts. One can compute that inductance fairly accurately without using a bridge/meter and then compare that to the leads of the device under test. The difference between the shorting lead and the device leads would then be the inductance of the device-under-test's body. If a few nanoHenries are involved and affect the circuit the device is to be used in, the frequency would be high enough that skin effect would be operative. Skin effect is the curious phenomenon where current flows in thinner and thinner volume spaces near the surface of a conductor. The thickness of the bridge/meter connections would only be advantageous at DC (no skin effect at all, only "straight" volumetric conductor bulk resistance) or lower frequencies (where the "skin" is quite deep in the conductor). An approximation of the RF resistance in Ohms/Inch-length is: R_rf = (2.61 x 10^-7 x Sqrt( frequency in Hz)) / (2 x (w + h)) for PCB foil and where w = trace width in inches, h = thickness of the PCB foil in inches (about 0.002 times "ounce" rating of foil thickness, give or take some). Reference: Nicholas Gray, Staff Applications Engineer, National Semiconductor, "Design Idea" insert in electronics trades for December, 2004. See also http://edge.national.com However, when push comes to shove on measuring things, a built-in inductance might be as much as 10 nanoHenries. At 100 MHz that inductance is equal to +j 6.283 Ohms. A precision resistor of 50 Ohms at DC would have an equivalent series magnitude of 50.39 Ohms, an increase over DC of only 0.79%. The end result to a circuit using that device wouldn't amount to very much change. If that inductance were (somehow) ten times higher to 100 nHy, then it would have +j 62.83 Ohms and the series magnitude would (of that 50 Ohm resistor at DC) be 80.30 Ohms. That WILL have some noticeable effect on a circuit at 100 MHz. The upshot of all this is that one considers the frequency(s) of operation first, then measures a part second to find out if the resulting part inductance will have any effect on things. In order to measure the part, the characteristics of the measuring instrument ought to be known so as to get a reasonably accurate measurement. Comparing the part to a wide conductor across the instrument connections as the reference inductance, the part is then measured with the wide conductor inductance subtracted from the mesaured series inductance. Some LC bridges measure admittance rather than impedance. If that's the case, then the complex admittance (an equivalent R & C in parallel) must be inverted to find the complex impedance which is a series R and L. A grid-dipper sort of measurement just won't yield any meaningful result for either Z or Y. |
It's not at all my intent to make up more work for you to do. What I was
mostly interested in was just how much the spiral construction of the resistor adds to its inductance -- is it or isn't it significant, or is the inductance mostly due to the component size and the leads? And it still seems to me that you could just measure a wrapped or painted resistor in the same fixture as you did the intact resistor. The difference in measured inductances should be the contribution of the spiral element. Although it would be interesting to see how small the inductance of a leaded resistor could possibly be, it's probably of more practical use to know the inductance of a leaded resistor with some realistic length of leads attached, which is I believe what you've already done. Anyone needing less inductance than that would be wise to abandon leaded resistors and go to SMD. Roy Lewallen, W7EL Ian White, G3SEK wrote: Paul Burridge wrote: On Thu, 2 Dec 2004 23:20:40 +0000, "Ian White, G3SEK" wrote: What does a typical "test jig" consist of Ian? That's one thing that's always puzzled me about these kind of measurements; particularly at UHF++. One that minimizes unwanted or uncontrolled lead lengths. In general, one that is based on solid lumps of metal and large, broad, low-inductance conducting surfaces. I had simply bent the resistor wires so that one end pushed into the centre of the VNA's N socket, and the other wire was literally tied onto the body of the socket. However, the measurement showed that most of the small inductance could be accounted for by those two wires - which you'd never leave as long as that in a practical layout. To home in on the inductance of the resistor body itself, I'd have to build a jig that allows the wire lengths to be reduced almost to zero. Harold W4ZCB sent a picture of something he uses, which is just a brass plate soldered to the back of an SMA connector. The Device Under Test is then soldered directly between the centre pin and somewhere on the plate. But I'm afraid my only visit to the workshop last weekend was to dump yet another cardboard box on the floor. |
Len wrote:
In order to measure any inductance component, the only requirement is to find the DIFFERENTIAL between a direct short across the bridge/RLC-meter connections and the device itself (in this case a resistor). Neil Hecht's excellent little LC Meter II does this automatically by the zeroing button that subtracts the shorting inductance from the device measurement, done arithmetically in the internal microcontroller's registers. I completely agree that all impedance measuring devices should be "zeroed" in this way. But the problem with the resistor wires that we're discussing here is *additional* to that. We want to know the inductance of the metal-film resistor body, with the wires cut very short as they would be for any application where low inductance is important. However, for convenience, my first measurements used almost the full length of the resistor wires to connect to the N socket of the VNA. It turned out that the total measured inductance is comparable to what you'd find from the wires alone, so the body inductance is very small (which is entirely consistent with the physical construction). The suggestion had been to determine the inductance of the resistor body by first repeating that original measurement, then applying conductive paint to short out the resistor body, and then measuring again. The body inductance would then be the difference between those two measurements. Unfortunately that would be a poorly designed experiment, because the resistor wires were bent around into a floppy loop whose size and shape - and therefore inductance - is not very well controlled. The very small inductance of the resistor body could easily become lost in variations caused by small accidental movements of the wires. It is also an experiment that cannot be repeated, because of the conductive paint. If I'd found time at the weekend, I would have made up a little plate like W4ZCB described. The N2PK VNA uses a three-step calibration with open, short and 50R standard loads. For this jig, I'd have had to start with the open-circuit connector spill, followed by a solder-blob short, and finally by the best solderable 50R load I could make (probably two 100R chip resistors in parallel). Then I'd have cut short the wires of the test resistor, and soldered that in place on the plate for the actual measurement. But unfortunately the whole weekend timed ou -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Len wrote:
All noted and good, Ian. The thread had become (in my estimation) one of those "intellectual exercises" that don't have much real practical significance. :-) "Small enough for VHF" is about as much as we need to know. Agilent (what HP test and measurement division became after Big Business of computer product take- over of HP) Or as someone said to me, "We're the world's #1 test equipment manufacturer, and they just gave our trademark away to the world's #3 computer manufacturer." I have some of the original free App Notes from HP (back when HP was HP...:-) and the going gets tough through all that four-port math. If someone else has gone the same route and put it all into some software/firmware to let the casual hobbyist go at it easily, then good all around. Yes, indeed someone has! -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
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On Tue, 07 Dec 2004 21:45:43 -0800, Roy Lewallen
wrote: It's not at all my intent to make up more work for you to do. What I was mostly interested in was just how much the spiral construction of the resistor adds to its inductance -- is it or isn't it significant, or is the inductance mostly due to the component size and the leads? And it still seems to me that you could just measure a wrapped or painted resistor in the same fixture as you did the intact resistor. The difference in measured inductances should be the contribution of the spiral element. Has anyone carried out any tests to compare conventional spiral elements with the 'doubling-back spiral' types that claim to be "non-inductive"? I just wonder if that doubling-back of the resistive element to 'eliminate' inductance is as effective as it's made out to be. -- "What is now proved was once only imagin'd." - William Blake, 1793. |
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