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#2
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On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article , (David Platt) wrote: When you lengthened the hairpin, you added inductance... probably too much, so you've not only cancelled out the capacitive reactance from the DE, but have left some excess inductance shunted across the DE. Backwards. Too much inductance to resonate with the effective (parallel) capacitance would have resonated with a somewhat smaller capacitance than you actually have. (The product of inductance and capacitance has to be the same, for a given resonant frequency.) So you can think of the actual capacitance present as consisting of however much would be needed to resonate with the (too large) inductance, in parallel with more capacitance which does NOT get cancelled out by the inductance. Result: the actual hairpin, plus the effective (parallel) capacitance the too short driven element presents, is capacitive, not inductive (in parallel with the desired 50 ohms). This is just the opposite from a series resonant circuit where too much inductance gives an overall inductive result. Otherwise, I agree with everything David Platt wrote. David, VE7EZM and AF7BZ No, David Platt is correct. You are assuming the effective capacitance is in parallel. The antenna equivalent circuit is a *series* RC. Put in too much *shunt* inductance and the combination looks inductive at the feed point. John KD5YI |
#3
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On 9/25/2013 4:27 AM, John S wrote:
On 9/24/2013 9:04 PM, David Ryeburn wrote: In article , (David Platt) wrote: When you lengthened the hairpin, you added inductance... probably too much, so you've not only cancelled out the capacitive reactance from the DE, but have left some excess inductance shunted across the DE. Backwards. Too much inductance to resonate with the effective (parallel) capacitance would have resonated with a somewhat smaller capacitance than you actually have. (The product of inductance and capacitance has to be the same, for a given resonant frequency.) So you can think of the actual capacitance present as consisting of however much would be needed to resonate with the (too large) inductance, in parallel with more capacitance which does NOT get cancelled out by the inductance. Result: the actual hairpin, plus the effective (parallel) capacitance the too short driven element presents, is capacitive, not inductive (in parallel with the desired 50 ohms). This is just the opposite from a series resonant circuit where too much inductance gives an overall inductive result. Otherwise, I agree with everything David Platt wrote. David, VE7EZM and AF7BZ No, David Platt is correct. You are assuming the effective capacitance is in parallel. The antenna equivalent circuit is a *series* RC. Put in too much *shunt* inductance and the combination looks inductive at the feed point. John KD5YI I'm wrong. Please disregard. Sorry. John |
#4
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On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article , (David Platt) wrote: When you lengthened the hairpin, you added inductance... probably too much, so you've not only cancelled out the capacitive reactance from the DE, but have left some excess inductance shunted across the DE. Backwards. Too much inductance to resonate with the effective (parallel) capacitance would have resonated with a somewhat smaller capacitance than you actually have. (The product of inductance and capacitance has to be the same, for a given resonant frequency.) So you can think of the actual capacitance present as consisting of however much would be needed to resonate with the (too large) inductance, in parallel with more capacitance which does NOT get cancelled out by the inductance. Result: the actual hairpin, plus the effective (parallel) capacitance the too short driven element presents, is capacitive, not inductive (in parallel with the desired 50 ohms). This is just the opposite from a series resonant circuit where too much inductance gives an overall inductive result. Otherwise, I agree with everything David Platt wrote. David, VE7EZM and AF7BZ Actually, thinking more about this, I believe a hairpin is a shorted transmission line. So, using a Smith Chart, I investigated a 25-j25 load and played with the chart to see what happened. It turns out that the feed impedance is indeed inductive if the stub is too long. So, I have now reverted to agreeing with Mr. Platt. I think that the stub (hairpin) will have no effect if its length is a quarter wave. Shorter than that, it becomes inductive. Very short and it is highly inductive. Longer is less inductive. See where this is going? Cheers es 73, John |
#5
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On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive. Longer is less inductive. John, the inductive reactance of an ideal shorted stub is a tangent function that increases from zero at zero length up to a maximum on the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance is proportional to the inductive reactance so - very short and it is slightly inductive (low reactance). Longer is more inductive (up to 1/4WL). What you say above seems to be referring to capacitive reactance in open-circuit stubs where very short is highly reactive and longer is less reactive.. |
#6
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On 9/26/2013 7:11 AM, W5DXP wrote:
On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote: Very short and it is highly inductive. Longer is less inductive. John, the inductive reactance of an ideal shorted stub is a tangent function that increases from zero at zero length up to a maximum on the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance is proportional to the inductive reactance so - very short and it is slightly inductive (low reactance). Longer is more inductive (up to 1/4WL). What you say above seems to be referring to capacitive reactance in open-circuit stubs where very short is highly reactive and longer is less reactive. Yep, you're right. I was only noticing that the length of the line was longer, but that is a shorter stub. My mistake again. |
#7
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In article ,
W5DXP wrote: On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote: Very short and it is highly inductive. Longer is less inductive. John, the inductive reactance of an ideal shorted stub is a tangent function that increases from zero at zero length up to a maximum on the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance is proportional to the inductive reactance so - very short and it is slightly inductive (low reactance). Longer is more inductive (up to 1/4WL). Agreed. So the inductive susceptance of a short shorted stub will be high, while the inductive susceptance of a longer (but less than a quarter wavelength) shorted stub will be low. Put that longer stub in parallel with a dipole a little bit shorter than a half wave long (and thus with a small amount of capacitive reactance and a large amount of capacitive susceptance) and the net result comes out capacitive. This is what I meant when I said that a shorted stub a bit too long to resonate the (too short) dipole makes the whole thing will come out capacitive, whereas if the stub is a bit too short for resonance, the whole thing will come out inductive. David, VE7EZM and AF7BZ -- David Ryeburn To send e-mail, change "netz" to "net" |
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