jeff millar wrote:
> John...
>
> The easy way to do this uses a computerized Smith chart program (free
> of course). It's got all the math buried inside and a nice interactive
> interface....Just drive the component values around until it matches
> and read off the results. No need to wrestle with the math and not
> really any need to understand how to design with Smith charts. In
> about 5 minutes I got the following...
>
> 144 MHz
> 14.8 pF tube and strays
> 128 degees long, 50 Ohm plate line
> 4.3 pF to ground tuning cap
> 3.15 pF output coupling cap
>
> Note that the tube capacitance and the output tuning/loading
> capacitance _both_ act to shorten the line, so it ends up only a bit
> more than a quarter wave (90 degrees) long. If you make the line
> longer, tuning cap gets smaller. If you make the line high impedance,
> it gets shorter for the same tuning cap.
>
> I used gsmc-1.1, but linsmith (Linux) or the RFDude's smith chart
> program (Windows) works about the same way. A google search will take
> you right to any of these.
>
> jeff, wa1hco
>
> David Kirkby wrote:
>
>> JMLTINC@aol.com wrote:
>>
>>
>>
>>> Hi again guys-
>>>
>>> Sorry if I was not clear about my question. Let me try it from a
>>> different angle. Specifically:
>>>
>>> How do you calculate the length for a 1/2wl plate line????
>>> That is, what is the formula?
>>>
>>> I already know:
>>> Freq = 144Mhz
>>> Zo = 87 ohms
>>> Ctotal (tube, strays, tune) = 74 ohms
>>>
>>> Thanks-
>>> John, N9RF
>>>
>>>
>>>
>>
>> I assume you mean Ctotal is 14.8pF, giving a reactance of 74 Ohms.
>> Anyway, forget the numbers - a general formula is better. I assume
>> the tube is at one end of a line, and the other end is a high
>> impedance too.
>>
>> The half-wave line is open circuited at one end, with the tube at the
>> other. But there are an infinite number of combinations of line
>> length that can be made to resonate, in different modes (0.5 lambda,
>> 1 lambda, 1.5 lambda etc), all open circuit at one end. Hence care is
>> needed when getting a formula that you get the right solution. I seem
>> to find a very different one from the results you give, but I'll
>> leave it for others to find a fault if I have made an error.
>>
>> I had a go at deriving a formula. I did not derive it from first
>> principles, but don't use an 'amateur knudges' that might be valid,
>> but might not be. Reducing my solution to its simplest useful form I
>> got:
>>
>> 2 Pi f Ctotal Zo == - Tan(2 Pi f L / c)
>>
>> where L is the length of the line you want to find in m.
>> c is the velocity of light 3 x 10^8 m/s
>> f is the frequency in Hz.
>> Pi=3.14159
>> Ctotal is the output capacitance in Farads.
>> Zo is the impedance of the transmission line.
>>
>> Attempting to take the ArcTan's of both sides never seemed to work to
>> well for me, although it would have in principle given a simpler
>> expression. But I'm pretty poor at Maths, so it's probably me -
>> either that or a bug in Mathematica. I might have a go in Matlab
>> later and see if I do any better with that, but I'm not such a
>> competent user of that.
>>
>> I understand when doing this you loose all except one solution, but I
>> seem to get a totally wild solution. Anyway, forgetting about trying
>> to invert fully, but instead doing a simple numerical solution (which
>> you could do on a calculator with a bit of thought about it), I tried
>> it in Mathematica.
>>
>> Just for the record, 'FindRoot' is a function in Mathematica that
>> looks a value of L such that the LHS == RHS. There are an infinite
>> number of solutions, so I tell it to look around 0.6m, as I know
>> there is one around there, which I *believe* is the right one, but
>> could be wrong.
>>
>>
>> FindRoot[ 2 Pi 144000000 14.8 10^-12 89 == - Tan [2 Pi L 144 10^6/ (3
>> 10^8) ], {L, 0.6}]
>> {L -> 0.752312}
>>
>> which indicates a solution for .752m. If you insist on working in
>> inches, I'll leave you the job of converting it!
>>
>> There are also solutions at -0.289m (not physically realisable, as
>> length <0) and at 1.794m, which is a higher mode. There's nothing
>> wrong with having negative solutions - mathematically they are valid,
>> but they are a bit hard to make.
>>
>> Proof of my formula - there may be an error here.
>> ****************************************
>>
>> The *general* equation for the input impedance of a lossless
>> transmission line of length 'L' , characteristic impedance Zo,
>> terminated in a load impedance of Zload is:
>>
>> Zin=Zo * (Zload cos (beta * L ) + j Zo Sin(beta * L))/(Zo Cos(beta *
>> L) + j Zload Sin(beta * L))
>>
>> where beta = 2*pi/lambda
>> j is the complex operator.
>> and c=3 10^8 m/s.
>>
>> That's valid for real and complex loads.
>>
>> If Zload =0 (short circuit, which it would be for a quarter wave
>> line), then that complex equation reduces to
>> Zin = j Zo tan (beta L) - this is just stated so you can redo it for
>> a quarter wave if you want.
>>
>> If Zload=Infinity (open circuit, which is what you will do for a
>> half-wave line), then that big complex equation reduces to
>>
>> Zin = - j cot[beta L] which simplifies things quite a bit.
>>
>> ( Note: cot(x) = 1/tan(x) )
>>
>> The reactance of the tube is
>>
>> Xc = -j/(2 Pi f Ctotal))
>>
>> But for resonance, Xc = - Zin, since you must balance the capacitive
>> (negative) reactance of the tube with an inductive (positive)
>> reactance from the transmission line. (The signs of the reactance
>> must be opposite).
>>
>> Hence -j/(2 Pi f Ctotal) == j Zo cot (beta L) for resonance
>>
>> imaginary parts cancel, so
>>
>> -1/(2 Pi f Ctotal) == Zo cot (beta L)
>>
>> Inverting.
>>
>> 2 Pi f Cotal Zo == - Tan(beta L)
>>
>>
>> using the fact beta = 2 Pi/lambda and 1/lambda = f/c, gives:
>>
>> 2 Pi f Ctotal Zo == - Tan(2 Pi f L / c)
>>
>> There might be an error in that lot, but I'm sure someone will pick
>> it to bits (I hope so anyway). It is of course a theoretical
>> computation, and ignores things such as the fact the transmission
>> line is short, so its not really too valid to give it a
>> characteristic impedance, when end effects are significant.
>>
>>
>>
>
>
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