[ARC5] [Milsurplus] 6AQ5 Tank

[jmfranke]

Exactly, he said:
>>>> I was looking at changing the tank to output at 3.9 MC,
>>>> but if Xc = 166 ohms is correct, the tank cap
>>>> at 3.9 MC would be 6.7 uFd.  That can't be right.

But we do agree on the best answer, "The best answer is to up the L and C by 
about 6x each."

Bestest,

John  WA4WDL


--------------------------------------------------
From: "J. Forster" <jfor at quik.com>
Sent: Wednesday, July 20, 2011 7:50 PM
To: "jmfranke" <jmfranke at cox.net>
Cc: "David Stinson" <arc5 at ix.netcom.com>; "ARC-5 List" 
<arc5 at mailman.qth.net>; <milsurplus at mailman.qth.net>
Subject: Re: [Milsurplus] 6AQ5 Tank

> Check the OP. He was questioning the huge C to resonate the same inductor,
> IMO.
>
> The best answer is to up the L and C by about 6x each.
>
> Best,
>
> -John
>
> ==============
>
>
>> But he wanted the Xc to be constant! To do that you have to increase C by
>> a
>> factor of 6. And if you increase C by 6 and L by 6, you get the factor of
>> 36.
>>
>> John  WA4WDL
>>
>> --------------------------------------------------
>> From: "J. Forster" <jfor at quik.com>
>> Sent: Wednesday, July 20, 2011 7:27 PM
>> To: "jmfranke" <jmfranke at cox.net>
>> Cc: "David Stinson" <arc5 at ix.netcom.com>; "ARC-5 List"
>> <arc5 at mailman.qth.net>; <milsurplus at mailman.qth.net>
>> Subject: Re: [Milsurplus] 6AQ5 Tank
>>
>>> If you want to change the frequency by a factor of 6, you need to change
>>> the LC product by 36. If the L stays constant, you have to increase the
>>> C
>>> by 36x -->  36 * 40pF = 1440 pF.
>>>
>>> Best,
>>>
>>> -John
>>>
>>> ================
>>>
>>>
>>>> John,
>>>>
>>>> Yes, you are correct about the coil. I was working the capacitance
>>>> issue
>>>> and
>>>> arrived at the same number as you for the capacitor. And, having
>>>> changed
>>>> the
>>>> capacitance by a factor of 6 the coil inductance would have to increase
>>>> by
>>>> a
>>>> factor of 6 for resonance. Working on maintaining the same reactance
>>>> values
>>>> for the coil and/or the capacitor should and does bring us both to the
>>>> same
>>>> point. Ignoring any change in resistive elements, Q would be maintained
>>>> by
>>>> maintaining the same reactance values as the frequency is changed.
>>>>
>>>> John  WA4WDL
>>>>
>>>> --------------------------------------------------
>>>> From: "J. Forster" <jfor at quik.com>
>>>> Sent: Wednesday, July 20, 2011 7:04 PM
>>>> To: "David Stinson" <arc5 at ix.netcom.com>
>>>> Cc: "ARC-5 List" <arc5 at mailman.qth.net>; <milsurplus at mailman.qth.net>
>>>> Subject: Re: [Milsurplus] 6AQ5 Tank
>>>>
>>>>>> Ok all you smart people-
>>>>>> I'm looking at a 6AQ5 Pierce crystal oscillator/tripler
>>>>>> with a plate tank circuit which outputs at 24 MC.
>>>>>> The plate tank capacitor is 40 pFd.
>>>>>> The capacitive reactance at 24 MC is only 166 ohms.
>>>>>
>>>>> You have to multiply by the Q. If the Q=100, it will look like 16.6K
>>>>>
>>>>>> This seems low for a tube tank.  Output capac.
>>>>>> of 6AQ5  is 6 pFd,  so that doesn't change the picture much.
>>>>>
>>>>> Agreed.
>>>>>
>>>>>> I was looking at changing the tank to output at 3.9 MC,
>>>>>> but if Xc = 166 ohms is correct, the tank cap
>>>>>> at 3.9 MC would be 6.7 uFd.  That can't be right.
>>>>>
>>>>> Let's see:
>>>>>
>>>>> The frequency scales as the SQ. RT. of LC
>>>>>
>>>>> 24 MHz / 3.9 MHz = about 6, so the C needs to be 6*6 *40 pF = 6*240 =
>>>>> 1440
>>>>> pF.
>>>>>
>>>>> This is probably still too large. I'd increase the number of turns on
>>>>> the
>>>>> coil by about 6 and the C to about 240 pF.
>>>>>
>>>>> Best,
>>>>>
>>>>> -John
>>>>>
>>>>> ================
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>> What's up with this?
>>>>>>
>>>>>>
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>>>>>
>>>>>
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