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Author Topic: Output Tank design  (Read 4292 times)
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Ed/KB1HYS
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« on: December 22, 2008, 04:47:28 PM »

In a PI output tank designed for 80m operation, if I just doble the size of the inductor say from 13 to 26uH will that be enough to get on 160? (all other things being equal. 
 I think there might be a problem with the Q with the new inductor???
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73 de Ed/KB1HYS
Happiness is Hot Tubes, Cold 807's, and warm room filling AM Sound.
 "I've spent three quarters of my life trying to figure out how to do a $50 job for $.50, the rest I spent trying to come up with the $0.50" - D. Gingery
W3FJJ
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« Reply #1 on: December 22, 2008, 04:57:49 PM »

In a PI output tank designed for 80m operation, if I just doble the size of the inductor say from 13 to 26uH will that be enough to get on 160? (all other things being equal. 
 I think there might be a problem with the Q with the new inductor???

I think so, but you'll also need to double the plate and loading capacitance also to maintain the same
Q of the circuit. ARRL (1986) handbook has a nice table of typical values for various plate impedances.
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W7XXX
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« Reply #2 on: December 22, 2008, 05:06:41 PM »

To resonate I am guessing you will need to add capacitance to the tuning capacitor, it all depends on what value you have now. The L/C ratio is of importance. Also you may have to add capacitance to the loading cap depending how heavy you want to load up. It is common to see an additional fixed capacitor added (switched in) to the tuning circuit in the old tube multiband 80-10 mtr rigs on 80 mtrs and if you are adding 160 to such a rig, you may need to add a switch to add more capacitance in the tuning circuit for 160 mtrs.
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Steve - WB3HUZ
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« Reply #3 on: December 22, 2008, 05:10:30 PM »

Just download the spreadsheet to calculate Pi and PI-L networks off the Web and run it. No guessing needed.
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steve_qix
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Bap!


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« Reply #4 on: December 22, 2008, 05:13:53 PM »

In a PI output tank designed for 80m operation, if I just doble the size of the inductor say from 13 to 26uH will that be enough to get on 160? (all other things being equal. 
 I think there might be a problem with the Q with the new inductor???

Well, in order to obtain resonance you will need to double both the inductor AND the tuning cap.  This is because the capacitance and inductance values are multiplied together, and then square-root'ed (there's a non-word!).  So, in order to halve the frequency, you have to quadruple the L * C value (the square root of which is then taken), multiplied by 2 PI - take the reciprocal  (1/xxxx), which yields the resonant frequency.   You could increase the inductance by 4x, and maintain the same capacitance, however the Q of the circuit will be lower.  This may or may not be a problem.

Regards,

Steve


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WD8BIL
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« Reply #5 on: December 23, 2008, 07:28:05 AM »

PI Net Calculator
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Ed/KB1HYS
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« Reply #6 on: December 23, 2008, 10:22:05 AM »

Buddly, that PI tank Calculator is NEAT. 

Looks like I'll be able to get 160 with the two caps I have and a 35uH inductor. with a little headroom to boot. 

Thanks for the assist.
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73 de Ed/KB1HYS
Happiness is Hot Tubes, Cold 807's, and warm room filling AM Sound.
 "I've spent three quarters of my life trying to figure out how to do a $50 job for $.50, the rest I spent trying to come up with the $0.50" - D. Gingery
Steve - WB3HUZ
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« Reply #7 on: December 23, 2008, 10:55:36 AM »

Google is your friend.   Cheesy
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WU2D
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CW is just a narrower version of AM


« Reply #8 on: December 23, 2008, 11:10:20 AM »

Its a tankless job - so tanks!
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These are the good old days of AM
WD8BIL
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« Reply #9 on: December 23, 2008, 02:33:01 PM »

It is, ED. I found it to be quite accurate too.
With the coil calculator I was able to pre-set the 80 and 40 meter taps on an amp I built. Didn't hafta move them at all. It tuned up first time.

Good luck.
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VE3GZB
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« Reply #10 on: December 25, 2008, 11:40:53 PM »

In a PI output tank designed for 80m operation, if I just doble the size of the inductor say from 13 to 26uH will that be enough to get on 160? (all other things being equal. 
 I think there might be a problem with the Q with the new inductor???

The formula below:



Notice that frequency is Inversely proportional to the Square Root of L*C. Therefore merely doubling the inductance will only provide a relatively small change in resonant frequency.
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