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what wattage value would I use for the resistors in this schematic

adrian21f

May 27, 2015
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The parts are: R1=10k, R2=2.2K, R3=100 C1=47uF/50V, C2=2.2nF, C3=10pF, SW=switch, B=Battery 9V.
The coil L should be closely wound 5 turns (start tunning with the closely wound distance of turns and then play with it and the C4 capacitor, to find the desired frequency )of enameled copper wire 1.5mm and internal coil diameter of 1cm (leads 2x20mm).
The variable capacitor C4 should be air trimmer capacitor rating 4 to 30 pf or the closest to this value.
As for the transistor Q its the 2N3553. Its a RF Power Silicon NPN Transistor in a TO39 type package designed to be used in amplifiers and oscillator applications in military and industrial equipment.
image.jpg
 

Fish4Fun

So long, and Thanks for all the Fish!
Aug 27, 2013
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Hey Adrian21f ! Welcome to the the forum!

So assuming the nominal circuit voltage is 9V * 3 = 27V, and the smallest resistor (r3) = 100 ohms a first order approximation would go something like this:

Code:
E = iR --> 27V = i * 100 ==> Imax = 0.27A = 270mA
P= iE -> 0.27A * 27V = 7.29W ==> r3 Power rating ~10W

Or, more simply:
P = E^2 / R

But obviously this is over-kill.....The circuit is some type of RLC resonant oscillator, and the actual current in r3 is dictated by the impedance of c4 and L at the tuned frequency, the DCR of L, ESR of c4 and the conductance of Q. You might have a look @ this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/serres.html#c1 , this http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html#c1 , and this http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html#c1 .

My gut feeling is that r3 should be rated @ ~ 1W to 3W and if I were building it in a hurry and wanted to be certain it would work w/o doing the math then I would likely go with 10 * 1/2W 1k in parallel, or 10 * 1/2W 10 ohm in series.....but that is CERTAINLY NOT because I did the math....

To be honest this looks a lot like a homework problem, so I hope if this is the case you will take the time to read through the links and actually do the math; regardless it would certainly be sloppy practice NOT to do the math.

Good Luck!

Fish
 

adrian21f

May 27, 2015
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May 27, 2015
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wow thanks... very helpful and easy to understand ill for sure take a look at those.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Just to add to the discussion, the average voltage across R3 is going to be about 0.6V lower than the dc bias point of the base. That's about 1/6 of the supply voltage, or about 4.5V.

I'll leave it as an exercise to the reader to determine if Colin's advice above is appropriate. :)
 

BobK

Jan 5, 2010
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If that was meant to be a hit a Colin, it doesn't work out that way.

4.5-0.6 = 3.9

3.9 ^2 / 100 = 0.15 W, so 1/4 W is fine.

Bob
 
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