L'Amp: A simple SIT Amp

Pass DIY Apprentice
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If you build Figure 11, with a 16 Ohm speaker, 1W THD goes to around .5%.

With the adjustable version (Figure 10) with a 50V volt supply, I found a nice sweet spot at around Id=2.42A and Vds=19.9V. There, the 1W THD was .069%, which is pretty nice. You'll get more gain, but less power, so your 16 Ohm speaker needs to be pretty efficient.

Now, the point of minimum distortion won't necessarily be your favorite spot. You may want to play with the setting a bit and see what you like best.

As for impedances much lower than 8 Ohms, I wouldn't use this little amp with 4 Ohm speakers.
 
juanitox,

Fig9 is just a test setup. Fig11 is probably the easiest to build because it does not require a separate bias supply or the input cap.

I am interested in how Fig11 would behave with R3 bypassed with a cap. I suspect that you get much of the advantage of Fig10, but without the added complexity of the external bias supply.
 
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juanitox,

Fig11 is probably the easiest to build because it does not require a separate bias supply.

I agree, Figure 11 is an easier first time build. Grade KD-33 is needed for that one. Other grades have a different Vgs.

About that input cap: Given enough leakage current, the input will be offset from zero, so eliminating the input cap isn't a slam dunk. Having tested only four parts and and not knowing what to expect from the universe of 2SK82s, I put one in the drawing.
 
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Hi Michael,

I am curious to know more about the effect of leakage current in this circuit. Specifically, what sort of non-zero voltage would be expected at higher leakage numbers?

Also, any thoughts on cap bypass for the source resistor?

I would like to try the circuit as drawn below.
 

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Pass DIY Apprentice
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and is 10000uF output cap the minimum value to get fullrange?

The cutoff frequency (-3dB) for the filter formed by the 10,000 uF output cap and an 8 Ohm speaker is:

Fc = 1/(2*PI*R*C(Farads)) = Hz
Fc = 1/(2*PI*8*.01) = 2Hz

Then the frequency response at 20Hz will be:

20*LOG10(1/(SQRT(1+(Fc/F)^2)) = dB
20*LOG(1/(SQRT(1+(2/20)^2)) = -0.04dB

You can do the math for a smaller cap and decide if it's suitable.

Also, the big resistor will work fine. You want 11-13 Ohms total. It will get very hot.

You could potentially string several 2 Ohm resitors in series and tap off to try different load values. But that might best be left for later.