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Started by Alexius II, September 22, 2010, 11:35:34 AM
Quote from: mensur on September 25, 2010, 06:08:07 PMUse MPF102's for both, voltage gain and as buffer.Use 680R for RS(voltage gain) without bypass cap, BTW. Cs reduce internal resistance of the channel, hence more gain.As for Rd use 2.2K.These are resistance values foe MPF102 FET (with -4.5V, 13.5mA Idss characteristics).From my cacs. it will bring you to -2.12V of Vg(humbacker sweetspot) and 3.46mA Id.Gain will be around 3 to 4x.As for buffer use same character. FET with 2.7K Rs, everything else remains the same.I don't like bipolars as buffers, you don't need much current there.
Quote from: Alexius II on September 26, 2010, 04:11:09 PMI was a kid and bit too young to remember anything from the old Yugoslavia, but they say that were the good times Anyway, thank you both. I went through some online theory pages and found a few other equations that helped me understand the few missing relations. I think I get it now... I hope :duhI will now go again through the calculations for my transistor:These are the equations I used for the gain stage:Vgs = -Id*RsVds = Vdd-Id*(Rd+Rs)Id = Idss*(1-(Vgs/Vp))^2 (the characteristic curve is drawn from this equation)gm0 = 2*IDSS/VPgm = gm0*(1-(VGS/VP))Av = (gm*Rd)/(1+gm*Rs) (if there is no Cs)My transistor:MPF102Vp = -4.5VIdss = 13.5mAI choose Vgs to be 1/2 of Vp, soVgs: 0.5*(-4.5V) = -2.25VId = 13.5mA*(1-(2.25V/4.5V))^2 = 13.5mA * 1/4 = 3.375mARs = Vgs/Id = 2.25V/3.375mA = 666ohm (closest is 680R)gm0 = 2*13.5mA/4.5V = 0.006mSgm = 0.006mS*(1-(2.25V/4.5V)) = 0.003mSThen I choose Rd. I understand that with larger Rd come larger voltage gain and output impedance + lower current.Rd = 2.2k (18V/2200R = 0.0081A = 8mA)Av = (0.003mS*2200R)/(1+0.003mS*680R) = 2.17 (= 6.7dB)Ok, I hope everything is OK this far.Then comes the buffer (aka "source follower" or "common drain amplifier") Only the Rs is necessary to bias it. I've read some theory about it, but I'm not sure. The way I understood it, you have to choose Rs in a way, that at the source you have half the voltage (Vcc). If this is correct, I then used the identical MPF102 transistor with calculated Id = 3.375mA (for Vgs = 1/2 Vp) inserted in the equation:1/2 Vcc = 9V = 3.357mA * RsRs = 9V / 0.003357A = 2680R = 2.7kIs this ok, or am I making things up? :loco