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21st October 2005, 08:57 PM  #11  
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21st October 2005, 09:00 PM  #12  
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22nd October 2005, 06:20 AM  #13  
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Join Date: Jan 2004
Location: Jakarta

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I've found a 6AU6 pentode to be easy to use and effective as a CCS, in the tail of a 6SL7 LTP which has its grounded grid at true ground (0v). My negative rail is 110v. I make sure the heater of the 6AU6 is at a safe voltage that doesn't exceed the rated heatercathode voltage limit 0f 100v. Current per triode in the LTP is 0.8mA, which means the plate current of the 6AU6 is 1.6mA. Voltage from 6AU6 plate to cathode is 112v, screen to cathode is 95v. If you look at the 6AU6 plate curves for screen voltage ~100v, you will see that they are very flat for such a low current. The plate current appears to be constant within 0.01%. For more current, e.g. as a CCS for a 6SN7 LTP, a negative rail of 150v would be preferable. 

22nd October 2005, 03:47 PM  #14  
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Join Date: Apr 2004
Location: Philly

Re: LongTailedPair gain > how to calculate?
Quote:
"Differential amplifier (phase splitter etc) made from a 12AX7. The plate load resistors are 150k each, and the next stage grid resistors are 220k. The cathodes are tied together and fed from a 2 mA current source. What is the gain from either input to either output? Solution: Effective Plate Resistance = 71k (plate resistance for these bias conditions) paralleled by 150k, paralleled by 220k or about 39k. Effective cathode resistance. The current source is assumed to have a very high resistance, so may be neglected. Each cathode is "looking at" the other cathode. The gm under these bias conditions is about 1400 umhos (710 ohms). Therefore the cathode resistance is 710 + 710 or about 1420 ohms. Stage gain = 39000/1420 = 27.5. Note that the differential gain (one input to both outputs) is double this (55), since equal and opposite output is provided by each plate. This method also works for sand state devices as well."
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