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Old 11th July 2012, 04:41 PM   #1
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Default A bit of DIY advice required

I attach below a copy of a Borbely headphone/line amp design. What I cant understand (and I am probably being stupid) is how the P Channel common source mosfet Q2 is fully biased into operation.

The class A standing current form the constant current source Q3 is 10mA. That gives 2v across the 200 ohm source resistor into Q2. That gives the source of Q2 source sitting at 24-2v=22v.

Now the ECC86 valve is operated at 2.35mA per side, which gives 1.17 volts across each 499 ohm resistor in the current mirror. Add 0.6V for a base emitter junction and I see the Gate of Q2 sitting at 24-1.17-0.6=22.23 volts.

So that gives a gate to source voltage of just 0.23volts, yet the 2sj79 takes around 2.5 volts to be fully switched on.

So my question, to anyone kind enough to help me is why does Q2 function. Is it correctly biased and if so, please explain to me how.

Many thanks
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Old 11th July 2012, 04:51 PM   #2
kevinkr is offline kevinkr  United States
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You've forgotten the triode end of things, the voltage on the collector of that mirror is determined by the operating point (dc condition) the 6GM8 diff pair establishes in order to close the feedback loop and drive the output to 0Vdc.
"To argue with a person who has renounced the use of reason is like administering medicine to the dead." - Thomas Paine
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Old 11th July 2012, 05:07 PM   #3
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The mistake is to consider only .6V in Q1. If the case, the transistor is saturated, and can't no longer act as wanted. (linear mode).
Osvaldo F. Zappacosta. Electronic Engineer UTN FRA from 2001.
Argentine Ham Radio LW1DSE since 1987.
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Old 11th July 2012, 05:30 PM   #4
jcx is online now jcx  United States
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there are spice models for tubes too - you can sim the whole amp, poke at the sim
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Old 12th July 2012, 06:52 AM   #5
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Thanks to those for their input - never thought to model the circuit. I will do that.
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