I suggest 2.2k as well as a 1uF capacitor from the base of the second transistor to negative rail.
I no have program to calculate R (connect to B-T4).so i will try jump B-T3 to ground by some resistance 1->2K then will know how much drop the voltage on B-T3...
Oh . if I not wrong .
R1 51K ( supply 100V) with R =1K i will get voltage at B-T3~1,9V with 2,2K get 4V
seem 1K is enough to keep stable for input stage or can reduce more .
R1 51K ( supply 100V) with R =1K i will get voltage at B-T3~1,9V with 2,2K get 4V
seem 1K is enough to keep stable for input stage or can reduce more .
There is no reason R515 needs to return to +51V (~100V) when ground (51V ~49V) is fine and that cuts the heat in R515 in half. The base resistor for T4 needs to be >0.7V/I(R515) = > 350. The forward voltage on two diodes should not exceed 1.4V so I don't know how you get 1.9-2.2V???
350R for the base resistor of the VAS CCS is not enough, it should be calculated such that it allow a low supply voltage
to render the amp fully functional.
Fi if the resistance that feed the primary CCS is say 33K tied to the ground and 51V rail voltage there will be about 1.5ma
that goes to the reference diodes, with a 2.2k resistance feeding the secondary CCS 250uA through the 33k resistance
are enough to get the pruimary CCS functional without the current being fully drained through the base emitter junction
of the secondary CCS T4, this way only 1.3 + 33(0.25) = 9.55V voltage rail is necessary for the amp differential to be fully functional.
to render the amp fully functional.
Fi if the resistance that feed the primary CCS is say 33K tied to the ground and 51V rail voltage there will be about 1.5ma
that goes to the reference diodes, with a 2.2k resistance feeding the secondary CCS 250uA through the 33k resistance
are enough to get the pruimary CCS functional without the current being fully drained through the base emitter junction
of the secondary CCS T4, this way only 1.3 + 33(0.25) = 9.55V voltage rail is necessary for the amp differential to be fully functional.
Where did you get 33k? The circuit from the OP shows (104-1.3)/51K = ~2mA. A 2W resistor as shown could be ~20mA (1W/50V/2k5), but ~4mA (220mW/50V/12K) from a 1/2W resistor would be more reasonable. 2k2 is probably OK but I would do some careful simulation to check for rail sticking and Miller effects on the phase margin etc.
Spice is a great tool and design tool. The only problem is the sim THD levels are often an order or two better then measured.
Some sim Amps thd 0.00004%
Some sim Amps thd 0.00004%
I was talking about the case where the resistance would be 33k and tied to GND instead of the +51V rail, it s useless
to pick the current on the positive rail since this will increase considerably the resistance dissipation for a same current.
Basically yes.
Not always ideal as mentioned.
But considering saturation problems mentioned.
The added resistor to be mandatory.
Usually solves the issue.
Since 1.2k already used in parts bom.
Might as well make value 1.2k instead of adding 1k value to list.
Impedance of simple diode current source usually about
250 to 300k
Since no current mirror in differential, and no beta enhancement
in second stage/ Vas.
More improved current sources have higher impedance of 1meg up to 3meg possible.
So can improve amplifier performance or lower distortion/ THD.
In this simple design increasing current source impedance likely offers no improvement.
So these simple 2 diode current sources work well for this design.
As first posted earlier, using 2 sets of diodes.
Often the first stage instead of using switching diodes.
Green Led is used instead for differential current source.
diode drop be closer to 1.8 volts instead of 1.1 volts.
1.8 volt Requires higher degen resistor value for same current.
So current source impedance will improve slightly.
Instead of being around 250 to 300k
Impedance will be closer to 750k
So if original schematic is used.
It solves the saturation concerns.
And instead of using switching diodes, use green Led instead for differential.
The current source impedance is increased to 750k and can sometimes slightly improve THD.
If use green led instead of switching diodes need calculate again Re T3 & T4 to get correct current for input stage & VAS . that why i still use 2 diodes and add 1 resistance 1K.. could be 2k2 as [B]wahab[/B] say . but some amp this resistance only 130 ohm .
Offcourse can improve circuit by add current mirror but it is other story .
Offcourse can improve circuit by add current mirror but it is other story .
This type of current source doesn't have the issue at all. When the second transistor saturates, the first (Q7) turns off. It doesn't draw excessive base current from the diodes.
Agree that . seem Resistance should be 1->2K2 with original circuit, and if reduce more need modify to reduce VAS ( see Sanyo chip amp stk 3102)
That doesn't help against the problem this thread started with. The current then flows through the Schottky diode instead of the base, but still pulls down the rail.
Seem , if I m not wrong that with using same voltage fixed by 02 diodes , when negative clip , T4 in saturation , it make drop voltage to T3 and reduce VAS drive also as a .clipping limitation .