Explain this preamp circuit

[Randall]

This is an Atomic reactor FR amp that I have never heard of. The pre/PI stage puzzles me. I assume Q1 is a clean gain stage, and that it passes through C2, but then I get lost. What is going on with the cathodes of V1 and the IC1? And how does V1 in this configuration create phase inversion? It seems quite unusual to me.

Reactor FR Schematic.pdf

[The Dude]

Yes, Q1 is just a gain stage. IC1 is a constant current regulator set up by R10- basically bias for the stage. I can't see much of an advantage to this circuit besides possibly some compression due to current clamping.

[Enzo]

JFET gain stage feeds through C2 straight into long tail pair phase inverter. In your typical amp the cathodes of the PI go through like a 470 ohm resistor to bias the tube grids. In this case they used a regulator to maintain steady current, but otherwise, just imagine a 470 ohm resistor in place of the regulator. LM334 is an adjustable current regulator.

Look at a Fender AB763 amp, like a Deluxe or something. This is really the same circuit. There is your 470 ohm resistor instead of the regulator IC serving the cathodes, there are the 1meg grid return resistors. Your tail resistor is 10k, while the Fender is 22k, but that is just a detail. The NFB comes in a slightly different place.

And your added a grid stopper to the input grid, at R9. Do you understand how the Fender circuit works? This works the same. The signal goes to the input grid, controlling current through the tube. WHich in turn affects the voltage across the tail resistor

[g1]

edit: Never mind, stated better above.

[loudthud]

The constant current source in the tail of the diff pair causes an equal signal (within resistor tolerance) on each plate. The usual 100K and 82K plate resistors are an attempt to equalize the two plate signals in the Fender circuit.

Note: the JFET is drawn with the arrow pointing the wrong way. The J270 is a P-channel device and the supply rail is -30V.

[Helmholtz]

And how does V1 in this configuration create phase inversion? It seems quite unusual to me.

As with all LTPIs, the cathode output signal of V1A feeds the cathode input of V1B (and vice versa for NFB).
So the plate signal of V1A is inverted, while the plate signal of V1B is in-phase with the V1A input.

[vintagekiki]

It's similar as MB Mk1
Instead of cathode resistor (1k5) it has a current diode regulator (1N5303)

bee.mif.pg.gda.pl/ciasteczkowypotwor/%23SM_scena/Mesa%20Boogie/Mesa_Boogie_Mark_I_Schematic.pdf

https://www.digitroncorp.com/Products-Overview/Current-Regulator-Diodes/1N5303

https://www.digitroncorp.com/getmedia/A26FAF4B-E2E6-4846-B564-E70F92E5F8E9/1N5283-1N5314-2c-1N7048-1N7055.aspx?ext=.pdf

About 1N5304 ¹⁾

Part number	NOM IP (mA) VS = 25V	MIN IP (mA) VS = 25V	MAX IP (mA) VS = 25V	Minimum dyn. Impedance VS = 25V ZS (MΩ)	Minimum knee Impedance VK = 6.0 V ZK = (MΩ)	Maximum limiting Voltage IL = 0.8 IS(min) VL (volts)
1N5304	1.80	1.62	1.98	0.420	0.074	1.75

[Randall]

Thanks as always gents!

[pdf64]

Why use the current source for the LTP bias, rather than the tail? Where its super high impedance would sort out the AC balance.

[Dave H]

Why use the current source for the LTP bias, rather than the tail? Where its super high impedance would sort out the AC balance.

It is in the tail It's ensuring a constant tail current through R6. I think R6 is redundant. It could be replaced by a short.

[loudthud]

Note that the LM334 only has a 40V breakdown Voltage, and requires a little over 0.8V to fully regulate current at 2mA.

[vintagekiki]

I think R6 is redundant. It could be replaced by a short.

No comment


http://www.valvewizard.co.uk/acltp.html

[Helmholtz]

The tube will force the voltage across the IC to be around 1.5V.

The 10k tail resistor lifts both cathode and grid voltages by a constant DC voltage of around 20V (constant current -> constant resistor voltage), while a normal tail resistor would carry some signal voltage. Not sure about the benefit.

[vintagekiki]

About "tail" resistor R6

https://www.tdpri.com/threads/how-the-long-tail-pair-phase-inverter-works.519077/

https://www.aikenamps.com/index.php/the-long-tail-pair

https://www.ampbooks.com/mobile/amplifier-calculators/long-tailed-pair/

[Dave H]

No comment


http://www.valvewizard.co.uk/acltp.html

Well, according to the wizard “The more (voltage) we allow for the tail, the larger the tail resistor will be and the better the balance” The balance is better because a larger tail resistor is a better approximation to a constant current sink. If you already have a CCS the resistor isn't needed. This has the advantage of there being no voltage drop across the tail resistor leaving more voltage swing available for the outputs. The voltage drop across the CCS will only be that necessary to bias the tubes to 1mA at the requires plate voltage.