question about B+ Voltage division and plate voltage

[pc1093]

Hi All,
I got 3 problem in a 6SN7-KT88 PP.
1. The B+ starts from 525v and after a 4.7K ohm resistor reduces to 480v.So the current is 0.0095A ((525-480)/4700). however ,from 480v drop to 370v, the resistor is not 11.4K ohm(110/0.0095).The resistor also is not 5.2K ohm (50/0.095)for 370v dropping to 320v.
The question is how to calculate the resistor value for B+ Voltage drop?

2.In the first 6SN7, it gives 320v on the anode resistor. So the current is 0.0068A(320/47000). I draw the load line on the 6SN7 plate  characteristics. and them how to find out the operating point even i know the cathode resistor is 470 ohm?

3. how to find the resistor value(39K ohm) for the Negative Feedback ?

thank you.

[PRR]

> how to find out the operating point even i know the cathode resistor is 470 ohm?

(I'll use "500" to ease the math.) (We should allow for the parallel path through 39k and the OT, but this makes little difference.)

If 0mA flows in 500r we have 0V. If 4mA flows in 500r we have 2V. If 8mA flows in 500r we have 4V. Plot those points. The intersection with the plate line is the operating point. About 5mA 100V.

No, I can not get exactly the numbers written on the plan. Your plate curves are not the actual tube used, 20% deviations are no big deal.