James Freeman
Member
- Joined
- Sep 8, 2021
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As my first (serious) post I want to gift an easy way to calculate negative feedback with a simple equation.
That's V=sqrt(P*R) to get voltage after the output transformer, then a simple voltage divider using the NFB and Tail resistors R2/(R1+R2), this is fed back into the phase inverter.
sqrt (Watt * Tap) * (Tail / (Tail + NFB))
Examples.
100W from 4Ω tap with 100k NFB resistor ('73 1959 Super Lead, 2203):
sqrt(100*4)*(4.7/(4.7+100)) = 0.89
50W from 8Ω tap with 100k NFB resistor (1987x)
sqrt(50*8)*(4.7/(4.7+100)) = 0.89
50W from 4Ω tap with 100k NFB resistor (2204):
sqrt(50*4)*(4.7/(4.7+100)) = 0.63
100W from 8Ω tap with 47k NFB resistor (1959 Plexi):
sqrt(100*8)*(4.7/(4.7+47)) = 2.57
100W from 16Ω tap with 82k NFB resistor (JVM410H):
sqrt(100*16)*(4.7/(4.7+82)) = 2.168
The correct NFB resistor for 2203 spec (0.89 as above) without changing the output tap is 205k.
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You can copy paste that equation into google and it will solve it for you.
Hope that saves you some time and puts decades of confusing forum posts to rest.
PS.
Sticky please.
That's V=sqrt(P*R) to get voltage after the output transformer, then a simple voltage divider using the NFB and Tail resistors R2/(R1+R2), this is fed back into the phase inverter.
sqrt (Watt * Tap) * (Tail / (Tail + NFB))
Examples.
100W from 4Ω tap with 100k NFB resistor ('73 1959 Super Lead, 2203):
sqrt(100*4)*(4.7/(4.7+100)) = 0.89
50W from 8Ω tap with 100k NFB resistor (1987x)
sqrt(50*8)*(4.7/(4.7+100)) = 0.89
50W from 4Ω tap with 100k NFB resistor (2204):
sqrt(50*4)*(4.7/(4.7+100)) = 0.63
100W from 8Ω tap with 47k NFB resistor (1959 Plexi):
sqrt(100*8)*(4.7/(4.7+47)) = 2.57
100W from 16Ω tap with 82k NFB resistor (JVM410H):
sqrt(100*16)*(4.7/(4.7+82)) = 2.168
The correct NFB resistor for 2203 spec (0.89 as above) without changing the output tap is 205k.
---
You can copy paste that equation into google and it will solve it for you.
Hope that saves you some time and puts decades of confusing forum posts to rest.
PS.
Sticky please.