Throwing out all the complications…

Assuming you are

__NOT__ powering actual speakers with the valve amplifier, but just want to take its valvish output and amplify that further with the Behringer, and reflecting on another poster's point that

*“valve amplifiers like at least a modest load”*, then a subtly different recommendation for resistors.

Let's have 3 output terminals, per your notes. 0 (

*“ground”*), 4 Ω, 8 Ω.

Hook RayMa's pair-of-resistors up, just as he says, to the 0 and 8 Ω output. Given that your valve amplifier is likely to be at

__least__ a 10 watt output (into 8 Ω) device, let's get a feeling for its output voltage.

There's a pair of fundamental electronics/electricity laws that come into play.

E = IR … Ohm's Law and

P = IE … Power product Law.

Using basic high-school algebra one can substitute 'E' in P=IE with 'IR' from the top equation, yielding

P = IE = I(IR) = I²R

Likewise, rearranging Ohm's Law

E = IR … rearranges to

I = E/R … then substituting I in P=IE

P = IE = (E/R)E = E²/R … and rearranging for E

E² = PR

E = √( PR )

Now we have a second quite-useful equation predicting the average voltage present on an 8 Ω load, having a certain power (10 watts)

E = √( PR = 10 × 8 )

E = √(80)

E ≈ 9 volts ('cuz 9×9 = 81, which is close enough)

Now, the INPUT to your Behringer amplifier likely won't want much more than 0 dBu, which is a funky unit if there e'er was one. Without resorting to more math, it is about 0.776 volts. So… the Behringer will be willing to take up to 0.776 volts

__average__ to produce

*“full rated output”* (which is my unresearched estimate!).

OK. We know the amplifier at 10 watts will be outputting about 9 volts from the 8 Ω winding. Kind of poetic. We want 0.776 volts.

0.776 = 9•x … where x = the divider ratio and • is multiply...

x = 0.776 ÷ 9

x = 0.086

This in turn drives the choice of resistors for the

__RayMa__ voltage reducer.

V_{out} = V_{in} • R_{zero-side} / (R_{top} + R_{zero-side})

V_{out} / V_{in} = x = R_{0} / (R_{0} + R_{8})

0.086 = R₀ / (R₀ + R₈)

0.086 (R₀ + R₈) = R₀

0.086 R₀ ⊕ 0.086 R₈ = R₀ … now get R₀ on the right side…

R₈ = (R₀ - 0.086 R₀) ÷ 0.086 … and factoring

R₈ = R₀ • (1–0.086)/0.086

R₈ = R₀ × 10.7

Isn't that cool? Choose an R₀, and R₈ pops right out with the 10.7 multiplier factor.

Now that we've been thru that, we can add ONE MORE RESISTOR to the circuit to give the valve amplifier a 'real load' of sorts.

It needs an 8 Ω load, on the 8 Ω tap. And one that can take the 10 or more watts of valve output. You could either buy a nice 25 watt wire-wound resistor for the job (which I recommend!), or a bunch of smaller 5 watt resistors to hook together.

Following this linkie …

25 W Wirewound Resistors | Mouser you will see many resistors that perfectly fit the bill for between $2 to $5 apiece. Cheap!

So, once you get one of these Big Boys, you hook it

__also to the 0 and 8 Ω__ terminals, so called

*“in parallel”* with the little-resistor-divider-RayMa-thingy.

Anyway, hope that helps.

Just Saying,

**Goat**Guy ✓