I need to take an audio signal from the output of the terminals of a valve amplifier, to send to the input of a Bheringher power amplifier.
I currently have that configuration (I have no other possibility) with a DaytonSPA250 board amplifier that has high and low level inputs. But I don't have the circuit of it.
I know that it is feasible then, (I understand that I should build a resistive voltage divider) but I have my doubts if it applies exactly the same to a valve amplifier or there would be some specific inconvenience.
Could someone guide me about it? Any link? Thanks in advance !
I currently have that configuration (I have no other possibility) with a DaytonSPA250 board amplifier that has high and low level inputs. But I don't have the circuit of it.
I know that it is feasible then, (I understand that I should build a resistive voltage divider) but I have my doubts if it applies exactly the same to a valve amplifier or there would be some specific inconvenience.
Could someone guide me about it? Any link? Thanks in advance !
First make sure the output terminal 0 Ohm (common) tap is grounded, some aren't.
If so, connect the resistor divider between the 16 and 0 taps as follows.
Connect a 1k to the 0 tap, and a 20k to the 16 tap. Connect the
free ends of the resistors together, and to the hot cable conductor.
Connect the ground shield of the cable to the amplifier 0 Ohm tap.
If the level is too high, increase the 20k.
If so, connect the resistor divider between the 16 and 0 taps as follows.
Connect a 1k to the 0 tap, and a 20k to the 16 tap. Connect the
free ends of the resistors together, and to the hot cable conductor.
Connect the ground shield of the cable to the amplifier 0 Ohm tap.
If the level is too high, increase the 20k.
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Thanks rayma!
, the amplifier has 4 and 8 ohms terminals.I just found this link, and it is as you say, I will have to calculate for an 8 Ohms terminal instead of 16 Ohms.
What about the indices of distortion, modification of the damping factor, etc., etc.?
There are no important variations in them to affect a Hi Fi chain?
Should I use some kind of special low noise resistance?
Voltage divider - Wikipedia
Hooking this up won't affect the amplifier/speaker (unless something shorts), it's a light load.
There's a chance you might get some hum in one of the speakers, but try it first and see.
Try the 20k first, and if it's too loud, try around 30k instead.
You can use any type of resistors you prefer or happen to have, and the wattage and value
are not critical at all. The divider makes up for the valve amplifier's, gain so the Behringer
won't be overloaded. Remember that the input to the Behringer will vary along with the input
to the valve amp.
There's a chance you might get some hum in one of the speakers, but try it first and see.
Try the 20k first, and if it's too loud, try around 30k instead.
You can use any type of resistors you prefer or happen to have, and the wattage and value
are not critical at all. The divider makes up for the valve amplifier's, gain so the Behringer
won't be overloaded. Remember that the input to the Behringer will vary along with the input
to the valve amp.
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Well, it's good news. I have not yet bought the Behringer, I am analyzing whether the connection is feasible and simple, I could even use 2 potentiometers instead of the resistors ... I will keep you informed, thanks again. !
PD: I don't like the design of the Inuke model, and my whole system is made of components with black fronts. So I would go for this one, sorry for the OT.
PD: I don't like the design of the Inuke model, and my whole system is made of components with black fronts. So I would go for this one, sorry for the OT.
Attachments
Prima Luna Dialogue Two
It is a modern amplifier that has no problems if you carelessly leave it without load, of course that should not last long, it is not a matter of tempting the devil .......
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I will study your resistance advice in parallel, I want to make it as accurate as possible, thanks.
I understand what you say, but it is the reality.
I have no other possibility because, as rayma pointed out earlier, it is that all components can be controlled through the volume of the Prima Luna amplifier. It has an output to record tapes, etc. (monitor) but is fixed level.
It is about achieving the correct overlapping point of the subwoofer amplifier (dedicated) with the valve amplifier (main), - crossing point, gain and phase - and then controlling everything from the PL volume.
Yes, this is very important, a valve power amp must be properly loaded with a dummy load if not connected to a loudspeaker.
Its the only safe way to interface to any valve amplifier as its isolated from deadly voltages by the transformer. Its probably the only part of an amp where the AC voltages are reasonable and at low impedance.
You can always add another transformer to do the job I suppose, but why not reuse the output transformer?
t is right. Dayton SPA250 is gone, Behringer NX4000D will enter service
I have not found a better way to provide the signal to the Bheringer than using a voltage divider from the terminals, as it is running smoothly in the Dayton. I can accommodate all connections and resistors inside a small box.
The PL weighs 30 KG, manipulating it is complicated. I could take a signal from the potentiometer, and place an RCA token somewhere on the back panel, but that would ruin it and ruin its resale value. I don't plan to change it, but you never know .....
I appreciate all the opinions again.
https://www.parts-express.com/pedocs/manuals/300-803--dayton-audio-spa250-manual.pdf
https://static1.squarespace.com/sta...1b631ba3f1a85b91/1469820793007/dia2manual.pdf
Attachments
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.
P = IE = I(IR) = I²R Likewise, rearranging Ohm's Law
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.
Vout = Vin • Rzero-side / (Rtop + Rzero-side)
Vout / Vin = x = R0 / (R0 + R8)
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 × 10.7Isn'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,
GoatGuy ✓
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 … Power product Law.
P = IE = I(IR) = I²R
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) I = E/R … then substituting I in P=IE
P = IE = (E/R)E = E²/R … and rearranging for E
E² = PR
E = √( PR )
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!).E = √(80)
E ≈ 9 volts ('cuz 9×9 = 81, which is close enough)
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. x = 0.776 ÷ 9
x = 0.086
Vout = Vin • Rzero-side / (Rtop + Rzero-side)
Vout / Vin = x = R0 / (R0 + R8)
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
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,
GoatGuy ✓
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I regret not having been more explicit, (I thought it was clear
) the main FR coverage of the system is provided with the Troels DTQWTII PL and cabinets.Although these cabinets were built with the hope that they will cover the spectrum from 20 to 40 hz (yes, general opinions say that there is not much music there, but it is a topic to discuss, and that I do not share) not really They do it.
So I decided to build two subwoofers, and I'm near the end of the adventure, see this link.
Eminence Delta Pro 18 A in prism sealed 150 liters
So, to cover the subfrequency zone, I use the current Dayton board amplifier.
But, I'm "opening the umbrella before it rains", and maybe I need more power.
The JBL vintage cabinets that I currently use as subwoofers are located very close to my listening position, in the near field.
But the new ones will go in a distant field several meters from my "sweet spot", they are corners and they will go on two front walls. Then perhaps
You need to go for an amplifier with more power.
I will only know when they are finished and installed, the room gain plays an important role here.
Thanks for the calculations, your dedication is greatly appreciated !!!
Attachments
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Thank you friend, but I think it will not be necessary to intervene on the OPT with other windings.
I am using an output from my valve amp to drive a subwoofer amp and it works fine.
I use a MiniDSP for crossover and EQ and so I built a little input board with a voltage divider as described above and installed it in a separate enclosure with the MiniDSP. You will need to match the level of the sub-amp with the main amp. If your sub amp has a level/volume control you're all good.
academia₅₀;5998387 said:I regret not having been more explicit … Thanks for the calculations, your dedication is greatly appreciated !!!
You're very welcome.
This may sound a little 'trivial', but a perfectly useful alternative would be to just use a 'pot'. Variable resistor. Use of a really nice high-grade unit, with multiple-contact wiper and so on, is highly stable when set. Use of a pot with an 'audio taper' gives it quite a large controllable divider-range.
Get out the drill, a half-inch bit, and an aluminum project box. You could put it together in less than an hour, I'd bet. Pass-thru device. Amp→box→speakers … with box→tapped-output as the side-effect.
One could also squash a quite-nice analog low-pass-filter in the box, with a few resistors and capacitors. Done that way, you could try all sorts of “what-if” experiments. Great learning experience.
Having a variable resistor also REALLY makes “what's the best value?” trivial. You can adjust it to your heart's content, and 'settle' on a setting. Then, if you want to commit that to hard-resistor values, just measure the two halves of the variable resistor. (hot to wiper, and wiper to ground). Done.
Remembering from all the exhausting math that it is the RATIO that is important (not the particular values), then relatiely standard, remarkably low-cost resistors would be procured and strapped into service.
Anyway…
Thoughts…
Good thoughts…
Just Saying,
GoatGuy ✓
What power does your amplifier deliver and what impedance terminal do you use? Maybe you could help me not to do the calculations.
PL has 38 watts per channel in UL
Thanks in advance
academia₅₀;5999343 said:
Sigh… you could do it the long way:
VRMS = √( P • R )
VRMS = √( 38 × 8 )
VRMS = 17.4 V
VIN = 0 dBu … = 0.775 V
k = VIN / VRMS
k = 0.775 ÷ 17.4
k = 0.0445
m = (1 - k) / k;
m = 21.5
RHOT = RGND • m
22.5 kΩ = 1.0 kΩ × 21.5 factor;
… or the short way with a single formula …VRMS = √( 38 × 8 )
VRMS = 17.4 V
VIN = 0 dBu … = 0.775 V
k = VIN / VRMS
k = 0.775 ÷ 17.4
k = 0.0445
m = (1 - k) / k;
m = 21.5
RHOT = RGND • m
22.5 kΩ = 1.0 kΩ × 21.5 factor;
m = ( √( PR ) - VAMP INPUT ) / VAMP INPUT
Where we have napkin-guessed that VAMP INPUT is about 0.775 volts som = ( √( 38 × 8 ) - 0.776 ) ÷ 0.776
m = 21.5
m = 21.5
Not that my line of thinking is being read… but there you are. Life is good. For me, audio band engineering is more a mathematical endeavor than a magical one. Still, there is a lot of magic in many of these fora!
Just Saying,
GoatGuy ✓
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