70v/100v to Line voltage divider

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

Studiogearlover

Well-known member
Joined
Mar 14, 2017
Messages
176
hello guys...could you please help me with this. I have a tube amplifier 25W with outputs of 5Ohms, 200 Ohms 70V and 400 Ohms 100v outputs.

I would like to convert any of the above to a line level in order to connect it to my mixer/AD converter.

The 10 to 1 voltage divider is I guess I need to follow but how do you calculate the resistors value for the above? What is the formula or calculation I need to follow? Please help me out with this.

Thanks  alot! Stay safe and healthy!

 
You'll need to load it first. Otherwise you could get a wacky frequency response. Do you have any high wattage resistors in your parts bin?
 
I think you need a dummy load.. if it's a guitar amp I heard some were equipped with the ability to drive another speaker (tweeter etc) and if it's an amp head it's designed specifically for this purpose.

Either way connecting a "powered" output into a line level device will break it.

If it's a tube guitar/bass amp I'd mic it if I were you but some prefer DI signal for reamping, although you get some tube sound from the amp circuitry with the dummy load.

For "real" DI you'd run the guitar through a box that splits the signal in two; the other comes straight from the pickups and the other goes to the amp which is miked and you'd multitrack and mix/reamp them in post (many newer transistor amps have a DI/thru output for this purpose)

The use of DI in recording is said to be useful if a multitrack is archived and remixed decades later as the sound is recorded "as is" without colouring it with amps, FX etc which makes it easier to reamp (re-record it from the desk/recorder through another amplifier)

As for the formulae for the R value it's hard to tell but a dummy load isn't much else than a power resistor in series, like 5W, your best bet is to test a few values (I guess line level in this context means 1Vrms max) but many consider resistors generating noise in audio circuits due to heat dissipation.

A "technical" term for dummy load is ballast resistor which is often used with LEDs, say you have a device with 24V power supply and want a 5V power on-LED so you'd put 1MOhm resistor in series first and downsize from there, ending up probably somewhere around 10-100kOhm.

To stay safe you could use two, one in + and other in - but someone more experienced could confirm this (as diodes such as LEDs don't need the resistor in minus because they're polarized and operate with DC)

Some power amplifiers have ground lifts which means switching off the minus but their use is seen as questionable (I've seen one actually, and afaik some DI boxes have ground lifts, I've never owned one though)

In theory adding one shunt (from + to -) acts as a dummy load too, in power supplies they're called load resistors (I once tested this with a diode bridge into a cap but I can't vouch for it)
 
Hello Thank you both for the suggestions......high wattage resistors...yes I have a few I think,  but thats easy to source..

This is a 25W EL34 tube amplifier with a microphone input and several taps of outputs....5 Ohms, 15 Ohms and 200 Ohms (70v) and 400 Ohms (100V)

I wonder what resistor \i would need to dummy load the 70V ? Must be massive right?

Attached what I thought it is the way to follow however I do not know the source of this and would that be a correct way to do it?

But still I dont know how to calculate the values for the resistors for 70V? is there no such calculation formula?

Many thanks again...

Of shall I purchase this instead?

https://www.canford.co.uk/Products/81651/58-443_RDL-TX-70A-AUDIO-TRANSFORMER-Speaker-level-input-25-70-100V-screw-terminal-I-O

 

Attachments

  • opt 1.JPG
    opt 1.JPG
    16.5 KB · Views: 14
Studiogearlover said:
Hello Thank you both for the suggestions......high wattage resistors...yes I have a few I think,  but thats easy to source..

This is a 25W EL34 tube amplifier with a microphone input and several taps of outputs....5 Ohms, 15 Ohms and 200 Ohms (70v) and 400 Ohms (100V)

I wonder what resistor \i would need to dummy load the 70V ? Must be massive right?

Attached what I thought it is the way to follow however I do not know the source of this and would that be a correct way to do it?

But still I dont know how to calculate the values for the resistors for 70V? is there no such calculation formula?

Many thanks again...

Of shall I purchase this instead?

https://www.canford.co.uk/Products/81651/58-443_RDL-TX-70A-AUDIO-TRANSFORMER-Speaker-level-input-25-70-100V-screw-terminal-I-O
25W into 5 ohms is 5Vrms. Thgis is already sufficient to drive a line input; in fact you need some attenuation.
You need to load the output with a 5 ohms 25+W and install an attenuator.
Making the attenuator 4:1 would create a nominal 1.25Vrms output, which is close enough to +4dBu.
The attenuator can be done with 1kohm + 330 ohms, or 500/330/500 if you want a balanced attenuator.
 
abbey road d enfer said:
25W into 5 ohms is 5Vrms. Thgis is already sufficient to drive a line input; in fact you need some attenuation.
You need to load the output with a 5 ohms 25+W and install an attenuator.
Making the attenuator 4:1 would create a nominal 1.25Vrms output, which is close enough to +4dBu.
The attenuator can be done with 1kohm + 330 ohms, or 500/330/500 if you want a balanced attenuator.

Thank you so much !

Could you please confirm if I understand you correctly:

1. dummy load needed for 5 Ohms with25W?
2. 1kOhms/ 330 Ohms for unbalance is the way how the attachment here is displaying the 10kOhm:1kOhm * so the suggestion is more like the 4; 1 right?

Would you put a 0,01uF capcitor there too or this is not needed? to act as a lo w ass filter? and a 0,25W resistor would be enough or this should be much larger resistors?

Thank you so much once again!  :)

 

Attachments

  • opt 2.JPG
    opt 2.JPG
    24 KB · Views: 12
Studiogearlover said:
Thank you so much !

Could you please confirm if I understand you correctly:

1. dummy load needed for 5 Ohms with25W?
2. 1kOhms/ 330 Ohms for unbalance is the way how the attachment here is displaying the 10kOhm:1kOhm * so the suggestion is more like the 4; 1 right?
that's correct. It's not critical at all. Usage will tell you if it needs to be altered.

Would you put a 0,01uF capcitor there too or this is not needed? to act as a lo w ass filter?
Why would you need to low-pass the signal? It's probably already low-passed by the period electronics.

[/quote] and a 0,25W resistor would be enough or this should be much larger resistors?5V into 1kohm is 25mW, so you're safe by a factor 10.
 
abbey road d enfer said:
that's correct. It's not critical at all. Usage will tell you if it needs to be altered.
Why would you need to low-pass the signal? It's probably already low-passed by the period electronics.

and a 0,25W resistor would be enough or this should be much larger resistors?5V into 1kohm is 25mW, so you're safe by a factor 10.

Thank you soo much. I am already on it...got a 330Ohm 1W and a 1kOhm 1 W resistor. I do not have a 25W + of the 5 ohm, will order that from Farnell i think....

Many thanks once again. !

Will take a pic wham I am done with this project :)

Have a great weekend.

 
Studiogearlover said:
Could you please confirm if I understand you correctly:

1. dummy load needed for 5 Ohms with25W?
Actually it should be more like 50W. Otherwise it could get quite hot with a lot of signal. Even at 50W it will get warm. If you really want to do this properly you should get a little die-cast aluminum enclosure and one of those aluminum encased resistors that have bolt flanges. Then bolt it to the inside through the enclosure into a heat sink which could just be a big chunk of aluminum or heat sink form Ebay. Those resistors are smaller for their wattage but only because they require a heat sink. And use a little heat sink mounting compound if you have it. It doens't have to be exactly 5 ohms though. You could use 3 small 2 ohm in series for example just to spread things out. Then just strap a 500 ohm log pot across that. Something cheap with a knurled shaft mounted to said enclosure would be fine. The power resistor will take most of the load (5 ohms vs 500 ohms). Google "Ohm's Law". To figure the power through the pot: 5V / 500 = 10mA * 5V = 0.05W. Conventional pots are usually like 0.25W so all good. Then just tap into the output from the wiper and ground. Crank the amp and adjust the pot to taste.

Studiogearlover said:
2. 1kOhms/ 330 Ohms for unbalance is the way how the attachment here is displaying the 10kOhm:1kOhm * so the suggestion is more like the 4; 1 right?
It might not be a bad idea to use a balanced attenuator like the second circuit here:

lpad.jpg


with the 5 ohm power resistor would be across the "In" terminals.

Although it wouldn't really be "balanced" since there is no signal on the ground ends. But if you do 500 ohms for the series resistors and 330 for the shunt, that would isolate the grounds of the two devices and so you would get some benefit of attenuating ground noise between the two devices and, perhaps more important, limiting current between grounds if there is a difference (which you should measure with a meter to make sure you don't have a bad ground loop - should be less then a few mV across the 500R of the ground side). You could make the shunt resistor a pot that is wired like a rheostat (meaning wiper is connected to one side so that it's a simple two terminal variable resistor). The pot could be like 500R.

Studiogearlover said:
Would you put a 0,01uF capcitor there too or this is not needed? to act as a lo w ass filter?
Probably not. The impedance of said attenuator is going to be a few hundred ohms so the capacitor would have to be little larger to even have an effect. And if you use a pot, the filter will shift up / down a little depending on the pot position so that's not really desirable. Google "rc filter calc".
 
I believe any winding could be loaded--might need to determine empirically what works best the 15 Ω tap could be loaded instead of 5. (In a paging sytem IDT there would be a dummy load.) Or even one of the CV taps--don't know what value to try there ,100Ω? -there should be some kind of load though to avoid damage to opt or el34s
 
shabtek said:
I believe any winding could be loaded--might need to determine empirically what works best
There's no "what works best".  The output needs to be loaded with a more or less correct impedance (tolerance is high there, typically +100/-50%.

Or even one of the CV taps--don't know what value to try there ,100Ω?
  25w@ 70V makes 196 ohms, 400 ohms @100V . It's all down to the availability of suitable resistors. However, since the 4 ohms winding is the one that will be used for outputting signal, I would favour loading that one.
 
Maybe I'm late but I would load 100V output, just to not leave open any secondary.  Also, if you will not play guitar in distorted mode, the power of the load resistor can be also lower because a microphone's signal effective power is low. 15-20W will suffice, IMO. Just to add that for 25W and 5ohms of the load,  the voltage is about 11V  so divider of 10:1 looks much  more suitable. However, I don't believe you will drive the amp to it's maximum so maybe already proposed 4:1 would be fine.
 
PRR said:
> I would load 100V output, just to not leave open any secondary.
Why??

PA tube output amps should withstand  significant mismatch of the output/load impedance at 70/100V output. 100V line can be fully loaded if the music is right, and minimally loaded when the party is over and all attenuators are set to off. But, if there is no load at all, at max output there is a voltage which can be significantly higher than 100V (25% or more) and some instability may occur (higher leak into feedback winding or so).  I modified in the past  similar amp (PA 35W, EL34) and had a lot of trouble with some partial oscillation until I loaded 100V output with few watts.
 

Latest posts

Back
Top