Advise for Active Summing Amp Neve BA283

[Golgoth]

Thanks a lot for the information! I have indeed contacted Dan and he's out of stock on these, my only option to get an original one is to buy a 1466 oscillator right now..

I'll try with Vintage King!

 

[avilen_tanaka]

OK, that is a reasonable specification. You do not say if the inputs and output are balanced so I will assume they are. You don't want pan pots so I will assume this is an 8 into 1 mixer. Here is one way to do it.

1. Unbalance the inputs with a transformer. In a Neve mixer this would have been done with a 31267 transformer wired 10K:600. You could use the equivalent Carnhill transformer wired the same way. The signal is now 12dB lower than the input level.
2. Use a 1K log pot for the volume control .This will be reflected via the transformer to the input as an impedance greater than 10K
3. Feed the pot wiper via a bus resistor to the bus. As this is a mono bus we do not have to worry about crosstalk. But we do want to minimise interaction between channel so we choose a bus resistor at least 10 times larger than the pot value. A typical Neve value would be 15K.
4. The bus loss is equal to the number of inputs i.e. 8 times or 18dB. This is the amount of gain we need to make up.
5. With only 8 inputs there is no need to use the Neve standard method of a transformer input to the bus amplifier. We can just wire the bus straight to the BA283 first stage and set its gain for 18dB.
6. The output of this stage can feed a 10K pot as the master volume. The pot wiper feeds the output stage of the BA283.
7. We still need to make up the 12dB loss from the input transformer which we do in the output stage.

Cheers

Ian

Thank you for your kind comments.
It took me a while to realize this, as my work is at its busiest time of the year. I apologize for that.
I will try to create one, based on your excellent comments.

 

[zone12bit]

OK, that is a reasonable specification. You do not say if the inputs and output are balanced so I will assume they are. You don't want pan pots so I will assume this is an 8 into 1 mixer. Here is one way to do it.

1. Unbalance the inputs with a transformer. In a Neve mixer this would have been done with a 31267 transformer wired 10K:600. You could use the equivalent Carnhill transformer wired the same way. The signal is now 12dB lower than the input level.
2. Use a 1K log pot for the volume control .This will be reflected via the transformer to the input as an impedance greater than 10K
3. Feed the pot wiper via a bus resistor to the bus. As this is a mono bus we do not have to worry about crosstalk. But we do want to minimise interaction between channel so we choose a bus resistor at least 10 times larger than the pot value. A typical Neve value would be 15K.
4. The bus loss is equal to the number of inputs i.e. 8 times or 18dB. This is the amount of gain we need to make up.
5. With only 8 inputs there is no need to use the Neve standard method of a transformer input to the bus amplifier. We can just wire the bus straight to the BA283 first stage and set its gain for 18dB.
6. The output of this stage can feed a 10K pot as the master volume. The pot wiper feeds the output stage of the BA283.
7. We still need to make up the 12dB loss from the input transformer which we do in the output stage.

Cheers

Ian

Hi. I've been thinking about building a summing mixer using neve ba283 modules for a long time. I've searched a lot of topics on the forum and I've really learned a lot. This topic is closest to what I want to build, so I'll allow myself to write here. I want to build an 8-channel line mixer with balanced inputs and 2 stereo outputs. I understand that I need 8 Carnhill VTB9046 line transformers, which will convert the balanced signal to unbalanced, so I'll have 8 unbalanced signals after the transformers. I'd like to add a volume potentiometer at this stage, to be able to adjust the signal that is sent to the ba283 after summing. I also need a switch at each input that will allow me to assign the channel to the Left/Center or Right output channel. In my case, a switch is enough, I don't need a potentiometer. Can I do it with a regular 3-position switch or a lorlin type? My next question is: can I use a 1k potentiometer as gain and 15k resistors at the output of each signal after the input transformer? I understand that the summed signal goes directly to two BA283 modules, there is amplification of the lost signal and symmetrization at the level of two Carnhill VTB1148 output transformers. As an output volume control, can I use a 10k stereo pot for a double BA283 module? I also plan to make stereo VU metering, I saw a neve VU buffer circuit diagram on the forum, will it be suitable for this project? If I could ask for a diagram of summing signals after input transformers with the gain option and a pan switch, I would be very grateful. Thank you in advance for your understanding, I really tried to find all the information and I think I found it, but I prefer to make sure. You have incredible knowledge on this topic, thanks!

 

[ruffrecords]

Basically you are correct. Here is the outline schematic of what you need:



You can use a three way switch as shown but it is possible to make one out of one particular type of DPDT centre off toggles switch.

Yes you can use a 10K log stereo pot for the output level control. This should go between the two stages of the BA 283. A VU buffer should be fine driven directly from the BA283 unbalanced output.

Cheers

Ian

 

[zone12bit]

Basically you are correct. Here is the outline schematic of what you need:

View attachment 135539

You can use a three way switch as shown but it is possible to make one out of one particular type of DPDT centre off toggles switch.

Yes you can use a 10K log stereo pot for the output level control. This should go between the two stages of the BA 283. A VU buffer should be fine driven directly from the BA283 unbalanced output.

Cheers

Ian

Thanks Ian for your reply and the switch diagram! I wonder if I really need 8 input transformers.. or would it be better to passively sum the 8 balanced inputs to a stereo bus, to two input transformers and two x BA283 modules and 2 output transformers? Would the PAN switches and gain knobs then be right after the input XLRs? And since the gain knobs will mix the incoming signal to the input transformers, would I be able to control the saturation of the individual channels? I am trying to understand the order of this audio path

 

[ruffrecords]

Thanks Ian for your reply and the switch diagram! I wonder if I really need 8 input transformers.. or would it be better to passively sum the 8 balanced inputs to a stereo bus, to two input transformers and two x BA283 modules and 2 output transformers? Would the PAN switches and gain knobs then be right after the input XLRs? And since the gain knobs will mix the incoming signal to the input transformers, would I be able to control the saturation of the individual channels? I am trying to understand the order of this audio path

It is possible to passively mix before the transformer and also do the pan switching but you will then be mixing balanced signals so your switches will need twice as many banks. New York Dave posted such a design here many years ago which I have attached. You then need a 10468 input transformer rather than a 31267 because you make up amplifier is basically Neve mic pre.

Cheers

Ian