Several questions: Radio Design Labs makes a 10K combiner/divider that sums or divides line level inputs/outputs. Why is 10K a useful resistance to achieve this as opposed to something else? The spec sheet doesn't give a ton of information regarding the schematic--just resistance/impedance. Below is a link to the product.


Are the two primary considerations of a combiner making sure your interfacing impedance to the next input is appropriate and that you're not feeding the outputs back into each other/loading down the outputs, whilst trying to combine them?

what's the reason for making as high-as-possible resistance?

  • \$\begingroup\$ No part numbers, no data, no links = no help. -1. \$\endgroup\$ – Transistor Apr 4 '16 at 13:36
  • 2
    \$\begingroup\$ I think the considerations you state are really what's behind the decision for as-high-as-possible resistances. \$\endgroup\$ – JimmyB Apr 4 '16 at 13:52
  • 1
    \$\begingroup\$ @-transistor--the manufacturer hasn't included what looked like relevant information since they basically just say 10k and give insertion loss. \$\endgroup\$ – inbinder Apr 4 '16 at 15:47

If you download the data sheet, you will find that the same product is available with different impedances.

You use the one whose impedance matches the audio system you are working with.

In response to comment: There are two ways to build a passive splitter/combiner.
One way is to use multiple coils on a single audio transformer core.
The other way uses resistors.

The datasheet specifies that the STD-10K uses a resistor network to combine the signals.

This guy built a resistor network combiner/splitter himself.

Here's the schematic: enter image description here

There's nothing magical about it. The signal goes through the resistors, and is summed where they all join together. The neat thing about it is that the summed signal also goes back out to any input that isn't driven by an input.

If you apply a signal to input 1 of that circuit, you will see the same signal on all of the other inputs. Conversly, if you input signals on seven of the inputs, then you will see the sum of them as the output on the unconnected input.

The only real trick is calculating the resistor values so that the desired impedance appears at all inputs.

  • \$\begingroup\$ @JRE--Are you able to explain why 10K or 600ohms allow for combining or dividing? \$\endgroup\$ – inbinder Apr 4 '16 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.