2

The audio source is an active mixer with an output impedance of 100 ohm, 25 mA output current and 3 V-pp max.

The inputs that would receive this source are (1) an ADC with an input impedance of 20 kohm and (2) another device with a "variable" input impedance of 2 kohm approx. I mean variable because it could be any device that receives audio I want to plug in (i.e. field recorder, camera, etc.), but as far as I know line-level input impedance in audio equipment never goes below that.

Can the split be done without an extra buffer? I suppose that the 100 ohm output impedance of the source can drive 20x the 20 kohm load (in theory). But how does the added 2 kohm load enter the equation? It is very important for this application that the ADC reading is not sensitive to plugging or unplugging the second device as well.

winny
  • 13,064
  • 6
  • 46
  • 63
Domingo
  • 61
  • 6

2 Answers2

5

It is very important for this application that the ADC reading is not sensitive to plugging or unplugging the second device...

The output impedance of the mixer is certainly low enough to drive both loads, but the signal voltage will vary slightly depending on the load. Furthermore, some devices may have an input coupling capacitor that charges or discharges when connected, causing a glitch or 'pop' when plugged in. If you want the ADC signal to be truly insensitive to whatever is plugged into the other output then you should put a buffer amp on the other output.

Bruce Abbott
  • 55,540
  • 1
  • 47
  • 89
2

For a 100 Ω source a 20 KΩ load will give you 99.5% of the original voltage.

$$ 1\mathit{V}~\cdot~\frac{20000\Omega}{20000\Omega + 100\Omega}~=~0.9950\mathit{V} $$
Adding 2 KΩ in parallel with the 20 K will give you 94.79%, so almost a 5% difference.

$$ \frac{20000\Omega~\cdot~2000\Omega}{20000\Omega~+~2000\Omega}~=~1818.18\Omega $$
$$ 1\mathit{V}~\cdot~\frac{1818.18\Omega}{1818.18\Omega + 100\Omega}~=~0.9479\mathit{V} $$

You could put a buffer in between the mixer and just the auxiliary load, and both your loads would get the 99.5% figure but the ADC would be largely unaffected by the aux load, or with a low enough output impedance buffer you could put both the ADC and auxiliary load on the buffer and get close to 100% on both inputs with the ADC very slightly affected by the aux load, but still better than not buffered.

For a 1Ω output impedance buffer with the 2000Ω load disconnected: $$ 1\mathit{V}~\cdot~\frac{20000\Omega}{20000\Omega + 1\Omega}~=~0.9999\mathit{V} $$
And with a 1Ω output impedance buffer driving the two loads in parallel: $$ 1\mathit{V}~\cdot~\frac{1818.18\Omega}{1818.18\Omega + 1\Omega}~=~0.9995\mathit{V} $$

If you're worried about isolation buffer just the aux load, if you're worried about accuracy buffer them both, if you want accuracy AND isolation use two buffers, one for each.

GodJihyo
  • 17,628
  • 1
  • 16
  • 45