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Mon Dec 13 17:20:44 EST 2010

## DI box

```I'm building a simple DI box buffer/mixer.  It reminds me again how
far removed I am from being an experienced analog circuit designer ;)

Goal: connect both bass and guitar to either a high-Z effects pedal
input (trivial) or a low-Z mixer input (impedance +- 10k).  Allow for

For the low-Z load, the main point is that the output impedance is
resistive over the region of interest while say 50% voltage loss is
not a problem.  This means an output impedance of about 10k is enough;
no need to go down to 1k.

Input impedance needs to be high.  This is straightforward: a 1M
biasing resistor to 1/2 VCC ref and a large enough coupling cap.  The
1/2 VCC ref can be a 100k/100k divider with 10uF buffer.

The output was where I'm puzzled.  Need to have both AC coupling to
remove the 1/2 VCC bias.  All the stomp box circuits I find online
seem to be designed for high input impedance, judging from the
couplings caps: 1uF on a 10kOhm input has a 3dB point of 15Hz.

So, to be careful, let's pick 10uF coupling.  This can come from the
10k linear mixing pot connecting the two buffer outputs.

10k
buf1 o--/\/\/\/--o buf2
^
|   10u        1k
\---|(---x---/\/\/\---o out
|
/
\ 100k
/
|
x------------o
|
GND

So, the 10u value is such that at a load of 10k, the impedance of the
cap is small.  At 20Hz this gives about 800 Ohm.

The influence of the mixer pot isn't so important: it's just signal
level attenuation (max 50% at 10k load).

The 1k is there to prevent short circuit, and the 100k is there to
bias the buffer capacitor at DC.

Take-away lesson: the expected load impedance determines the size of
the output coupling cap.

```
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