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I successfully built an active DI box using this circuit (image posted below), everything went fine and I'm happy of the result.

Parts List

R1, R2 - 1k
R3, R5, R6, R7, R8, R11, R12 - 10k
R4 - 1M
R9 - Jumper
R10 - 100 ohms
R13 - 10k reverse log potentiometer
C2 - 1uF
C4 - 100pF
C1, C5, C6, C9 - 10uF
C7, C8 - 47uF
IC1 - NE5532

V+ is 9 Volts

Can anyone help me understand how this circuit arrangement provides differential signals and impedance balancing? Briefly, can anyone explain me how this circuit works in details?

op amp based balanced transmitter

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    \$\begingroup\$ active d.i. box? \$\endgroup\$ – Leon Heller Apr 28 '15 at 12:40
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    \$\begingroup\$ en.wikipedia.org/wiki/DI_unit apparently? \$\endgroup\$ – scld Apr 28 '15 at 13:12
  • \$\begingroup\$ If a person wanted to add an unbalanced buffered send to this circuit for a 1/4" output to amp monitor, would it be as simple as splitting the output from IC1a and would that have any negative effects? \$\endgroup\$ – Jonathan S. Fisher Mar 3 '16 at 6:18
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The main amplifier is IC1A. R1+R13 and R3 set the gain, C1 and C2 are DC blocking caps, and most of the remaining mess of resistors and capacitors is for providing a low-noise DC bias at (presumably) 1/2 the supply voltage. The circuit around IC1B takes the output of the main amp and inverts it. The output of IC1B is then just the output of IC1A flipped around the DC bias point, so together they make a differential signal. C5 and C6 are DC blocking caps, this time for the output.

DC blocking the signal on the way in and then on the way out again decouples the DC operating point of the circuit from the input and output levels, which are both centered around ground. Since you don't care about frequencies below 20 Hz, it's perfectly fine to pick a arbitrary DC (0 Hz) level internally.

My only quibble with this circuit is R2 and C4. Although you didn't show the values, I'm expecting C4 to be much less than C7. R2 and C4 probably filter high frequencies beyond what the amp is intended to handle (20 kHz for "HiFi"), but that could have been accomplished by tying the bottom end of C4 to ground instead of the internal bias supply.

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    \$\begingroup\$ Be aware that capacitors C5 & C6 are a problem - many inexpensive mixers have only a global phantom-power On-Off switch and those capacitors will fail if this unit is connected to a mic input that has phantom-power turned ON. The normal fix for that is to have back-to-back caps of double the desired value. Voltage rating of the outside caps has to be 63V or more. \$\endgroup\$ – Dwayne Reid Apr 28 '15 at 15:00
  • \$\begingroup\$ Hello Olin, many thanks for your time, everything is much clearer to me now. Still have some obscure points i'd like to solve: R4 is a 1Meg resistor, what's the purpose of it? Has it something to do with Ic1a input impedance or with the DC offset on it? \$\endgroup\$ – Tonno Subito Apr 30 '15 at 14:38
  • \$\begingroup\$ @Tonno: R4 feeds the DC bias point onto the positive input of IC1A. It is large to not get in the way of the AC signal. \$\endgroup\$ – Olin Lathrop Apr 30 '15 at 14:49
  • \$\begingroup\$ C5 + C6, are they facing the wrong direction to block 48v? \$\endgroup\$ – Jonathan S. Fisher Mar 3 '16 at 6:56
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IC1 is an "Operational Amplifer." Essentially an op-amp is a mathematical amplifier. The purpose of all amplifiers are to take a very weak signal and make it stronger, able to "drive" more energy into something, like a speaker. When we say "impedance matching" we mean just this: using something like an amplifier to match an input signal to whatever it is being fed to.

If an op-amp's -input were grounded, and a signal applied to the +input, a larger but identical +output signal would be generated, just like a traditional amplifier. So that is called a non-inverting configuration, since the same signal comes out.

Op-amps have a -input though, which does the reverse: if the +input was grounded, and the signal applied to -input, the opposite signal comes out. So that would be an inverting configuration. This is very useful for many things.

In this schematic, IC1a takes the instrument input (+input) and outputs essentially the same thing to Out1. That is a non-inverting configuration. Out1 is also coupled to the -input of IC1b, which is the inverting configuration, so the opposite signal comes out of Out2.

So Out1 and Out2 are differential outputs, because opposite signals are coming out of them. That's exactly what an XLR cable wants, and due to the physics behind sending signals along cables this way, it is very low-noise.

All of the other components are just for setting the voltage levels right (search for 'resistor divider'), blocking DC voltages (capacitors), and other signal-conditioning needs.

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Olin Lathrop provided you most of the information that you need. There are a couple of other nuances that may not be apparent:

1) Capacitor C4 provides capacitive loading for the signal source. 100 pF might be a little bit high - you may have to experiment with it.

It's also going to the wrong place - ideally, the bottom end of C4 should be going to Ground. This reduces noise pickup that slips through the bias voltage divider.

C4 is important because the normal cable and amplifier input impedance and capacitance load the source signal. It's not as much problem with inductive pickups (guitar, etc) but can be a real problem with crystal (piezo) pickups.

The reason that it has so much effect is that the normal cable has a fairly large capacitance. The frequency response of the pickup is altered to have more high-frequency present when the pickup is unloaded. The load capacitance then rolls that high-frequency increase off so as to approximate a flat response.

When you shorten the cable to go to a DI box instead of an amplifier, the capacitance drops. This leads to a rise in the high-frequency response. C4 compensates for that.

2) R2 provides some measure of transient protection should a crystal (piezo) pickup be dropped or otherwise receive a significant impact. These pickups can have an amazingly-high transient output voltage under these conditions. Very fast rise-time but short duration.

Many acoustic instruments use crystal pickups. There are some very-nice sounding crystal pickups available.

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  • \$\begingroup\$ 'C4 is important because the normal cable and amplifier input impedance and capacitance load the source signal'. Isn't this way a total higher load? 'The frequency response of the pickup is altered to have more high-frequency present when the pickup is unloaded'. Where and in which way this happens? 'When you shorten the cable to go to a DI box instead of an amplifier, the capacitance drops'. Why this happens? Many thanks for your time too, i appreciate \$\endgroup\$ – Tonno Subito Apr 30 '15 at 14:43

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