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Hex Inverter vs Hex Schmitt-trigger Inverter
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author Hex Inverter vs Hex Schmitt-trigger Inverter
dr.earlfunkton
Hey y'all!

So i've come across a bit of confusion and need some IC wizards to help.

Im building a midi thru box based on the MidiSizer schematics. I accidentally ordered a SN74LS14N in SMD package instead of DIP package. While trying to just roll with it, I broke one of the legs off from over bending. Before I order the right format of this trigger, I looked through my stash of IC's to see if I had a viable replacement on hand. I found an SN7404N in DIP package.

The 7404N is a Hex Inverter while the 74LS14N is a Hex Schmitt-Trigger Inverter. They seem to have the same pinout and do the same function, so should I be able to substitute the 7404N in this circuit? Is a Schmitt-Trigger used just for the hysteresis aspect?

Thanks in advance!

Schematic: https://midisizer.files.wordpress.com/2013/01/midithru1.pdf

SN74LS14N Datasheet:http://www.ti.com/lit/ds/symlink/sn74ls14.pdf
SN7404N Datasheet:http://www.ti.com/lit/ds/symlink/sn7404.pdf
latigid on
The function of a Schmitt trigger is to provide sharp transitions with a noisy or "slow"-moving input signal. The MIDI thru signal here is quite well buffered so I would expect it to work with a normal inverter.
ixtern
In many MIDI designs "analog" 6N138 optocouplers are used for the input. In such circuit Schmitt trigger is rather necessary after optocoupler. 6N137 is a "digital" optocoupler.
MIDI output is another case. Almost any buffers/inverters may be used which can handle required current but at least in the past open collector one's were preferred.
MIDI organization specifies +5V supply and 2x 220 Ohm resistors for the output loop. Optocoupler input diode forward voltage is 1.4V (typ.) so required (sink) current will be:
(5V - 1.4V)/440 Ohm = 3.6 / 440 = 0.0082A = 8.2 mA.
Graham Hinton
ixtern wrote:
MIDI organization specifies +5V supply and 2x 220 Ohm resistors for the output loop. Optocoupler input diode forward voltage is 1.4V (typ.) so required (sink) current will be:
(5V - 1.4V)/440 Ohm = 3.6 / 440 = 0.0082A = 8.2 mA.


MIDI is a 5mA current loop. You forgot the 220R resistor in the input circuit.

Some interfaces now use a 3.3V supply and lower resistors, but they still have to drive 5mA into a standard input.
ixtern
Graham Hinton wrote:


MIDI is a 5mA current loop. You forgot the 220R resistor in the input circuit.

Some interfaces now use a 3.3V supply and lower resistors, but they still have to drive 5mA into a standard input.

You are right.
dr.earlfunkton
Graham Hinton wrote:
ixtern wrote:
MIDI organization specifies +5V supply and 2x 220 Ohm resistors for the output loop. Optocoupler input diode forward voltage is 1.4V (typ.) so required (sink) current will be:
(5V - 1.4V)/440 Ohm = 3.6 / 440 = 0.0082A = 8.2 mA.


MIDI is a 5mA current loop. You forgot the 220R resistor in the input circuit.

Some interfaces now use a 3.3V supply and lower resistors, but they still have to drive 5mA into a standard input.


So the power supply in this schematic is supplying 5v at around 5mA, and is being fed probably a 9v from a wall wort of some sort, so could I just not build this power supply and just use a 5v 5mA supply as my VCC?
ixtern
dr.earlfunkton wrote:
Graham Hinton wrote:
ixtern wrote:
MIDI organization specifies +5V supply and 2x 220 Ohm resistors for the output loop. Optocoupler input diode forward voltage is 1.4V (typ.) so required (sink) current will be:
(5V - 1.4V)/440 Ohm = 3.6 / 440 = 0.0082A = 8.2 mA.


MIDI is a 5mA current loop. You forgot the 220R resistor in the input circuit.

Some interfaces now use a 3.3V supply and lower resistors, but they still have to drive 5mA into a standard input.


So the power supply in this schematic is supplying 5v at around 5mA, and is being fed probably a 9v from a wall wort of some sort, so could I just not build this power supply and just use a 5v 5mA supply as my VCC?

You could. But not 5mA. You should include other components. Notice that 7404 are not very efficient buffers as every buffer input needs 1.6 mA in low state. LED takes about 10mA (depends on type). Add this and 20 mA should be the minimum for Vcc.
MIDI input current loop (5mA) is driven from external MIDI source and is separated from buffers so not counts into Vcc total current.
Graham Hinton
ixtern wrote:
Notice that 7404 are not very efficient buffers as every buffer input needs 1.6 mA in low state. LED takes about 10mA (depends on type). Add this and 20 mA should be the minimum for Vcc.


Wrong again. Each MIDI output draws 5mA in the on state, so that is 25mA alone. Add 13mA for the LED and a bit more for the ICs and 50mA is the minimum.

dr.earlfunkton wrote:
So the power supply in this schematic is supplying 5v at around 5mA, and is being fed probably a 9v from a wall wort of some sort, so could I just not build this power supply and just use a 5v 5mA supply as my VCC?


You can use a 5V DC regulated PSU, but you are unlikely to find one <200mA and the regulation is iffy at that end of the market. An on board small regulator is the best way of cleaning up some of the dubious external PSUs out there.

A 6N137 needs a lower pullup resistor, 1k2 is recommended, and it needs a decoupling capacitor close to pins 5 and 8. See a data sheet for the recommended layout.
You don't need Schmitt trigger buffers and if you use a 7417 or 7407 non-inverting open collector buffer you can have six MIDI outputs.
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