[sdiy] MIDI phantom power...over 5 pin MIDI connector ?

[rsdio at audiobanshee.com]

Hi A.

5 V into 660 ohms is not 7.5 mA, because the opto-isolator LED accounts for an additional voltage drop. The 6N138 has a typical forward voltage of 1.3 V, but it can be as high as 1.7 V or 1.75 V depending upon temperature or current. Thus, 5 V into 660 ohms with a 6N138 in the current loop works out to be 5.6 mA typically, or exactly 5 mA at high temperatures. They probably planned it that way so that 5 mA would be the minimum, but it would only go only about 10% above that.

3.3 V wouldn't really produce 5 mA, but more like 3 mA (due to the LED drop) or even as little as 2.4 mA.

Those 20 mA ratings are Absolute Maximum, not recommended for normal use. Usually, a data sheet will recommend typical values after the absolute maximums. In the specific example of the 6N138, the device operates on as little as 0.5 mA, so there's not really much need to exceed the standard MIDI value of 5 mA. MIDI is already designed to send 10 times the minimum current required for operation.

But, you do have a point, A. We need to ask JP what he really wants to solve here. There might be a way to solve the problem without "more power."

Brian Willoughby
Sound Consulting


On Sep 9, 2015, at 11:40 PM, eidorian at aladan.net wrote:
> I just re-read the original post, and it might be that JP isn't asking how he can get power from a MIDI cable, he might actually be asking a subtly different question: how can he *send* more power over the two MIDI pins (presumably without affecting the ability of those pins to transmit MIDI to older devices at the same time).
> 
> If it really is the case (that you have control over the MIDI-out) then maybe you could run the current loop at higher than 5mA.  I looked at data sheets for PC900, 6N137, 6N138 and 6N139 and they all have various maximum rated If(avg) of 20mA or greater, so something like 15mA should be safe.  Much more than that and you run the risk of burning out MIDI-in opto-couplers on older gear.
> 
> http://www.midi.org/images/midihw.gif
> 
> 15mA from a 5V supply would require replacing each of the MIDI-out 220 ohm resistors with 56 ohm.
> 
> Interestingly, 5V into 660 ohms is already 7.5mA.  But 3.3V into 660 ohms is 5mA!
> 
> But yeah, there's no fun to be had doing this.  Any device that requires your "high powered" MIDI-out ports for power is going to damage your "normal" gear if you ever plug it in accidentally :-(
> 
> Cheers,
> A.
> 
> 
> On 2015-09-10 15:11, rsdio at audiobanshee.com wrote:
>> Hi JP,
>> 
>> I would strongly recommend against designing any product that pulls
>> power from the MIDI cable. MIDI simply wasn't designed for that, and
>> if you try to take advantage of the vague power that's there, you'll
>> still run into many devices which don't work. The link that A provided
>> below has long lists of incompatible MIDI devices (because MIDI
>> Solutions is deviating from the standard by pulling power where it's
>> not supposed to be used).
>> 
>> That said, the best clues (if you're going to do this anyway) come
>> from the MIDI Specification.
>> 
>> You can see that pin 4 of all MIDI outputs is supposed to be tied to
>> 5 V through a 220 ohm resistor. I believe that A is wrong in saying
>> that you can get only 5 mA, because the 5 mA limit is the amount of
>> current that is supposed to flow when a MIDI bit is being transmitted.
>> That 5 mA flows out of pin 4 of the MIDI Out, in pin 4 of the MIDI In,
>> through the optoisolator, out pin 5 of the MIDI In, and in pin 5 of
>> the MIDI Out. That makes the total circuit.
>> 
>> However, if you make a different circuit path for current to flow,
>> then you can potentially pull much more than 5 mA from the +5 V supply
>> of the MIDI Out device. This current would flow out of pin 4 of the
>> MIDI Out, in pin 4 of your MIDI In, through your charge pump, and then
>> out pin 2 of your MIDI In, before finally flowing in pin 2 of the MIDI
>> Out. Of course, the more current you pull, the lower the voltage - but
>> that shouldn't be a problem for your charge pump. At 5 mA, the voltage
>> you see will be 3.9 V, but if you double the current to 10 mA then
>> you'll still have 2.8 V. I've used the MAX1595 charge pump (not for
>> MIDI, but for USB-MIDI) in products, and it works with as little as
>> 1.8 V on input. That would allow you to pull 14.5 mA, but there would
>> only be 9.5 mA available when the MIDI signal is using its 5 mA.
>> 
>> Note that you might cause problems for the MIDI Out device by pulling
>> more than 5 mA, because their 5 V supply may have been designed to
>> only provide 5 mA and no more. If there isn't any headroom for extra
>> current, then your device could overheat the MIDI Out circuits and
>> cause them to fail. This is the biggest reason why I argue against
>> designing anything like this.
>> 
>> MIDI Solutions suggests that customers who make their own power
>> supplies connect ground to pin 2 and power to both pins 4 and 5. This
>> suggests that MIDI Solutions is also pulling current from pin 5. Note
>> that this will only work if the buffer circuits in the MIDI out are
>> not open collector. MIDI Solutions is probably assuming a typical pair
>> of inverting gates which drive both high and low logic levels. This
>> would indeed provide some power to your charge pump. However, any MIDI
>> device that uses an open collector output will only be driving logic
>> low levels, and not logic high levels. This is perfectly valid for
>> MIDI, but not so helpful when abusing MIDI for free power. I assume
>> that most of the power is coming through the resistor at pin 4 because
>> it's always available, whereas the Voltage on pin 5 depends upon the
>> logic level of the data and the type of driver.
>> 
>> I think you'd need a couple of diodes to combine the power from both
>> pins 4 and 5 to your charge pump, but those diodes would rob more
>> voltage. You can select LDO (low-dropout) diodes to minimize the
>> losses there. Or, you can just ignore the power that may or may not be
>> coming in on pin 5.
>> 
>> Finally, you're depending upon the MIDI Out cable shield for ground.
>> Personally, I think it's a bad choice to assume that the shield will
>> have a low resistance to ground. There are many ways to shield a
>> cable, especially when you are aiming for low noise. A good designer
>> of MIDI audio gear might choose to put a 1 Mohm resistor in parallel
>> with a 1 uF capacitor, and only connect the MIDI Out shield to ground
>> through that filter network. This will make their MIDI audio device
>> much less prone to picking up noise from the cable shield, and it's
>> perfectly legal in terms of normal MIDI functionality. If you pay
>> careful attention to the MIDI Specification, a different GND symbol is
>> used for the opto-isolator versus the symbol for cable shield
>> connections. This means that it's perfectly legal if they do not
>> connect directly. By the way, this means that MIDI Solutions is wrong
>> when they claim that MIDI devices that aren't compatible with their
>> power supply "don't conform to the MIDI specification" - the truth of
>> the matter is that MIDI Solutions has not studied the original MIDI
>> Specifications carefully enough.
>> 
>> There you go. I think that's plenty of clues.
>> 
>> 
>> On Sep 9, 2015, at 5:57 PM, eidorian at aladan.net wrote:
>>> 
>>> Also see http://www.midisolutions.com/faqs.htm
>>> 
>>> Maybe contact those guys and see if they're willing to send you a service manual?
>>> 
>>> Cheers,
>>> A.
>>> 
>>> On 2015-09-10 09:43, jpdesroc wrote:
>>>> Does anybody have schematics of an easy way to send
>>>> power using the 2 MIDI pins in a 5 pins cable ?
>>>> I don't want to use any of the 3 remaining wires of the cable
>>>> for universal use on ANY MIDI cables..
>>>> I know this company uses that technology:
>>>> 
>>>> http://www.midisolutions.com/prodmrg.htm
>>>> 
>>>> but I can't find any circuit ideas that uses it..
>>>> I know I'll need some kind of charge pump design to get something like 5 volts
>>>> out of the MIDI (idle or in use) signal.
>>>> 
>>>> Any clues ?
>>>> 
>>>> JP