Phantom power problem in Direct Injection Box

[ChanchoPancho]

Hey, I hope you can help me with this.

The thing is I have a DI box which can run its power from either battery or mixer's phantom.
When I run the power from battery everything works like it supposed to, but when I turn the phantom on,
the signal level drops a lot.

I opened the box and i found every component looks fine, every solder and PBC track is Ok as well.
I started measuring supply voltages and found that with phantom power Vcc is like 2,42v and Vref 1,21.
I also measured Phantom voltaje in the xlr cable and it has the 48v in pins 2 and 3 (1 is 0v), but when
i plug it in the voltage drops downs to 19v in each pin (2 & 3)

The Di box is a Samson S direct plus (stereo)

I forgot to mention that both left and right channels have the same problem.

The schematic is close to the one in Rod Elliott's web:

http://sound.westhost.com/project35.htm

I hope you can give me a hand. If you need it I'll figured out the correct schematic.
Thanks in advance,
FBP

[Gus]

What preamp are you using for the phantom power?  P48 Phantom should supply up to 10ma
read the Implementation variations section
http://en.wikipedia.org/wiki/Phantom_power

Can you try a different phantom supply maybe the one you are using does not supply enough current

http://www.samsontech.com/site_media/legacy_docs/S-direct_plus_ownman.pdf
 does not appear to have the current required at the last page

48VDC - 19VDC = 29VDC  Phantom resistors are often 6.81k  using ohms law 29VDC/6810= 4.25ma to 2 and also to 3 for a total of 8.25ma.

[Minion]

The Voltage will drop when you plug it in ...... Phantom power has 6.8k resistors on the 2 power legs (they are in paralell with each other) .... when Current is drawn the voltage will drop because of these resistors ....... If it also runs off of a battery then you know that it isn"t useing 48v ........

[ChanchoPancho]

Thanks for the replies guys.

Gus, the power source I'm using to prove it is a behringer mixer (xenyx 1202 fx)
I read its manual and it doesn't say anything technical about its phantom source, just that it is 48v.
I measured the current draw (in pin 1) and it is 7,3mA.
A friend had connected the direct box to another mixer and same level drop happened
i'll try to connect it to other mixers this week.

I'll leave you with the schematic I draw after hours of looking the pcb and measuring with the multimeter.


Later i'll post some voltage measures i made.

[PRR]

It's a stereo box. Are you feeding Phantom to _both_ XLRs, or just one?

Though some ruff calcs suggest it should run on just one Phantom, and that 7.3mA is the right answer for true 48V through 2*6.8K to a 10V-12V Zener.

[ChanchoPancho]

Are you feeding Phantom to _both_ XLRs, or just one?

I'm feeding it with just one XLR. Since you asked that, I measured with both xlr to see what happens.
With one XLR it draws 7,63 mA, and with both its 0,33 mA

I measured some voltages too.
test points were:
1. XLR pin 2 (same voltage as in pin3)
2. Node connecting R13, R31, R22 & C15. (same voltage as node connecting R18, R32, R23 & C16.)
3. Node connecting R31, R32 & D2
4. Node connecting D2, Led1 & Led2
5. Vcc

test point	Left	Right
1	19.1	19.1
2	16.4	16.5
3	5.1	5.1
4	4.4	4.4
5	2.4	2.4

This was done connecting just 1 XLR
The same test was done with both XLR:

test point	Left	Right
1	18.9	19.0
2	16.6	16.6
3	6.4	6.4
4	5.7	5.7
5	3.7	3.7

Also measured the zener:
Pin 2 XLR to an 1k resistor to the zener and then the zener back to XLR pin 1.

What the hell is wrong here?
Phantom voltage was 32.6 and zener voltage was 30.6 so i guess it is a 30v zener.
(Zener reads c30et)

What the hell is wrong?!?!

[gritz]

I fear that the power supplies on the mixer(s) you're trying it with just can't supply the current. The voltage drops across the 680R resisors indicate that it's drawing nearly 8mA - that's is maybe a milliamp or two more than a couple of the 072s that I have here draw, but it's still about_right. This problem may be down to the fact that phantom power was originally designed to supply the milliamp(ish) that condenser mikes draw. The antiphase opamps are also having to drive a fairly low impedance (about 1k5 in parallel with whatever the DI box is plugged into) so current draw would only get worse under large signal conditions.

All your numbers look right, so I can't think of another explanation right now.

[PRR]

Re-draw just the power system, including Phantom, combining redundant resistors.

1 19.1V - XLR pin 2 (same voltage as in pin3)
2 16.4V - Node connecting R13, R31, R22 & C15.
3  5.1V - Node connecting R31, R32 & D2
4  4.4V - Node connecting D2, Led1 & Led2
5  2.4V - Vcc



The quickie analysis shows 1.1mA "missing" at the XLR. I neglected a couple 100K which account for 0.4mA. It is possible the C15 C16 caps are leaking, but also very possible that +/-10% resistor tolerances account for it. In particular if the 1202's "P48" is really 45V, it is within P48 tolerances and would make everything add-up.

The JRC072 is presumably a TI TL072 sold by JRC and presumably eats the same power. TI specs 1.4mA nominal 2.5mA Max per amplifier. Since four amps are powered, it could eat 10mA. Another 0.5mA in the 100K bleeders and 0.6mA in the rail-splitter, 11mA is possible.

> power supplies on the mixer(s) you're trying it with just can't supply the current.

That was my thinking also. However I now think this particular DI will not power-up correctly with any P48.

P48 is 44V-52V through 3.4K. 11mA in 3.4K is 37.4V drop. If we start from true 48V that leaves 10.6V. From 44V we have 6.6V. Then knock-down for D2 and the LEDs, say 2.5V. We come near 4V. Which is not quite what you have, but close-enough for thought.

Maybe the designer believed the "1.4mA" nominal spec. This might even be almost-justified in a large console with hundreds of '072s: the tolerances might balance-out. With just two chips probably from the same lot and wafer, max-spec '072s seem to suck-down the voltage more than this plan can stand.

i.e. Sampson sold you a lame box.

Since you already tore into it, you can't send it back.

If the chips are socketed (ha!) you can try other '072s. Odds are it won't get worse and may get better. However I am fond of JRC chips: they were about the last Mostly-Audio chip company and their chips sometimes sound better in ways the Big Chip Makers wouldn't hear.

If chips are socketed and you only need one channel, pull the unused chip. Half the power drain. Still pretty lame.

If you can live without the LEDs, shorting them gains another 2V.

If you wanna hack: replace R21 680r with two 330r resistors in series. Tie their center junction to the R31 R32 junction. Now the inside-the-box loss is 165 ohms plus 2.7V of diode-drop. However a pair of worst-spec '072 chips is still going to drop too much.

Change the chips or use a battery.

[gritz]

Yeah, I just went with the ballpark current figure from the drops across the 2 x 680R resistors.

I suppose that the designer just made the assumption that 48V it a stiff 48V and never had cause to talk to the guys who design the mixers...

I was thinking maybe some well specced low current opamps (and losing R21 in the process) would help, albeit at a price. As PRR points out, shorting the leds would gain another 2V for free.

If you only use it occasionally then maybe stick with battery power...

[PRR]

> suppose that the designer just made the assumption that 48V is a stiff 48V

He should know it isn't "stiff": there's 3.4K in series.

My bet is that the designer used the "Nominal" supply current spec. But even then it's way too close to the edge.

1.4mA*4 is 5.6mA, plus misc is 7mA. In 3.4K+1.9K that makes 37V drop. Pretending 48.0V source we are left with 11V. Take out 2V+ for diodes, it is under 9V. So even with Nominal chips and Phantom applied it will tend to run on the battery!

I do suspect that several people touched the design between initial beauty and final production run. The LEDs sure look tacked-in. The original chips may have been something else but audio-product factories tend to put TL072 everywhere.

TL062 will work fine here at a fraction of the supply current.

[JRay]

TL062 will work fine here at a fraction of the supply current.

I have also had excellent results with Texas Instr TLE2022 in simple low power op-amp audio circuits. Should drop in OK in this circuit too.

Ray

[PRR]

> Texas Instr TLE2022

Thanks. A fine choice, maybe better.

[ChanchoPancho]

Hey guys thaks for the replies.

If the chips are socketed (ha!) you can try other '072s...

Of course it is soldered directly. Why put a couple of sockets if is cheaper to solder the chips directly? Damn economics!
Well the thing is i tried to desolder one of the chips and was impossible. When i first opened the di box I saw one resistor(R28) a little dirty/burned or whatever so I tried to get it out. It was so hard that I finally broke it. Then i realized it was working good in the first place, it was just dirty. With that experience i tried to desolder the zener diode to get the voltage rating by its reading and to see if it was good or not. After like a half hour I finally got it out and realized it was also OK.
I have my share of experience desoldering electornic stuff, but this is the first time I have such problems getting out simple (2 leg) components.

The quickie analysis shows 1.1mA "missing" at the XLR. I neglected a couple 100K which account for 0.4mA. It is possible the C15 C16 caps are leaking, but also very possible that +/-10% resistor tolerances account for it. In particular if the 1202's "P48" is really 45V, it is within P48 tolerances and would make everything add-up.

Resistors are metal film 1/8w 1% (brown tolerance band) so i don't think the excess current draw is a matter of R's tolerance.

If you wanna hack: replace R21 680r with two 330r resistors in series. Tie their center junction to the R31 R32 junction. Now the inside-the-box loss is 165 ohms plus 2.7V of diode-drop. However a pair of worst-spec '072 chips is still going to drop too much.

With this mod R31 & R32 would be redundant. Won't it be better to get them out?

What is the function of R13/R18 and R21?
Are phantom voltage attenuators?

If you wanna hack: replace R21 680r with two 330r resistors in series. Tie their center junction to the R31 R32 junction. Now the inside-the-box loss is 165 ohms plus 2.7V of diode-drop. However a pair of worst-spec '072 chips is still going to drop too much.

This will be my last resource to repair it. I'm also considering getting the diodes out.

I had just measured the current draw from the battery and it is 30mA . I think this is too much, maybe a leaking Cap...
I think the only way to know that is desoldering... dammit there's a lot to try: C9, C8, C1 & C11.
I'm going to do that tomorrow since it's almost 5 am.

Thanks for the replies again guys!
talk to you tomorrow.

FB

[PRR]

> I tried to get it out. It was so hard that I finally broke it.
> i tried to desolder the zener ... After like a half hour I finally got it out
> R31 & R32 would be redundant. Won't it be better to get them out?

They do no harm, and clearly these parts DON'T want to come out.

> What is the function of R13/R18 and R21?

Opamps don't like to drive low-impedance mike inputs.

Guitar level is higher than mike level (and the second opamp gives effective 2X gain).

The resistor pad lowers the level while increasing the impedance seen by the opamp.

[PRR]

> current draw from the battery and it is 30mA

OK, something IS wrong. Worst-case opamps 10mA, another 1mA here or there.

Look for caps which should have DC voltage across them, but don't.

Use skinny-nose wire cutter to snip the V+ pin of each opamp. If current drops a little, you can solder-bridge the pin back together. If current drops a lot, snip all 8 legs (easier than desoldering) then drop a DIP8 socket on the stubs and solder. Use jeweler's screwdriver to beat the legs and be sure the joints will hold up.

[ChanchoPancho]

Look for caps which should have DC voltage across them, but don't.

Doing that the multimeter will give me the same result in every cap in parallel, won't it? I think I'll better desolder them, it'll be a little easier to grab than an 1/8W resistor.

Use skinny-nose wire cutter to snip the V+ pin of each opamp. If current drops a little, you can solder-bridge the pin back together. If current drops a lot, snip all 8 legs (easier than desoldering) then drop a DIP8 socket on the stubs and solder. Use jeweler's screwdriver to beat the legs and be sure the joints will hold up.

Good idea! I'm not at home at the time, but tomorrow will, so the i'll try everything. thanks again.

[ChanchoPancho]

Well I did the disoldering of caps and Ic's and found out that they weren't the problem.
My next guess would be the voltage divider resistors (R14 & R16) so I got out R14 and found out that
it wasn't the problem neither. If nothing of the above components are wrong, the it must be one of Vref.
I revised R7 & R27 and realized that R7 was the problem. I got it out, resolder all the other things back again and measured the current draw: 7,6 mA. 

I measured again all the phantom voltages for anyone who have problems with this in the future...
test points were the same as before:
1. XLR pin 2 (same voltage as in pin3)
2. Node connecting R13, R31, R22 & C15. (same voltage as node connecting R18, R32, R23 & C16.)
3. Node connecting R31, R32 & D2
4. Node connecting D2, Led1 & Led2
5. Vcc

test point   L/R
1   21.2
2   18.9
3   8.6
4   7.9
5   5.9

Thanks for all the help guys, regards 
ChanchoPancho.

[PRR]

> R7 was the problem

What was wrong?

I don't see how R7 could cause excess power current.

[ChanchoPancho]

When I got out the caps and ic's power supply (pin current draw was 28 mA. I concluded that any current draw must come from the Vref circuit. I got out R14 and current draw was still very high, hence the problem may be in R7,  R27 or the "+" pin from the signal inverter from ic's. I took a shot with R7 and when I got it out the current draw when down inmediatly. I measured its value with a multimeter and was ok. I soldered everything back (except for R7) and current draw was around 7 mA. Then soldered R7 back and everything work ok.

[ChanchoPancho]

One thing I forgot to mention is that R7 is the equivalent resistor of R27 in the left channel. It can be infered from my previous post, but I make the distinction anyway.