How to find the correct power supply for a device with no markings

Thread Starter

Bassizer

Joined Aug 12, 2011
8
Hello,

I have an instrument preamp for which I am trying to identify the proper power supply. There are several factors that are complicating the matter: A: there are no markings on the outside of the device B: the manufacturer has gone out of business C: there is significant reason to suspect that the manual may be incorrect.

I do know that it uses a dual output + and - power supply and that it is either +/-9V DC or +/-12 V DC.

So my question is whether or not one can identify which one it is by looking at the actual circuit inside the unit. It doesn't appear to be overly complicated inside. My thought was to look for voltage regulators and find out what voltage they would be looking for, but I'm not seeing any.

On a couple different places I see markings for +/- 9 volts, but I also know that the unit is supposed to be able to supply 12 volts of phantom power and I'm not sure if there is a way to dish out 12 volts of phantom power from a 9 Volt supply.

If I posted pictures, might someone here be able to help nail it down?

Any help would be appreciated!
 

flat5

Joined Nov 13, 2008
403
I would look for the biggest electrolytic capacitors and note the plus & minus signs on them.
If you see that many of them share a path to the power supply + & - that is a big clue.
There is likely two sets to find from what you describe.

Yes, good pics will help the experts here to inform you.
 

Thread Starter

Bassizer

Joined Aug 12, 2011
8
OK here are some pics. I hope they are clear enough:

Here is the power input section. The 2 big caps are both labeled 6800μF 16V. Both of the red things say WIMA 0.1 63-A ND. There is an AC IN and an AC OUT because you can daisy chain units together with one power supply. The glare is obscuring the text" -VCC and +VCC ".



Here is a shot of the bottom of the same board:


Here is the remainder of the rear board:


And the bottom:


Here are some pics of the boards on the front of the unit. Notice all the references to + and - 9V. If the supply is +/-12V, can you see anything that would step the voltage down to 9V for these leads? The wires are blocking the text " -9V Ch1 and +9V Ch1 ".





I just reached my picture limit. Hopefully they are clear enough. I realize most of the Channel 2 board is obscured by the upper, Channel 1 board. If you need to see the lower board let me know and I can get some pics of it. Or if anything isn't clear, I'll take better pics.

Thanks!
 

praondevou

Joined Jul 9, 2011
2,942
+VCC and -VCC seem to be the same thing as +9Vx and -9Vx since the VCCs come from the board where the input connector is, then seem to go to SW5 and after SW5 they are called +9V and -9V, right? Then the 9V are being distributed to the rest of the circuit.

Now why would the input be called AC? I don't see any rectifier diodes, so it has to be DC.
If you want to be sure that input is 9V, measure resistance from one of the two input terminals, press SW5, resistance then should be near zero. + should be where the C20 or C18 + pole is connected to the input connector. follow the traces. If the input is a dual supply C20 and C18 are probably in series and their middlepoint would be ground.

Nice pictures btw
 
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praondevou

Joined Jul 9, 2011
2,942
In parallel you may also post a picture of the front and the back of the amplifier which may help to identify it via image search.
 

Thread Starter

Bassizer

Joined Aug 12, 2011
8
Thanks for your replies!

I think it says "A/C IN" referring to the A/C adapter. On the back of the device it says "A/C Adapter In and Out". You're right it has to be DC.

Pardon my lack of knowledge here, but I'm not quite clear on where to make the resistance measurement.

The A/C adapter input jack is a 3.5mm TRS female connector. I'm pretty sure the tip is negative. I know where each of these pins meets the board and I know where the + of C18 meets the board (it does appear that C18 and C20 are in series and that the center point is ground.)

With that being said, can you tell me where exactly should I connect my meter to check for near 0 resistance? Or between what 2 points?

Many thanks!
 

praondevou

Joined Jul 9, 2011
2,942
The A/C adapter input jack is a 3.5mm TRS female connector. I'm pretty sure the tip is negative. I know where each of these pins meets the board and I know where the + of C18 meets the board (it does appear that C18 and C20 are in series and that the center point is ground.)

With that being said, can you tell me where exactly should I connect my meter to check for near 0 resistance? Or between what 2 points?

Many thanks!
Refer to the attached drawing.
Try to name the two caps. Which one is which?
Press the power button and measure from the input jack to +9V and then from the input jack to -9V. Should have zero resistance. Put your findings in the drawing and post it.

Ground should be everywhere on the boards, so when you measure from the middlepoint of the capacitors to let's say the point where C15 and C16 (near U5) are connected you should also measure zero resistance. I imagine these ceramic caps to be bypass caps for the ICs (which could be Opamps). Verify also where these caps are connected to at U5 and post your findings. You may have seen that all of the 8-pin ICs have them. (Their capacitance is quite small though)
 

Attachments

Thread Starter

Bassizer

Joined Aug 12, 2011
8
I put my findings in the drawing. See attachment. Please let me know if this isn't what you were looking for.

I apologize, but I'm not sure what you mean when you say, "Try to name the two caps. Which one is which?" Are you referring to C18 and C20 or to the conductors of the input jack? I'm assuming the latter.

With SW5 pushed in, I measured from the tip of the jack (-) to all of the points marked -9V (not -VCC) and had 0 resistance. I also measured the ring of the jack (+) to all points marked +9V and also found 0 resistance. When I depressed SW5, I lost continuity.

I measured between the center point of C18 and C20 to both of the pins of C15 and C16 and measured anywhere from 19KΩ to 80KΩ. I also tried using another ground point and had the same results.

Here is how C15 and C16 are connected to U5:


How is this looking so far?
 

Attachments

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thatoneguy

Joined Feb 19, 2009
6,359
When measuring resistance, if there is an electrolytic/large value capacitor in the circuit, the resistance will appear to "climb" to a certain value, that value usually being incorrect, but lets you know there is a cap in the way.

What is the output of the preamp supposed to be? 1V peak to peak?
 

thatoneguy

Joined Feb 19, 2009
6,359
Ok

5th image down, where Vcc goes to the power switch. There is a power LED there, with a resistor inline with it to limit current. What is the value of the resistor between power and the LED?

Colors are fine, like brown red brown or whatever.

Ohms law and assuming 20mA LED current should let you know if it was designed for 9V or 12-13.8V
 

bertus

Joined Apr 5, 2008
22,270
Hello,

The text A/C is confusing as the input voltage seems to be + and - 12 Volts.

REAR PANEL FEATURES: Effects Send and Return jacks for each channel and blended mono signal; Stereo Mix In jack; Tuner Send; Chan. 1, Chan. 2, and Mono XLR outs; Ground lift; Individual Unbalanced Line Outs; A/C adapter In and Out jacks.
This comes from this archived page:
http://web.archive.org/web/20090106074358/http://www.raven-labs.com/pages/products/usip/usip.html

See this page for more info:
http://web.archive.org/web/20090204070442/http://www.raven-labs.com/mainframe.html

A/C ADAPTER IN/OUT JACKS: Insert the A/C adapter that came with your unit into the IN jack of your preamp first, then plug the adapter into the wall socket or power strip second. This will avoid accidentally shorting out part of the mini stereo plug to ground when first inserted. You can use the OUT jack to power (daisy chain) other equipment from Raven Labs. To do this, use a mini stereo to stereo patch cord. Since the adapters output is + and - 12 volts DC, the cable will not induce humm into other equipment.
Bertus
 
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KJ6EAD

Joined Apr 30, 2011
1,581
The Raven Labs mixers and preamplifiers were intended to be run from two 9V batteries or a dual ±9V wall wart supply. The power requirement is obviously minimal so you could use any regulated dual ±9V supply if one were available (unlikely). It would be much easier to buy two matching single 9V supplies and wire them togther. The connections to a ⅛" stereo plug are: tip -9V, ring +9V and sleeve ground. You can minimize the wiring effort and avoid having to cram 4 wires into a connector by reversing the polarity of one of the supplies and using a purchased stereo to mono adapter as shown below. Mark the left and right plugs and jacks with colored shrink tube or a paint mark.

If the unit you have is the Universal Stereo Instrument Preamplifier (USIP), it was intended to be run from a dual ±12V wall wart supply. The supply wiring was the same as for the 9V setup. The multiple silkscreened 9V references were probably due to the fact that the USIP reused many of the same PCBs that were in prior units. It's doubtful that 12V would have harmed any of them and that 9V was chosen as the original power source simply because the designer intended that they be operable from batteries.

Various findings for reference below.
Raven Labs MDB-1 powered by two 9V batteries:
http://www.harmonycentral.com/blogs/News-RavenLabs/1998/09/14/raven-labs-debuts-with-the-mdb-1.

Various Raven Labs models:
http://www.musictoyz.com/studio/rack/rave.php.

A manual for a PMB-II:
http://web.archive.org/web/20051024164338/http://www.raven-labs.com/pages/products/pmb1/pmb1om.html.
 
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praondevou

Joined Jul 9, 2011
2,942
assuming 20mA LED current should let you know if it was designed for 9V or 12-13.8V
How would I know that they chose 20mA? :)

You know now that the DC that comes in they later call "9V". So why would they do this if it weren't supposed to work with 9V. I suggest you try it with 2x 9V power supply wiring the 3.5mm according to the last diagram you posted (polarity).
 

Thread Starter

Bassizer

Joined Aug 12, 2011
8
What is the output of the preamp supposed to be? 1V peak to peak?
I have no idea what the output voltage should be. All I can find in the manual is the maximum input voltage:

"Maximum input level for both Channels 1 and 2 is 1.5 Volts RMS"

Does this tell you anything?

What is the value of the resistor between power and the LED?
Colors are fine, like brown red brown or whatever.
I'm not sure in which direction to read the colors, but they appear to be either:

Brown Black Black Red Brown or Brown Red Black Black Brown

But when I measure across the resistor, I get 10KΩ...any ideas???

The multiple silkscreened 9V references were probably due to the fact that the USIP reused many of the same PCBs that were in prior units.
I don't think this is the case because all of the boards in the USIP are too big to fit inside any other Raven Labs device. The USIP is a rack mount unit and everything else they made was much smaller. All of the boards in the USIP pretty much span the width of the device. It appears to me that the designer intended 9V markings.

Further, because there is no resistance between the jack and the points marked 9V, doesn't that dictate that the supply should be 9V?

KJ6EAD, thank you very much for the nice diagram (the tip should be - fwiw). However, the problem isn't procuring a ±9V supply. I have one. it was sent to me by RL after they ceased production. The problem is that shortly after receiving the supply I left the country for several years and never used it. Well now I'm back and I want to use it. But in reading the manual I discovered the ±12V reference in the manual and hence my LONG quest to find the CORRECT voltage.

Does anyone here think it would be harmful to try it with the ±9V adapter that I have?

Finally, assuming the ±12V reference in the manual is incorrect and it SHOULD be ±9V, is there anyway the device could produce 12V of phantom power from a 9 volt power supply?
 
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praondevou

Joined Jul 9, 2011
2,942
Further, because there is no resistance between the jack and the points marked 9V, doesn't that dictate that the supply should be 9V?
That's what I thought too.

Does anyone here think it would be harmful to try it with the ±9V adapter that I have?
No, I don't think it would be harmful. Worst case it could be clipping some signals.

Finally, assuming the ±12V reference in the manual is incorrect and it SHOULD be ±9V, is there anyway the device could produce 12V of phantom power from a 9 volt power supply?
It doesn't look like this from the pictures.
 

KJ6EAD

Joined Apr 30, 2011
1,581
Power it with a 12V split supply and be done with it. There's nothing in the box but a bunch of connectors, switches, potentiometers and op amps. They can all handle ±12V. If you power it with ±9V, nothing bad will happen and it will have 9V phantom power which will also probably be enough since other Raven Labs products did just that.

Here's the USIP manual you have but neglected to post:
http://web.archive.org/web/20081017182516/http://www.raven-labs.com/pages/products/usip/usipom.html.
It mentions 12V in 3 places, twice in reference to phantom powering and once in relation to the power supply. If you want to believe that all 3 are errors, you may, but then the whole manual becomes suspect.

By the way, it's much more useful to supply what information you have in your first post as background instead of in the 15th as the basis for doubt.
 
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Thread Starter

Bassizer

Joined Aug 12, 2011
8
Power it with a 12V split supply and be done with it. There's nothing in the box but a bunch of connectors, switches, potentiometers and op amps. They can all handle ±12V. If you power it with ±9V, nothing bad will happen and it will have 9V phantom power which will also probably be enough since other Raven Labs products did just that.
Thanks for the info. At least now I know I can operate it without doing harm.

Here's the USIP manual you have but neglected to post:
http://web.archive.org/web/20081017182516/http://www.raven-labs.com/pages/products/usip/usipom.html.
It mentions 12V in 3 places, twice in reference to phantom powering and once in relation to the power supply. If you want to believe that all 3 are errors, you may, but then the whole manual becomes suspect.
If you think it would have been helpful to post the manual first, then point taken. FWIW, I deliberately did not post the manual because I wanted this to be a purely objective discussion based solely on the circuit itself. I didn't want it turn into a debate on the validity of the manual. I figured I would have to make up my own mind on whether the manual was correct or not.

But since the manual is now part of the discussion, I completely agree about the 12V references in the manual, and it makes it hard to dismiss that. However, if you plug a 12V supply into the device, all of the points marked "9V" on the circuit boards will now be getting 12V, which then makes these "9V" markings suspect, doesn't it? Shouldn't they have marked the boards "12V" instead?

I know you said it doesn't really matter and you can operate the device either way, but I am just trying to understand this the best I can.

If after trying the dual 12V adapter option, this appears to work better than my ±9V adapter, how hard would it be to either find a ±12 adapter or to make one? From my searching they seem to be basically non-existent, but I thought people here might know more on this.

Thanks again!
 

praondevou

Joined Jul 9, 2011
2,942
Before you connect the TRS connector make sure you measured correctly.

KJ6EAD found out that: "The connections to a ⅛" stereo plug are: tip +9V, ring -9V and sleeve ground."

You indicated tip being -9V and ring being +9V. Inverted polarity may damage your equipment.

A dual 12V power supply you could make out of a 24V power supply. There are also plenty of small dual power supplies on the market, even though the stable regulated power supplies are not the cheapest, so maybe you buy an unregulated with a higher voltage and put two voltage regulators at the output.
 

MrChips

Joined Oct 2, 2009
30,706
Power it with a 12V split supply and be done with it.
Yeh, I think we were splitting hairs here. There's only 3V difference between 9V and 12V. In this particular circuit I think 12V would do no harm.

However, if you plug a 12V supply into the device, all of the points marked "9V" on the circuit boards will now be getting 12V, which then makes these "9V" markings suspect, doesn't it? Shouldn't they have marked the boards "12V" instead?
Not really. We don't know the history in the product development. It is not unusual to change design specs after the board was printed.

Just go with the +12V/-12v supply.
 
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