How to add DC jack to a Uniwah?

[Paul Marossy]

I picked up a Uniwah that someone put a 9V DC jack on, but the circuit operates on 4.5V. It was a total hack with the wiring anyway and am happy that I never even tried using it. It needs 4.5V because it also uses a #47 6.3V bulb. I don't want to run it on 9V in any case.

So I thought, OK, I'll just put a 5V voltage regulator on it and knock down the output voltage 1/2 volt with a resistor. Now I want to be able to switch between battery power and wall wart power. When I got into it I realized that there appears to be no way to do this because the circuit is either connecting directly to the batteries or the output of the voltage regulator.

So am I just up the creek on this idea? Or is there a way to do this that I haven't thought of?

[ElectricDruid]

What's that thing Duck always says? Something abut schematics? hohoho

Honestly, without seeing it, I'm just guessing. And that's not likely to help you much.

[Paul Marossy]

What's that thing Duck always says? Something abut schematics? hohoho

Honestly, without seeing it, I'm just guessing. And that's not likely to help you much.

Schematic: http://www.diyguitarist.net/Images/UNIWAH_SCHEM.JPG

[ElectricDruid]

I don't see any problem with it. Leave the batteries where they are, but add a switched DC jack so that they're only connected when there's no power supply plugged in. And I'd have thought 5V was close enough for a circuit like that (especially if someone's succesfully managed to run it on 9V), but you *could* tweak the 100R resistor if you wanted. I wouldn't bother, since it'll also affect the batteries.

One thing's weird though. That looks like a positive ground circuit, but with NPNs? What am I missing?

PS: Love the germanium diode as reverse power supply protection. Imagine having so many of them so cheap that you think that's a good idea! 

[duck_arse]

hohoho, like I allays say.

wired as per the schem, the lamp will light, but the circuit won't produce a peep - the reverse polarity diode will get in the way of all those volts.

[Paul Marossy]

I don't see any problem with it. Leave the batteries where they are, but add a switched DC jack so that they're only connected when there's no power supply plugged in. And I'd have thought 5V was close enough for a circuit like that (especially if someone's succesfully managed to run it on 9V), but you *could* tweak the 100R resistor if you wanted. I wouldn't bother, since it'll also affect the batteries.

One thing's weird though. That looks like a positive ground circuit, but with NPNs? What am I missing?

PS: Love the germanium diode as reverse power supply protection. Imagine having so many of them so cheap that you think that's a good idea! 

Oh yeah, forgot to mention that part of the schematic is drawn incorrectly - it's neg gnd circuit.

Problem is that the 9V goes to the regulator, the output of which goes to the circuit, so it has to go THRU the regulator to the circuit. Won't work for use with batteries, because they have to go thru the regulator to get there. In battery mode the batteries connect directly to the circuit. So it seems to me you'd need to have parallel runs to the circuit board, which creates a situation where the wall wart will be connected to the batteries at the circuit board and the circuit will just get straight 9 volts because one of those wires are in parallel with the regulator, which defeats the purpose of the regulator. I was wondering if maybe you could get around that with diode but that also seems problematic because of the internal circuitry of the voltage regulator.

I drew this up to demonstrate what I'm talking about:


Seems to me the only real way to make this work is to add a SPDT switch to select between DC jack or batteries, and the DC jack just becomes strictly a wall wart connector. I mean it could be done that way but it would be difficult to implement due to limited room to do all this. I also thought of just using a voltage divider but that has its own set up problems. 

[R.G.]

You could use a transistor to sense whether there is 9V on the DC jack, and use that to work the switching from battery to jack+regulator. The switching could easily be a $3 DPDT relay (Panasonic, Kemet, at Mouser) or a couple of transistors, bipolar or MOSFET.

Look at this one:

The resistors may need tweaked, as I don't know how much current the effect draws, especially with that 6.3V bulb as a volume device. If you were to measure the voltage across that 100 ohm resistor, I could make better guesses.

[ElectricDruid]

Problem is that the 9V goes to the regulator, the output of which goes to the circuit, so it has to go THRU the regulator to the circuit. Won't work for use with batteries, because they have to go thru the regulator to get there.

Aaah, ok. Yeah, now I get it. Thanks Paul. I was thinking you were going to just stick a 5V power supply into it. My bad.

If the battery was 9V too, then no problem. But switching between a lower-valued battery or a regulated supply is going to need either a completely different power jack from normal or some intelligence internally like RG suggested.

[Paul Marossy]

Problem is that the 9V goes to the regulator, the output of which goes to the circuit, so it has to go THRU the regulator to the circuit. Won't work for use with batteries, because they have to go thru the regulator to get there.

Aaah, ok. Yeah, now I get it. Thanks Paul. I was thinking you were going to just stick a 5V power supply into it. My bad.

If the battery was 9V too, then no problem. But switching between a lower-valued battery or a regulated supply is going to need either a completely different power jack from normal or some intelligence internally like RG suggested.

I like RG's idea but there's not really enough room for all that stuff in the space that I have to work with. 

[R.G.]

I like RG's idea but there's not really enough room for all that stuff in the space that I have to work with. 

Hmmm. Three TO-92s, three resistors, and a Schottky diode are too big? Possibly with the resistors cord-wooded and the parts all hot glued together?

[Paul Marossy]

I like RG's idea but there's not really enough room for all that stuff in the space that I have to work with. 

Hmmm. Three TO-92s, three resistors, and a Schottky diode are too big? Possibly with the resistors cord-wooded and the parts all hot glued together?

Well you mentioned a relay above... I think could be done the way you propose in that schematic. Still is tight due to how the plastic enclosure is designed. The batteries are in between the jacks and the circuit, so there's not a lot of space to work with as the jacks are in their own little compartment. I am under the impression that the bulb alone is 150mA, so I need to use LM7805 type regulator. The circuit always gets power thru that 100 ohm resistor (two tranistors) and the lamp is switched on/off separately via a very strange DPDT push button switch with 12 pins.

[R.G.]

Hmm. (again!)
Can you get a current reading of the batteries in both A and B positions?

Digging into the schemo a bit, I'm guessing that the circuit operates by wobbling a magnetic core into and out of the inductor coil to change its inductance for the wah, and has a light shield moving between the lamp and LDR for volume change.

Depending on how much taste you have for adventure, you might consider replacing the lamp with an LED and resistor. That alone would cut the current from ~150ma to 20ma (using the 1/7th the current for same light approximation for LEDs).

[Paul Marossy]

Hmm. (again!)
Can you get a current reading of the batteries in both A and B positions?

Digging into the schemo a bit, I'm guessing that the circuit operates by wobbling a magnetic core into and out of the inductor coil to change its inductance for the wah, and has a light shield moving between the lamp and LDR for volume change.

Depending on how much taste you have for adventure, you might consider replacing the lamp with an LED and resistor. That alone would cut the current from ~150ma to 20ma (using the 1/7th the current for same light approximation for LEDs).

Sorry for late reply... been very busy at work designing HVAC for nutraceutical manufacturing plant. I think was actually the most complex commercial project I have worked on. Client was also a P.I.T.A. which didn't help me much.

Anyway, now that I finally have my head above water again, it measures 0.8mA in wah mode and 112mA in volume mode. I had already thought about an LED replacement but I'm not sure how the photocell would like that... I guess if the light level is correct it shouldn't care? I have a feeling that's probably not as straightforward as it sounds. We have 6.3V bulb running on 4.5V. Would a LED designed for 6.3V even light up on 4.5V?

EDIT: Yes, it has a variable inductor that works same was as the Kay Wah Wah. There is a flat spring which has a disc attached to it and one end of it is at a tilt to the other part. When that is moved closer the winding/cup-core assembly it increases the inductance.

[R.G.]

Sorry for late reply... been very busy at work designing HVAC for nutraceutical manufacturing plant. I think was actually the most complex commercial project I have worked on. Client was also a P.I.T.A. which didn't help me much.

Condolences - making a living can sure get in the way of the important stuff. I know all too well.

Anyway, now that I finally have my head above water again, it measures 0.08mA in wah mode and 112mA in volume mode. I had already thought about an LED replacement but I'm not sure how the photocell would like that... I guess if the light level is correct it shouldn't care? I have a feeling that's probably not as straightforward as it sounds. We have 6.3V bulb running on 4.5V. Would a LED designed for 6.3V even light up on 4.5V?

.08ma ? Wow. A zener would be fine with an 80 microamp load. The batteries are for the bulb. It's a flashlight. 8-)

Photocells have a spectral response, and are more sensitive to some frequencies of light than others. But they are otherwise simple - more or less light means less or more resistance. So the amount of light is what matters, wobbled a bit by the color. CdS is most sensitive in the greens, but has response down in the reds. That bulb puts out most of its light in the red, so even a red LED might be fine. LEDs light up in general at voltages lower than 4.5V for red, orange, yellow and green "simple" LEDs. Blue and UV+phosphor LEDs are higher, but that's not what you'd use. The voltage above the LED forward voltage is used for a resistor or current source to set the LED current, and then the light output varies with the current. LED voltages of 1.4 to 2 are very common.  So yeah, there's enough voltage. A resistor and an LED would do it.

My concern is the mechanism of varying the light to the photocell. I'm guessing that they put some kind of moving shutter between the bulb and the photocell, since the schematic doesn't show any electronic way to vary the bulb's current. Incandescent lamps are light-promiscuous - they spit light everywhere. LEDs are more directed. Their light comes out in a narrow cone. Depending on the LED and the mechanical setup, you might have to tinker the LED to be more spread out or be further away so the shutter works well.

[amptramp]

Anyway, now that I finally have my head above water again, it measures 0.08mA in wah mode and 112mA in volume mode. I had already thought about an LED replacement but I'm not sure how the photocell would like that... I guess if the light level is correct it shouldn't care? I have a feeling that's probably not as straightforward as it sounds. We have 6.3V bulb running on 4.5V. Would a LED designed for 6.3V even light up on 4.5V?

An LED provides light output proportional to the current through it.  A red LED would have less than 2 volts across it and a green LED would have a little more than 2 volts across it.  I have measured 1.88 volts for red and 2.05 volts for green in circuits I have built.  You then have to add a current-limiting resistor sized so that the difference between power supply and LED voltage divided by the current you want to run is the resistance.  It may not favour green - if you use red LED's, you might be able to get away with two in series.  Two greens in series would be 4.1 volts and your resistor would be small - possibly too small to keep current within limits over the range of parts tolerance, temperature and supply voltage variation.

[R.G.]

Yeah, resistors can only do so much. The fact that it's on batteries also causes some grief.
In the past, I've gone to something like a simple active current limiter. Two garden variety PNPs and a couple of resistors would do it.  This would keep the current fixed even with declining battery voltage. This can be set up to only take as little as a diode drop plus a couple of tenths of a volt.
It is more complex, a little, but gives good control when there's not much voltage for a resistor to be effective over voltage variation.

[Paul Marossy]

Anyway, now that I finally have my head above water again, it measures 0.08mA in wah mode and 112mA in volume mode. I had already thought about an LED replacement but I'm not sure how the photocell would like that... I guess if the light level is correct it shouldn't care? I have a feeling that's probably not as straightforward as it sounds. We have 6.3V bulb running on 4.5V. Would a LED designed for 6.3V even light up on 4.5V?

An LED provides light output proportional to the current through it.  A red LED would have less than 2 volts across it and a green LED would have a little more than 2 volts across it.  I have measured 1.88 volts for red and 2.05 volts for green in circuits I have built.  You then have to add a current-limiting resistor sized so that the difference between power supply and LED voltage divided by the current you want to run is the resistance.  It may not favour green - if you use red LED's, you might be able to get away with two in series.  Two greens in series would be 4.1 volts and your resistor would be small - possibly too small to keep current within limits over the range of parts tolerance, temperature and supply voltage variation.

Yeah... I dunno how well that would work. I suspect these drop in replacement LED "light bulbs" likely have current limiting resistors inside of them already. That's why I was wondering if it would work at all. Maybe so but it has to be a certain minimum brightness.

What about a dimmed down white or blue LED? I think they would be closer to the brightness of the flashlight inside the pedal. This a tough one to work on because of how it's designed - moving parts inside and not much physical room for any kind of modifications to be done to it.

Also another obstacle with the whole idea is how to mount an LED. It needs to be in a very specific location for the volume mode to work well. When this pedal is working properly with all of the out of spec parts replaced, the volume pedal function is actually nice. I've never actually used the volume function on any of other wah-volume pedals, with the exception of the DOD FX-17, which I use ONLY as a volume pedal.

[amptramp]

Anyway, now that I finally have my head above water again, it measures 0.08mA in wah mode and 112mA in volume mode. I had already thought about an LED replacement but I'm not sure how the photocell would like that... I guess if the light level is correct it shouldn't care? I have a feeling that's probably not as straightforward as it sounds. We have 6.3V bulb running on 4.5V. Would a LED designed for 6.3V even light up on 4.5V?

An LED provides light output proportional to the current through it.  A red LED would have less than 2 volts across it and a green LED would have a little more than 2 volts across it.  I have measured 1.88 volts for red and 2.05 volts for green in circuits I have built.  You then have to add a current-limiting resistor sized so that the difference between power supply and LED voltage divided by the current you want to run is the resistance.  It may not favour green - if you use red LED's, you might be able to get away with two in series.  Two greens in series would be 4.1 volts and your resistor would be small - possibly too small to keep current within limits over the range of parts tolerance, temperature and supply voltage variation.

Yeah... I dunno how well that would work. I suspect these drop in replacement LED "light bulbs" likely have current limiting resistors inside of them already. That's why I was wondering if it would work at all. Maybe so but it has to be a certain minimum brightness.

What about a dimmed down white or blue LED? I think they would be closer to the brightness of the flashlight inside the pedal. This a tough one to work on because of how it's designed - moving parts inside and not much physical room for any kind of modifications to be done to it.

Also another obstacle with the whole idea is how to mount an LED. It needs to be in a very specific location for the volume mode to work well. When this pedal is working properly with all of the out of spec parts replaced, the volume pedal function is actually nice. I've never actually used the volume function on any of other wah-volume pedals, with the exception of the DOD FX-17, which I use ONLY as a volume pedal.

You are not looking for a LED light bulb, like the replacements you can get for typical dial lamps in radios and stereos.  They will definitely have a resistor in series because they are used where the source is an AC voltage, not a fixed current.  As R.G. said, most photocells are CdS with a peak response in the green, so the extra voltage drop you get with blue or white is a waste of power.  You can mount a green LED and a resistor or current source as a complete circuit element into a light bulb socket.

[Elektrojänis]

Totally outside of the solutions you are discussing... How about something like this:



Sorry for the rough sketch. I drew it on some random piece of paper at the gym. There also might be some aspect I didn't think of that makes it a bad idea or simply non functional.

One caveat I can see is, that if you are going to use a regular barrel jack for the DC, you will need to wire it center positive. This is because the switched part of those jacks is usually the outside contact. (I hope I remember that right.) That goes against the common center positive Boss "standard".

Edit: I left out the input and output caps for the 7805 too. Those should be added ofcource.

[Elektrojänis]

That goes against the common center positive Boss "standard".

And I meant it is against the common center negative Boss "standard". Duh!