"Little Angel" - Super Simple PT2399 Mini Chorus

[anotherjim]

Remove it.

[jfrabat]

Ok.  I will try it.  So if I am reading the diagram right, that would take the transistor out of the equation, right?  OK, I will try. 

Now, quick question, to make this a richer learning experience; explain to a noob in simple words what is the purpose of that transistor (to understand why I am taking out of the loop).  In my overdrive, I use them for boost (or so I understood, anyway!), but this pedal uses an OpAmp, so it should not be the source of boost, right?

[EBK]

It's being used as a delayed switch here, I believe.

[anotherjim]

Yes it's time delay switch using the transistor as the "switch"

On power up the 100R should be disconnected from the PT2399. The transistor should be turned off initially and that means there is no path to 0v through the transistor collector-emitter for the 100R, effectively switching it out of circuit. When the 47uF capacitor charges up sufficiently (the time delay), the transistor turns on because there is enough voltage to "forward bias" it's base-emitter junction. In your case, this appears to happen about 0.5V.

This was the easiest schematic to find -  ignore the red modification notes.

Now, the time delay is how long it takes the 47uF to charge via the 68k resistor. This must be long enough for the PT2399 internals to get running normally with only the 2k2 resistor setting the delay time. Then the 100R can be switched in, making the PT2399 delay shorter to suit the chorus effect.

If you think about it, the time delay due to the 47uF capacitor only happens IF the capacitor is completely discharged before power is switched on. I'm not sure that this is always happening. Indeed the 47uF may keep a small charge for a very long time after 9v power is removed.

A cure might be to fit another 1N4148 diode in series with the transistor base. This would raise the time it takes for the capacitor charge to turn the transistor on AND negate the effect of any small charge remaining in the capacitor from the last time power was applied. I also wonder if it wouldn't be better for the existing 1N4148 diode that is meant to discharge the capacitor to have it's cathode connected to the 9v supply instead of 5v. I suspect it will be able to discharge the capacitor faster that way.

This is a bit technical, but MOS devices like the PT2399 tend to have very high resistance to the power supply once the supply voltage falls to about 1v, because none of the MOS transistors in it can be turned on at that low a voltage, so there isn't much resistive loading on the 5v supply that could discharge the 47uF capacitor below that 0.5v level. It could perhaps have 0.4v on it for quite some time.

[jfrabat]

You know, now that you mentioned the transistor, it got me wondering; I know that transistors can get damaged by heat when soldering.  I did not socket my transistors.  Could it be that I damaged them while soldering?  Should I try removing the transistor and soldering a socket and THEN trying the pedal?

[EBK]

You know, now that you mentioned the transistor, it got me wondering; I know that transistors can get damaged by heat when soldering.  I did not socket my transistors.  Could it be that I damaged them while soldering?  Should I try removing the transistor and soldering a socket and THEN trying the pedal?

I'd say yes.  We haven't tried everything until we tried that, right?

[EBK]

I'm looking at that transistor arrangement.....
Does that cap really have a discharge path other than through the transistor?  I don't see the diode doing anything here.

Update: ok, I see that the 68k resistor is connected to the +5v node, which would eventually drop to ground potential upon power down, providing another discharge path.  I can also see how the diode can speed up the discharge once it conducts, decreasing this effective discharge path resistance.

[jfrabat]

OK, just to be clear then, the best way to proceed for me now would be remove the transistor (2N2222) and solder the socket and try the pedal without the transistor, or do I need to take out the 100R resistor as well?

[EBK]

OK, just to be clear then, the best way to proceed for me now would be remove the transistor (2N2222) and solder the socket and try the pedal without the transistor, or do I need to take out the 100R resistor as well?

Without the transistor, the 100R is already effectively disconnected.  Leave it on your board.

[jfrabat]

OK.  Will report tonight on the results.

[EBK]

Indeed the 47uF may keep a small charge for a very long time after 9v power is removed.

Upon removing the 9V supply, the 7805 provides a resistive path to ground through its output terminal.  (The PT2399 possibly also provides a discharge path to ground.)
The charge would neutralize fairly quickly.

[anotherjim]

There is a resistive path to ground in the 7805 output, but I don't know what value.
I'm still not sure having the timer threshold voltage as low as the base-emitter forward voltage be a good thing. 0.5V seems a touch low for a silicon small-signal transistor base-emitter junction, maybe that particular transistor is damaged/leaky.

[duck_arse]

if we were to fit the reverse cap-discharge regulator diode (no. I dunno what to call it, the cap flyback preventorator?) across the regulator out to in, like in the datasheet, wouldn't that provide a [known] discharge for the offending cap?

[EBK]

I don't think we really care how fast it discharges, as long as it does.

More important to us is the charge time, which must be more than 400ms, I believe.

[anotherjim]

Capacitor charge should reach 0.5v in about 350ms. That situation is complicated by the conduction "knee" of the base-emitter, which starts to take some charging current away from the capacitor before it gets clamped at the forward voltage drop - so maybe it can take 400ms. It's bit tight though isn't it?

The problem anyway with that discharge diode is that it will be "off" until the 5V supply falls sufficiently below the capacitor voltage. If the capacitor voltage is 0.5v and the diode forward voltage is 0.6v, then the 5v supply needs to fall to -0.1v, which can't happen. It's possible that the diode never conducts.

The discharge path is then via the 68k in series with whatever else goes to 0v.
If it was only 68k, and the 5v supply falls to 0v instantly (it won't), at 5 seconds after switch off the capacitor voltage is down to 0.1v. Not 0v - so I think you can fool the circuit if power goes on and off quickly enough, which is easily done when testing.

[jfrabat]

Well, guys, we have still not found the problem.  Took out the transistor tonight, but no change.  Still only sending the dry signal through.

[maiko]

Sorry to hear that
i hope you wont mind  ive taken it upon myself to check your layout against the JMK pdf all seems to
be ok but can you verify if your sockets for your ic on the solder side has continuity
to your ic legs.

If the jmk board is a dual sided board there are some soldering on your component
side that appears to be incompletly soldered.  maybe you can check that speed lug 3
in your picture there is cleary a trace from that to a 220k resistor but there doesnot
appear to be solder on the component side of the board. check and verify if there is continuity
from that to the resistor

[jfrabat]

i hope you wont mind  ive taken it upon myself to check your layout against the JMK pdf all seems to
be ok but can you verify if your sockets for your ic on the solder side has continuity
to your ic legs.

I dont mind AT ALL!  IN fact, thank you!  It would not be the first time I misplace components or use the wrong ones!  This one I have tripple checked hunting down whatever it is that it is making it not work!  I will test continuity tomorrow (I'll be busy playing the guitar this evening!)

If the jmk board is a dual sided board there are some soldering on your component
side that appears to be incompletely soldered.  maybe you can check that speed lug 3
in your picture there is clearly a trace from that to a 220k resistor but there does not
appear to be solder on the component side of the board. check and verify if there is continuity
from that to the resistor

I will check all soldering where there appears to be some suspect contact on the top of the board, and re-solder those that are suspect.  I will also install the socket for the transistor (I simply uninstalled it last night, got frustrated, and left the whole thing as it was!).

Felipe

[anotherjim]

You know how those PT2399 are prone to stay locked up. Did you test the chip in your working delay or discharge any static after removing the transistor?

[jfrabat]

No, I did not...  Let me try that first.