LED fades out when input cable is removed instead of turning off immediately...

[mrsage]

I've been over my wiring a thousand times and I can't figure out what's going on here...

When I plug a cable into the input jack, the LED turns on as expected. However, instead of flipping off instantaneously when the plug is removed, the LED fades out gradually over a period of about 5 - 10 seconds.

I'm using a standard negative-ground power jack. I have one lug connected to the PCB, the next lug connected to the battery(+) terminal, and the final lug connected to the Battery(-) terminal and the ring of the input jack (so that inserting the input plug connects that to ground). The board ground is connected to the sleeve of the input jack, which is grounded on the enclosure.

That should work, right? To answer my own question, I have another circuit next to me that I wired up exactly the same way, and the LED flips off instantaneously when the input plug is removed...


To make things even stranger, I'm also getting "phantom" voltages when I use my DMM. If I measure the voltage going to the (+) terminal on the LED, I get a little over 2 volts reading on it. Strange, huh? This is when the pedal is unplugged, no battery, and nothing on the input jack.

Finally, the strangest thing. When I attach the battery to the battery clip, I sometimes get continunity between the ring and the sleeve on the input jack. Even when nothing is plugged into the input jack. The only thing I've done is plug a battery in there.

Again, I've been over all my wiring, and I can't seem to find anything I've done wrong. I switched out the input jack for a different one (thinking maybe I had a faulty one that was resulting in some continuity betweent he sleeve and the ring). I also swapped out the DC jack for a different one, thinking perhaps it was a problem there. That didn't work, nor did swapping out the battery clip.

I can't seem to figure out what's going wrong here! Any ideas? Things I may need to check?

The circuit itself works fine and sounds the way it should...I'm just concerned that I'm either draining battery power if I leave a battery plugged in (because of that weird continuity between the ring and sleeve) and/or I'm leaving the circuit powered at all times.

Help!

[brad]

Perhaps it's taking a while for the power decoupling cap to discharge?

Only thing I can think of 

[mrsage]

Perhaps it's taking a while for the power decoupling cap to discharge?

Only thing I can think of 

Well, that was my first thought, but it worries me that I'm getting continuity between the ring and sleeve of the input jack when I connect either the power supply or the battery...even when there's nothing plugged in there!

That makes me worry that power is draining or something.

I can't find any reason at all why those two spots should suddenly be connected!

[Peter Snowberg]

circuit? pictures?

[mrsage]

It's a Red Llama circuit.

My camera is pretty bad, but hopefully these will help a little...

In this first one you can see the input jack. Tip is connected to the 3PDT switch, sleeve is connected to the board ground and (via 1Meg resistor) to the input, ring is connected to the power jack.

On the power jack, one lug is connected to the ring and to the Battery(-), next lug is connected to Battery(+), and the last lug is connected to 9V+ on the board.

[mrsage]

[A.S.P.]

a switching jack?

[A.S.P.]

ahh: that one (top) jack`s contact is very close to the pot`s lug...
(when you plug s.th. in, it might short...)

[Peter Snowberg]

Great work on that build!

My 2 cents: The led is being powered by the decoupling cap like Brad guessed too. It drains slowly until it gets below the Vf of the LED where it just about stops leaking because it's also below the Vf of the inverter chip. At that point no light, but your meter is going to get VERY confused if you add that charge to the voltage it's already using to measure the resistance bewtween the probes.

Everything is probably just fine.

If you want to check for sure, just pop out the battery and check for a change.

[mrsage]

ahh: that one (top) jack`s contact is very close to the pot`s lug...
(when you plug s.th. in, it might short...)

It's not quite as close as it looks in the picture...

Thanks for the suggestion, though!

[A.S.P.]

if not, Peter is always very close, with his thoughts 

[mrsage]

Great work on that build!

My 2 cents: The led is being powered by the decoupling cap like Brad guessed too. It drains slowly until it gets below the Vf of the LED where it just about stops leaking because it's also below the Vf of the inverter chip. At that point no light, but your meter is going to get VERY confused if you add that charge to the voltage it's already using to measure the resistance bewtween the probes.

Everything is probably just fine.

If you want to check for sure, just pop out the battery and check for a change.

Thanks for the compliment! I just ordered some PCBs from ExpressPCB, and they came out pretty nice.

Well, the voltage definitely changes when I pop the battery out, but it stays right around 2 - 2.5V (around 5V when the jack is plugged in and the battery is connected).

How do I check the current draw of the circuit to see if it's drawing power even when there's no plug in the input jack?

[Peter Snowberg]

You need a meter that measures current in the mA range.

Snap the 9V into one of the two snaps in the holder and touch the probes to the other side of the battery and the other side of the snap, with the meter set to measure mA and the probes plugged into the appropriate holes for current measurement.

[mrsage]

You need a meter that measures current in the mA range.

Snap the 9V into one of the two snaps in the holder and touch the probes to the other side of the battery and the other side of the snap, with the meter set to measure mA and the probes plugged into the appropriate holes for current measurement.

Okay...I did that and I'm not getting a reading (set the probe to DCA in the 20m and 200m ranges).

How can I test to see what the circuit draws when it's engaged? I tried doing that same thing with a plug in the input jack, but I still got no reading...

[Caferacernoc]

I have a DOD BIFET preamp pedal that does the same thing. I pull out the input cable and the LED slowly fades out over like 5 seconds. Nice boost pedal by the way but I don't care for the "bypass". Works good if you plan to leave it on all the time, though.

[vanhansen]

My Marsha Valve build does the exact same thing.  It's the big power filter cap draining.  No biggie.  After unplugging the input jack, put your meter probes on the circuit power and ground rails.  If you get zero, the battery is disconnected.

[mrsage]

I have a DOD BIFET preamp pedal that does the same thing. I pull out the input cable and the LED slowly fades out over like 5 seconds. Nice boost pedal by the way but I don't care for the "bypass". Works good if you plan to leave it on all the time, though.

Yeah, the BiFet is a pretty nice booster.

I'm surprised someone hasn't cloned it yet.

I've got a schematic somewhere for one of the other DOD FET boosts, but not the BiFet.

[MartyMart]

My MXR Micro amp clone was doing this, I plugged in after a week or so and the battery was
drained.
Turned out that the input jack was grounding on the box ( damm 1590B BimBox )
There's only a 2mm "window" for drilling error on the sides using switchcraft open sockets, too
much either way and the socket will ground !!
I changed to the plastic square type and it's fine now

MM.

[petemoore]

  To light an LED you need a source of power, if the *power source is disabled, the only other power source would be in the 'other battery'...the BFC, it's charge storing capabilities are plenty to keep an LED lit for a 'while'...it's typical to see an LED stay lit, fade and go out after a 'few' [up to about 15 seconds, not much more I'd guess].

[Sir H C]

I don't know the circuit but it is possible that the cap stops draining at about 2.5 volts because that is where everything is shut down and removes current paths for the cap to discharge.  Pull the battery and see if that voltage remains there, if so, then it is just that the circuit is now high impedance and no problems.  Remember 2.5 V = 3 diode drops, which is probably a minimum path in the ICs and stuff, so that could well be what is happening.