An observation of power supply behaviour

[Bunkey]

An Observation of Power Supply Behaviour





So I was getting fed up of replacing batteries in a quirky boost pedal I built recently...
It features a bright LED which seems to be rather power hungry and was giving me less than 30 hours play time before the battery supply voltage dropped below an acceptable level.

I found a 9v AC adapter hidden in a bundle of wires - I think it belonged to an old Sony wireless audio peripheral - and replaced the jack with a battery clip so I could just connect it to the battery clip in the pedal.





The pedal is of minimalist design, it was intended solely for battery operation and doesn't feature any supply filtering or the like. It has a switched input jack.

When switching the power off (at the wall) I observed the voltage slowly dissipate through the LED until it reaches ~2.5v where it seems to hover.
I imagine if I unplugged the input jack to open the circuit at the same time, the supply would remain hovering at a 9v potential - Coming back to it today, 18 hours after doing something similar, it was still sitting at 3.5v.

Would it be preferable then to include in the design of circuits something like a 10M resitor between the supply terminals to discharge the circuit as a precaution if used with power adapters that don't seem to discharge themselves when switched off..?

[Rob Strand]

I'd expect the bias resistors (on the base) to discharge any capacitance.   However, the Bass-fuzz doesn't have those bias resistors so when the supply reaches the sum of the diode and base-emitter voltages the supply drain will drop to almost zero and the supply capacitors will only discharge via leakage.  Good caps might take a long time to discharge.

As a general comment I wouldn't add a discharge resistor unless there was a problem with the circuit being left at some intermediate voltage.   Adding the 10M resistor would of course drain any caps.

Switch-mode type AC adaptors  often have a small load resistor that will eventually drain the caps but the old transformer type AC adaptors rarely have any such load resistor.   That's just how they are.  (In some scenarious the charged caps could cause a spark hazard.)

[Bunkey]

I'd expect the bias resistors (on the base) to discharge any capacitance.   However, the Bass-fuzz doesn't have those bias resistors so when the supply reaches the sum of the diode and base-emitter voltages the supply drain will drop to almost zero and the supply capacitors will only discharge via leakage.  Good caps might take a long time to discharge.

As a general comment I wouldn't add a discharge resistor unless there was a problem with the circuit being left at some intermediate voltage.   Adding the 10M resistor would of course drain any caps.

When the input jack is unplugged the circuit itself is floating so there's nowhere for the risidual charge to drain.

I guess it's a non-issue but it just seems like good design etiquette to not have residual charge in a circuit when the circuit is technically powered off.

It was a concern here because I regularly handle the circuit to swap the clipping diodes - but then I'd be unplugging the adapter anyway & any charge in the in/out caps is going to remain whether I'm using a battery or mains supply...


Considering the design of the fuzz however (I'm calling it the 'fusscake' btw - a nod to it's origins - but it's as much a Bazz Fuss as a Woolly Mammoth is an NPN FuzzFace & I don't want people to think I've spent 9 months repackaging a Bazz Fuss and called it my own pedal  ).. considering it's design, it's possible for someone to short two points together if they were doing something careless; so I'm more concerned about that.

Assuming then that it is preferable to add a drain resistor in this case - There's no downside you can think of I might be overlooking besides 0.9uA added current draw?

[merlinb]

Your bias resistors will already drain the power supply. Adding 10M doesn't really do anything extra. But it won't hurt to add it if you want to.

[Bunkey]

My fault for not drawing the schematic like this in the first place. The potential is being measured across the supply points on the pedal (obviously I don't have the 10M in place yet).

I believe you've answered my query though thank you