Dying battery… diode switch, lm317, pot wired as voltage divider. Which and wh

Started by Locrian99, April 03, 2025, 12:01:27 AM

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Locrian99

So I'm looking at adding a dying battery type thing to a fuzz face, or really just anything right now though a fuzz face.

I was reading over at jack ormans site on the led switch, and the lm317 set up.   Def lean towards the pot rather than the switch.   But what I don't see any mention of is a simple voltage divider wired pot with a variable resistor wired pot in series to accomplish the resistance.   This seems like the easiest option to me.   10k pot wired as a voltage divider, 10k resistor between lug 1 and ground to set your minimum voltage at 10k, then just put a pot in series to give the extra current resistance. 

Is there a reason the other options are better?

Also is there a reason to put this before or after say a 1044 charge pump.   

Thx!

PRR

> over at jack ormans site on the led switch, and the lm317 set up.

Do you mean muzique.com ? That's a big site. I did not find "dying".

Patents On Parade has a plan with a regulator, switches, and resistors (no LED). Is that where you are looking?

http://www.muzique.com/lab/patent2.htm   Fig A7
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Locrian99

Quote from: PRR on April 03, 2025, 01:14:38 AM> over at jack ormans site on the led switch, and the lm317 set up.

Do you mean muzique.com ? That's a big site. I did not find "dying".

Patents On Parade has a plan with a regulator, switches, and resistors (no LED). Is that where you are looking?

http://www.muzique.com/lab/patent2.htm   Fig A7

Yes that is what i was referring to. 

Rob Strand

When a (9V) battery dies the *open circuit* voltage drops and the impedance goes up.

Here's some measured results done by Jack,
https://www.muzique.com/lab/batteryz.htm

So while the load causes a voltage drop across the battery impedance the actual voltage does drop.

If you have a circuit which pull very little current the drop across the impedances isn't the main cause of voltage drop, the drop in the open circuit voltage does more.

So when you see low current circuit with large-ish series resistances to simulate a flat battery.  It only works because the small current and large resistor cause a similar voltage drop, and that the circuit still works with a high impedance supply.

If you had a higher current circuit them the large resistor wouldn't work.  To the same degree of approximation we could choose a lower valued series resistor to get the same drop as the low current circuit.

Where the resistor approximation doesn't work is when the added impedance in the supply rail causes different parts of the circuit to interact via the supply rail.   If we control the voltage directly we don't need such a large resistance.   The impedance of the supply can be low and the weird interactions are avoided, or are less likely.
[FYI: some of the Zvex pedals do this.   It's deliberate and what some people want.  I call it changing the circuit more than battery control.]

So in general you might find the direct voltage control is less problematic.    You could argue we should be adding some series resistance, as per Jack's results,  we be more accurate emulation of the battery.   In fact the impedance really needs to increase as the voltage is reduced.

The 10k pot will work.   It does provide variable open circuit voltage.  It does provide increasing impedance as the voltage is reduced from full to say half.   It's not bad.   It would probably work for most transistor circuit but perhaps wouldn't work so great for opamp based circuits.


Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Locrian99

Quote from: Rob Strand on April 03, 2025, 03:48:10 AMThe 10k pot will work.   It does provide variable open circuit voltage.  It does provide increasing impedance as the voltage is reduced from full to say half.   It's not bad.   It would probably work for most transistor circuit but perhaps wouldn't work so great for opamp based circuits.




Well the good news is I just want to use this in a fairly straightforward FF circuit.   It sounds to me like the pot would be better placed after the charge pump I am using to get my -9v.    Do you think the impedance from the pot will make the series resistor unnecessary?    Inwas going to experiment with a 500r pot wired as a variable resistor in series with the voltage divider. 

Thx again.

Rob Strand

Quote from: Locrian99 on April 03, 2025, 11:34:58 AMDo you think the impedance from the pot will make the series resistor unnecessary?    Inwas going to experiment with a 500r pot wired as a variable resistor in series with the voltage divider. 

The natural resistance of the pot is nearly there.   Arguably a good battery has an impedance of 1 to 10 ohm, so  adding that small resistance on full isn't going to do much in most cases.

Rememeber the lower the impedance of the pot the more current is drawn from the supply: 9V/10k => 900uA but 9V/500R => 18mA!
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Rob Strand

I've got these measurements for very flat old batteries:

Open circuit voltage and DC output resistance.

Voc [V]   Ro [ ohm]
2.3   60000
5.2   3100
6.45   1000
7.77   210
7.36   151
7.74   20

See AMZ page for the case when the battery isn't so flat.

When the open circuit voltage is less than say 5V the circuit current is going to pull down the actual battery voltage quite a bit.   The actual voltage is going to be very circuit dependent - perhaps a motivation to control the voltage and not try for crazy high output impedances.

Also when you have such a high DC output impedance the circuit could misbehave.  A real battery has an AC output impedance as well.  If you wanted the dying battery circuit to emulate such high DC resistance you would need to add another circuit (cap + resistor) to emulate the AC impedance, which won't be as high as the DC impedance.  If you don't the pedal circuit won't behave like it does with a real dying battery.   The patent page on the AMZ site show the AC impedance circuit.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Mark Hammer

I distinguish between how a "dying" carbon-zinc battery behaves, and mere lower supply voltages.

The whole "dying battery" thing came out of Eric Johnson's use of germanium Fuzz Faces.  The thing is that when carbon-zinc batteries are in their twilight years, they have the characteristic of being able to deliver current for very brief spurts, before needing to catch their breath.  When used with low current-draw fuzzes, and such, this can yield a sort of compression or sag, where the fuzz responds to pick attack, but can't get enough current from the battery to provide much in the way of dynamics.  This is NOT the same as what happens with an alkaline 9V, or a regulated power supply turned down to a stable 8V,7V, or other.  It is also unlikely to occur if the pedal requires more current than a simple germanium fuzz.  Those puppies are generally down in the <7ma range.  If you can find any player touting the virtues of a dying carbon-zinc battery with a drive pedal that normally draws >12ma, I'd be VERY surprised.

That is NOT to say that running a pedal off a stable supply OTHER than 9V won't yield anything interesting or musically valid.  It just won't sound or "breathe" the same as an actual carbon-zinc 9V that easily runs out of steam.

So if you're simply curious about how a given circuit sounds with a different supply voltage, have at it, and have fun.  But if you're pursuing emulation of a dying carbon-zinc battery without having to use a battery, it's going to take a lot mere than simply lowering the V+ with a regulator.

antonis

I do presume that we all agree in "dying battery" effect is mainly power matter rather than voltage or current one (i.e. ideal voltage or current source)..

i.e.: Neither a 6V(say) nor a 10mA(say) stiff PS can simulate a dying battery..
(the former keeps providing constant voltage independently of current demand where the later keeps providing constant current independently of voltage variations..)
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Mark Hammer

Yes.

I think the mistaken inference that some make is that:
a) the same battery behaviour would be true of any type of battery (carbon-zinc, alkaline, lithium, NiMH, etc.);
b) since the desired character only occurs after the battery has dropped down a volt, or 2 or 3, from peak "fresh" state, that it is the voltage that matters, rather than voltage being a marker for something more relevant.

PRR

Quote from: antonis on Today at 11:44:27 AMI do presume that we all agree in "dying battery" effect is mainly

I don't think "we all agree" what "dying battery" is.

I have a suspicion it is more in the ears and the hands (or just "all in your head") than in the pedal.
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