I modified the "Road Rage" charge pump circuit, thoughts?

[bean]

....even the "good" supplies aren't putting out exactly 9v. My One Spot is a steady 9.42v......

We say "9V" so we don't get a 1.5V or a 67V battery. No battery is ever what it says on the pack. A "12V" car battery floats at 12.6V, is typically cycled 13.8V down below 12.0V.

The One Spot probably aims for "typical battery voltage", NOT "marketing number".

The fresh "1.5V" zinc cell was close to 1.56V (around 1960). Reliable enough to calibrate meters. (They keep dinking the chemistry and it may not be 1.56V now.) So a six-pack would be 9.36V.

Your Spot showing 9.42V is certainly possible for somebody's "9V batt", is well within 1% of my old-old reference.

Well, sure. But the point I was making is about why it might be a good idea to have a voltage divider in this application instead of relying on the actual supply voltage. I think most people here realize 9v doesn't generally mean "or possibly 1.5v"

PS I'm being cheeky here

[niektb]

What I did in my first pedal was this, I did however omit it in the later pedals because I use them always in context of a pedalboard (and without option for battery) meaning that the input jack is always plugged in:


So basically I have 2 seperate grounds (one for the FET and a green LED that indicates power status) and the main ground. They are connected to the ring and sleeve of your input jack (meaning they will be shorted together as soon as you plug something in) and TADA! It works!

[ElectricDruid]

If you're running at 18V, what's the need for the 4.5V "midpoint" supply, when you've already got a 9V actual midpoint?

There are a couple reasons I can think of:
1) You rarely get 18v once a load is on the charge pump circuit. It does work best if you use low ESR caps on the pump circuit but generally you are going to end up with something between 17-18v depending on the current demand.
2) Even in best case scenario (you have a steady 18v out of the pump circuit) even the "good" supplies aren't putting out exactly 9v. My One Spot is a steady 9.42v with or without any current draw on it.

I agree, Brian, but I don't think it's that crucial. Ok, maybe we've got a 17V rail, and maybe our incoming power is 9.5V. So things are a little bit asymmetrical, but we've still gained a ton of headroom, which is presumably the general idea. I'm not going to stress about a volt here or there.

Maybe there's a reason not to have virtual ground the same as the power supply input, as well? That's above my pay grade.

Yeah, this does sound like a risk, I agree, although its above my pay grade too. Still, worth a shot, right?!

[Boner]

So you guys are saying using 9v as a virtual ground in a 0/9/18 volt configuration might be asking for trouble, but a -9/0/9 and using actual groundy-ground as a virtual ground for audio signals would be more ideal?

That makes sense.

*edit*

but then if you're using JFET boosters, which I do all the time, wouldn't ground in this case now be -9 if you wanted 18 volts of headroom?

Could you use 18 volts referenced to zero (true zero) for jfet boosters and whathaveyou AND -9/9 with virtual ground being true zero for op amps and anything else requiring a virtual ground?

Wouldn't that solve any weird problems associated with using a power rail as a virtual ground?



*double edit*

came up with that after some spice tinkering. I'm embarrassed to admit that I can't get a BJT pair like you guys were talking about to work. However this seems to work....


Left out the 15pF timing cap that goes between pin 2 and 7 to make it more legible. This cap raises the clock freq to help with noise.

D1 protects from input voltages exceeding 12v. Q13 turns the circuit on when a jack is plugged into the output socket. Q9 turns on only for positive voltages (reverse protection). D16 protects Q9 from static discharge which could hurt it. All that nonsense ensures safe power coming into the rest of the circuit that is the charge pump..... Thoughts?

[ElectricDruid]

So you guys are saying using 9v as a virtual ground in a 0/9/18 volt configuration might be asking for trouble, but a -9/0/9 and using actual groundy-ground as a virtual ground for audio signals would be more ideal?

That makes sense.

Yeah...be cautious of stuff that sounds like it makes sense but which no-one can actually justify. It might just sound plausible, but actually be nonsense.
I'm not offering any justification.

Voltage is all relative, so there is no difference between +9, 0v, -9V and +18V, +9V, 0V. They are literally the same thing. The only question is how you derived those voltages, and how much noise that gives you on one wire over others and whether your circuit might be more sensitive to noise on one than another. The more I think about it, the less I find it makes sense to think it makes any odds.

[bean]

One advantage to bi-polar circuits is they do reduce part count. Also (since I've taken a lot influence from Peter @ VFE) you can ditch the 100n decoupling caps on the +9 and -9 at the supply end and instead place them at critical points on the audio circuit. So, for example, a circuit utilizing a dual op-amp with a bi-polar supply: place the 100n decoupling caps right at pins 4 and 8 as close to the IC as you can get them. This is really good practice as far as circuit design.

[niektb]

Like I tried to explain in my last post, you can ditch the whole Q13 thingy if you connect the lower leg of R61 to the jack sleeve (I used that in my pedals and it absolutely works )

[Rob Strand]

Like I tried to explain in my last post, you can ditch the whole Q13 thingy if you connect the lower leg of R61 to the jack sleeve (I used that in my pedals and it absolutely works )

Then the power doesn't turn off when you pull the jack - it powers through the MOSFET body diode.

If you don't use batteries you might not notice the problem because there's nothing to go flat.   What you will notice with a power supply is the LED still goes off and on with footswitch when the Jack is out.

[niektb]

Like I tried to explain in my last post, you can ditch the whole Q13 thingy if you connect the lower leg of R61 to the jack sleeve (I used that in my pedals and it absolutely works )

Then the power doesn't turn off when you pull the jack - it powers through the MOSFET body diode.
[...]

The ground is disconnected...?

[Rob Strand]

The ground is disconnected...?

The point of having the solid-state switch was so the power ground didn't pass through the input socket.   The charge-pumps have current pulses on the input supply the solid-state switch stops those currents passing through the input socket altogether.   A big input cap, like C29 on boner's schematic, certainly helps reduce those currents but if you want to be 100% sure those current are gone then it's best to add the solid-state switch.

It's a bit of a value judgement which way to go.    The current pulses are likely to be more problematic on higher gain pedals than unity gain pedals.

I'm assuming boner wanted a solid-state switch as his original ckt has one (Antonis and I thought it was for reverse protection but that got cleared up in the thread).


To boner:

One thing bad about the input zener protection is if the supply is reversed polaritied the zener will forward conduct short circuit the power-rail just like a SI diode.   That completely defeats the purpose of having the graceful disconnection of MOSFET reverse polarity ckt.   One fix would be to add a silicon diode in series with the zener.

[Boner]

So you guys are saying using 9v as a virtual ground in a 0/9/18 volt configuration might be asking for trouble, but a -9/0/9 and using actual groundy-ground as a virtual ground for audio signals would be more ideal?

That makes sense.

Yeah...be cautious of stuff that sounds like it makes sense but which no-one can actually justify. It might just sound plausible, but actually be nonsense.
I'm not offering any justification.

but man if you think about it like an electron it makes total sense man.

ditch the 100n decoupling caps on the +9 and -9 at the supply end and instead place them at critical points on the audio circuit. So, for example, a circuit utilizing a dual op-amp with a bi-polar supply: place the 100n decoupling caps right at pins 4 and 8 as close to the IC as you can get them. This is really good practice as far as circuit design.

Thanks! 100n the norm?

add a silicon diode in series with the zener.

Good idea... like this? Wouldnt the Si diode short as well?

[Rob Strand]

Good idea... like this? Wouldnt the Si diode short as well?

The diode needs to go the other way, so when the input voltage is reversed the diode and the zener are not conducting.    When the input voltage is over voltage the zener kicks and the si conductors.    That would increase the effective zener clamp voltage to 0.6+12 = 12.6V.

So with that mod, the circuit would have these two undesirable features,

- For the reverse-voltage case technically the circuit still has a problem since the BE junction of the switch transistor Q13 gets reverse biased when the input is more than say 8V (depending on the transistor).

- For the over-voltage case the zener is subject to large currents and is likely to blow-up.   The zeners impedance isn't that low so under high faults the zener voltage could be raised well above the labelled zener voltage.   Zeners aren't that tough.  Some of the TVS devices are a little tougher.

Zener tolerances mean the zener could strongly conduct at a lower voltage than "expected".   A 10% tolerance zener might have considerable conduction at 12V*0.9 =10.8V.    Add the Si diode and that's 11.4V.

The voltage clamp boils down to choosing a voltage high enough not to kick-in undesirably and clamping the voltage to a low enough voltage to protect the circuit.    Many circuits will handle much higher voltages than 10V but those charge pump devices have a tendency to fry quite close to their absolute max input voltage.

Suppose we allow the circuit to suffer under high input voltages.  "Less angry" alternative would be to move the zener to the output size of the switch.   In this position you probably don't need to add the silicon diode.   What you would do is choose  R37 to be large as possible so that it limit the fault current  through the zener,  Ifault = Icollector =  hFe * Ibase.    It's still not great since the switch requires R37 to be chosen based on 9V and the fault current will be determined based on Ibase at the maximum input voltage (say 40V?).   It's still an "angry" method of protection just that it's not as angry as the straight zener and the "anger" is shared between the zener and the transistor Q13.


There are more graceful over-voltage protection schemes.   They require more parts but not a lot more.

Image:


Link
https://www.electricaltechnology.org/wp-content/uploads/2019/11/Over-voltage-protection-using-Zener-Diode.png

From,
https://www.electricaltechnology.org/2019/11/simple-overvoltage-protection-circuit-using-zener-diode.html

There's no right answer with this stuff.   It's a matter of how far you want to go.   For larger more expensive products a fully protected supply input doen't make a big impact on the cost or the size of the product.    But for a pedal, anything from no protection to fully protection is up for grabs.   I guess it is a matter of knowing where you want to drawn the line.

[PRR]

There are more graceful over-voltage protection schemes.   They require more parts but not a lot more.
Image:

Is that really a good example for our purposes?

At a glance my so-called brain thought it looked like a cut-out. The Idiot gives the same trend. From small to 5.1V, out=in. Past 5.1V, out=zero. Since we can conveniently find regulators to hold a 5V output up to 35V input, I am not sure how this is better.

[Rob Strand]

Is that really a good example for our purposes?

At a glance my so-called brain thought it looked like a cut-out. The Idiot gives the same trend. From small to 5.1V, out=in. Past 5.1V, out=zero. Since we can conveniently find regulators to hold a 5V output up to 35V input, I am not sure how this is better.

The goal is to cut out, it disconnects the high voltage.

Sure a regulator will work, in fact that site I linked has such an example, but the regulator introduces a significant drop-out.   Also with a high input voltage the regulator would have to handle the corresponding heat dissipation.

The cut-off method saves the day and we don't have to deal with any of the regulator issues.

We could design a universal input product but that's going to be a lot more work.   Probably a universal input switchmode.  Then we could create many output rails.     Many products with universal main input products are like that but it's probably raising the bar too high for circuit complexity.