Cant wrap my head around combining a bipassable 9V battery with a 9V regulator..

[Boner]

First, what do you think of the following supply schematic? Its meant to take an unregulated BOSS supply, hence ground being on the pin and not sleeve.




Second, I have no idea how to throw in an optional 9v battery so that when the power jack is inserted, the battery is bipassed. I was trying to avoid using a TRS input jack that turns the pedal on and off.... simply because I dont have any 
Does that make it impossible?

[FiveseveN]

Why not put the regulator in the PSU?

[thermionix]

Second, I have no idea how to throw in an optional 9v battery so that when the power jack is inserted, the battery is bipassed. I was trying to avoid using a TRS input jack that turns the pedal on and off.... simply because I dont have any 
Does that make it impossible?

Usually in a negative ground pedal the + of the battery is switched at the DC jack, to cut the battery when outside power is used, and the - of the battery is run to the ring of a TRS jack to cut the battery when the cable is unplugged.  Without a TRS jack you could add a switch to turn the battery off.

Is it a bad idea to run a 9v battery into a 9v regulator circuit?  I have no idea, I haven't yet learned about regulators.  Assuming it is okay, you could switch battery + at the DC jack like is commonly done.  Otherwise, maybe build the regulator as a separate box, to run between your Boss supply and pedal(s).

[R O Tiree]

The 78LXX series need to be supplied with about 2V more than the stabilised voltage you require.  So, if you plug a battery into that 78L09, it will only give you 7V or so, and it may or may not be stable.  It might be worth installing a "Batt/Ext" switch, but that could get knocked quite easily in your gig bag and drain your battery.  Maybe use a recessed slider switch?

...Or you could spring for a "proper" power supply, centre pin negative (which is the de facto standard for this stuff, as it makes life much easier to wire up).

[Phoenix]

Current boss power supplies (the PSA) are regulated. The ACA power supplies from Boss were unregulated, but they were discontinued in 1997.
As has been mentioned, 7809 regulators have a dropout voltage of 2V at minimum (more at higher currents), so will not function correctly without at least 11V input, which is higher than even the unregulated ACA power supplies will provide.

What exactly are you trying to solve with the regulator? If you let us know what you're trying to do we can give you better advice.

[Ice-9]

Yeah, the BOSS power supplies that you can buy now are regulated and quite good quality, or use a 1-Spot. If it ain't broke, don't fix it.

[Boner]

Holy schnikeys am I glad I asked these questions!! I had no idea newer boss PSUs were regulated... I have two and they are unregulated.
That makes things way easier and understandable (I think)

What about this??



U9 is the input jack, stereo so there is: tip, going to the stomp switch; ring goiing to negative of the battery and sleave going to ground.
U12 is output, here stereo only because I wanted to use the same footprint.
'input' is just a generic placeholder for an effect input
'output' is just a generic placeholder for an effect output

Hows the conditioning part between D1 and C25??

Thank you all for the awesome help! Greatly appreciated

[Rob Strand]

Usually in a negative ground pedal the + of the battery is switched at the DC jack, to cut the battery when outside power is used,

If you use a negative regulator then the +rail can we wired in the usually way.    The pin/centre of the DC jack can then run to the regulator -ve input without going anywhere else.   The -ve output of the regulator then goes to ground.   

Everything seem good so far *BUT*  if you draw this out for the two cases of PSU power and battery power you will see under battery power the Battery is feeding the output of the regulator.     

Is this a problem?   Well there a number of possible problems and they depend on very specific details of the regulator.
1)  If a 9V regulator has an output protection zener there's a good chance it will short across a fresh battery (which can be as high as10.5V).
2) You can fry the regulator as it doesn't like power fed into the output.  In many cases the addition of a diode like D3 in the OP's original schematic (but pointing the other way) can stop this.
3) Feeding power into to the output may increase the battery drain, maybe by a little, maybe a lot.

One way to clobber all this is to put a diode (say a Schottky diode) in series with the output of the regulator but that will further increase the dropout.  It will also decrease the output voltage and make the regulation poorer; for effects ripple rejection is often more important then regulation. Another option is to increase the regulator voltage to compensate for the added diode - the we need more input voltage.   Anyway not the best solution.

If we look at one of the old Boss pedals like the BF-2 we can see it does in fact use a negative regulator.  A simple shunt regulation with an 11V zener.    So Boss have already done this!

This will probably work but the dropout's not great.
http://3.bp.blogspot.com/-cj-YAcRGUm4/U0bo6Vby3DI/AAAAAAAABI8/NVqZNGJNpmg/s1600/negative.jpg

At the end of the day you need a negative regulator that doesn't misbehave with power fed into the output.

[Rob Strand]

What about this??

It unnecessarily creates a lot of drop on the battery path.

Just copy the Boss circuit from the BF2.  You don't even need the zener if you just want a filter.

[Edit: see
https://www.hobby-hour.com/electronics/s/schematics/boss-bf2-flanger-schematic.gif
Left-hand side R56, D1, D10.

Tune R56 so the circuit sees 9V when the adapator is plugged in.
]

[Boner]

What about this??

but at the benefit of creating voltage reversal protection, right?


*edit*

tinkered a bit with it....




That 9.1 zener will protect against both over-voltage AND reverse voltage right?

[thermionix]

That 9.1 zener will protect against both over-voltage AND reverse voltage right?

I've been told that it's a bad idea.

[Boner]

ohhhhhhhh that makes sense.

What about replacing it with a 1N4xxx series diode then?

[Rob Strand]

What about replacing it with a 1N4xxx series diode then?

That will work.
You are best putting the diode (or zener) on the other side of the 100ohm as it limits the fault current.

The zener needs careful attention to the zener voltage and the maximum power it dissipates in fault and normal conditions.