2 power (supply) questions + one more

[Athin]

Hi,
   I want to power an effect with +-4.5V using a 9V battery. I suppose using 2 100Ohm resistors in series and two 100uF caps in parallel should be enough [ground would be connected between the resistors and the battery to either side of the divider]. The question is will I need DC-blocking caps to ground [ie. will there be any DC current flow form one effect to another via ground?]. I got a bit confused, cause ground is a bit imaginary - just a reference point. The thing is that how are the electrons to know I mean depending on how you look at it - I either have a +-4.5V powered effect or a 9V powered effect with ground at +4.5V. Someone - relp?
   Another question [request] - I've received about 20 7660's [as a 'present'] and thought to make some use of them. I've read here that the max1044 is far better, but I'd like to play around with the 7660's [they must have >>some<< value, other than scarecrows for centipedes that is...]. To get them to work above audio frequencies I'd need an external oscillator - will a 555 do [got a few of those too, any simpler solutions]? I suppose 60kHz will do. Any other problems I might bump into whilst 'playing' with the 7660's ? maybe some [not necessarily dirty] tricks?

[puretube]

if you use those little block 9V batteries like in most "B*ss" pedals, they will be empty in far less than an hour, if you use a 100 Ohm divider...

somebody else will link you to an active schem for dual supply at low current draw (R.G.?)

[Athin]

I've seen the schems at geofex.com [with the maxim 1044] . So are you saying 10k resistors?

[Torchy]

If you've been to geofex check the Neutron schem (pdf download) which uses a 7660 to give +/- 9V supply.

[Athin]

tanx  

[Lonestarjohnny]

Not to mess up your thread, But, That's Funny ! " Hey Mom, Guess what I found this Morning, What Son ? I found a Centipede ! Yes ? What about it ?
Mom, it was Makin Love to my 1044's  :shock: , Son, You need to Discuss that with your Father ! "  
Good Thread ! LMAO !
JD

[niftydog]

I mean depending on how you look at it - I either have a +-4.5V powered effect or a 9V powered effect with ground at +4.5V.

that won't bother the effect, because in either config the effect "sees" 4.5V accross it's terminals.

Just don't connect it's ground to the "real" ground!

But, you should connect it from the real ground to the 4.5V, that way, all the grounds are at the same potential.

The question is will I need DC-blocking caps to ground

no. if you think about it, only negative curreent can flow "from" ground. And if you're effect is well de-coupled, and powered from positive supply, then you should have no worrys.

(lets not get started with the "conventional versus actual current flow" discussions ok!?!)

No matter what resistors you choose, there's always going to be current draining from the battery.

Small resistors drain the battery fast.

Large resistors limit the current available from the 4.5V AND they waste more and more voltage as the current goes up; resulting in something other than 4.5V!

But, it will work. Just remember to disconnect the battery from this circuit when you're not using it!

there is some explanation in the FAQ of this site.

There is also a way of using op-amps to provide more current whilst not loading down the voltage divider resistors.

[puretube]

using op-amps

that link is exactly what I meant - go for it!

[Athin]

thanx for the in-depth explanation...
the opamp thing looks neat, I think I'll go for it.

[Athin]

Just don't connect it's ground to the "real" ground!

But, you should connect it from the real ground to the 4.5V, that way, all the grounds are at the same potential.

Wait - are you saying not to connect the real ground [the ground on the cable/jack] to the '-' of the batery but to the middle of the divider? the potential [the one form 'infinity' to 'here' would be different = current flow?]  :?: I understand that to get rid of the noise form the cable I have to get the signal to the ground (even if it is virtual), but a direct connection, not via 4.7uF caps like on the neutron (thx again Torchy) ? I think I think too much

[Athin]

I don't feel like opening another thread for this (esp that there's already been a discussion), so I'll just post it in here. I want to get the eighteen, professor tweed and english channel into one box and hook 'em up to the amp simulation part of SansAmp. Because there are quite a lot of knobs on those and most of them do the same thing [volume, tone] I thought I could use the same pot for all three circuits. I've read [here] that you can wire a resistor in series with the input and then wire a PNP to ground. The base voltage can be regulated via one pot and the circuits are isolated, but the BJT in that configuration will asymmetrically clip the signal. This could be  solved by using a symmetrical circuit with an NPN and negative control voltage, but there is no point. So I thought JFET, and it works [in my head], as on the schematic. I want the control voltage to be positive, so I'd have to use a PJFET.

So how (or rather why is it impossible to do it simple) do I get the thing to work using a NJFET? For a NJFET to work I'd need a negative control voltage, or, I'd need the NJFET's drain to be more positive, but if I get the NFET drain more positivecurrent flows 'backwards'.
I already know that digital pots won't work, LED/LDR is either too bulky and/or too expensive [normal pots come out cheaper], BJT's clip. Is there any reasonable way out.... I mean correct me if I'm wrong, but the input signal swings aren't big enough to distort a JFET, and if it comes out better I'll use NJFETs and a 7660 for negative or just PJFETs.

I'm not sure about the resistors [the 470k's are typical, the red one is to get current flow, I dunno about R1, I suppose it should be (the value of the pot I want to replace) minus (minimum JFET resistance) ].

pS a reply to the previous post would be nice


http://kosi.website.pl/vcr.gif

Edit: the picture added
Edit2: yes, ok, I know, copy-paste the link into a new window for the image....

[niftydog]

er... no. I've confused you unnecessarily, sorry.

connect the ground on the cable/jack to the battery - terminal as always.

if you want to connect up something to 4.5V, connect between the ground and the middle of the divider.

[Athin]

ahhh... now it makes sense... but if I fry anything, or anything blows up it's still your fault  :lol: