Splitting 9Volts to +/- 4.5v?

[jwhitmore]

There's an answer to this question in the FAQ [1], question "Power Supply - What is V.R., V.B, VREF, 1/2V+ etc....?". Basically you connect the input voltage, (9v) across two equal value resistors in series. You can now make ground or the reference voltage the center of the two resistors in series. That point has a voltage of 4.5V relative to the -Ve input of the power supply but if you use that point as ground you have +4.5 and -4.5 as the two terminals of the input power supply.

I'm happy with that, but unsure of the value of the resistors being used and why. In the example from [1] 2 10K resistors were used in series, which means with nothing else connected you have 9V/20,000 Amps being drawn in idle, or 0.45mA. It's easy to deal with the series resistors in isolation but if you add circuits to the +ve and -ve side of the reference voltage you're putting another effective resistance in parallel with the two 10K resistors, so you no longer have two equal value resistors in series, but 10k || PosR? and 10k || NegR?

And what does all this mean if you have a circuit that's hungry and wants 200mA as opposed to a circuit that only needs 50mA. Should the two 10k resistors be changed to higher/lower value resistors, depending on power hungry circuits?

This all probably changes depending on what you're circuit looks like. I'm only talking about +/-4.5 as I want to use an opamp and maybe in that case a opamp has a very high input resistance, but then does it depends on the configuration of the Amp? Inverting/Non Inverting, soft/hard clipping.

I'm trying to think of a +/- power supply which I can just use without worrying about all this and wondering what should I use, or is there a simple IC that you put in 9V and get the +/- voltages and not have to worry about resistances in parallel or configurations of Amps. Maybe you don't have to worry about all that and two 10k resistors will just always work? But if it's engineering there's probably trade offs involved

Might go and look at digikey or something. There might be a part like a 7805 which is two LDOs glued together

[1] https://aronnelson.com/diywiki/index.php/Frequently_Asked_Questions_(DIY_FAQ)#POWER_SUPPLY

[GibsonGM]

Welcome to the forum!   I'm no guru, and someone will have a fuller explanation shortly I bet, but here is what I understand:

The 2 R's are creating a bias voltage for an active device such as a transistor or opamp...the current required of this network is extremely low, uA.   So you should never have a situation where that divider can't provide it...you can APPROACH this situation in something like an MXR Dist +, which uses 1M to 'starve' the opamp a little, but it still functions.   

Why the values are chosen (10k here, 22k there) can be found by looking up opamp circuit design files on the net.  Texas Instruments has many of them freely available...they usually specify "greater than 10k" and so on, to ensure that the device will have a 'stiff voltage' at the associated input.   A typical rule for dividers is to have 10x the current in the divider as the load will draw...(which holds true for most applications, to your question about varying current draw - you'd design for the higher...and have to take into account dissipation).

As for other devices being driven by the VR....no...don't think that's too good, LOL.    This isn't for +/- power, it's just for biasing.   Adding another opamp bias or something to that network is fine since the current demand is so micro.   mA level draws on it WILL skew it around, you're correct.   The cap that is found at VR helps keep the reference stable, as well.   If I was going to use something like this to POWER chips, it would be a separate, independent thing that takes into account the power dissipation, and would not be centered on a 9V supply....+/- 4.5V aren't much to work with. 

For actual +/- supplies, one has to look into other things.   A positive and negative voltage regulator is one way.  Simple dividers like this CAN work if you design for the load (10x...)...perhaps noise is of concern this way, too.  Hope this helps a little, I'm not entirely sure what you're looking for.  If not, keep asking and someone with much more in depth info will add to this basic info 


Download and enjoy!  I'm pointing you to section 4, re. single supply applications...you may be happy to have this as you go along

https://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf

[duck_arse]

hello jwhitmore, hello Mike (Sir).

the reference supply is only for looking at, it is not to provide POWER!! for that you want a low-impedance power supply, so as to supply whatever the circuit draws, without blinking. opamps and jfets and some transistor circuits can bias nicely by looking at a Vref.

the lm386 [I think] is an IC that you can wire in to provide half supply voltage automatically without needing to think about it too much.


[no mention of backwards this post.]

[amz-fx]

Rail splitters and virtual grounds:

https://tangentsoft.net/elec/vgrounds.html

regards, Jack

[jwhitmore]

Thank you all for taking the time to respond. There's interesting information in there to work through. One thing that's relevant is what circuit is this aimed at and to be honest it's not. I'm working on a modular system for myself which will modularise everything which can be. So audio jacks, foot switch, controls, (pots and toggle switches) will all be in a modular board that I can plug a vero board onto and make whatever takes my fancy. Don't want to have to start at the beginning every time. This has to fit into a standard enclosure as well.

Now I'm coming at this from an embedded programming background so playing about with the Spin FV-1 Reverb DSP. So my baseboard will have footprints for 5v and 3v3 but I'll only populate what I need, and pass what I need up to the vero board/strip board. But I'd like to play with analog effects as well.

Thanks again for all the help

[garcho]

If you'll only ever have 5V for the supply, an LT1054 can give you +/-5V, but you have to account for switching supply noise and total current draw (~100mA). What's powering the DSP? Digital and analog need to learn how to play nice together.

[jwhitmore]

On the Spin FV-1 at present I only have a dev board which for the moment I'm powering both the Spin FV-1 and a microcontroller from a bench top power supply set to 3.3v. I've not been too demanding as yet, and currently only getting a feel for what I want to do with it. When it comes to actually putting this into pedal and powering uC and FV-1's Analog and Digital I'm not sure what I'll have to do. Have you any suggestions? I'd be hoping to power all 3v3 electronics from a single linear DC/DC Converter, but might need to carry two DC/DC's one for digital and one for Analog reading between the lines. I have the SPIN dev board and I think that powers from a single Linear, but then again that's just a dev board as well and not production ready.

[garcho]

When it comes to actually putting this into pedal and powering uC and FV-1's Analog and Digital I'm not sure what I'll have to do.
...linear DC/DC Converter...

Firstly know what will power this, then figure out the circuitry. Standard guitar power supply (9VDC ~50-500mA), USB, internal transformer and mains, etc. What's the plan?

As far as regulators linear or switching, that will also depend on knowing more details. Is 100% of the analog circuitry on the development board? As Mr Arse pointed out, the v/2 is a reference, not a supply. Unless you need voltage beyond 9, there's not much of a need for bipolar supplies in guitar pedals unless you're using ICs that require it. "Pro" or studio grade audio gear has to worry about noise and distortion that we don't have any business being concerned with, luckily.

just a dev board as well and not production ready

And why would that matter for you?

[jwhitmore]

    just a dev board as well and not production ready

And why would that matter for you?

If I got something I wanted to use in the wild, put in a case, and stick it on a pedal board it'd have to be powered from a 9V supply and be robust. Early days