limiting power from 9v or adapter to the board to 4.5v

[jschwalls]

what i want to do is get 4.5 volts to the fuzz instead of the 9 volts from battery or the adapter......how would i limit the incoming voltage to 4.5 volts ?
i image it would be easy but i am new at this game....

how would i go about doing this.......

my reason is simple... i heard about one of the guys in the Allman brothers draining his 9 volt batteries down to half life 4.5 volts then running them in his fuzz.... he said it squashed the load or something like that.....

this is a request from a friend of mine....he knows Greg Allman and got the story from him.... Greg lives here and visits the store often.. His cousin is the owner...

thanks
JON

[darron]

hey. if you want the dying battery effect, i can think of a couple of ways to do it.


if you want the exact (very close) 4.5 volts, then make a voltage divider out of two resistors. get two resistors of about 100k, connect in series:

(+) positive > 100k resisor > 100k resistor > (-) negative

then, the voltage that you will get in the middle of the two resistors will be half of the supply voltage, about 4.5 volts. so tap off between the two resistors which are sitting between positive and negative and you will get the 4.5 volts.


another way to get the dying battery effect with a bit more control would be to use a potentiometer in series with the power supply. at one end you will have zero resistance and the full 9-ish volts, at the other end the resistance will increase and the voltage will proportionately lower.

i just checked google for a diagram, this came up first
http://www.beavisaudio.com/Projects/DBS/

[theblueark]

You could use a voltage divider, 2 resistors of the same size in series from 9V to ground. The voltage from in between them would be 4.5V.

For battery sag simulation zvex recommends in an old post just using a 5k pot in series with the effect.

I don't think just giving the effect 4.5V will get you the results you want, or if the effect will have enough to work at all. But do try both and report the results 

Edit: We posted at almost the same time saying almost the same thing 

[blanik]

the 5k pot works fine and you can fine tune the sound, just a couple of millimetres in one direction or the other creates a big change but i don't know if this trick would world on something else than a Germanium fuzz...

[miqbal]

Lot of schematics use coding like VCC/2, 1/2VCC, or some times VSS. They all mean the same: half of voltage source, i.e.: 9V/2=4.5V.

EDIT: if the source voltage other than 9V, just divide the source voltage by 2.

[zachary vex]

Zachary Vex says, if you want to get all the way down to 4.5 volts and simulate a dead battery, the first thing to do is to actually kill an old battery (short it out for a few minutes) and get it down to 4.5 volts or whatever, and LISTEN to the fuzz with that battery to see if you actually like the sound.

If you don't like the sound, of course, you don't have to go any further.

If you do like the sound, a series pot of about 50k to 100k should be enough to reduce the voltage down somewhere near there with most fuzzes.  Just connect one end of the pot to the battery and the wiper to the circuit, and adjust it to reach nirvana.  Maybe.

Let's do a thought experiment here.  Just exactly what is the output resistance of a dying battery?  It still has some voltage, but its output series resistance keeps increasing as it dies, because of chemical pollution of the elements inside.  How do you determine just how much series resistance a dying battery (or even a new battery) has?  It's easy to figure out.  Much like figuring out the "output impedance" of a given amplifier, you simply put a load on it (put a resistor across the terminals) and keep trying different values until the voltage of the battery drops to exactly half of the voltage it has with no resistor across its terminals.  When the battery reads exactly half of its normal voltage, the resistor you've put across it is exactly its output series resistance.  Why?  Because half of the voltage is across the external resistor, and half is across the internal chemical resistance of the battery, which means the two resistances are equal!

You can use a similar trick for measuring the output impedance of an amplifier, if you have a high enough wattage resistor or wirewound pot to put across the output of the amp.  Just connect a tone generator to the amp at low to moderate volume with a speaker cab, then switch over to your adjustable resistance load and measure the ac voltage across the load, then adjust the resistance of the load until you get the maximum output wattage (you'll have to square the voltage and divide by the resistance of your load, E^2/R.)  That's the output "impedance" of your tube amp, where it can deliver the maximum power.  You may be surprised to discover that your Marshall or Fender amp has a completely different output impedance than its rating!

[jschwalls]

VERY COOL....
thanks guys...... i will start with the resistor trick first... space is an issue...... unless a mini pot on the board would do the trick....

jon

[foxfire]

you sure can use a trim pot...

[petemoore]

  LM386
  power it with 9v to V+ pin
  Ground V- pin
  output pin is now a nice clean 1/2v.

[darron]

  LM386
  power it with 9v to V+ pin
  Ground V- pin
  output pin is now a nice clean 1/2v.

a feature of the 386 being auto-biassing? that's a cool tip, a little bit more effort though maybe? i suppose you came across that accidentally too (:

[PerroGrande]

Trimpot between +9v and ground. 

Wiper connected to non-inverting input of an op-amp capable of sourcing enough current for your needs. (TS922 can source 80mA for example)

Connect op-amp output to inverting input (as in a buffer).

Output pin will be at a voltage that matches where the trim pot it set (within the range of the op-amps ability to swing to the poles).

[petemoore]

  Takes a socket and board space, otherwise is 3 connections, I don't think there's a fewer connections option available.

[Sir H C]

You can also add a capacitor (say 100uF) to the output of the divider to help with some of the spiking, put a separate resistor/pot in series with this cap to simulate sag independent of the voltage being driven.

If you need to do high current with the divider, you can put a NPN bipolar transistor's base to the divider, collector to V+ and the emitter to the circuit, this will be about .7 volts below the dividers voltage, but can handle a lot more current, so you can try voltage starving without current starving.

[sevenisthenumber]

try this. I built this idea out in a small box and I use it alot on my distortions and overdrives. The lower the voltage the more gain in most circuits. Takes all of 15min to build. Try a lower K pot. (2K - 50k) . Or you can put it in the pedal itself!   Works great on the dynacomp....
http://www.beavisaudio.com/Projects/DBS/

[sed]

Check this out. Haven't tried it myself but maybe it works. 

http://www.muzique.com/news/?p=60#more-60

[John Lyons]

You can also add a capacitor (say 100uF) to the output of the divider to help with some of the spiking, put a separate resistor/pot in series with this cap to simulate sag independent of the voltage being driven.

If you need to do high current with the divider, you can put a NPN bipolar transistor's base to the divider, collector to V+ and the emitter to the circuit, this will be about .7 volts below the dividers voltage, but can handle a lot more current, so you can try voltage starving without current starving.

Do you mean to say that the 100uf cap would go to ground from the 1/2 voltage point as in a typical 4.5v Vref and then a resistor can be added between cap and ground as a sag simulation?
Or is the 100uf cap in series with the 1/2 voltage point in the divider and then the resistor in series with the cap to simulate sag?
Just trying to clarify.

Thanks

John

[amz-fx]

More info about voltage sag:

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

regards, Jack