RM potential divider beginners query.

[coxy46]

On the RM circuit, can the two resistors that form a potential divider before the transistor, (normally 470k and 68K) be substituted for different values of the same ratio?
i.e 470k/68k = 4.5
So can say 100k/22k ( =4.5 )be substituted without any noticeable effect?
Just that i have never seen it any different to 470K/68K.

If not what difference does lowering of highering the resistors (keeping the ratio the same) do to the end result/sound?

[petemoore]

  I think it's pretty much like that, maybe there's a 'curve' I don't know about, ie when altering resistor values for bias at what point does the ratio not apply exactly...
  I've made the 68k a 100k pot before, and a 500k pot for the 470k, when made smaller enough, high end first, and general loading, 'damp it down'.
  I did one I think with 100k and 1meg where the [68k and 470k were]...something like that...as long as the bais is good it's good...

[davebungo]

Like most things it is a compromise.  Ideally we would want to use 470Meg and 68Meg, in order to reduce current consumption to a minimum, but the impedance of this is so high that if any current is required from the network (say into a BJT), it would not be available.  Also, the use of larger values can increase noise if used before high gain stages, so this is another consideration.  The answer lies somwhere inbetween, so the resistance values used can provide the required bias current at the required bias voltage whilst keeping the drain on the battery to a minimum.

Thevenin's theorem cropped up the other day, and it is useful here also (sorry to harp on).  A resistive divider as follows:

          Vbatt
        ___
          |
          R1
          |
          |------Vbias
          |
          R2
          |
        ___
       0V

can be viewed more simple as its Thevenin equivalent.  It is easy to see what the result will be by attaching a load resistance (i.e. your circuit) to this network.

          ______Rth_____
          |                      
          |                      ^
          |                       |
          Vth                   Vbias
          |                       |
          |
          ___________
 
      Rth = R1//R2 = R1 X R2 / ( R1 + R2 )
      Vth = Vbatt X R2 / ( R1 + R2 )

This assumes that the battery has a relatively low impedance itself.