Below is the screen capture of the +DC supply to a +-DC supply as found in Weekly_Update_2007-08-30. Please note that the input is Channel 1 and the output is Channel 2.
Input = 20.2V
Ouput = 10.0V
Figure 1 (Schematic to the left is the +DC supply to a +-DC supply)
Figure 2 (Channel 1 = Input = 20.2V, Channel 2 = Output = 10.0V)
The figure below is the second half of the circuit attached NOT as it is in the above schematic but as it is originally. The only difference is that the ground for U1 above for the power is the new ground formed by the +-DC supply. Please note that the input to the +DC supply was 20V.
Figure 3 (Output voltage by the DC motor power amplifier with the +-DC=20V supply)
I then made the +DC supply a 15V wall wart rated for 1.67A. The input still had an amplitude of 7.2V but the output was clipped because though 15V/2 is 7.5V there are voltage drops in the Darlingtons thus we will need to either decrease the input voltage or increase the source voltage. See Figure 4 below for the input and output with the use of the wall wart. Please see Figure 5 below for the output where the input has been reduce so there is no clipping of the output signal.
Figure 4 (+DC = 15V DC Channel 1 = Input = 14.6Vpp Channel 2 = Output = 10.8Vpp)
Figure 5 (+DC = 15V DC Channel 1 = Input = 14.4Vpp Channel 2 = Output = 10.8Vpp)
By looking at Figure 5 above you can see that the maximum amplitude you can put in the amplifier with a wall wart of size 15V is about 10.4V/2 or 5.2V. It is important to note that the motor that we are using is rated for use at between 4.8V and 7.2V. The preference would be to have the wall wart be able to supply about 20V but the current is also important, and because the motor is rated for 4.8V it might be fine to use a 15V supply as long as it has a relatively large current rating, in this case 1.67A will probably be sufficient.