From your description you talk about using a 3V
battery. In your calculations for #3 you use 4.8
volts for the PIC. Do you have two supplies. If so
are the grounds common? If not tell me what you do
have 'cause the calculation are different.

Don't forget the 1.4V base emitter voltage drop
for a darlington.
Resistor    = (PIC Output Voltage-Vbe) / BEI
              = (4.8 -1.4)Volts / 0.12mA
              = 28.3K

Working the equation backwards with the 40K
resistor you calculate and a Hfe of 5K you still
should get 0.425 amps. Maybe not enough to start
the motors but enough to run them. What is the
starting current required?

- -Mark
>>> myke predko <myke@PASSPORT.CA> 12 July 1996
11:31 am >>>
Hi Folks,

- -snip- -

I originally tried to copy the circuit presented
in the "Runabout Robot" in
Electronics Now, but I've reached a stumbling
block.  The motor driver is an
"H" transistor Driver (a PNP 2N4403 from 3 Volts
(battery) to a motor
terminal to an NPN 2N4401 dropping the circuit to
Ground).  The Transistor
bases are connected to a PIC driver via a 750 Ohm
resistor.

- -snip- -

3.  What PIC Pin to Transistor Base Resistor Value
should I use?  Assuming
an Hfe of 5K for the Darlington NPN Array and
current requirements of 600 mA
(for some guardband):

  C-E Current = Hfe x B-E Current ("BEI")
  600mA       = 5K (for example) x BEI
  BEI         = 600mA/5K
              = 0.12 mA

  Resistor    = PIC Output Voltage / BEI
              = 4.8 Volts / 0.12mA
              = 40K

So, from these calculations, a 40K resistor should
give me approximately 600
mA through a Darlington Pair NPN Transistor driven
by a PIC Pin.  Is this
correct?




Do you ever feel like an XT Clone caught in the
Pentium Pro Zone?