>>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 <
.....mykeKILLspam
@spam@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?
>
>
>
