I've found that lately that using transistors for interfacing is critical.
In the past I've used optoisolators and open collector IC's for higher
voltage interfaces. But it's clear that understanding how plain old NPN
transistors do their work is important.

So today's question is how to use an NPN transistor to switch 24V
(arbitrarily picked) with a PIC output pin. From what I've read if you
do the standard:

        +24
         |
       Load
         |
         C
        /
PIC-R--B
        \
         E
         |
        GND

Then the transisor won't be saturated because the 5V output from the PIC
isn't high enough relative to the collector voltage.

Now there are several alternatives:

1) Stick an OC buffer between the PIC and the resistor and add a pullup
resistor between +24 and B. Then the PIC can pull the base low and the
pullup will pull the base to +24, causing saturation.

2) Do the same with an optoisolator.

3) Replace the NPN with a logic level power MOSFET. The MOSFET is completely
on with an input of 4V. And it has such a high impeadence that essentially
the PIC is isolated from the output circuit.

4) Use a relay, but most times that's what I trying to drive with this circuit
as it is. I have cascaded a 5V relay with a higher voltage one in a pinch.

Now Bob Blick has an HBridge motor drive that uses a darlington configuration
with no pullup between the high voltage and the base of the drive transistor.
Here's the schematic:

http://www.saber.net/~bblick/bob/projects/hbridge/hb01sch.gif

What I don't understand is why the drive transistor in any quadrant would
completely saturate. From what I've been reading transistors are current
devices. So maybe the switch transistors provide enough current to get
the drive transistors to saturation, without having to go all the way up
to the drive transisor's collector voltage.

Would anyone care to explain? And let me simplify this to an example. What
happens if I extend the example above:


                  +24
                   |
                 Load
                   |
                   C
                  /
         +-------B
         |        \
         C         E
        /          |
PIC-R--B         GND
        \
         E
         |
        GND

Now it looks to me like this won't work because there's no mechanism to
provide positive voltage to the cascaded base so the second transisor will
never conduct.

Would it be sufficient to pull the second base up to +24 with a pullup
resistor? While the first transisistor won't saturate, the second one will.
Right?

I need this circuit to drive a high voltage relay. I already have the
second transistor in place and I was driving it with a RS-232 output which
goes to +12. I want to replace it with a PIC output.

This is an instance where I'm a computer guy trying to understand some simple
EE principals.

Any suggestions?

BAJ

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