At 10:57 PM 6/30/2005, Russell McMahon wrote:
>>I think you're wrong, Russell.  The input transistors are operate as 
>>constant current sinks so long as they don't saturate.
>
>It depends how you run your H Bridge.
>My concept of H Bridge driving is that you run everything saturated by 
>default unless the deign specifically prevents that happening. (My design 
>(circuit soon, fully described in prior posts) so far in fact doesn't 
>always run the bottom switch pair in saturation but this is a fully 
>defined state).

Ahh - I think this is the sticking point.  Bob's circuit does *not* want 
the input transistor saturated.  Yes - both output devices should be 
saturated.  But by operating the input transistor as a constant current 
sink, the current you have chosen to be fed into the bases of both output 
transistors doesn't change (much) even as Vbatt changes.

>This pulls the top driver base down to V+-Vbe2. The input driver tries to 
>saturate and its collector is essentially at 5-Vbe1. As it's hard 
>connected to the base of the upper driver which is also essentially 
>saturated then 5-Vbe1 ~~= V+-Vbe2.

Yep - when Vpic = Vbat, the input transistor is heading towards 
saturation.  I say "heading towards" because I expect the processor pin 
voltage to start sagging (PIC FET RDSon).  When Vbatt actually goes below 
Vpic, you start to have problems.  Problems because you begin to draw 
significant current from the processor pin.

>When V+=6.5v and driver v = 5v then we have about 6.5-5-0.7-.0.8 = OKish.

Here's where you miss the beauty of Bob's circuit.  With Vbatt > than Vpic, 
the input transistor is NOT saturated.  Its in constant current mode.  That 
current is (aprox) (Vpic - 2Vbe) / emitter resistor..

Try it and see!  Model the RDSon of the PIC as a 100R resistor.  Put a 
voltmeter across the driver transistor emitter resistor and vary Vbatt 
while keeping Vpic constant.  So long as Vbat > Vpic, the voltage will stay 
reasonably constant.

Also note that this talk of constant current applies to only the driver 
transistor.  Its emitter resistor is chosen to ensure that the output 
transistors are saturated.

Of course, subsequent messages from you have shown that saturated output 
transistors isn't what you are looking for.  But I wanted to clear up the 
misconception about the circuit that Bob showed.

dwayne


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Dwayne Reid   <dwayner@planet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
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